Specific enthalpy is the measure of the total energy of a unit mass. It is defined as the sum of specific internal energy and flow work across the boundary of the system.
Units of Specific Enthalpy
The unit of specific enthalpy (h) is kJ/kg.
Specific Enthalpy equation
The equation of specific enthalpy is
h = u + Pv
Where,
h = Specific Enthalpy
u = Specific Internal Energy
P = Pressure of the system
v = Specific volume of the system
Specific Enthalpy formula
h = u+Pv
h = cp (dT)
Where,
cp= specific heat capacity
dT = Temperature difference
Specific Enthalpy of dry air
It is defined as the product of the specific heat capacity of air at constant pressure and dry bulb temperature
h = cp (T)
Cp: Specific heat of air at constant pressure
Cp(AIR) : 1.005 kJ/kg-K
T: Dry Bulb Temperature
Specific Enthalpy of ethanol
The specific enthalpy of ethanol (C2H5OH) is 2.46 J/g℃
Specific Enthalpy of water at different temperatures
Specific enthalpy of water (hwater) is given by the product of the specific heat capacity of water Cwater and the temperature. At ambient conditions (Pressure 1 bar), water boils at 100℃, and the specific enthalpy of water is 418 KJ/Kg.
Cwater = 4.18 kJ/kg K
Specific enthalpy of liquid water at atmospheric pressure under condition and different temperature has been illustrated below:
Fig 1: specific enthalpy of liquid water vs. Temperature
Enthalpy equation specific heat
Enthalpy is defined as the total energy content of a system. It is expressed as product of mass, specific heat and change in temperature of system.
H = m Cp (Tf – Ti)
Where,
H = enthalpy
Cp =specific heat capacity at constant pressure
m = mass of the system
Ti = Initial temperature
Tf = final temperature
Specific Enthalpy of air
It is defined as the summation of specific enthalpy of dry air and specific enthalpy of moist air.
h = 1.005*t+ω (2500+1.88 t)
h = enthalpy of moist air kJ/kg
t = Dry Bulb Temperature in ℃
ω = specific humidity or humidity ratio in kg/kg of dry air
Specific humidity is defined as the ratio of the mass of water vapour per Kg of dry air in a given volume and given temperature.
Specific enthalpy of air table
Variation of thermodynamic properties of air with respect to temperature at atmospheric pressure condition have been provided below.
Fig 2: Thermodynamic property of liquid gas (image credit :thermopedia)
Specific Enthalpy of liquid water
A Phase diagram of water plotted between temperature and specific entropy illustrate the enthalpy of water at a different state.
Saturated dry steam curve separates super-heated steam from the wet steam region, and saturated liquid curve separates sub-cooled liquid from the wet steam region.
The point where both saturated vapour and saturated liquid curve meets is known as the critical point. At this point water, directly flashed off to vapour.
Note: At critical point, The latent heat of vaporization is equal to zero.
At critical point degree of freedom is zero.
Critical point pressure for water is 221.2 bar
Critical point temperature of the water is 374℃
The line 1-2-3-4-5 represents a constant pressure line.
Fig 3: Phase diagram representation on T-S curve
Subcooling: It is the process of decreasing the temperature at constant pressure below the saturated liquid.
Specific enthalpy of liquid water is the difference of enthalpy of water at the saturated liquid line (2) and specific enthalpy of water in sub cool region (1). Unit of specific Enthalpy (h) is kJ/kg.
h1 = h2 – c p(liquid) (T2 – T1)
Where,
h1 = enthalpy of water in sub cool region
h2 or hf = enthalpy of water at saturated liquid curve
Cp (liquid) = 4.18 kJ/kg (specific heat capacity of water)
T2 = Temperature of liquid at saturation point
T1 = Temperature of liquid in sub cool region
Specific enthalpy of steam
Specific enthalpy of the steam at any arbitrary point (3) in the wet region is given by sum of specific enthalpy at saturation liquid curve at constant pressure and product of dryness fraction and difference of enthalpies at saturation liquid curve and saturation vapour curve as same constant pressure.
h3 = hf + X(hfg)
h3 = specific enthalpy of steam in wet region
hg = specific enthalpy of steam at saturation vapour line
hf = specific enthalpy of steam at saturation liquid line
hfg = hg – hf
Wet Region : It is the mixture of liquid water and water vapour
Dryness Fraction (X): It is defined as the ratio of the mass of water vapour to the total mass of the mixture. The value of dryness fraction is zero for saturated liquid and 1 for saturated vapour.
X = mv/(mv+ml)
Where mv = mass of vapour
ml = mass of liquid
Specific enthalpy of superheated steam
Super heating: It is a process of increasing the temperature at constant pressure above saturated vapour line.
h5 = h4 + cp(vapour) (T5 – T4)
Where,
h5 = specific enthalpy of steam in super heated state.
h4 = specific enthalpy at saturation vapour curve.
Cp = heat capacity at constant pressure
T4 = Temperature at point 4
T5 = Temperature at point 5
Specific Enthalpy on steam table
Steam table contains thermodynamic data about the properties of water or steam. It is mainly used by the thermal engineers for designing heat exchangers.
Some frequently used values on the steam table has been shown below.
Pressure based saturated steam table (Image credit : www.tlv.com)
Enthalpy and Specific Enthalpy
Enthalpy (H): It represents the total heat content of the system.
The mathematical expression is
H = U + PV
H = Enthalpy of system
U = Internal Energy of system
P = Pressure
V = volume
Change of enthalpy (dH) is defined as the product of mass, specific heat capacity at constant pressure and temperature difference between two state.
dH = mCp(dT)
m = mass of the system
Cp = heat capacity of fluid
dT = change in temperature
SI unit of Enthalpy is kJ
Specific Enthalpy and heat capacity
Specific enthalpy (h) is defined as the summation of specific internal energy and flow work.
The mathematical expression is given by
h = u +Pv
u = specific internal energy
Pv = flow work
SI unit of specific enthalpy kJ/kg
Specific heat capacity (Cp) of water is defined as the amount of heat required to raise the temperature of 1 kg of water by 1 K. For ex specific heat capacity of water is 4184 J/kg-K.
cp = specific heat capacity.
SI unit of specific heat capacity is kJ/kg-K.
Specific enthalpy of combustion
It is defined as the enthalpy change when a substance reacts vigorously with oxygen under standard conditions. It is also known as “heat of combustion”. The enthalpy of combustion of petrol is 47 kJ/g and diesel is 45 kJ/g.
Specific Enthalpy of evaporation
It is defined as the amount of energy that must be added to 1 kg of a liquid substance to transform it completely into gas. The enthalpy of evaporation/vaporization is also known as latent heat of vaporization.
Specific enthalpy of evaporation of steam
The heat energy required by the water at 5 bar pressure to convert it into steam is basically less than the heat needed at atmospheric conditions. With the increase of steam pressure specific enthalpy of evaporation of steam decreases.
Specific Enthalpy of moist air
Specific enthalpy of moist air is given by
h = 1.005*t+ω (2500+1.88 t)
h = enthalpy of moist air kJ/kg
t = Dry Bulb Temperature in ℃
ω = specific humidity or humidity ratio in kg/kg of dry air
Specific Humidity (ω) is defined as the ratio of the mass of water vapour per Kg of dry air in a given volume and given temperature.
Specific enthalpy of saturated steam
The specific enthalpy of a saturated steam at corresponding temperature and pressure is 2256.5 kJ/kg. It is represented by hg.
Specific enthalpy of saturated water
The specific enthalpy of saturated water at standard atmospheric conditions is 419kJ/kg. It is generally represented by hf.
Specific enthalpy of water vapour
At standard atmospheric conditions,i.e 1 bar pressure, water starts boiling at 373.15K. The specific enthalpy (hf)of water vapour at saturated condition is 419 kJ/kg.
Absolute Specific Enthalpy
The enthalpy of the system is measured of total energy in the system. It cannot be measured in absolute value as it depends on change in temperature of the system and can only be measured as the change in enthalpy. For ideal gas, Specific enthalpy is the function of temperature only.
Acrylic Acid Specific Enthalpy
Acrylic acid is used in many industrial products as raw material for Acrylic Easter. It is also used in manufacturing polyacrylates. Specific enthalpy of formation of Acrylic Acid is in the range of -321± 3 kJ/mole.
FAQ/Short Notes
1. Specific Enthalpy of Helium:
Specific heat of helium is 3.193 J/g K. Latent Heat of vapourization of Helium is 0.0845 kJ/mole.
Heat of vaporization of Helium
Fig 5: The heat of vapourization of helium (Image credit: people)
2. Can specific enthalpy be negative?
Yes, the enthalpy of formation of ethanol is negative. Enthalpy of formation is defined as the energy removed during the reaction to form compound from elements under standard conditions. The higher negative the enthalpy of formation, the more stable the compounds is formed.
3. Specific enthalpy vs specific heat capacity
Specific enthalpy is the total energy of a unit mass or defined as the sum of specific internal energy and work done across the boundary of the system.
Specific heat capacity is defined as the heat required to raise the temperature of 1 kg of water by 1 K.
4. Specific enthalpy vs specific heat
The heat interaction per unit mass at constant pressure (Isobaric process) is known as specific enthalpy.
5. Air specific enthalpy vs temperature
Specific enthalpy of air is defined as the product of heat capacity of air at constant pressure and change in temperature whereas the temperature is an intensive property of the system by virtue of which heat transfer takes place.
6.Mass enthaply vs specific enthalpy
Mass enthalpy or enthalpy is defined as the total energy content of the system . Its unit is kJ.Specific enthalpy is defined as total energy content of the system per unit mass. Its unit is kJ/kg.
7.Difference between Enthalpy and Entropy
Enthalpy is defined as the total heat content of the system where as the entropy is defined as the total randomness of the system.
8.Why does specific enthalpy of steam on steam tables begin to decrease after about 31 bar?
The liquid and vapour phases of a substance are indistinguishable from each other. If we consider the internal energy of the steam, it should decrease with enthalpy, But as the random vibration of molecules is hindered by other molecules due to increase in pressure.m which results in decrease of specific volume, thereby decreasing internal energy. As the specific enthalpy is defined as the sum of specific internal energy and flow work on boundary of the system, the specific enthalpy also decreases.
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Gauge Pressure can be defined as the pressure that is relative to the atmospheric pressure. For pressures that are above atmospheric pressure, gauge pressure is taken to be positive while for pressures below atmospheric pressure, gauge pressure is noted to be negative. Gauge pressure is referenced to be zero at atmospheric pressure.
For example, while filling air in a flat tyre, the air is inside the tyre is filled in terms of gauge pressure and the atmospheric pressure is observed to be zero. This is because the tire gauges are designed to operate at 0 atmospheric pressure.
What is the relationship between Gauge pressure and True pressure?
Gauge Pressure can be formulated as the difference between absolute pressure and atmospheric
Pabs = Pg + Patm
where Absolute Pressure is denoted as Pabs, Atmospheric pressure as Patm and Gauge pressure as Pg
If the tire gauge reading is observed to be 36 psi (pounds per square inch), then the Absolute Pressure will be sum of the atmospheric pressure (which is a constant, i.e., 14.7 psi) and gauge pressure reading
i.e. Pabs = Pg + Patm
= 36 psi + 14.7 psi
= 50.7 psi
What is the gauge pressure of the trapped air?
The gauge pressure of air trapped in a vessel or a tube can be measured using a manometer. A manometer is a U- shaped tube often filled with mercury as a fluid to measure pressure. The difference in the height of fluid (i.e., mercury) is used for measuring the gauge pressure.
For example, the gauge pressure can be measured using a U- tube with one end exposed to the atmosphere and a balloon connected to the other end. The absolute pressure is greater than the atmospheric pressure by an amount hρg which is taken to be the gauge pressure.
Pressure Gauge is a tool used for measuring the pressure exerted by a fluid which can be liquid or gas, per unit area which is expressed in terms of Newton per square meter or pounds per square inch.
Liquid Glycerine is often used in pressure gauges due to its excellent vibration dampening properties at room temperatures. They usually operate in the temperature range between -200 C and + 600C. There are other fluids which are used as liquids in pressure gauges depending on the application, but the most promising liquid is glycerine.
Working principle of Pressure Gauge
Pressure gauges work using principle of Hook’s law which states that the force required to compress or expand a spring depends on the distance i.e., F = kx, where k is the spring constant, x, the distance to which the spring is compressed or expanded, and F is the forced applied.
When pressure is applied on an object, there exists an inner pressure force and an external pressure force. Further, the pressure exerted in a Bourdon tube will be more in the inner surface due to the smaller surface area compared to the outer surface
Pressure Gauge calibration is the comparison of values of the unit that is being tested to the values that are measured from an accurately calibrated device. The pressure gauge is usually used for calibrating and tuning fluid flow machines. The fluid flow machines would be unreliable if not calibrated using a pressure gauge. Pressure gauges are calibrated according to the National Standards (NMISA).
These types of pressure gauges are used for measuring relative pressure in the range of 0.6 to 7000 bar. They belong to the category of mechanically driven pressure measurement devices as they do not require electrical energy to power.
The Bourdon Tube Pressure Gauge has an oval-shaped cross-sectional area with tubes that radially packed. The pressure exerted by the measuring source creates a motion on the other end of the tube which is not clamped. This motion that is created on the other end of the tube is taken to be the pressure which is measured. A C- shaped Bourdon tube can be used for measuring pressures up to 60 bar. Bourdon tube packed with windings of exact angular diameter i.e., helical tubes are used for measuring high pressures that exceed 60 bar.
Bourdon tube pressure gauges are manufactured according to set standard of EN 837-1. There are Bourdon tube pressure gauges which are filled liquid, these types of gauges are used for critical applications where the readings to need to be accurate and precise.
What is Oil Pressure Gauge?
Oil pressure Gauges can be categorised into mechanical gauges and electrical gauges
Mechanical Pressure Gauge
This gauge measures the pressure of oil at the end of pipe connecting the pump and the filter. To measure the pressure, an oil take-off pipe taps on the engine block. The needle movement in the dial indicates the measured pressure.
By tapping of the engine block, oil is sent to the gauge by using a copper or plastic bore. The pipe is arranged in such a way that it will be exposed to minimum damage to prevent leakage of the engine oil. The gauge is composed of a coiled tube which is termed as bulb. The open end of the bulb is connected to outer casing of the gauge.
The oil fed that is fed into the supply pipe is at almost the same pressure as when the oil leaves the engine. Under this pressure, the bulb tries to maintain its position, in doing so the needle in the dial moves and indicates the pressure. The higher the pressure, the larger will be the degree of movement of the needle.
Electrical Pressure Gauge
This gauge measures the pressure of oil at the end of pipe connecting the pump and the filter. To measure the pressure, a screwed sensor taps on the engine block. The needle movement in the dial indicates the measured pressure.
Working Principle
This type of pressure gauge is powered by electric current which is supplied from one of the wires that is present in the dashboard. The current that is supplied through the wire passes through a coil which is wound with a wire in the needle’s pivot. A magnetic field is produced which causes the needle to move within the dial depending on the measured pressure.
The extend to which the needle moves or the reading it shows depends on the current that flows through the gauge. The contributing factor is the resistance offered by the gauge wire that is earthed in the engine block using the sensor. All gauges are illuminated for the ease of reading the measurement at night.
It is an instrument used for measuring pressure as well as pressure differences. This pressure gauge was modelled and developed by Dwyer which has currently set standards for the pressure gauges used in industries. It is primarily used for measuring positive and negative pressure i.e., vacuum.
Working Principle
A magnehelic pressure gauge is composed of a diaphragm that is sensitive to pressure changes. The dial of the pressure gauge responds based on the pressure applied. The appropriate positioning of the instrument is required for proper functioning of this pressure gauge. It should be placed at the right level and in a vertical position or else the diaphragm will give inaccurate readings as it will sag.
Industries and labs had been using conventional analogy tire gauges for measuring pressure since ages. But ever since the discovery of digital instruments has led to the use of digital pressure gauges which provide the most accurate reading. It is easy to operate a digital tire gauge, that is to switch on the gauge and position it on the valve stem to get the corresponding reading.
This type of pressure gauge is used for measuring both positive and negative pressures in a vacuum. Few examples where compound pressure gauge is employed are
for leaking testing in pressure lines,
for measurement of low pressures, and
for pressure measurements in test chamber
Its capable to measure positive and vacuum pressure only for pressures below 200 psi.
Working Principle
The compound pressure gauge consists of a sensor that is capable of measuring both positive as well as negative vacuum pressures. The zero pointer of the instrument is referenced at ambient pressure. The gauge consists of a vent hole which allows to compensate for the changes in atmospheric pressure.
These are pressure gauges that can measure the pressure from fluid and provide direct reading of the pressure measurement unlike Analog pressure gauges which require an operator to manually read the positioning of the needle in the dial for the respective pressure reading.
Psi in a pressure gauge is pounds per square inch which is the unit for the measured pressure. It is the pressure exerted by one pound force over an area of one square inch
A supply pressure gauge helps in determining the amount of air or water or fuel in a tank. Air brake vehicles are usually provided with a supply pressure gauge to measure the amount of air in the tank. For vehicles with dual air brake system, there is a pressure gauge for every half section of the system
Column of water is sometimes used for measuring pressure. A non- SI unit for measuring pressure is inch of water and can be defined as the pressure that a column of water that is 1 inch height exerts under standard conditions.
A pressure canner is a vessel that is fitted with a lid that has a dial or weighted gauge that regulated the steam that builds up inside. The steam that is build up inside the vessel is released when the pressure exceeds the limit the vessel can handle. Further, the steam that is build up inside is hotter than boiling water. Dial gauge regulators are found in older types of pressure canners. The dial displays the exact pressure build up inside the canner.
A pressure canner is a vessel that is fitted with a lid that has a dial or weighted gauge that regulated the steam that builds up inside. The steam that is build up inside the vessel is released when the pressure exceeds the limit the vessel can handle. Further, the steam that is build up inside is hotter than boiling water. Weighted gauge regulators are made up of disc like pieces that must be placed on the vent pipe with preferred choice of weight and like the one-piece regulator, this regulator makes a rocking sound.
This type of pressure gauge is used as a diagnostic tool to ensure that the fuel pressure in the engine is maintained and is running at good performance levels. They also help to prevent any kind of damage that might occur due to pressure build up on the fuel pump or on the injector
This pressure gauge is used for measuring absolute pressure of the fuel-air mixture contained in the intake manifold. The diaphragm in the manifold pressure gauge is used for measuring the absolute pressure. The accurate power configuration and settings for an aircraft engine is obtained using the manifold pressure.
The normal oil pressure gauge reading when an engine is running should be between 25 and 65 psi. When the pressure gauge reading is higher than 80 psi, then there is problem of high oil pressure which needs to be dealt with.
Photohelic pressure gauge is a Magnehelic pressure gauge equipped with a switch to adjust between high and low gas pressures. It is an advanced version of Magnehelic pressure gauge that helps in saving money with reduced usage of compressed air and provides a longer life for the pressure gauge.
High pressure required for waterjet cutting is smoothened out by using pressure gauge snubber. These high-pressure fluctuations are created by reciprocating pumps and controlling these fluctuations help in extending the life of the pressure gauge and reducing the calibration time. These gauges are preferred over a valve due to their small orifice which reduces the cases of clogging.
A pressure gauge snubber consists of a pressure vessel with a capillary that has a small bore. The pressure is accumulated in the gauge and the built-up pressure is smoothened out thereby reducing the fluctuations. The gauge is equipped with a steel filter at the inlet to the capillary to avoid dirt from entering or clogging the bore.
This transducer converts pressure into an electrical signal. The principle behind the working of a strain gauge pressure transducer is piezo resistance i.e., the change in resistance value with respect to the physical deformation or changes caused to the material when exerted by pressure. This transducer is when wired to a Wheatstone bridge can convert small changes in resistance to electrical signals corresponding to the pressure exerted.
The accuracy class in pressure gauges help in determining the permissible percentage of error. The accuracy classes for pressure gauges are 0.1, 0.25. 0.6, 1, 1.6, 2.5 and 4. The gauges with pointer stop are the range of 10 to 100%.
Pressure gauges have removable rings which are termed as bayonet ring. A bayonet is an indentation on the outer surface of the ring. Usually, a bayonet ring has up to five indentations. The rings help in holding a gasket and window. The dial can be found on removing the gasket and window. This ring is mostly seen in pressure gauges where the operator must access the adjustable pointer.
Thin walled cylinder with convolutions and metal as material of construction, are Bellow pressure gauge they are closed at one end while the other end is open and can move about. On applying pressure to the sealed end, the bellows will compress and move upwards. The rod in between the bellows and the transmission system will also move up and initiate the movement of the pointer. They can provide longer stroke length and exert greater forces. These bellows are fabricated using different materials depending on the application.
The deflection that is produced can be expressed as below
In case of high pressure, the entire disc will be blown and break into piece to release the build-up pressure. To protect the gauge from breaking or from being blown up, pressure gauge blows out protection is provided. An advisable design for protecting the pressure gauge from the over pressure is by separating the front and back part of the pressure gauge using a solid wall. Using such a design, the front part will not be affected though the back part will blow out thereby providing protection to the pressure gauge.
A differential pressure gauge helps in measuring the differences in two measured pressure. They are usually used for measuring pressure levels in closed tanks, over pressure in room and for controlling pump stations.
A pressure element divides the two chambers in a differential pressure gauge. If the pressures in the two chambers are the same, then there occurs no difference in the pressure element. On the other hand, if there exists difference in pressure between the two chambers, then the pressure element displaces, and mechanical movement indicates the pressure difference value.
The Brayton Cycle, also known as the gas turbine cycle, is a thermodynamic cycle used in power generation and jet engines. It is named after George Brayton, an American engineer who patented the first version of the cycle in 1872. The Brayton Cycle is widely used in gas turbines, which are commonly found in aircraft engines, power plants, and even some automobiles.
Definition of the Brayton Cycle
The Brayton Cycle is a closed-loop thermodynamic cycle that consists of four main components: a compressor, a combustion chamber, a turbine, and a heat exchanger. It operates on the principles of the ideal gas law and follows a series of processes to convert thermal energy into mechanical work.
The cycle begins with the compressor, which takes in ambient air and compresses it to a higher pressure. This compressed air then enters the combustion chamber, where fuel is injected and ignited. The resulting high-temperature and high-pressure gases expand, driving the turbine. The turbine extracts energy from the expanding gases, converting it into mechanical work to drive the compressor and any external load, such as an aircraft’s propeller or a power generator.
The exhaust gases from the turbine then pass through a heat exchanger, where they transfer some of their heat to the incoming air before being expelled to the atmosphere. This heat exchange process increases the overall efficiency of the cycle by preheating the air before it enters the combustion chamber.
Diagram of the Brayton Cycle
To better understand the Brayton Cycle, let’s take a look at a simplified diagram of the cycle:
As shown in the diagram, the cycle consists of four main processes:
Process 1-2 (Isentropic Compression): The compressor takes in ambient air at point 1 and compresses it to a higher pressure at point 2. This process is isentropic, meaning there is no heat transfer or change in entropy.
Process 2-3 (Constant Pressure Heat Addition): The compressed air enters the combustion chamber, where fuel is injected and ignited. This process occurs at constant pressure, resulting in a significant increase in temperature.
Process 3-4 (Isentropic Expansion):The high-temperature and high-pressure gases from the combustion chamber expand through the turbine, driving it and producing mechanical work. This expansion process is also isentropic.
Process 4-1 (Constant Pressure Heat Rejection): The exhaust gases from the turbine pass through a heat exchanger, where they transfer some of their heat to the incoming air. This process occurs at constant pressure, reducing the temperature of the exhaust gases before they are expelled to the atmosphere.
P-V (Pressure-Volume) and T-S (Temperature-Entropy) diagrams are commonly used to visualize the Brayton Cycle. These diagrams provide a graphical representation of the cycle’s processes and help in analyzing its performance.
In the P-V diagram, the vertical axis represents pressure, while the horizontal axis represents volume. The cycle’s processes are represented by lines on the diagram, allowing us to see how pressure and volume change throughout the cycle.
On the other hand, the T-S diagram plots temperature against entropy. It helps us understand the heat transfer and energy exchange that occur during the cycle. The T-S diagram shows the cycle’s processes as curves, allowing us to analyze the changes in temperature and entropy.
Both diagrams provide valuable insights into the performance of the Brayton Cycle, allowing engineers to optimize its efficiency and power output.
In the next sections, we will explore the important relationships within the Brayton Cycle and answer some frequently asked questions about this thermodynamic cycle.
Steps of the Brayton Cycle
The Brayton Cycle is a thermodynamic cycle that is commonly used in gas turbine engines and power generation systems. It consists of four main processes that work together to produce power efficiently. Let’s take a closer look at each step of the Brayton Cycle.
Process 1-2: Reversible Adiabatic Compression
In this first step of the Brayton Cycle, the air is drawn into the compressor, where it is compressed to a higher pressure. The compression process is adiabatic, meaning that no heat is added or removed from the system. As the air is compressed, its temperature increases. This step is crucial as it prepares the air for the subsequent combustion process.
Process 2-3: Constant Pressure Heat Addition
After the air is compressed, it enters the combustion chamber, where fuel is injected and ignited. The high-pressure air from the compressor mixes with the fuel, and combustion occurs. This process is carried out at a constant pressure, allowing for efficient heat transfer from the combustion products to the working fluid. As a result, the temperature and pressure of the working fluid increase significantly.
Process 3-4: Reversible Adiabatic Expansion
Once the air-fuel mixture has undergone combustion and reached its maximum temperature, it enters the turbine. In the turbine, the high-pressure, high-temperature gases expand, driving the turbine blades and producing useful work. The expansion process is adiabatic, meaning that no heat is added or removed from the system. As the gases expand, their temperature and pressure decrease.
Process 4-1: Constant Pressure Heat Rejection
In the final step of the Brayton Cycle, the low-pressure gases from the turbine enter the heat exchanger, where heat is rejected to the surroundings. This process occurs at a constant pressure, allowing for efficient heat transfer. As the gases cool down, their temperature and pressure decrease further, preparing them to re-enter the compressor and start the cycle again.
By following these four processes, the Brayton Cycle can continuously produce power in a gas turbine engine or power generation system. The cycle is highly efficient, as it maximizes the conversion of heat energy into useful work. The thermal efficiency of the Brayton Cycle can be improved by increasing the pressure ratio and temperature ratio, which can be achieved through design modifications and advanced technologies.
In summary, the Brayton Cycle is a fundamental thermodynamic cycle used in gas turbine engines and power generation systems. It consists of four main processes: reversible adiabatic compression, constant pressure heat addition, reversible adiabatic expansion, and constant pressure heat rejection. Each step plays a crucial role in the overall efficiency of the cycle, allowing for the continuous production of power.
Brayton Cycle Refrigeration
Introduction to Brayton Refrigeration Cycle
The Brayton Cycle, also known as the gas turbine cycle, is a thermodynamic cycle that is widely used in power generation, jet engines, and gas turbines. It consists of four main components: compressor, combustion chamber, turbine, and heat exchanger. The cycle operates on the principle of converting thermal energy into mechanical work.
In the context of refrigeration, the Brayton Cycle can be modified to create a refrigeration cycle known as the Brayton Refrigeration Cycle. This cycle utilizes the same components as the traditional Brayton Cycle but with a different configuration. Instead of producing work output, the goal of the Brayton Refrigeration Cycle is to remove heat from a low-temperature reservoir and reject it to a high-temperature reservoir.
The Brayton Refrigeration Cycle is commonly used in cryogenic applications, such as liquefaction of gases and air separation. It offers several advantages over other refrigeration cycles, including high efficiency, compact size, and the ability to achieve very low temperatures.
Inverted Brayton Cycle
The Inverted Brayton Cycle, also known as the Brayton Heat Pump Cycle, is a variation of the traditional Brayton Cycle. In this cycle, the roles of the hot and cold reservoirs are reversed compared to the Brayton Refrigeration Cycle. The goal of the Inverted Brayton Cycle is to absorb heat from a low-temperature reservoir and reject it to a high-temperature reservoir, thus providing heating instead of cooling.
The Inverted Brayton Cycle finds applications in heat pumps, where it can be used for space heating, water heating, and industrial processes. It offers advantages such as high efficiency, low operating costs, and the ability to provide both heating and cooling.
The Joule Brayton Cycle, also known as the simple Brayton Cycle, is the basic form of the Brayton Cycle. It operates on the principle of constant pressure combustion and is commonly used in gas turbine engines. The cycle consists of a compressor, combustion chamber, turbine, and heat exchanger.
In the Joule Brayton Cycle, air is compressed by the compressor, then heated in the combustion chamber where fuel is burned, resulting in a high-temperature, high-pressure gas. This gas expands through the turbine, producing work output, and then passes through the heat exchanger to reject heat to the surroundings. The cycle is then repeated.
The Joule Brayton Cycle is widely used in power generation, where it converts the energy of a fuel into mechanical work to drive a generator. It offers high thermal efficiency and is capable of generating large amounts of power.
Reverse Brayton Cycle
The Reverse Brayton Cycle, also known as the Brayton Cryocooler Cycle, is a modification of the traditional Brayton Cycle that is used for cryogenic cooling applications. In this cycle, the roles of the hot and cold reservoirs are reversed compared to the Brayton Refrigeration Cycle. The goal of the Reverse Brayton Cycle is to absorb heat from a high-temperature reservoir and reject it to a low-temperature reservoir, thus achieving cryogenic temperatures.
The Reverse Brayton Cycle finds applications in cryogenic systems, such as cooling of superconducting magnets, infrared detectors, and medical imaging devices. It offers advantages such as high cooling capacity, compact size, and the ability to achieve very low temperatures.
In conclusion, the Brayton Cycle and its variations play a crucial role in various industries, including power generation, refrigeration, heating, and cryogenics. Each variation of the cycle offers unique advantages and is tailored to specific applications. Understanding the principles and applications of the Brayton Cycle is essential for engineers and researchers working in these fields.
The Brayton Cycle and Rankine Cycle are two thermodynamic cycles commonly used in power generation and propulsion systems. While both cycles involve the conversion of heat into work, they differ in several aspects.
Brayton Cycle
Rankine Cycle
Used in gas turbine engines and jet engines
Used in steam power plants
Operates on an open cycle
Operates on a closed cycle
Uses a compressor, combustion chamber, and turbine
Uses a pump, boiler, and turbine
Utilizes a gas as the working fluid
Utilizes a liquid (usually water) as the working fluid
Higher thermal efficiency
Lower thermal efficiency
Higher power-to-weight ratio
Lower power-to-weight ratio
Differences in Heat Addition and Rejection
One of the key differences between the Brayton Cycle and Rankine Cycle lies in the way heat is added and rejected. In the Brayton Cycle, heat addition occurs in the combustion chamber, where fuel is burned, and the resulting high-temperature gases expand through the turbine, producing work. The heat rejection takes place in the heat exchanger, where the exhaust gases transfer their heat to the surroundings.
On the other hand, the Rankine Cycle involves heat addition in the boiler, where the working fluid is heated by the combustion of fuel. The high-pressure liquid then expands through the turbine, generating work. Heat rejection occurs in the condenser, where the working fluid is cooled and condensed back into a liquid state.
Handling of Low-Pressure Gas
Another notable difference between the Brayton Cycle and Rankine Cycle is the handling of low-pressure gas. In the Brayton Cycle, the low-pressure gas is discharged directly into the atmosphere after passing through the turbine. This open cycle allows for continuous operation without the need for a condenser.
In contrast, the Rankine Cycle is a closed cycle, which means the low-pressure liquid is pumped back to the boiler to be reheated and undergo the cycle again. This closed-loop system requires the use of a condenser to cool and condense the working fluid back into a liquid state before it is pumped back to the boiler.
Overall, while both the Brayton Cycle and Rankine Cycle are thermodynamic cycles used for power generation, they differ in terms of their applications, working fluids, heat addition and rejection processes, and handling of low-pressure gas. Understanding these differences is crucial in designing and optimizing power generation systems and propulsion systems for various applications.
Brayton Cycle Explained
The Brayton cycle, also known as the gas turbine cycle, is a thermodynamic cycle used in power generation and jet engines. It consists of four main components: a compressor, a combustion chamber, a turbine, and a heat exchanger. In this section, we will explore the different aspects of the Brayton cycle, including its ideal form, derivation and analysis, regeneration, and modifications for actual applications.
The ideal Brayton cycle is a theoretical model that assumes perfect conditions and no losses. It consists of two reversible adiabatic processes and two isobaric processes. The cycle starts with the compression of air by the compressor, followed by the addition of heat in the combustion chamber. The high-pressure and high-temperature gases then expand through the turbine, producing work output. Finally, the gases are cooled in the heat exchanger before returning to the compressor.
The thermal efficiency of the ideal Brayton cycle can be calculated using the temperature and pressure ratios. The temperature ratio, denoted by T3/T2, represents the ratio of the turbine inlet temperature to the compressor inlet temperature. The pressure ratio, denoted by P3/P2, represents the ratio of the turbine inlet pressure to the compressor inlet pressure. The thermal efficiency, denoted by ηth, is given by the formula:
ηth = 1 – (1 / (P3/P2)((γ-1)/γ))
where γ is the specific heat ratio of the working fluid.
Derivation and Analysis of Brayton Cycle
To derive the Brayton cycle, we consider the first law of thermodynamics and apply it to each component of the cycle. By assuming ideal gas behavior and neglecting kinetic and potential energy changes, we can derive the expressions for work and heat transfer in each process. This allows us to analyze the performance of the cycle and calculate important parameters such as the work output and heat input.
The analysis of the Brayton cycle involves evaluating the net work output, thermal efficiency, and specific work output. These parameters depend on the pressure ratio, temperature ratio, and specific heat ratio of the working fluid. By varying these ratios, we can optimize the cycle for different applications, such as power generation or aircraft propulsion.
Regeneration is a technique used to improve the thermal efficiency of the Brayton cycle. It involves recovering some of the waste heat from the exhaust gases and using it to preheat the compressed air before it enters the combustion chamber. This reduces the amount of fuel required to reach the desired turbine inlet temperature, resulting in higher thermal efficiency.
In a regenerative Brayton cycle, a heat exchanger, known as a regenerator, is placed between the compressor and the combustion chamber. The regenerator transfers heat from the hot exhaust gases to the cold compressed air, increasing its temperature. This preheated air then enters the combustion chamber, where fuel is added and combustion occurs. The rest of the cycle remains the same as the ideal Brayton cycle.
Actual Brayton Cycle and Efficiency Modifications
In real-world applications, the Brayton cycle deviates from the ideal model due to various losses and inefficiencies. These include pressure losses in the compressor and turbine, heat losses to the surroundings, and combustion inefficiencies. To account for these factors, modifications are made to the ideal Brayton cycle to improve its efficiency and performance.
One common modification is the use of intercooling and reheating. Intercooling involves cooling the compressed air between stages of the compressor, reducing its temperature and increasing its density. Reheating, on the other hand, involves adding heat to the gases between stages of the turbine, increasing their temperature and expanding them further. These modifications help to mitigate the effects of irreversibilities and improve the overall efficiency of the cycle.
Another modification is the inclusion of a bypass system, commonly used in aircraft engines. This allows a portion of the compressed air to bypass the combustion chamber and directly mix with the exhaust gases, reducing fuel consumption and increasing thrust.
In conclusion, the Brayton cycle is a fundamental thermodynamic cycle used in gas turbines and jet engines. Understanding its ideal form, derivation, regeneration, and modifications is crucial for optimizing its performance and efficiency in various applications. By continuously improving and refining the Brayton cycle, engineers can enhance power generation, propulsion systems, and other industrial processes.
Frequently Asked Questions (FAQ) about the Brayton Cycle
How to Increase Efficiency of Brayton Cycle
The efficiency of the Brayton cycle, also known as the gas turbine cycle, can be improved by implementing certain measures. Here are some ways to increase the efficiency of the Brayton cycle:
Increasing the Pressure Ratio: The efficiency of the Brayton cycle is directly proportional to the pressure ratio. By increasing the pressure ratio, the cycle can extract more work from the same amount of heat input, resulting in higher efficiency.
Increasing the Temperature Ratio: Similar to the pressure ratio, increasing the temperature ratio also improves the efficiency of the Brayton cycle. This can be achieved by using more efficient combustion techniques or by utilizing advanced materials that can withstand higher temperatures.
Utilizing Regenerative Heating: In a regenerative Brayton cycle, a heat exchanger is used to preheat the compressed air before it enters the combustion chamber. This reduces the amount of heat required in the combustion process, resulting in improved efficiency.
Optimizing the Compressor and Turbine Design: The efficiency of the compressor and turbine plays a crucial role in the overall efficiency of the Brayton cycle. By optimizing the design and using advanced materials, the losses in these components can be minimized, leading to higher efficiency.
Application of Brayton Cycle
The Brayton cycle finds its application in various fields, including power generation and jet engines. Here are some key applications of the Brayton cycle:
Gas Turbines: Gas turbines are widely used in power generation, aviation, and industrial applications. The Brayton cycle forms the basis of gas turbine engines, where the combustion of fuel produces high-temperature gases that drive the turbine, generating power or thrust.
Jet Engines: Jet engines, commonly used in aircraft, also operate on the Brayton cycle. The incoming air is compressed, mixed with fuel, and ignited in the combustion chamber. The resulting high-velocity exhaust gases propel the aircraft forward, providing thrust.
Power Generation: Gas turbine power plants utilize the Brayton cycle to generate electricity. The combustion of fuel in the gas turbine produces high-pressure and high-temperature gases that drive the turbine, which is connected to a generator, converting mechanical energy into electrical energy.
Brayton Cycle Problems and Solutions
While the Brayton cycle offers numerous advantages, it also presents some challenges. Here are some common problems encountered in the Brayton cycle and their solutions:
Compressor Surge: Compressor surge occurs when the flow rate through the compressor decreases abruptly, leading to a disruption in the cycle’s operation. To prevent compressor surge, anti-surge control systems are employed, which regulate the flow and maintain stable compressor operation.
Combustion Instability: Combustion instability can cause fluctuations in the flame, leading to reduced efficiency and increased emissions. Advanced combustion techniques, such as lean premixed combustion, are employed to mitigate combustion instability and improve overall performance.
Heat Exchanger Fouling: Fouling of the heat exchanger surfaces can reduce the efficiency of the Brayton cycle. Regular maintenance and cleaning of the heat exchanger surfaces help prevent fouling and ensure optimal heat transfer.
Power Calculation and Compressor Efficiency
Calculating the power output and compressor efficiency is essential to assess the performance of the Brayton cycle. Here’s how these parameters are determined:
Power Calculation: The power output of the Brayton cycle can be calculated using the equation: Power Output = Mass Flow Rate * Specific Work Output. The mass flow rate is the rate at which air passes through the cycle, and the specific work output is the work done by the turbine per unit mass of air.
Compressor Efficiency: Compressor efficiency is a measure of how effectively the compressor compresses the air. It is calculated as the ratio of the actual work done by the compressor to the ideal work done. Compressor efficiency can be improved by optimizing the compressor design and reducing losses.
Comparison of Simple and Regenerative Brayton Cycles
The Brayton cycle can be implemented in two configurations: simple and regenerative. Here’s a comparison between the two:
Parameter
Simple Brayton Cycle
Regenerative Brayton Cycle
Heat Exchanger
Not present
Present
Preheating of Compressed Air
Not applicable
Achieved through a heat exchanger
Efficiency
Lower efficiency compared to regenerative cycle
Higher efficiency due to preheating of compressed air
The Brayton cycle forms the basis of gas turbine engines used in power generation and aviation. Here’s how the Brayton cycle is implemented in gas turbines:
Compressor: The incoming air is compressed by the compressor, increasing its pressure and temperature.
Combustion Chamber: The compressed air is mixed with fuel and ignited in the combustion chamber, resulting in the release of high-temperature gases.
Turbine: The high-temperature gases expand through the turbine, driving its blades and extracting work to generate power or thrust.
Exhaust: The exhaust gases, after passing through the turbine, are expelled into the atmosphere, completing the Brayton cycle.
Gas turbines offer high power-to-weight ratios, making them suitable for applications where weight and size are critical factors, such as aircraft propulsion and mobile power generation.
In conclusion, the Brayton cycle, with its various applications and potential for efficiency improvements, plays a vital role in power generation and aviation. Understanding the key concepts, challenges, and solutions related to the Brayton cycle is essential for optimizing its performance and exploring future advancements in this thermodynamic cycle.
Frequently Asked Questions
Q: What is the Brayton cycle?
A: The Brayton cycle, also known as the gas turbine cycle, is a thermodynamic cycle used in power generation and jet engines. It consists of four main components: a compressor, combustion chamber, turbine, and heat exchanger.
Q: What are the steps involved in the Brayton cycle?
A: The Brayton cycle involves four steps: compression, combustion, expansion, and exhaust. During compression, the air is compressed by the compressor. In the combustion step, fuel is added and ignited in the combustion chamber. Expansion occurs as the high-pressure gas passes through the turbine, generating work output. Finally, the exhaust step involves releasing the remaining gas to the environment.
Q: How does the Brayton cycle work in refrigeration?
A: The Brayton cycle can be used in refrigeration systems by reversing the direction of heat transfer. Instead of generating power, the cycle absorbs heat from a low-temperature source and rejects it to a high-temperature sink, providing cooling.
Q: Why is 1 Decembrie not considered in the FAQ terms?
A: The term“why not 1 Decembrie” is not relevant to the topics of Brayton cycle, gas turbine cycle, or power generation. Therefore, it is not included in the FAQ terms.
Q: What is the difference between the Brayton cycle and the Rankine cycle?
A: The Brayton cycle is an open cycle used in gas turbines, while the Rankine cycle is a closed cycle used in steam power plants. The Brayton cycle uses air or gas as the working fluid, while the Rankine cycle uses water or steam.
Q: What are the working principles of the Brayton cycle?
A: The working principles of the Brayton cycle involve compressing the working fluid, adding heat through combustion, expanding the fluid to generate work, and then exhausting the fluid. This cycle enables the conversion of thermal energy into mechanical work.
Q: Can you explain the Brayton cycle in more detail?
A: Certainly! The Brayton cycle starts with the compression of air by a compressor, increasing its pressure and temperature. The compressed air then enters the combustion chamber, where fuel is added and ignited, resulting in a high-temperature gas. This gas expands through the turbine, producing work output. Finally, the exhaust gas is released, and the cycle repeats.
Q: What is the role of gas turbines in the Brayton cycle?
A: Gas turbines are the key components of the Brayton cycle. They consist of a compressor, combustion chamber, and turbine. The compressor compresses the air, the combustion chamber adds fuel and ignites it, and the turbine extracts work from the expanding gas.
Q: How does the pressure ratio affect the Brayton cycle?
A: The pressure ratio, defined as the ratio of the compressor outlet pressure to the inlet pressure, affects the performance of the Brayton cycle. A higher pressure ratio leads to increased thermal efficiency and work output, but it also requires a more robust and efficient compressor.
Q: How is the thermal efficiency of the Brayton cycle calculated?
A: The thermal efficiency of the Brayton cycle is calculated as the ratio of the net work output to the heat input. It can be expressed as the difference between the compressor and turbine work divided by the heat input from the combustion chamber.
What are the examples of Quasistatic processes in our daily lives ?
Why a reversible process is necessarily a Quasi static process ?
Since the pressure is uniform in the quasi static process, how can there be any work done?
Quasi-static process definition
It can be defined in simple words process happening very slowly, and all state passed by this process is in equilibrium.
The meaning of the word “Quasi” is almost. The static means the thermal properties are constant concerning time. All the reversible processes are quasi. The slow rate of the process is the main characteristic of the this process.
Non quasi-static process
It is not realized for any finite difference of the system. Most of the processes happening around us (in nature) can be termed as non quasi-static process.
It helps analyze. It is primarily studied in books and references. We already know the introductory study of thermodynamic starts with quasi processes. We can readily notice work PdV in this diagram. The curve of non quasi looks half-circle type. The quasi-static method is represented by a straight line.
Difference between quasi-static and reversible process
We can define a reversible process as if the system restores its initial or starting stage and there is no effect of the process on the surrounding.
In a reversible process, the process follows the same path in the forward and reverse functions. There is no impact of the system on the surrounding. Ideally, this type of process can not be possible due to friction.
It can be defined in simple words process happening very slowly, and all state passed by this process is in equilibrium.
There is no friction present in this process. So, we can say that ideally, the processes are reversible.
There is no entropy generation in both processes. We can make any process reversible if we continue the process at a prolonged rate.
Example of quasi-static process
We can consider the static compression process as an example of the quasi-static process. In this process, the system’s volume will change very slowly, but the pressure of the system remains throughout the process.
Compression process with cylinder and piston is shown in figure below,
Compression process of quasi-static process
Characteristics of quasi-static process
It is a thermodynamic process where the process occurs at a very slow rate. We can say the process occurs at near to rest condition.
Every point or stage in the this process is considered in equilibrium conditions.
We can say that control on the quasi process is effortless. In the non quasi-static process, the control can be challenging compared to ideal quasi. The reason behind it is the speed of the process.
It is a thermodynamic process in which the time taken for the complete process will be infinite.
It is highly efficient as there is no loss in this process. There is no friction or heat generation due to friction. In the case of a non quasi process, friction is present, which is ultimately loss so less efficient than quasi.
This process is reversible in nature.
Device working on quasi-static process produce maximum work
Common quasistatic processes
Ideally, the quasi reversible process can not possible practically. There is always some loss in any system. With some assumptions, we can consider some processes as quasi processes.
Ideal Gas processes at a slow rate.
Compression process at a prolonged rate
Reversible Processes.
Growth of tree
Huge Temperature Reservoir
Condition for an ideal gas in a quasi-static adiabatic process
If we consider the quasi adiabatic process, there is some condition to be satisfied. If ideal gas is compressed from state 1 to state 2, then
P1 and V1 Is the initial condition of the system,
P2 and V2 Is the final condition of the system,
The condition for the system is,
We can write this condition for both condition as below,
Conditions for quasi-static process
It is a thermodynamic process where the process occurs at a very slow rate. We can say the process occurs at near to rest condition.
Every point or stage in this process is considered in equilibrium conditions.
We can say that control on this process is very easy. In the non quasi-static process, the control can be challenging compared to quasi. The reason behind it is the speed of the process.
It is a thermodynamic process in which the time taken for the complete process will be infinite.
This process is highly efficient as there is no loss. There is no friction or heat generation due to friction. In the case of a non quasi process, friction is present, which is ultimately loss so less efficient than quasi.
This process is reversible in nature.
Device working on this process produce maximum work
Difference between quasi-static and non quasi-static process
It can be defined in simple words that it is the process happening very slowly, and all state passed by this process is in equilibrium.
This process is always reversible in nature.
There is no friction or loss present.
It is not realized for any finite difference of the system. Most of the processes around us (in nature) can be termed a non quasi-static process.
The non quasi process is always irreversible.
There is always friction and loss present in the system.
We can write relation for entropy generation,
Where dS denotes entropy change in system
The entropy change in the system can be positive, negative, or zero.
Heat transferred in an infinitesimal quasi-static process
Heat transfer equation for this ideal process can be written in following foam for calculation,
Here
dQ = Heat transfer
Cv= Constant volume heat capacity
n= no. of moles of substance
R= ideal gas constant
Cp= Constant pressure heat capacity
V = volume,
dV = Differential volume
P= Pressure,
dP = Differential pressure
Importance of quasi-static process
It is proposed in 1909 as a ” quasi-static process.” It is an essential process in the field of thermodynamic for analysis. It is providing maximum output work in the system. Though this process is ideal, this process in the various study is vast.
In this process, the system remains in equilibrium for infinitesimal time. The reasons behind its importance in the field of engineering are
1. This process is easy for analysis
2. Any device working on this process produces maximum work. There is no loss of any energy.
Non quasi-static process example
Every process in nature is a non quasi-static process,
Those processes do not occur at a prolonged rate. You can consider any processes non quasi-static process.
Fast Heat transfer,
Fast compression,
Expansion,
Non-quasi-static cyclic process
It is not realized for any finite difference of the system. Most of the processes around us (in nature) can be termed a non quasi-static process.
We can readily notice the curve in the diagram given below. As we know, the non quasi-static process does not return with the same path. The backward process is always with a different direction to be considered a cyclic process.
Quasi-static process diagram
The diagram for both of the process shown below for the expansion process.
Quasi-static and Non Quasi-static Diagram
Quasi-static process entropy
We can write relation for entropy generation,
Where dS denotes entropy change in system
The entropy change in the system can be positive, negative, or zero.
Quasi-static process equation
It can be derived for various processes in thermodynamics. The equation for different process with the constant property is given below,
It can be stated as the force applied very slowly on the system. Due to this force, the system deforms very slowly with infinite time. This type of force can be defined as a quasi-static force.
Is reversible process quasi static?
This process is always reversible.
There is no friction or loss present.
It is not realized process for any finite difference of the system. Most of the processes around us (in nature) can be termed a non quasi-static process.
Is adiabatic process quasi static ?
An adiabatic process is a process with no heat transfer. It is also considered as an isentropic process means constant entropy of the system.
There are some conditions of the process to be quasi.
If the adiabatic process occurring at a very slow rate, then it can be considered as quasistatic adiabatic process
What are the examples of Quasi static processes in our daily lives?
It is an ideal process in nature; still, the process that occurs very slowly can be considered as quasi.
Growth of tree,
Why a reversible process is necessarily a Quasi static process?
This process is always reversible in nature.
There is no friction or loss present. There is no heat loss at all in this process
It is not realized for any finite difference of the system. Most of the processes around us (in nature) can be termed a non quasi-static process.
Since the pressure is uniform in the quasi static process, how can there be any work done ?
If the pressure is constant in any system with the this process, the work done can be given by the following equation,
A Coupling is a connecting device, which connects two rotating shafts. A coupling is used for power transmission and torque transmission. Connected at the ends of the shafts, there is the possibility of failure or slippage depending on the torque limit of the shafts.
Image credit: Occupational Safety and Health Administration part of the U.S. Department of Labor, Rotating coupling, marked as public domain, more details on Wikimedia Commons.
Applications:
⦁ The purpose of coupling is to transmit power and torque.
⦁ Transportation of shafts becomes easier by dismantle and assembling the shafts by the use of a coupling.
⦁ Connect the driving part to the driven element.
⦁ To reduce transmission shocks.
⦁ Protects the system.
There are some shafts that are manufactured separately and still can be joined together by the use of coupling:
A rigid coupling is a coupling device used to connect shafts that are perfectly aligned, or there is no misalignment in the shafts in all directions.
These type of couplings are mostly used in vertical actions in the system. Rigid couplings transmit rotational as well as an axial motion to the two connected shafts rotating at certain rpm. Rigid couplings transmit power and torque between the shafts and between the two systems only if the shafts are appropriately aligned. Example: Vertical pump.Example: electric motor.
A rigid coupling is connected from the equipment shaft to the motor shaft.
The coupling shafts transmit axial thrust.
Configurations: Split configuration: Split along the centerline. Flanged configuration: The two couplings and the flanges are bolted together. Flanged rigid couplings use adjusting plates that are utilized to set up proper vertical position. .
Rigid couplings types:
The flanged type rigid coupling The clamp-type rigid coupler The sleeve type rigid coupling
Flange rigid coupling:
Flange coupling is the device that is used to connect to shafts if both the machine shafts are properly aligned to each other. Flange coupling is used where free access is available for both the shafts.
It is mostly used coupling, and the couplings flanges and the shafts are bolted together.
Advantages of flange coupling:
Flange coupling is less expensive as compared to the other type. Less space is required for installation. Interchangeable.
Material used :
Flanged couplings are constructed using various materials, including grey cast iron, malleable iron, carbon steels and carbon steels series ranging from 1035 to 1050.
Rigid couplings can be manufactured from most metal materials. This allows Rigid couplings to be used in many applications and variable conditions. It can transmit more power.
MATERIAL AND ITS PROPERTIES:
The manufacturing process used to manufacture flanged coupling is the casting process, as it contains recess and projection. The flange coupling is commonly made from grey cast iron those which are characterized by graphite microstructure, causing a fracture to the material to appear as grey.
Cast iron is the most commonly used material due to its casting properties having less tensile strength than compressive strength. Alloys of iron contain carbon and silicon 2.5-4% and 1.3%, respectively.
Cast iron experiences less solidification shrinkage.
Silicon is corrosion resistant, and in the casting process, it leads to an increase in fluidity and offers good weldability.
Advantages of rigid coupling:
⦁ Rigid couplings can be used in complex motion systems.
⦁ Rigid coupling provides more torque between the shafts.
⦁ It is also useful for better positioning as it has High torsional stiffness.
⦁ Easy availability.
⦁ Cost-efficient.
⦁ It has precision with zero backlashes.
⦁ Rigid couplings are used for the proper alignment and rigid connection.
⦁ Easy to assemble and disassemble.
⦁ Easy for the maintenance operations
Specifications: Rigid flange couplings
⦁ Rigid couplings are stiff connected, and it do not absorb vibrations leading to the possibility of replacement of the coupling due to wear on the parts are not properly aligned. ⦁ It requires routine check-up for wear and alignment check. ⦁ Apply lubrication regularly.
Difference between rigid and flexible couplings:
Rigid coupling is the coupling device used to connect shafts, and the connection between the shafts is the rigid connection where the two shafts are closely connected, whereas, in the case of flexible coupling, the connection between the two shafts is the flexible connection.
Flexible coupling provides the connection between the shaft components, which assemble the fixed coupling with some amount of loose connection. This gives some misalignment between the shafts.
Rigid coupling gives the smooth transmission of torque between both the shafts and the components, whereas in the flexible coupling, only the metallic type flexible coupling has a large capacity than other flexible couplings, and there is the possibility of some torque loss during the operation.
Flange coupling adapter:
A flange coupling adapter connects the end of the ductile iron pipe to the flanged pipe, valve or fitting.
Design procedure for flange coupling:
Assembly of muff coupling:
A hollow cylinder is attached at the ends of both the shafts using the sunk key. The hollow cylinder is called the sleeve. Torque and power are transmitted through shafts using these hollow cylinders.
First, it is transmitted from the first shaft to the sleeve; From the sleeve, it is transmitted to the key. Then from the key, it is again transmitted to the hollow cylinder(sleeve).
It is easy to manufacture and design and difficult to assemble and disassemble. The material used: Cast iron The factor of safety =6-8 (on the ultimate strength) It is required to have more axial space and less radial space dimensions.
Sleeve standards :
Sleeve outer dia. D = 2d + 13 Length of the sleeve, L = 3.5d d= diameter of the shaft.
Design of Shafts:
Shaft design is based on the torsional shear stress.
For torque transmission, shear stress T is given by,
Where, T = Torque acting on shafts, J = shaft polar moment of inertia, r = d/2
Allowable shear stress=[τ] determines the dimensions of the shaft.
Sleeve Design:
D = 2d + 13 L = 3.5d,
Consider a hollow shaft, The torsional shear stress in the sleeve is calculated,
Design of Key:
Cross-section of the key selected corresponding to the shaft dia and key dimensions. cross-sections of the keys: Square and rectangular Length of the key in each shaft,
Shear and crushing stresses,
shear stress, where, w= Width of the key. h= height of the key.
Clamp coupling:
Clamp coupling is compression coupling or also called split muff coupling. Split coupling is coupling in whose sleeves are split into two halves along the plane passing through the shaft axis.
These split sleeves are attached using bolts and placed in the recesses. Assembling and disassembling is easy for the clamp coupling. Clamp coupling balancing is difficult for high speeds and shock loads.
Design of Bolts:
Bolts design is based on torque transmission.
Let [σt] = permissible tensile stress,
dc = diameters of bolts, n = number of bolts
Clamping force of each bolt,
clamp force is applied equally on each shaft.
Frictional Torque,
Flange Coupling:
Flange coupling is the coupling device consisting of two flanges that are keyed to the shafts. The flanges are joined together using the bolts on a circle concentric to the shaft.
Power transmission is from the driving shaft to the flange on the driving shaft with the help of the key and from the flange on the driving shaft to the flange on the driven shaft using the key again.
For the proper alignment, projection and recess is used with the flanges Inner hub, flanges and protective circumferential flanges – Protected type flanges Flange coupling design dimensional proportions:
Flange coupling: The outer diameter of hub, D = 2 Bolts diameter, D1 = 3 d Flange diameter, D2 = 4 d Hub length L = 1.5 d tf = 0.5 d tp = 0.25 d
Design of Hub:
A hollow shaft is considered, with the inner diameter = diameter of the shafts, Outer diameter= 2* inner diameter. For torsional shear stress.
Where, T = In designing Hub required Twisting moment (or torque) J = shaft’s Polar moment of inertia ( axis of rotation) r = D/2
Design of Flange:
The hub is for the toque transmission through the bolts, The flange is subjected to the shear.
Tangential force, Shear stress,
Design of Bolts:
Let n be the total number of bolts. Force acting on each bolt, where D1 is the pitch circle diameter of bolts. Area resisting shear,
where, dc = core diameter of bolts Shear stress,
Area under crushing Crushing stress,
Bolts are subjected to both shear and crushing stress, Due to the transmission of torque, the force acts perpendicular to the bolt axes.
Types of flange coupling as follows: ⦁ Protected type flange coupling ⦁ Marine flange coupling. ⦁ Unprotected type flange coupling.
Unprotected type flange coupling:
In unprotected type flange coupling, No of bolts used= 3-6 The keys are attached at the right angle along the circumference of the shafts dividing the keyways.
Unprotected flange coupling and cast iron flange coupling dimensions:
d= diameter of the shaft, then D = 2 d Hub length, L = 1.5 d, flange, D2 = D1 + (D1 – D) = 2D The thickness of flange, tf = 0.5 d Number of bolts = 3,
Flanges are attached using the bolts.
Protected type flange coupling:
A protective circumferential rim is used. The rim covers the nut and the bolt.
It consists of the the following protective procedures:
Perform visual inspections, Check signs of wear or fatigue, Clean couplings regularly and change the lubricants regularly. Maintenance is required in operating conditions and adverse situations.
Advantages of the protective type flange coupling:
It can transmit high torque.
It is simple to construct.
Easy to assemble and disassemble
Marine flange coupling:
This is a type of coupling where the flanges are attached to the shafts using the tapered headless bolts. thickness, t=d/3, bolts, D1=1.6d, D2=2.2d,
Advantages:
It is cheap.
It is simple in structure.
More efficient.
Maintainance is not required.
Disadvantages:
1.It cannot be de-engaged in motion. 2.This type of coupling cannot transmit the torque between the shafts that are not linear.
Checking the coupling balance:
Balancing requires cost and it is difficult to balance. The amount of the coupling unbalance can be tolerated by the system. The analysis gives the detailed functions and the characteristics of the system and the connected machines.
Rigid flange couplings are less expensive than the flexible couplings. Rigid type couplings have rigid connections so they are torsional stiff and does not give access to any misalignment between the shafts. Due to the thermal effect, parts have misalignment during the operation, and both the shafts are physically aligned.
Rid couplings are couplings having rigid connections. It does not absorb vibrations leading to the replacement of the parts. Due to wear on the parts, misalignment occurs. The operators require routine maintenance and checking of the parts for wear and alignment.
Flanged pin bush couplings:
Flanged pin bush coupling is also called as bush pin type coupling.
This coupling works as protective type flange coupling with better modifications.
This coupling device has pins, and it is used to work with coupling bolts.
The material used: Rubber The rubber bushing can absorb vibrations and shocks during its operations.
Flange compression coupling:
The flange compression coupling of the coupling device.
Flange compression couplings have two cones which is used to place over the shafts.
The shafts should be coupling shafts.
The hollow cylinder is a sleeve that is used to fit over the cones.
Sleeve coupling flange:
Sleeves are attached to the shafts.
To lock the coupling in position, two threaded holes are provided.
Split flange coupling:
Split flange coupling is the coupling device the sleeves are split into two halves made up of the cast iron. These split parts are connected using mild steel bolts.
Advantages of the split flange coupling:
Easy assembling and dismantling without changing the position of the shafts. It can be used to connect two shafts of heavy transmission at moderate speed.
FAQS:
Flange coupling is what type of coupling:
Rigid type coupling.
Flange coupling specifications. Explain.
⦁ It should be easy to assemble or disassemble. ⦁ Flange coupling should transmit torque and power. ⦁ Maintain proper alignment. ⦁ Minimize the shock loads transmission.
Requirements to ensuring of the shaft alignment before attaching the fixing bolts:
⦁ If it is easy to connect or disconnect the coupling. ⦁ No projecting parts There should be less misalignment in running operation, leading to maximum power transmission.
Why is the key used in protective type flange coupling?
Keys are used to preventing rotational motion. The surfaces of the shaft and hubs parts provide cut to mount the keys, joints.
Why did recess provide in flange coupling?
To provide the clearance in the flanges, recess is provided. The flanges are tightly fitted with the use of bolts using the torque to be transmitted.
The minimum number of bolts required in flange coupling:
Four, six, or up to 12 bolt assemblies.
What is the grade of cast iron used to make rigid type flange coupling?
Bulk modulus is the ability of the material to be resistant to compression. It is the volumetric elasticity and is inversely proportional to Compressibility. The object having incompressibility deform in all directions when the load is applied from all directions.
Bulk modulus is the volumetric stress over volumetric strain.
The ratio of the increase in pressure relative to the decrease in volume.
Hydraulic fluid incompressibility is the Compressibility resistant property of the material. Hydraulic fluid gets affected by the applied pressure. As the applied pressure increases, the volume of the body decreases.
The Bulk modulus of elasticity:
The modulus of elasticity of liquid varies depending on the specific gravity and the temperature of the liquid. K is always constant within elastic limit of the material. This is the Bulk modulus of elasticity.
K=Volumetric stress/volumetric strain
The sign indicates the decrease in volume.
Volume modulus is associated with a change in volume.
Compressibility is calculated as reciprocal of incompressibility. Compressibility represented as, Compressibility=1/K SI unit: m^2/N or Pa^-1. Dimensions of Compressibility: [M^-1L^-1T^2]
Derivation of Bulk modulus of elasticity:
Bulk fluid modulus is the ratio of the change in pressure to change in volumetric strain. \\frac {-\\delta V} {V}=\\frac {\\delta P} {K} δV: change in volume δp: change in pressure V: actual volume K: volume modulus δp tends to zero V=1/density
The Bulk modulus of incompressible liquid:
The volume of incompressible fluid does not change. As the force is applied, the change in volume is zero due to the volumetric strain of the incompressible fluid is zero.
Temperature dependence:
The modulus of incompressibility evolves due to the volumetric stress evolves periodically.
It is coupled to shear modulus, Assume constant Poisson’s ratio.
The time-dependent modulus is represented as,
Elastic constants relationships:
Relationships between Poisson’s ratio, Young’s modulus and shear modulus with bulk modulus:
For an incompressible fluid, the maximum limit of poisson’s ratio be 0.5. For K to be positive μ should be always than 0.5. n = 0.5. 3G = E. K = ∞. E= 3K(1-2 μ) E= 2G(1+μ) 2G(1+μ)=3K(1-2 μ)
Distinguish between young’s modulus and bulk modulus:
Young’s modulus is related with longitudinal stress and longitudinal strain of the body.
Incompressibility is the form of volumetric stress and volumetric strain. Bulk modulus exists in solid, liquid and gas, whereas Young’s modulus exists in only solids. Young’s modulus gives the change in length of the body, whereas Bulk modulus gives the change in volume of the body.
Distinguish between Shear modulus and Bulk modulus :
Bulk modulus is the form of volumetric stress and volumetric strain. It involves the effect of the applied pressure. as the pressure increase , the volume of the body decreases. This gives the negative sign to the ratio of the stress to strain. The ratio is associated with the volume of the body. In case of shear modulus, shear modulus is the form of shearing stress and shearing strain. It involves the effect of shear stress on the body. It is the response to the deformation of the body. The ratio is associated with the shape of the body.
where, T=shear stress gamma=shear strain Incompressibility exists in solid, liquid and gas, whereas shear modulus exists in only solids.
Isentropic Bulk modulus:
Incompressibility of the body at the constant entropy is called isentropic bulk modulus. The ratio of change in applied pressure to fractional volume change in the body due to the pressure change is a form of the isentropic incompressibility.
Isothermal Bulk modulus:
When the temperature is constant throughout the incompressibility is called isothermal bulk modulus. The ratio of change in applied pressure to fractional volume change in the body due to the pressure change is a form of the isentropic incompressibility
Negative Bulk modulus:
Why negative:
Bulk modulus has a negative sign because of the decrease in volume due to an increase in pressure.
Adiabatic Bulk modulus:
Adiabatic Bulk modulus is the ratio of the pressure to change in fractional volume in the adiabatic process when there is no heat exchange with the surrounding.
Dimensional analysis is the process of solving a physical problem by reducing no relevant variables and appealing it to the dimensional homogeneity. Processing: Experimental data interpretation Solve physical problems Presentation of equations Establish relative importance Physical modelling
Bulk modulus,
P= pressure = [M L-1 T-2] V=volume= L3 dP=change in pressure= [M L-1 T-2] dV=change in volume= L3
1) A solid ball has initial volume v; it is reduced by 20% when subjected to volumetric stress of 200N/m^2.Find the Bulk modulus of the ball.
Solution: V1=v, Volumetric strain= Final volume to initial volume *100 Volumetric stress related to volumetric strain=200N/m^2 K= (volumetric stress/Volumetric strain) = (200/0.02) =10^4N/m^2
2) The initial pressure of the system is 1.0110^5Pa. The system undergoes a change in pressure to 1.16510^5Pa. Find out the incompressibility of the system.
Solution: P1=1.0110^5Pa, P2=1.16510^5Pa, At 20°c change in volume=20% Bulk modulus=-dP/(dV/V) =- (1.01×10^5−1.165×10^5)/0.1 =1.55*10^5Pa.
3)5 litres of water is compressed at 20atm.Calculate the volume change in water.
Given:
K of water =20*10^8 N/m^2
Density of mercury=13600 kg/m^3 g=9.81m/s^2
Normal atm.=75cm of mercury
Original volume=5L=510^-3 m^3 Pressure dP=20atm=207510^-2136009.8 Solution: Volumetric stress= pressure intensity=dp K = dp/(dv/v)
Change in volume=dpV/K =5*10^-6 m^3 =5 cc.
Frequently asked questions:
What is the Bulk modulus of granite? 50Gpa.
Can incompressibility be negative: No.
Bulk modulus formula speed: The speed of sound depends on the Bulk modulus and density,
Bulk modulus of air at 20 c: Density of air at 20°C =1.21kg/m^3 Speed of sound=344m/s So, K can be calculated from the above formula, K=143186.56N/m^2 Hence, K=0.14Mpa
Flexural modulus and incompressibility: Bulk modulus is the volumetric elasticity and is inversely proportional to the Compressibility. The object having incompressibility deform in all directions when the load is applied from all directions. Flexural modulus is the ability of the material to resist bending. Flexural modulus is the ratio of stress to the strain in flexural deformation.
Modulus of elasticity and incompressibility:
Modulus of elasticity is the ability of the material to resist deformation elastically when applied to external forces. Modulus of elasticity occurs under the elastic deformation region in stress-strain curve. Incompressibility is the volumetric elasticity and is inversely proportional to the Compressibility. The object having volume modulus deform in all directions when the load is applied from all directions
What material has the highest bulk modulus values? Diamond.
Why is the value of K maximum for a solid but a minimum for gases? Incompressibility is the resistance to compression of the substance. High pressure is required to compress the solid rather than compressing a gas. Hence the modulus of solid is maximum, and that of gas is low.
If Young’s modulus E is equal to the incompressibility K then what is the value of Poisson’s ratio:
K=E/3(1-2u) K=E 3(1-2u)=1 1-2u=1/3 u=1/3 So the value of Poisson’s ratio=1/3.
With increase in pressure, Does compressibility decrease or increase ?
As the pressure increases, volume of the body decreases. Decrease in volume gives rise to increase in incompressibility. Incompressibility is the ability to resist the compression of the body. So as it increases the compression of the body decreases. Hence the compressibility of the decreases.
What is the effect of the temperature increase?
As the temperature increases, The resistance to compression decreases. As the ability of compression of the body decreases, the bulk modulus decreases, leading to an increase in compressibility.
When the incompressibility of a material becomes equal to the shear modulus what would be the Poisson’s ratio:
2G(1+u)=3K(1-2u) as G=K, 2(1+u)=3(1-2u) 8u=1 u=1/8 Hence the value of Poisson’s ratio=1/8.
What will the velocity of sound in water m/s be if the volume modulus of water is 0.2*10^10 N/m 2:
c=2*10^6m/s.
To compress a liquid by 10% of its original volume, the pressure required is 2*10^5 N /m^2. What is the K (modulus of the liquid)?
Fourier’s law of conduction heat transfer can be states as below,
“The heat transfer rate passes from the material or specimen is directly proportional to the cross-sectional area (perpendicular area) from which heat is passing through, and temperature difference along the end surfaces of the material.”
Fourier’s law of heat conduction
We can write this statement mathematically as,
Where,
q = heat transfer rate in watt (W or J/s)
K = Thermal conductivity of material or specimen (W / m K)
A = Cross-sectional area from which the heat is passing through in m2
dT = Temperature difference between the hot side and cold side in K ( Kelvin )
dx = Thickness of material in m (thickness between hot side to cold side)
Most important: Here in the equation, the negative sign indicates that the heat always flows in the direction of decreasing temperature.
Fourier’s law equation
The equation of heat conduction law is as derived above. It is widely used to solve problems on heat conduction and analysis. The fundamental of the equation remains the same, but the parameters will be changed upon shape and situation of object.
Fourier’s law spherical coordinates
The heat conduction law applied to cylinder and equation is given as below,
Here, at any location the area
,
r is radius of considered cylindrical portion,
Rectangular, cylindrical and spherical coordinates Image Credit Book Cengel and Ghajar
Fourier’s law cylindrical coordinates
The heat conuction law applied to cylinder and equation is given as below,
at any location the area A = 2πrL,
r is radius of considered cylindrical portion,
Fourier’s law experiment
Conduction heat transfer is occurred by microscopic diffusion and collisions of molecules or quasi-particles inside an object because of a temperature difference. If we see microscopically, then diffusing and colliding any material includes molecules, electrons, atoms.
Typically, metals have free electrons mobility inside an object. This is the reason behind its good conductivity.
Consider two-block A and B,
Block A is very hot
Block B is cold
Experiment for Fourier’s law of heat conduction
Suppose we join these two blocks and insulate all other outer surfaces. The insulation is provided to reduce surrounding heat loss from the block. You can quickly get the idea that the heat energy will flow from hot block to cold block. The heat transfer will continue until both of the blocks attain the same temperature (temperature equilibrium).
It is one of the method of heat transfer in both blocks. It is conduction heat transfer mode. Using the equation of heat conduction law, we can calculate the heat transfer with this experiment. It is very informative and important practical to be performed in heat transfer lab ( Mechanical engineering and Chemical engineering)
Fourier’s law history
Fourier started his work to express conduction heat transfer in 1822. He has also given the concept of Fourier series and Fourier integral. He was a mathematician. His law on conduction is well known on behalf of his name, “Fourier’s law of heat conduction.”
Fourier’s law units
Fourier’s law of heat conduction is stated for heat transfer. So, we can consider the unit of heat transfer for it. The unit of heat transfer is the watt ( J/s) W.
Fourier’s law assumptions
There are some assumptions made for Fourier’s law of heat conduction. The law only applicable if following conditions will be followed and satisfied.
Fourier’s law of heat conduction example
There are many examples of law of heat conduction in day-to-day life. Some examples are discussed below.
There is hot coffee inside the mug. Now you know that heat will be transferred from the hot side to the cold side. Here, the heat transfer occurs from the inner wall to the outer wall of the mug. It is conduction heat transfer and based on Fourier’s law of heat conduction.
We can consider the wall of our house as for example.
If there is internal heat generation in the rod, heat will flow in the interior portion to outer surfaces.
You can touch any electrical and electronics equipment. You will get realize some heat. These all devices can be the example of Fourier’s law.
Fourier’s number
It is a dimensionless number derived by a non-dimensionalization heat conduction equation.
Fourier’s number is denoted by Fo
Where,
L is plate length (Diameter in case of the cylinder) in m
K is the coefficient of gradient transport
T is time in s
Fourier’s law flux
According to heat conduction law law,
The heat flux can be defined as the heat flow per unit area in unit time is directly proportional to the temperature difference between the hot and cold side (Temperature gradient.)
Heat flux
The heat flux can be defined as the heat flow per unit area in unit time is directly proportional to the temperature difference between the hot and cold side (Temperature gradient.)
Heat flux equation
The equation for heat flux is given as below,
Where,
q- is heat flux in w / m2
K is thermal conductivity in w / m K
ΔT /ΔX is a temperature gradient,
Heat flux units
The unit of heat flux is w / m2
FAQs
What is Fourier’s law
“The rate of heat transfer through the material or specimen is directly proportional to the cross-sectional area from which heat is passing through, and temperature difference along the end surfaces of the material.”
We can write this statement mathematically as,
Where,
q = heat transfer rate in watt (W or J/s)
K = Thermal conductivity of material or specimen (W / m K)
A = Cross-sectional area from which the heat is passing through in m2
dT = Temperature difference between the hot side and cold side in K ( Kelvin )
dx = Thickness of material in m (thickness between hot side to cold side)
Most important: Here in the equation, the negative sign indicates that the heat always flows in the direction of decreasing temperature.
What are the assumptions of Fourier s law of heat conduction?
There are some assumptions made for Fourier’s law of heat conduction. The law only applicable if following conditions will be followed and satisfied. Fourier’s law of heat conduction can be compared with newton’s law of cooling and fick’s law of diffusion. The assumptions are different in every law.
Conduction heat transfer will take place under steady-state conditions of an object.
The flow of heat should be unidirectional.
The temperature gradient will not be changed, and the temperature profile should be linear.
The internal heat generation should be zero.
The bounding surfaces should be adequately insulated.
The material should be homogeneous and isotropic.
What is proof of the Fourier s law of heat conduction and the negative gradient?
The proof of Fourier’s law of heat conduction is already given in topic “Fourier’s law.”
The negative gradient is used because the heat always flows in decreasing temperatures.
This question is very important for interview because interviewer always try to check your fundamental knowledge.
How does Fourier’s law of heat conduction contradict the theory of relativity?
Fourier’s law contradicts the theory of relativity due to its instantaneous heat propagation through heat diffusion. If we consider time-dependent heat diffusion with a partial differential equation, The growth of heat flux will be with relaxation time. This time is in order of 10-11. Heat propagation takes infinite time in nature. The relaxation time is negligible.
If we eliminate relaxation time, then the equation becomes Fourier’s law of heat conduction. It is violating the popular theory of Einstein (theory of relativity). The velocity of light in a vacuum is 2.998 * 108
How is the physics behind Fourier’s law different from the one behind Newtons Law of cooling
As we are already knowing, Fourier’s law is used for conduction heat transfer, and Newton’s law of cooling is used for convection heat transfer. Suppose you have a question that why two different laws are required for the heat transfer rate analysis. The reason behind it is modes of heat transfer are different from individual physics.
Conduction heat transfer is occurred by microscopic diffusion and collisions of molecules or quasi-particles inside an object because of a temperature difference. If we see microscopically, then diffusing and colliding any material includes molecules, electrons, atoms. They transfer kinetic and potential energy microscopically to each other. This energy is known as internal energy in the object. The law states conduction heat transfer is Fourier’s law.
Convection heat transfer in any object can be defined as heat transfer from one molecule to another by moving fluids or flow of fluid. Newton’s law of cooling defines convection heat transfer.
The physics used for the individual process is different. Hence, the governing law for an individual is different.
What are the similarity between Newtons law of viscosity, Fourier’s law of heat conduction, and Fick’s law of diffusion?
It is the analogy between these equations.
Fourier’s lawof Heat Conduction
It states the conduction heat transfer process. The equation can be written as below,
The equation for heat flux is given as below,
Where,
q- is heat flux in w / m2
K is thermal conductivity in w / m K
ΔT /ΔX is a temperature gradient,
Fick’s law of Diffusion
It is used to describe and state the mass transfer process. The equation for mass transfer can be written as below,
(dC/dx) is gradient of concentration
D is transport property diffusivity
Newton’s law of Viscosity
It is used for momentum transfer and widely used to study viscosity of any fluid.
Here, (du/dx) is the velocity gradient
μ is the viscosity of fluid
Thus, you can analyze three different laws straight away about these equation’s relativity.
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“Thermal stress is the stress in the material due to the temperature change and this stress will lead to plastic deformation in the material.”
Thermal stress equation | Thermal stress formula:
The stress induced due to temperature change: σ=Eα∆T It is documented that changes in temp will cause elements to enlarge or contract and if increment in the length of a uniform bar of length L and ∆L is the change f length because of its temp has been changed from T0 to T then ∆L could be represented as ∆L = αL (T – T0) where α the coefficient of thermal expansion.
Thermal stress unit:
SI unit: N/m^2
Thermal hoop stress:
Stress generated for thermal change. Let us assume a thin tire having diameter ‘d’ has fitted on to the wheel of diameter ‘D’. If the temp of the tire has been changed in such a way that the diameter of the tire is increased and it has become equal to the diameter of the wheel and if temp of the tire is decreased to original, the tire diameter tries to return to its original dimension and because of this process a stress has been generated in the tire material. This stress is an example of Thermal hoop stress. so, the temperature difference=t degree. thermal strain=D-d/d Hoop stress= e. E Hence, Hoop stress=(D-d).E/d
The objective of the thermal analysis is to study the behavior of material after applying thermal loading and thermal stress. To study heat transfer within an object or between objects and thermal analysis is utilized for temp measurement, thermal gradient, and warmth flux distributions of the body.
Types of thermal analysis:
There are two sorts of thermal analysis:
Steady-state thermal analysis:
Steady-state thermal analysis aims to seek out the temperature or heat flux distribution in structures when an equilibrium is reached.
Transient thermal analysis:
Transient thermal analysis sets bent determine the time history of how the temperature profile and other thermal quantities change with time Also, thermal expansion or contraction of engineering materials often results in thermal stress in structures, which may be examined by conducting thermal-stress analysis.
Thermal stress importance:
Thermal Stress Analysis is essential to determine the thermal stresses due to temperature changes in structures. We can proceed to
Solve Equation K. T = q ⦁ To obtain the temperature change fields initially apply the temperature change ΔT as initial strain ⦁ The stress-strain relations due to temperature change were determined by first using 1D case materials. The thermal strain (or initial strain): εo = αΔT
Case study with ANSYS Workbench:
Material: Aluminum k = 170 W/(m · K) ρ = 2800 kg/m3; c = 870 J/(kg · K) E = 70GPa; v = 0.3 α = 22 × 10–6/°C Boundary conditions: Air temperature of 28°C; h = 30 W/(m2 · °C). Steady state: q′ = 1000 W/m2 on the base. Initial conditions: Steady-state: Uniform temperature of 28°C.
Start ANSYS workbench
Create a steady-state thermal analysis system:
Add new material: provided with all given data.
Launch design modeler program.
Create body
Launch the steady-state thermal program
Generate mesh
Apply boundary conditions.
Solve and retrieve results.
Thermal analysis of water-cooled engine:
The following steps are followed after finalizing the engine specification.
Design of water-core and head-core system.
Design of liner system. (Based on its parameters like bore, stoke and thickness etc.)
Design of water pump and installation.
Cooling system design and it’s subsystems such as radiators, fans, oil-cooler design.
Aspects of thermal analysis of engine block:
Cylinder head valve bridge water velocities (design of cross section in headwater core).
Piston and valve cooling aspect analysis.
Liner cavitation analysis.
Cylinder head gasket design analysis.
Thermal stress weathering:
Thermal stress weathering is the thermal fracture is a mechanical breakdown of rock due to thermal expansion or contraction caused by the change in temperature.
Effects of thermal stresses in weld joints: Thermal stress in welding and in bonded joints:
The temperature of the body is uniformly1 increased, The normal strain of the body is, x = y = z = α(T) Here, α is the co-efficient of thermal expansion. T is the temp variation. The stress is represented as σ1 =− E =−α(T)E in a similar means, if a consistent flat plate is restrained at the sides and also subjected to a constant temp rise. σ2 =− α(T)E(1−ν) The stresses σ1, σ2 are called thermal stresses. They arise due to a natural process during a clamped or restrained member.
Thermal stress equation for cylinder| Thermal stress in thick walled cylinder:
The process of heat treatment is used to decrease the residual thermal stresses in the materials. First, the part needs to be heated at 1100-1200degree F, leading to relief of the stresses and hold it there for an hour per inch of thickness, and then left to chill in tranquil air at temperature.
Thermal Expansion:
When a solid material experiences an increase in temp or temperature difference, the volume of the structure of solid material increases, this phenomenon is acknowledged as thermal expansion and this volume increment will lead to an increase in stress of the structure.
Coefficients of Thermal Expansion:
(Linear Mean Coefficients for the Temperature Range 0–100°C):
Aluminum: 23.9(10)−6 Brass, cast: 18.7(10)−6
Carbon steel: 10.8(10)−6 Cast iron: 10.6(10)−6
Magnesium :25.2(10)−6 Nickel steel: 13.1(10)−6
Stainless steel: 17.3(10)−6 Tungsten: 4.3(10)−6
Thermal stresses in composite bars formula: Thermal stress in compound bars:
Compound bars and composite bars, when undergoing temperature change, tend to contract or expand. Generally thermal-strain is a reversible process so material will return to its actual shape when the temp also decreased to its actual value, though there are some materials that does not behave according to thermal expansion and contraction.
Bars in series:
Thermal stress and strain: Thermal stress and strain definition:
The stress produced due to change in temperature is known as thermal-stress. Thermal stress=α(t2-t1).E The strain corresponding to thermal stress is known as thermal strain. Thermal strain=α(t2-t1)
Engine, radiator, exhaust, heat exchangers, power plants, satellite design, etc.
Thermal residual stress:
Differences within the temperatures during the manufacturing and dealing environment are the most explanation for thermal (residual) stresses.
Thermally induced stress
σ=E ∆L/L
Thermal stress calculation in pipe:
Pipes expand and contract due to variable temperatures. The coefficient of thermal expansion shows the rate of thermal expansion and contraction.
Factors affecting thermal stress:
Temperature gradient.
Thermal expansion contraction.
Thermal shocks.
Thermal stress is dependent on the thermal expansion coefficient of the material and if change of temp is more, then the stress will be more too.
Modulus of Elasticity in thermal expansion:
If the bar is prevented from completely expanding within the axial direction, then the typical compressive stress-induced is σ=E ∆L/L where E is the modulus of elasticity. So the thermal stress needed is, α = –αE (T – T0) In general, in an elastic continuum, the natural process is non-uniform throughout and this is a function of time and space usually. therefore the space coordinates (x, y, z), i.e. T = T(t, x, y, z).
Limitations of thermal stress analysis:
The body into account could also be restrained from expansion or movement in some regions, and external tractions could also be applied to other regions and stress calculation under such circumstances may be quite complex and difficult to compute. This also having following case is constrained.
Thin circular disks with equal temp difference.
Long circular cylinder. (This could be hollow and solid)
Sphere having radial temperature variation. (This could be hollow and solid)
Straight beam of arbitrary cross section.
Curved beam case.
Thermal stress problems and solutions:
1) A steel rod of length 20m having temperature 10-degree Celsius. Temperature is raised to 50 degrees Celsius. Find the thermal stress produced. Given: T1=10, T2=50, l=20, α=1210^-6, E=20010^9
Thermal stress=α(t2-t1).E
=1210^-6(50-10)20010^9
=9610^6 N/m^2.
FAQ/Short Notes:
What is the effect of thermal stresses ?
This has a significant effect on the materials and can lead to fracturing, and plastic deformation depends on the temperature and material type.
Which material can be used as thermal insulator and why?
Cellulose. Because it blocks air better than fiberglass and has low thermal conductivity.
What are the three most common types of heat stress ?
Types of heat stress commonly used:
Tangential
radial
axial.
How to calculate thermal stresses in glass?
Thermal stress in glass is varied at different temperatures.
Thermal stress and deformation:
Thermal deformation is the property of a substance to expand with heating and contract with cooling, normally kind of deformation because of temp change and this is stated by linear expansion coefficient α. α=ΔL/L×Δt Here, ⦁ α is that the linear expansion coefficient of a substance (1/K). ⦁ ΔL is that the expansion or contraction value of a specimen(mm). ⦁ L is the actual length. ⦁ Δt is the temperature difference measured in Kelvin or degree Celsius. The higher the thermal expansion coefficient, gives higher the value of thermal deformation.
Thermal stress weathering:
Thermal stress weathering is the thermal fracture a, mechanical breakdown of rock due to thermal expansion or contraction caused by the change in temperature.
What is the formula for thermal expansion stress and strain?
Thermal stress formula:
α(t2-t1). E
Thermal strain formula:
α(t2-t1).
What is the relationship between thermal stress and thermal strain?
Thermal stress and Thermal strain in 2D-3D cases: Temperature changes do not yield shear strains. In both 2-D and 3-D cases, the entire strain is often given by the following vector equation: ε = εe + εo And the stress-strain relation is given by σ = Eεe = E(ε − εo).
Which parameters have to be defined for isotropic materials for structural and thermal analysis in ANSYS?
Isotropic Thermal Conductivity
Material
Heat transfer coefficient
If strain causes stress, then in free thermal expansion why is stress absent even though there is thermal strain:
Stress is the internal resistance when applied to an external load. When the material is undergone any load or force, the material tries to resist the force leading to stress generation. If the material is undergoing free thermal expansion, the material won’t experience any internal stress leading to no stress generation.
What are some examples of thermal expansion in everyday life?
What is the application of thermal diffusivity in real world?
⦁ Insulation.
Does Hooke’s law fails in case of thermal expansion?
Hook’s law applies to a thermal expansion only when there is a restriction to the object undergoing thermal stress. If there is no applied stress, there won’t be any expansion and Hook’s law states that stress is directly proportionate to strain.
Why does copper have such a low thermal expansion ?
If the coefficient of thermal expansion is nearly equal for both steel and concrete, then why is a concrete structure considered a better firefighter If the coefficient of thermal expansion is almost equal for both steel and concrete, then why is a concrete structure considered a better firefighter: A concrete structure has low thermal conductivity and does not heat up quickly. Hence If the coefficient of thermal expansion is nearly equal for both steel and concrete, then why is a concrete structure considered a better firefighter.
Why we do static structure buckling modal thermal nonlinear fatigue based on stress and strain in Ansys?
It is a finite element method. To predict the exact and accurate strength of the structures, nonlinear analysis is performed. It takes into the changes in the parameters as the load is applied.
What does thermal capacity mean?
The thermal capacity of the material is the amount of heat required to change the material temp by unit mass of material.
What is the difference between the thermal expansion coefficients of steel and copper?
Thermal expansion coefficients 20 °C (x10−6 K−1) copper=17 steel=11-13.
What is the use of thermal conductivity?
Thermal conductivity is that the ability of an object to conduct heat. It measures the amount of heat that transfers through the material.
Do any materials have a zero thermal expansion coefficient?
There exist few materials which have zero thermal expansion coefficient. Mesopores.
Hooke’s law| Hooke’s law for thermal stress:
σth = Eϵth If the material is undergoing free thermal expansion, the fabric won’t experience any internal stress leading to no stress generation.
What is thermal shrinkage in concrete:
When the hot concrete cooled down at ambient temperature, the volume of the concrete reduces; this process is called thermal contraction or thermal shrinkage in concrete.
What is the best simulation and analysis software for mechanical engineering mainly structural analysis and dynamic analysis thermal not required?
Ansys, Nasttan, Abaqus, 1-deas NX, etc.
Thermal stress-strain: Why the bar does not bend when it is heated from the bottom with one end alone fixed:
Thermal stresses in cantilever beams:
Case1: Fixed free bar: If a rod is heated through temp rise, the rod will tend to expand by amount εo=αLΔT, if the rod is free at other ends, undergoes thermal expansion ε=αΔT, ε = εo, εe = 0, σ =E(ε- εo)=E(αΔT- αΔT)= 0 That is, there is no thermal stress in this case.
Case2: Fixed-fixed bar If there’s a constraint on the right-hand side, that is, the bar cannot expand to the proper, then we have: ε = 0, εe =−εo σ=E(ε-εo)=E(0- αΔT)= = −αΔT, σ = −EαΔT Thus, thermal stress exists.
Shear strains do not change only normal strains change.
If the temperature changes, the size of the body changes, though it will not change the shape of the body. So, considering this fact, the shear-strain of the body does not change.
Thermal diffusivity is defined as the ratio of conducted heat to heat stored in material per unit volume.
Unit of Thermal diffusivity
The unit of thermal diffusivity is given as m2/s
Thermal diffusivity formula
The equation of thermal diffusivity is given by,
α = k/ρ
Where,
α is thermal diffusivity,
k is thermal conductivity (w/mK)
? is the density of the material (kg / m3)
Cp is the specific heat (J/ kg k)
Thermal diffusivity of water
The thermal diffusivity of water changes with temperature and pressure. If we consider atmospheric pressure condition, then the thermal conductivity values with temperature are given as below table.
The thermal diffusivity of air with temperature change is shown in the above table. Generally , the thermal diffusivity of gas is more than liquid in practice. we will study is more in next topic.
Thermal diffusion
Thermal diffusion of substance is the relative motion of the molecules due to temperature gradient.
Thermal diffusivity of aluminium
Thermal diffusivity of aluminium material is given as 9.7 * 10-5 m2/s
Flash method Thermal diffusivity
The flash method is used to determine the thermal diffusivity of the material. The short duration radiant energy pulse is passed through the sample. The laser or light flash lamp source is used for radiant energy. The piece will absorb the emitted energy. The process is repeated for the sample. Due to this emitted radiation, there is a temperature increase of the material sample. The infrared temperature detector records this increase in temperature.
The duration of the measured signal is calculated. The thermal diffusivity will be found from the following equation.
α = 0.1388/l2(t2)
=
Where L is the sample thickness,
t/2 is the half time,
we can find thermal diffusivity, specific heat, and density using the flash method.
The schematic diagram of the flash method is shown in the figure below
The thermal diffusivity can be measured using the flash method as discussed above. in this method, the short energy pulse is radiated one end, and temperature rise is calculated on the other end.
Thermal conductivity and thermal diffusivity
To differentiate between thermal conductivity and thermal diffusivity, consider two materials having the same thermal conductivity but with different thermal diffusivity. Both will permit the same rate of heat flow in the steady-state condition. At the start of the heat transfer process, the material with higher thermal diffusivity will reach a steady-state first compared to other material since it retains less heat energy. Heat energy penetrates fast through this material, but after getting a steady-state, the rate of heat flow will be the same. Also, remember that the material having less thermal diffusivity takes more time to reach the steady state.
Thermal diffusivity measurement techniques
There are mainly three types of thermal diffusivity measurement techniques.
Flash method
Thermal wave interferometry
Thermographic method
Thermal diffusivity of asphalt
The thermal diffusivity of the asphalt (Ah-70) is 0.123 mm2/s,
Asphalt (Ah-90) 0.128 mm2/s
Thermal diffusivity of rubber
The thermal diffusivity of the rubber material is in the range of 0.089-0.13 mm2/s
Thermal diffusivity values
Thermal diffusivity values for various materials are given in the table below. The values are changes with properties like temperature. These values are given for standard temperature and pressure.
The highest thermal diffusivity is of pure silver 165.63 mm2 / s
Thermal diffusivity of sand
Thermal diffusivity of dry sand varied from 0.6 * 10-7 to 7.0 * 10-7 m2/s.
Thermal diffusivity heat transfer
There are three modes of heat transfer conduction, convection and radiation. Heat conduction is dependent on main two properties. One is thermal conductivity and thermal diffusivity. Thermal conductivity is well-known property, but thermal diffusivity is not well known. It defines the rate of heat transfer through a given medium.
The rate of heat transfer is faster is the thermal diffusivity is higher. Thermal diffusivity is balancing between the medium of heat transfer and heat storage.
The Thermal diffusivities for gases are generally
The thermal diffusivities of gases substance are found more than liquid substance
Thermal diffusion coefficient
It is one of physical parameter which describes the dependency of mass diffusion flow of the mixture. In other words, the thermal diffusion coefficient is the ratio of a temperature gradient to the absolute temperature.
Thermal diffusion meaning
Thermal diffusion of substance is the relative motion of the molecules due to temperature gradient.
Glass Thermal diffusivity
The thermal diffusivity of glass is 0.34 * 10-6 m2/s at normal atmospheric condition.
Stainless steel Thermal diffusivity
The thermal diffusivity of stainless steel at 100 °C is 4.55 *10-6 m2/s
Thermal diffusion ratio
The thermal diffusion ratio is the ratio of the thermal diffusion coefficient to the concentration coefficient.
FAQs
Thermal diffusivity of gas vs liquid
The thermal diffusivities of gases substance are found more than liquid substance
Which material has the highest Thermal diffusivity
The most increased thermal diffusivity is of pure silver 165.63 mm2 / s
Application of Thermal diffusivity
The conduction heat transfer in any apparatus requires the study of thermal diffusivity. The industries are using the analysis of thermal diffusivity to optimize the heat transfer rate.
If we take a particular example, then insulation is one example. In insulation, the thermal diffusivity of the material is minimum so that it can resist maximum heat flow.
We are using computers, laptops and other electronic gadgets. Do you know what the method to extract heat from devices is? Yes, it’s Heat sinks.
Heat sink requires higher thermal diffusivity to transfer faster heat from any gadgets.
An increase in heat transfer in any electronics degrades its performance. The higher thermal diffusivity material should be used to improve its performance in that case.
Thermal diffusivity of concrete
The thermal diffusivity of the concrete is 0.75 *10 -6 m2/s
What is the physical significance of thermal diffusivity?
Thermal diffusivity can be defined as the ratio of the thermal conductivity of the substance to the heat storage capacity of the substance.
The ratio defines the generated heat gets diffused out at a specific rate. The higher value of thermal diffusivity indicates that the time required for heat diffusion is less. The study of the equation of thermal diffusivity can be possible by the higher value of thermal conductivity or the lower value of heat capacity.
Thermal diffusivity is helpful for more intransient heat transfers. In steady-state heat transfer, the thermal conductivity is enough to study.
Why is the thermal diffusivity of gas greater than liquid even though the thermal conductivity of the liquid is greater than gases?
Thermal diffusivity means the ability of a material to transfer heat and store the heat at an unsteady state. A faster heat transfer can be possible if the thermal diffusivity is higher. The lower thermal diffusivity of material means the storage of heat in it.
Gas possesses low volumetric heat capacity because of low density. Due to low volumetric heat capacity, the value of thermal diffusivity is high.
Liquid possesses a high heat capacity compare to gas; hence thermal diffusivity is lower in the liquid.
What is the order of thermal diffusivity for solid, liquid, and gas?
The order of thermal diffusivity in solid, liquid and gas as shown below,
Gas > Liquid > Solid
What is the difference between momentum diffusion and thermal diffusion?
Momentum diffusion
It can be considered the kinematic viscosity of the fluid, i.e. the ability of the fluid to flow the momentum. Momentum diffusion is occurred by shear stress in a fluid. Shear stress causes a random and any direction movement of molecules.
Thermal diffusion
It can be defined as thermal conductivity divided by the multiplication of density and specific heat capacity (when the pressure is constant). It measures the heat transfer rate for a given material from the hot side to the cool side. It is predictive analogous to whether a given material is “cool to the touch.”
How is the Prandtl number related to kinematic viscosity and thermal diffusivity?
The Prandtl Number is dimensionless. It can be given as the ratio of momentum diffusivity (it is kinematic viscosity as explained above) to thermal diffusivity.
It can be formulated in equation as,
Pr = v/α
Pr = Prandtl number
V= momentum diffusivity ( m2/s )
α = Thermal diffusivity ( m2/ s )
MCQs
Thermal diffusivity is the _________
(a) Dimensionless parameter (b) Function of heat (c) Physical property of the material
(d) All of the above
Thermal diffusivity of a material is __________________?
(a) directly proportional with thermal conductivity (k)
(b) inversely proportional with the density of a material
(c) inversely proportional with specific heat
(d) all of the above
(e) none of the above
Find the wrong statement: Specific heat of a material ______________.
(a) Constant for a material (b) Heat capacity per unit mass
Principal Stress is the maximum and minimum stresses derived from normal stress at an angle on a plane where shear stress is zero.
How to calculate principal stress ?
Principal stress equation | Principal stress formula: Maximum and Minimum principal stress equations:
Principal stress derivation | Determine the principal planes and the principal stresses
Normal stresses:
Differentiate,
“p” represents the principal plane.
There are two principal stresses, one at angle and other at Maximum and Minimum principal stresses:
R=
substitute in equation 1:
substitute value of R
Maximum and minimum normal stresses are the principal stresses:
The state of Stress:
The principal stress is the reference co-ordinate axes to the representation of the stress matrix and this stress components are the significance of the state of stress could be represented as,
Stress tensor:
Principal stresses from stress tensor and stress invariants |principal stress invariants
There are three principal planes at any stressed body, with normal vectors n, called principal directions where the stress vector is in the same direction as normal vector n with no shear stresses and these components depend on the alignment of the co-ordinate system.
A Stress vector parallel to the normal unit vector n is specified as,
Where, \\lambda represents constant of proportionality.
The principal stress vectors represented as,
The magnitude of three principal stresses gives three linear equations. The determinant of the coefficient matrix is equal to zero and represented as,
Principal stresses are the form of normal stresses, and the stress vector in the coordinate system is represented in the matrix form as follows:
I1, I2, I3 are the stress invariants of the principal stresses, The stress invariants are dependent on the principal stresses and are calculated as follows,
The principal stresses equation for stress invariants:
Principal stress trajectories | Principal directions of stress
Stress trajectories show the principal stress directions and their varying magnitude of the principal stresses.
Von Mises is the theoretical measure of the stress yield failure criterion in ductile materials. The positive or negative sign depends on the principal stresses. Principal stresses Boundary conditions:
Theories of failure give the yield stresses of the components subjected to multiaxial loading. Further, when it is compared with the yield point of the components shows the margin of the safety of the component.
Maximum principal stress is considered for brittle elements such as casting components (i.e., clutch housing, gearbox, etc.) Von-mises stress theory is based on shear strain energy theory is suggested for ductile materials like aluminum, steel components.
Why von mises stress is recommended for ductile and Principal Stress for brittle materials?
Failure of brittle materials used to uni-axial test is along a plane vertical to the axis of loading. So, the failure is because of normal stress generally. Out of all theories of failure, principal stress theory is based on normal Stress. Hence for brittle materials, principal stress theory is recommended,
Ductile materials fail at 45 degrees inclined at the plane of loading. So, the failure is due to shear stress. Out of all theories of failure shear strain energy or von-mises theory and maximum shear stress theory is based on shear stress. By comparison, von mises gives better results. Hence for ductile materials, von mises theory is recommended.
Absolute principal Stress | Effective principal Stress:
The principal stresses are based on maximum Stress and minimum Stress. So, the range of the Stress is between the maximum and minimum Stress, (stress range is limited and less) and might lead to higher fatigue life. So, it is important to find out the effective principal Stress that gives the maximum value out of the two over the given period of time.
What is Maximum Normal Stress theory?
This states that brittle failure occurs when the maximum principal Stress exceeds the compression or the tensile strength of the material. Suppose that a factor of safety n is considered in the design. The safe design conditions require that.
Maximum principal stress equation
Where σ1, σ2, σ3 are three principal stresses, maximum, minimum, and intermediate, in the three directions, Sut and Suc are the ultimate tensile strength and the ultimate compressive strength, respectively.
To avoid brittle failure, the principal stresses at any point in a structure should lie within the square failure envelope based on the maximum normal stress theory.
Maximum principal stress theory |Maximum principal stress definition
consider two-dimensional stress state and the corresponding principal stresses such as σ1 >σ2 >σ3 Where σ3=0, σ2 may be compressive or tensile depending on the loading conditions where σ2 may be less or greater than σ3.
According to maximum principal stress theory, failure will occur when σ1 or σ2 =σy or σt The conditions are represented graphically with coordinates σ1,σ2. If the state of Stress with coordinates (σ1,σ2 ) falls outside of the rectangular region, failure will occur as per the maximum principal stress theory.
Mohr’s circle principal stresses
Explain the Mohr’s circles for the three-dimensional state of stress:
Consider a plane with a reference point as P. Sigma is represented as normal Stress and tau by the shearing stress on the same plane.
Take another plane with reference point Q representing sigma and tau as normal stress and shear stress, respectively. Different planes are passing through point p, different values of principal and shear stress.
For each plane n, a point Q with coordinates as shear stress and principal Stress can be located.
Determine the normal and shear stresses for point Q in all possible directions of n.
Obtain three principal stresses as maximum principal Stress, minimum principal Stress, and intermediate principal Stress and represent them in ascending order of the values of the stresses.
Draw three circles with diameters as the difference between the principal stresses.
The shaded area region is the Mohr’s circle plane region.
The circles represent the Mohr’s circles.
(σ1-σ3) and the associated normal stress is (σ1+σ3)
There are three normal stresses, so are three shear stresses.
The principal shear planes are the planes where shear stresses act and principal normal stress acts at a plane where shear stress is ‘0’ and shear stress act at a plane where normal principal stress is zero. The principal shear stress act at 45° to the normal planes.
The shear stresses are denoted by And the principal stresses are denoted by
Third principal stress
3rd principal Stress is relative to the maximum compressive stress due to the loading conditions.
3D principal stress examples:
For three-dimensional case, all three planes have zero shear stresses, and these planes are mutually perpendicular, and normal stresses have maximum and minimum stress values and these are the normal stresses that represent the principal maximal and minimal stress.
These principal stresses are denoted by, σ1,σ2, σ3. Example: 3D Stress in hub-a a steel shaft is force-fitted into the hub. 3D Stress in machine component.
Principal deviatoric stress:
Principal deviatoric stresses are obtained by subtracting mean Stress from each principal Stress.
Intermediate principal Stress:
The principal Stress, which is neither maximum nor minimum, is called intermediate Stress.
Principal stress angle | Orientation of principal stress: θP
Orientation of the principal Stress is computed by equating shear stress to zero in x-y direction at the principle plane rotated through an angle theta. Solve θ to get θP, the principal stress angle.
Important Frequently Asked Questions (FAQs):
Maximum principal stress theory is applicable for which material?
Answer: Brittle materials.
What are the 3 principal stresses?| What is maximum and minimum principal stress ?
Maximum principal Stress | Major principal stress: Most tensile (σ1) Minimum principal Stress | Minor Principal Stress: Most compressive (σ3) Intermediate principal Stress (σ2)
Principal Stress vs normal Stress:
Normal Stress is the force applied to the body per unit area. Principal Stress is the stress applied to the body having zero shear stress principal Stress is in the form of normal Stress giving maximum and minimum stresses on the principal plane.
Principal Stress vs Bending Stress:
Bending Stress is the Stress that occurs in the body due to the application of a large amount of load that causes the object to bend.
Principal Stress vs axial Stress:
Axial Stress and principal stress are the parts of normal stress.
What is the significance of principal Stress?
Principal Stress shows the maximum and minimum normal stress. Maximum normal Stress shows the component’s ability to sustain the maximum amount of force.
What are the principal stresses in a shaft with torque applied ?
The normal Stress is maximum or minimum, and shear stress is zero.
when shear stress=0,
Important Principal stress problems:
1) A rectangular stress vector having shear stress in XY direction of 60Mpa and normal tensile Stresses of 40Mpa.How to find principal stresses ?
Solution: Given: Principal stresses are calculated as,
σ1=100Mpa
σ2=-20Mpa
2)What are the coordinates of the center of Mohr’s circle for an element subjected to two mutually perpendicular stresses, one tensile of magnitude 80MPa and other compressive of magnitude 50MPa?
σx = 80 MPa, σy = -50 MPa Co-ordinates of center of Mohr’s circle =[ ½( σx + σy),0] = [(30/2),0] = (15,0)
3)A body was subjected to two mutually perpendicular stresses of -4MPa and 20MPa, respectively. Calculate the shear stress on the plane of the shear.
σx+σy /2= -4+20/2 = 8Mpa Radius= σ1-σ2/2 = 20-(-4)/2 = 12 where σx,σy are principal stresses at pure shear stress,σn=0 shear stress= squareroot12^2-8^2= 8.94Mpa.
4) Application of principal stress | Find the principal stresses for the following cases.
i)σx=30 Mpa, σy=0, \\tau=15Mpa.
solution:
σ1=36.21Mpa
σ2=-6.21Mpa ii)σx=0,σy=80MMpa, \\tau=60Mpa.
σ1=97Mpa
σ2=12.92Mpa
iii)\\tau=10Mpa, σx=50Mpa,σy=50Mpa.
σ1=60Mpa
σ2=40Mpa
5) The maximum principal Stress is given 100 Mpa, and the Minimum principal Stress is 50 MPa. Calculate the maximum shear stress and the orientation of the principal plane using Mohr’s circle.
Given: Maximum principal stress=100Mpa(compressive) Minimum principal stress=50 Mpa(compressive) Solution: Maximum shear stress is the radius of the Mohr’s circle, then we can write as follows.
R=
2θ = 90, from the maximum principal stress direction. So, the orientation at that point is θ = 45 from the maximum principal stress direction.
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