Engineering

Topic of Discussion: Thermal insulation

Thermal insulation definition

When two objects are in thermal contact with each other or under the influence of radiation, the process of depletion of heat transfer among the entities is known as thermal insulation. It is quite the opposite of what thermal conductivity can be defined as. Essentially, an object with very low thermal conductivity can be regarded as a well-insulated material. 

Thermal insulator

While thermal insulation is the process of depletion of heat transfer, thermal insulators are materials that employ the insulation process. It prevents heat energy from being transferred from one object to another. This can be viewed in detail from a thermodynamics perspective while comprehending heat energy principles and more.

Heat Insulation

It is a form of energy that depends upon another factor called temperature. The transfer of energy in the form of heat from one body to another result in a temperature difference. Heat usually flows from a hotter to a colder body. It plays a significant role in the principles of thermodynamics. If a body is cold, it means that heat is removed and not coldness added, which brings about a fun fact about this form of energy.

Heat can be transferred by three different means.

1. Conduction
2. Convention
3. Radiation

Conduction is the energy transfer process between two objects where the medium of exchange is through direct contact. At the same time, convection is the transfer of energy through the motion of matter, using air as a medium. Radiation is the transfer process that happens without any medium but with the aid of electromagnetic waves.

The three equations concerning the three forms of heat transfer are as follows,

Conduction: Q = [k · A · (Thot – Tcold)]/d

Convection: Q = hc · A · (Ts – Tf)

Radiation: P = e · σ · A · (Tr4 – Tc4) (Using Stefan-Boltzmann law)

Examples of the modes of transfer that can be found in our everyday life under conduction can be as simple as accelerated vibrating molecules in the hand when in contact with a hot coffee mug. This means that the hand has heated up where the transfer of energy took place through direct contact. 

A typical example of convection would be refrigeration, where the food items kept in the fridge would essentially get cold through convection of air and other coolants. 

Radiation is the mode of transfer through a void, such as the heat from the sun that reaches the earth.

Why thermal insulation? Its purposes and requirements

The objective of thermal insulation is to moderate the temperature in something as small as an individual house to as complex as a nuclear reactor. Thermal insulation is to fortify the constructional elements against damage caused by moisture or thermal impact on the component. The wear on the object or the part can be decreased during winter with thermal insulation, which serves the purpose of energy conservation. At the same time, during the summer, overheating is significantly depleted. 

Advantages of thermal insulation

Thermal insulation creates an optimum environment that keeps the surroundings warm in the winter and chill during the summer, enabling a comfortable living and operating. Due to the demand for a comfortable lifestyle environment, thermal insulation greatly enhances energy conservation and maintenance costs. It also helps prevent the deposition of moisture on the interior walls of a room or a container that can be caused due to the effect of temperature and humidity.

Thermal insulation materials

1. Fiberglass
2. Polyurethane foam
3. Cellulose
4. Polystyrene
5. Mineral wool

Thermal insulation Fiberglass: 

it is the most common and frequently used for thermal insulation method in modern-day houses. It is derived from finely woven silicon, recycled glass fragments, and sand particles containing glass powder. 

Fiberglass or glass wool is generally used as an acoustic insulation material, an indoor material applied under pitched roofs or wooden floors. Since fiberglass loses their value insulation when in contact with damp or moisture, they are mostly seen inside homes and not outside.

Insulation values of the material are given by,

• Density = 25 kg/m³
• Heat storage capacity = 800 J/kgK
• Fire class => A2, S1, d0 (extinguish by self and low flame ability)
• λ= 0.032 to 0.040 W/mL-K
• Diffusion resistance: 1

Cellulose:

This type of thermal insulation method is considered one of the most eco-friendly processes in the modern-day. Cellulose comprises 70-80% recycled denim, paper or cardboard in the form of loose foam heavily treated (15% volume) with (NH₄)₂SO₄, boric acid or borax. It is considered the best form of thermal insulation against fire resistance solutions that are essentially used to moderate heat loss and gain noise transmission. 

Properties of cellulose, 

• Thermal conductivity = 40 mW/m·K 
• R value = R-2.6 to R-3.8 per 100mm 
• Density = 57 kg/m3

Mineral wool:

The glass wool or mineral wool is used widely for its functional properties, easy purchase, and simple handling. Mineral wool comprises spun yarn manufactured from melted or recycled glass or stone (rock wool). Rock wool is made from basalt, where the threads are combined in a unique way for a wooly structure to be formed for insulation. Hereafter the wool is compressed into mineral batts or boards that can be purchased off the market for insulation purposes. 

Mineral wool is generally used to Insulate cavity walls, exterior walls, partition walls, and stored floors. They are also extensively applied in industrial applications like machines, air conditioners, etc. 

Properties: 

• λ= 0.03 W/mK to 0.04 W/mK
• Density= 30-200 kg/m³
• R= 0.035 W/mK

Polystyrene:

This is also commonly known as styrofoam, is a waterproof thermoplastic foam that insulates temperature and sound very effectively. They come in two types: EPS (expanded) and XEPS(extruded), differing in cost and performance. They possess a very smooth surface of insulation not found in any other types, usually created into cut blocks, making it very ideal for insulation. The foam is sometimes flammable and requires a coating of Hexabromocyclododecane (HBCD), a fireproofing chemical. 

Its significant advantages are that it possesses magnificent cushioning properties, lightweight in nature, low thermal conductivity, and absorbs very little moisture, mostly 98% air and 100% recyclable. 

Properties: 

• R= 4-5.5
• Density= 0.05 g/cm3
• λ= 0.033 W/(m·K)
• Refractive index= 1.6

Polyurethane foam:

It is the most abundant and exceptional form of thermal insulation utilizing non-chlorofluorocarbon (CFC) as a blowing agent to decrease the ozone layer’s damage. They are low-density foams that consist of low conductivity gas in their shells that can be sprayed onto the insulated areas. 

They are lightweight in relativity and weigh almost 2lb/ft3. They are also fire-resistant and used on surfaces like brick blocks, concrete, etc., by direct fixation. It is also used in the case of unfinished masonry by cutting the foam into the desired shape and size. The foam is then covered with constructive adhesive, pressing it against the masonry surface and seal the joints between the sheets with the expanding foam. 

Properties:

• λ= 0.022 W/mK to 0.028 W/mK
• Density= 30 kg/m³ to 100 kg/m³
• R= 6.3/ inch of thickness

Types of thermal Insulation

Blanket: Batt and Roll Insulation

The most well-known and broadly accessible sort of insulation is Blanket insulation, which comes in Batts or Rolls. It comprises flexible fiber, fiberglass’s. Batts and Rolls are also finished from mineral-wool, plastic, and natural fiber, such as cotton and sheep’s wool. The Blanket insulation is most likely to be used in unfinished walls, floors, and ceilings and these insulation could easily be fitted in-between studs, joists, and beams. This insulation type is highly used since it is suited for standard stud and joist spacing that is comparatively free from different obstruction. This type is also relatively expensive in comparison to the others.

Concrete Block thermal Insulation

Concrete block insulation is incorporated in several ways, like adding foam bead or air into the concrete mixture to get the desired R-values. Concrete Block insulation is widely used for unfinished walls, including foundation walls, and is also used prominently for construction and renovation. Installation requires specialized skills like stacking, insulating concrete blocks without using mortar, and surface bonding. The cores are insulated to achieve R’s desired values, which helps us moderate temperatures as well.

Insulating Concrete foam 

The material used in the making of this is foam boards or foam blocks. This type of insulation is highly used to complete unfinished walls, as well as foundation walls for new construction. They are incorporated as a part of the building assembly also. This category of insulation is highly in use for construction. Since they are built into the home’s walls, it increases the thermal resistance.

Rigid fibrous or fiber insulation

Fiberglass and mineral wool are used to assimilate fiber insulation. Rigid fibrous insulation is highly used in regions that withstand high temperatures and often used for ducts in unconditioned spaces. Fiber insulation is established by HVAC contractors, usually manufactures the insulation and install them onto vents. These are mainly utilized for the reason of its ability to withstand high temp.

Structural insulated panels (SIPs)

This is primarily foam-board or liquid-foam insulation core and straw-core insulation. They are incorporated in unfinished walls, ceilings, floors, and the initial construction of roofs. They are implemented by construction workers who fit SIPs together to form walls and roofs. The perks of using this type of insulation provide consistent and higher insulation compared to traditional insulation. SIPs take a limited amount of time to implement.

Thermal insulation in the nuclear sector

The generic idea of a nuclear power plant is that it is utilized to generate electricity with nuclear fission. 

Nuclear reactor cores serve a particular purpose in energy-releasing enormous amounts of heat and work output. The containment of the nuclear reactor in an container is a large space incorporating the nuclear steam supply system (NSSS).

 The NSSS has a reactor, valves, pipe, pumps, and other various components and equipment. The NSSS produces a very substantial net positive heat load. Insulation on hot pipe and equipment inside the reactor has one objective: to control containment cooling loads. Containment cooling is performed to remove that heat linked directly to a water body (river, lakes, etc.) or vapor compression cooling such as air conditioning. The nuclear plants’ technical specifications will be alarmed if heat source release heat more than the cooling rate in standard.

Thermal properties of insulation

There are particular primary considerations to be chosen during the selection process of insulation. These properties vary with the material selected, ranging from wool to nuclear reactor thermal insulation. The difference in the insulation types’ thermal properties makes a difference in the amount of efficiency, performance, and sustainability. 

The various properties to be considered are: 

Emissivity (E):

A material written as ε is defined as the ratio of energy radiated by the material to the energy emitted by a black body at a similar temperature. In layman’s terms, it is useful in emitting energy as thermal radiation such as infrared energy.

Thermal conductance (C):

It can be termed by the unit temperature difference between two bodies that infer the time rate of a steady-state heat flow through a unit area of the given material. 

Temperature limits:

upper and lower levels of temperature should be satisfied by the materials chosen for insulation. 

Thermal resistance (R-Value): the temperature difference between two surfaces induces a unit heat flow rate through the objects’ unit area (K.m²/W). 

Thermal transmittance (U):

through an assembly, heat flow’s overall conductance is coined as thermal transmittance. 

Thermal conductivity (k-value): 

Where, L= thickness of the material, (m)

T= temperature, (K)

q = heat flow rate, (W/m²)

To learn more about thermodynamics click here!

Contents: Moment Area Method and Macaulay’s Method

• Macaulay’s Method Definition
• Macaulay’s Method for slope and deflection
• Macaulay’s Method example 1: Slope and Deflection in a Simply supported Beam for Uniformly Distributed Load
• Macaulay’s Method example 2: Slope and Deflection in an Overhanging Beam
• Moment-Area Method
• Moment Area Theorem
• Example related to Moment Area Method
• Bending Moment by parts
• Applying Moment Area method on overhanging Beam with Uniformly distributed Loading for finding slope and deflection
• Maximum Deflection due to unsymmetrical Loading
• Q & As on Macaulay’s Method and Moment Area Method

Macaulay’s method

Mr W.H Macaulay devised Macaulay’s Method. Macaulay’s Method is very efficient for discontinuous loading conditions.

Macaulay’s Method (the double integration method) is a technique used in structural analysis to determine the deflection of Euler-Bernoulli beams and this method is very useful for case of discontinuous and/or discrete loading condition.

Macaulay’s Method for Slope and Deflection

Consider a small section of a Beam in which, at a particular section X, the shearing force is Q and the Bending Moment is M as shown below. At another section Y, distance ‘a’ along the Beam, a concentrated load F is applied which will change the Bending Moment for the points beyond Y.

Between X and Y,

And Beyond Y

For the slope at Y, equating [5] and [2] we get,

But at Point Y, x = a

Substituting the above equation in [5]

Also, for the same deflection at Y equating (3) and (6), with (x=a) we get

On solving these equations and substituting value of C3

Substituting in equation [6] we get,

By further Investigating equations [4], [7] and [8] we can conclude that the Single Integration Method for obtaining Slope and deflection will still be applicable provided that the term F(x-a) is integrated with respect to (x-a) and not x. Also, the term W(x-a) is applicable only for (x>a) or when (x-a) is positive. Thus, these terms are called Macaulay terms. Macaulay terms should be integrated with respect to themselves and must be neglected when they are negative.

Thus, the generalized equation for the whole Beam becomes,

Macaulay’s Method example 1: Slope and Deflection in a Simply supported Beam for Uniformly Distributed Load

Consider a simply supported beam with uniformly distributed loading over the complete span. Let weight  acting at distance a from End A and W2 acting at a distance b from end A.

The Bending Moment Equation for the above beam can be given by

The U.D.L applied over the complete beam doesn’t require any special treatment associated with the Macaulay’s brackets or Macaulay’s terms. Keep in mind that Macaulay’s terms are integrated with respect to themselves. For above case (x-a) if it comes out negative then it must be ignored. Substituting the end conditions will yield the values of constants of integration in the conventional way and hence the required value of slopes and deflection.

In This case the U.D.L starts at point B the bending moment equation is modified and the uniformly distributed load term becomes Macaulay’s Bracket terms.

The Bending Moment equation for the above case is given below

Integrating we get,

Macaulay’s Method example 2: Slope and Deflection in an Overhanging Beam

Given below is the overhanging beam in Fig. (a), we are need to calculate

(1) the equⁿ for the elastic curve.

(2) the mid-values in-between the supports and at point E (indicate whether each  is up or down).

To determine the bending moment for the above beam the equivalent loading is used, is given below as Figure (b). In order to use Macaulay’s Bracket in the Bending Moment equations, we are required to extend each distributed load to the right end of the beam. We extends the 800 N/m loadings to point E and eliminates the un-necessary portion by applying an equal and opposite loadings to C-E. The global expression for the bending moment represented by free-body diagram in figure(c).

Substituting M into the differential equation for the elastic curve,

Integrating it,

Again, Integrating it,

At Point A, the deflection is restricted due to simple support at A. Thus, at x = 0, y=0,

Again, at Point D the deflection is restricted due to simple support at D. at x = 6 m, y = 0,

When we substitute the values for P and Q to Eq. (a), we get

This is the Generalized equation to find deflection over the complete span of overhanging Beam.

In order to find the deflection at a distance of 3 m from the left end A, Substitute the value of x =3 in Eq. (b),

The equation of elastic curve so obtained is given by,

The negative sign of the value indicates that the deflection of the beam is downward direction in that region.
Now finding the Deflection at the extreme of the Beam i.e., at Point E
Put x = 8 m in eq. [b]

Again, the negative sign indicates the downward deflection.

Moment Area Method

In order to determine the slope or deflection of a beam at a specified location, the moment area method is considered most effective.

In this Moment Area Method, the bending moment’s integration is carried out indirectly, using the geometric properties of the area under the bending moment diagram, we assume that the deformation of Beam is below the elastic range and this results in small slopes and small displacements.

The First theorem of Moment Area method deals with slopes; the second theorem Moment Area method deals with deflections. These Two theorems form the basics of the Moment Area Method.

Moment Area Theorem

First – Moment Area Theorem

Consider a beam segment which is initially straight. The elastic curve AB for the segment taken into consideration is shown in fig (a). Consider two cross-sections of the beam at P and Q and rotate them through the angle dϴ relative to each other also separated by the distance dx.

Let’s assume the cross sections remain perpendicular to the axis of the beam.

dϴ = Difference in the slope of curve P and Q as depicted in Fig. (a).

From the given geometry, we see that dx = R dϴ, where R is the curvature radius of the deformed element’s elastic curve. Therefore, dϴ = dx/R, which upon using the moment-curvature relationship.

Integrating Eq.(a) over the segment AB yields

The left-hand side of Eq. (b) is the change in the slope between A and B. The right-hand side represents the area under the M/EI diagram between A and B, shown as the shaded area in Fig. (b). If we introduce the proper notation , Eq. (b) can be expressed in the form

This is the First theorem of Moment Area Method. The First theorem of Moment Area method deals with slopes

Second – Moment Area Theorem

Let  t (B/A) be the vertical distance of point B from the tangent to the elastic curve at A. This distance is called the tangential deviation of B with respect to A. To calculate the tangential deviation, we first determine the contribution dt of the infinitesimal element PQ.

We then use integration for A to B dt = t (B/A)  to add all the elements between A and B. As shown in the figure, dt is the vertical distance at B between the tangents drawn to the elastic curve at P and Q. Recalling that the slopes are very small, we obtain from geometry,

Where x’ is the horizontal distance of the element from B. Therefore, the tangential deviation is

Putting value dϴ of  in Equation [a] we get,

The right-hand side of Eq. (c) represents the first moment of the shaded area of the M/(EI) diagram in Fig. (b) about point B. Denoting the distance between B and the centroid C of this area by ,  we can write Eq. (c) as

This is the second theorem of moment area method. The second theorem Moment Area method deals with deflections.

Bending Moment by parts

For studying Complex applications, the evaluation of the angle ϴ (B/A)  and the tangential deviation  can be simplified by independently evaluating the effect of each load acting on the beam.  A separate Bending Moment diagram is drawn for each load, and the slope is obtained by algebraic summation of the areas under the various B.M.Ds. Similarly, the deflection is obtained by adding the first moment area about a vertical axis through point B. A bending-moment diagram is plotted in parts. When a bending-moment is drawn in parts, the various areas defined by the BMD consists of shapes, such as area under 2nd degree curves, cubic curves, rectangles, triangles, and parabolic curves, etc.

Steps to draw bending moments by parts

• Provide appropriate fixed support at a desired location. Simple supports are usually considered to be the best choice; however, another type of support is used depending upon the situation at hand.
• Calculate the support reactions and assume them to be applied loads.
• Draw a bending moment diagram for each load. Follow proper sign conventions while drawing bending moment diagram.
• The slope is obtained by algebraic summation of the areas under the various B.M.Ds.
• the deflection is obtained by adding the first moment area about a vertical axis through point B.

Applying Moment Area method on overhanging Beam with Uniformly distributed Loading for finding slope and deflection

Consider a Simply Supported overhanging beam with uniformly distributed loading from A to B and C to D as Shown below [ . Find slope and deflection by Using Moment Area method.]

From a free-body diagram of the beam, we determine the reactions and then draw the shear and bending-moment diagrams, as the flexural-rigidity of the beam is constant, to calculate (M/EI) diagram we need to divide each value of M by EI.

Drawing Shear Force and Bending Moment Diagram for the given beam

For Reference tangent: since the Beam is symmetric along with its loading with respect of Point C. The Tangent at C will act as a reference Tangent. From the diagram above

Thus, tangent at E can be given by,

Slope at E: according to M/EI diagram and applying the First Moment area method as discussed above we get,

Similarly, for A2

From equation [1] we get,

Deflection at Point E can be calculated by using Second moment area method

Similarly,

But we know that

Maximum Deflection due to unsymmetrical Loading

When a simply supported beam carries an unsymmetrical load, the maximum deflection will not occur at the centre of beam and required to be identify the beam’s K-point where the tangent is horizontal in order to evaluate the maximum deflection in a beam.

1. We start with finding Reference tangents at one of the supports of the beam. Let ϴa be the slope of the tangent at Support A.
2. Compute the tangential deviation t of support B with respect to A.
3. Divide the obtained quantity by the span L between the supports A and B.
4.  Since the slope ϴk=0, we must get,

Using the first moment-area theorem, we can conclusively predict that point K can be found by measuring an area A

By Observation we conclude that the maximum deflection y (max) = the tangential deviation t  of support A with respect to K (Fig. a) and we can determine y(max)  by calculating the first moment area between Support A and point K with respect to the vertical axis.

Question and Answer of Macaulay’s Method and Moment Area Method

Q.1) Which method is useful in order to determine the slope and deflection at a point on a Beam?

Ans: Macaulay’s Method is very efficient for this case.

Q.2) What does Second Moment Area Method states?

Ans: The Second Moment Area method states that,” the moment of Bending moment diagram B.M.D between any two points on an elastic line divided by flexural rigidity (EI) is equal to the intercept taken on a vertical reference line of the tangent at these points about the reference line.”

Q.3) Calculate the deflection of the beam if the slope is 0.00835 radians. The distance from the free end to the center of gravity of bending moment is 5 m?

Ans: The deflection at any point on the elastic curve is equal to Mx/EI.

But we know that M/EI is slope equation = 0.00835 rad.

So, Deflection = slope × (The distance from the free end to the center of gravity of bending moment

Deflection = 0.00835*5 = 0.04175 m = 41.75 mm.

To know about Strength of material(click here)and Bending Moment Diagram Click here.

List of content

• Physical properties of the fluid
• Specific weight
• Density
• What is the requirement of indicating the density value of petrol and diesel?
• What happens if the density of petrol or diesel will get change?
• Specific gravity
• Specific volume
• Compressibility and Bulk modulus
• Viscosity
• Newton’s law of viscosity
• Kinematic viscosity
• Effect of temperature on viscosity
• Questions & Answers
• Multiple choice questions

Physical properties of fluids

The properties can describe the physical condition of any fluid. It is essential to understand various properties of fluid before analyzing the fluid flow problem.  The properties can be defined as the physical characteristic which indicates its state

Properties of fluid are broadly divided into two parts

Intensive property: it is a property whose magnitude is not dependent on mass. For example, pressure, temperature, mass density, etc.

Extensive property: it is a property whose magnitude is dependent on mass. For example, weight volume, mass, etc.

Specific weight

Specific weight is defined as weight per unit volume

W=w/v

Here w is the weight of the fluid,

V is the volume of fluid.

As we know, the body’s weight is the force of the body to center of the earth.

It is expressed as the multiplication of the mass of the body and gravitational acceleration. The value of g is measured at sea level 9.8 m/s²

Weight is a force, so the unit of weight is Newton (N). The unit of volume is m³

Hence, the unit of specific weight is N/m³

The specific weight of water is 9810 N/m³ at standard pressure 760 mm of  Mercury and temperature of 4°C.

Specificc weight of seawater is 10000 -10105 N/m³.

The higher value of specific weight in seawater is due to dissolved salt and solid particulate matter. The specific weight of Mercury is 13 times greater than water. The air has specific weight around 11.9 N/m³ (at temperature 15°C and standard atmospheric pressure).

Since the specific weight is dependent on gravitational acceleration, its value changes with gravity.

Density

Density, the symbol of density, is rho (?). The standard definition of density is mass per unit volume.

In other words, we can say that it is a matter (amount) of fluid storage in the given volume.

ρ=m/V

Here, m is mass of fluid, V indicates the volume of fluid,

We know that the unit of mass is in kg and the unit of volume is in m³

So, The unit of density is taken in kg/m³

The mass density of water 15.5°C is 1000 kg/m³

The mass density of air is 1.24 kg/m³ at standard temperature 20°C and normal atmospheric pressure.

I have one practical question for you. You are frequently visiting the petrol pump filling petrol in your bike or car. You have noticed that the density of petrol or diesel is indicated on display. Now understand my question carefully,

What is the requirement of indicating the density value of petrol and diesel? What happens if the density of petrol or diesel will get change?

Think about it as an engineer and find an answer.

Specific gravity

Specific gravity is well-defined as the ratio of mass density or specific weight of the fluid to mass density or specific weight of the standard fluid.

Here, the standard fluid (sf) for liquid is water at 4°C and the standard fluid for gases is air at 0 °C.

As we can see, the specific gravity is the ratio of the same property, so the specific gravity is unitless.

There is no dimension of specific gravity..

The specific gravity of Mercury (Hg) is usually 13.6 times higher than water. It means that Mercury is 13.6 times heavier than water.

Specific volume

The specific volume is reciprocal of mass density

It can be defined as the ratio of volume and mass

v=V/m

Practically specific volume is a more useful incompressible fluid study.

The unit of specific volume is m³/kg.

Compressibility and bulk modulus

The study of fluid mechanics includes compressible and incompressible fluid.

Compressible fluid means it will get a contract when pressure is applied, and removal of pressure it will get expand.

Compressibility is the an essential property of the fluid. It is the ability of fluid to get change volume under pressure. The equation of the Coefficient of compressibility is given as,

Here, the dp change in applied pressure and dV is a volume change.

Here, the -ve sign indicates an increase in pressure results reduction in volume.The Coefficient of compressibility is symbolized by Β_c.

Generally, In this measurement compressibility of fluid is represented by its bulk modulus of elasticity and the bulk modulus of elasticity is taken as reciprocal of the Coefficient of compressibility.

Viscosity

Viscosity can be defined as it is property of fluid by whic it exerts resistance to flow.

Practically if we take an example, fluid is flowing over any solid surface or planer surface. The velocity of the fluid is considered negligible (zero) at the solid surface boundary, and velocity is found increasing far away from the solid surface boundary.  The fluid Layers offers resistance to each other. It is one type of friction between fluid layers.

Suppose we observe the velocity profile in the fluid layers.  The velocity is found lesser near to the solid surface. The velocity is found greater at the outer layer, far from the solid surface boundary. This happens because of internal resistance, and it is known as viscous resistance. All real fluid possesses viscosity. As we know, that ideal fluid does not have viscosity.  Some examples of highly viscous fluid are glycerine, tar,   and molasses, etc.

The fluids with lower viscosity are air, water, petrol, etc.

Newton’s law of viscosity

Let’s, consider two adjacent layers at distance dy,

Layer 1 velocity is u,

Layer 2 velocity is u+du,

The top layer is flowing with velocity u+du.  The top layer offers resistance to the lower layer with exerting force F.  The lower layer also provides resistance to the top layer with equal and opposite force F.  These two opposing forces generate shear resistance. 

 It is denoted by τ shear resistance. It is proportional to the velocity gradient.

If we remove the proportional limit, can we have to put one constant?

Here, the constant of proportionality or proportionality factor is μ

It is acknowledged as the Coefficient of viscosity. The value of Coefficient of viscosity is dependent on the type of surface and surface roughness.

This equation is widely known as Newton’s law of viscosity.

There is some observation based on this law. These observations are useful to study viscosity and velocity distribution.

Shear stress is the maximum velocity gradient is high.

When the velocity gradient is zero, the shear stress is also zero.

The value of shear stress is maximum at the boundary, and it will simultaneously decrease from the boundary.

The unit of viscosity can be formulated from Newton’s law of viscosity..

Here, N/m² istaken as Pascal (Pa). Sometimes, the Coefficient of dynamic viscosity is taken in poise (P).

1 Poise = 0.1 Pa*s

Dynamic viscosity of water is  1  centipoise (cP)=  10^-3 N s/m²

Dynamic viscosity of air is 0.0181 centipoise =0.0181 *10^-3 N s/m²

Water is 55 times denser than air.

The given value is at standard temperature 20°C and atmospheric pressure.

Kinematic viscosity

The kinematics viscosity is well-defined as the ratio of dynamic viscosity and density.

The unit of kinematics viscosity is formulated as,

v=μ/ρ

As we know, that metric does not involve any force or energy, so the unit of kinematic viscosity only of length and time.

This unit is commonly known as stokes.

The kinematics viscosity of water is 10 raise to minus 6 meter square per second

The kinematic viscosity of air is 15

The value is at standard temperature of 20°C and atmospheric pressure.

The kinematics viscosity of air is 15 times higher than water.

Effect of temperature on viscosity

The effect of the temp. value of viscosity is different in liquid and gas.

If we consider the fluid is a liquid value of dynamic viscosity is decreasing with an increase in temperature

Suppose the fluid is gas; the value of viscosity is increasing with an increase in temperature.

Let’s see why

In liquid, the molecules are more closer as compare to gases.

Viscosity is act mainly due to molecular cohesion. The molecular cohesion is decreasing with increasing temperature.

 Empirical relation is developed to explain the variation in viscosity due to temp.

For liquid:

Here, μ is the viscosity at the desired temperature t°C.

 μ₀ is the viscosity at 0°C

A, B are the constant, and their value is dependent on the used liquid.

For water μ₀= 0.0179 poise, A= 0.03368, B= 0.000221

For gases:

Here, μ_t is the viscosity at desired temperature t°C.

 μ₀ is the viscosity at 0°C

α,β are the constant and its value is dependent on used gas

For air. μ₀=1.7*10^-5 Ns/m², α=0.56*10^-7, β= 0.1189*10^-9

Questions & Answers

What is an intensive property?

It is the property of fluid whose magnitude is not dependent on mass or matter.

What is the weight of the body? Is it one type of force?

Yes, Weight is force. The weight of the body is the force of the body to the center of the earth.

Why is specific gravity unitless?

Specific gravity is the ratio of density of the fluid to density of the standard fluid. It means that ratio of the similar types. So there is no unit of specific gravity.

Which type of study requires the use of specific volume?

The study of compressible fluid requires the use of specific volume property.

What is compressibility?

Compressibility is the important property of fluid. It is ability of fluid to get change volume under pressure.

What is meaning of negative sign in the equation of compressibility?

The negative sign indicates increase in pressure results decrease in volume.

Enlist the observation based on newton’s law of viscosity.

Shear stress is maximum velocity gradient is high

When the velocity gradient is zero the shear stress is also zero

Value of shear stress is maximum at the boundary, and it will simultaneously decrease from the boundary.

Define kinematic viscosity. Why is the unit only include length and time dimensions?

The kinematics viscosity is represented as the ration of dynamic viscosity and density. We know that kinematic does not involve any force or energy, so the unit of kinematic viscosity only of length and time.

What is the effect of temp. on gaseous fluid?

If the fluid is gaseous, then the value of viscosity is increasing with an increase in temperature.

Give some examples of highly viscous fluid.

Examples of highly viscous fluid are glycerin, tar, and molasses, etc.

What are the values of constants in correlation for “effect of temperature on viscosity of gases  ?

μ₀ is the viscosity at 0°C

α,β are the constant and its value is dependent on used gas

For air. μ₀=1.7*10^-5 Ns/m², α=0.56*10^-7, β= 0.1189*10^-9

Multiple Choice Questions

Which one of the following is extensive property?

a) Pressure         b) Mass density                                c) Volume           d) Temperature

Give the unit of specific weight.

a) N/m                  b) N/m²                                c) N/m³                                d) m/N

What is the value of specific weight of seawater (at standard condition)?

a) 10000 -10105 N/m³     b) 20000 -20105 N/m³     c) 1000 -1105 N/m³          d) None of above

How many times is Mercury heavier than water?

a) 11                      b) 12                      c) 13                       d) 14

What is the density of water at 15.5°C in kg/m³

a) 994                    b) 1000                 c) 1500                  d) 846

The specific gravity is ratio of mass density of fluid to mass density of_______

a) Compressible fluid                     b) Incompressible fluid                  c) Standard fluid                               d) None

The specific volume is reciprocal of__________

a) Specific weight             b) Viscosity                         c) Mass density                 d) Specific gravity

The bulk modulus of elasticity is reciprocal of___________

a) Coefficient of viscosity   b) Coefficient of performance              c) Coefficient of compressibility                 d) None

Viscosity can be defined as resistance to ________

a) Fluid flow       b) Current flow                                 c) Temperature flow      d) Pressure

What is the unit of kinematic viscosity?

a) N/m                  b) m/s                   c) m³/s                  d) m²/s

If fluid is liquid then the value of dynamic viscosity will ________ with increase in temperature of liquid.

a) Increase          b) Decrease                        c) be constant                   d) None of this

The molecular cohesion is decreasing with________ temperature.

a) Increase          b) Decrease                        c) Remain constant                         d) None

Conclusion

This article is the concept of various properties and their relation. The properties like specific weight, mass density, specific gravity and specific volume are defined with the unit. The concept of viscosity and newton’s law of viscosity are described in detail with its equations. The most important phenomenon, the effect of temperature on the fluid’s viscosity, is discussed to make the concept easier to understand.

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Contents: Crankshaft

• What is Crankshaft?
• Material and manufacture of Crankshafts
• Crankshaft diagram
• Crankshaft Design Procedure
• Crankshaft Deflection
• Crankshaft Deflection Curve Plotting
• Marine Crankshaft Failure Case Study
• Failure Analysis of Boxer Diesel Crankshaft: Case Study
• Crankshaft Fatigue Failure Analysis: A Review
• Failure of Diesel Engine Crankshaft: A Case Study

What is crankshaft?

“A crankshaft is a shaft driven by a crank mechanism, involving of a series of cranks and crankpins to which the connecting rods of an engine is attached. It is a mechanical part able to perform a conversion between reciprocating motion and rotational motion. A reciprocating engine converts reciprocating motion of a piston to the rotational form, although in a reciprocating compressor, it translates opposite way means rotational to reciprocating forms. During this change in-between two motions, the crankshafts have “crank throws” or “crankpins” additional bearing surface which axis is offset from the crank, to which the “big end” of the connecting rod from each cylinder is attached.”

A crankshaft can be described as a component used to convert the piston’s reciprocating motion to the shaft into rotatory motion or vice versa. In simple words, it isa shaft with a crank attachment.

A typical crankshaft comprises of three sections:

1. The shaft section that revolves inside the main bearings.
2. The crankpins
3. The crank arms or webs.

This is categorized in two types as per position of crank:

1. Side crankshaft
2. Centre crankshaft

The crankshaft can be further categorized in Single throw crank-shafts and multi throw crank-shafts depending on the no. of cranks in the shaft. A crankshaft which possesses only center crank or one-sided crank is entitled as single-throw crankshaft. A crankshaft with 2 or multiple center cranks or ‘2’ side cranks,  ‘1’ on each end is recognized as “multi-throw crankshafts”. The Side crank configuration includes geometric simplicity, are comparatively simple to be manufactured and assembled. They can be used with simple slide-on bearings and are relatively cheaper than Center crankshaft.

The center crank configuration provides better stability and balancing of forces with lower induced lower stresses. Their manufacturing cost is high, and a split connecting rod bearing is required for assembly. Applications which require multiple pistons working in phase, a multi-throw crankshaft can be developed by placing several centers cranks side-by-side, in a specified sequence, along a common centerline of rotation. The throws are rotationally indexed to provide the desired phasing.

Multi-cylinder internal combustion engines such as Inline and V- series Engine utilizes Multi-throw crankshaft. All types of crankshafts Experience dynamic forces generated by the rotating eccentric mass center at each crank pin. It is often necessary to utilize counterweights and dynamic balancing to minimize shaking forces, tractive effort and swaying couples generated by these inertia forces.

Material and manufacture of Crankshafts:

The crankshaft often experiences shocks and fatigues loading condition. Thus, the material of the crankshaft must possess more toughness and better resistance to fatigue. They are usually product of carbon steel, certain steel or cast-iron materials. For Engines used in industry, the crankshafts are generally generated from carbon steel such as 40-C-8, 55-C- 8 and 60-C-4.

In transport engine, manganese steel i.e., 20-Mn-2, 27-Mn-2 and 37-Mn-2 are commonly utilized to prepare the crank shafts. In aero engines, nickel-chromium steel such as 35-Ni-1-Cr-60 and 40-Ni-2-Cr-1-Mo-28 are generally utilized for manufacturing the crankshaft. 

The crank shafts are commonly finished by drop forging or casting process. The surface hardening of the crankpin is finished through the case carburizing process, Nitriding or induction hardening process. The selected Crankshaft materials must meet both the structural strength requirements and the bearing-site wear requirements.

In the typical crankshaft application, soft, ductile sleeves are attached to the connecting rod or the frame, so the crankshaft material must have the ability to provide a hard surface at the bearing sites. Many materials may meet structural strength requirements, but providing wear resistance at the bearing sites narrows the list of acceptable candidates.

Because of the asymmetric geometry, many crankshafts have been manufactured by casting or forging a “blank,” to be finish-machined later. Built-up weldments are used in some applications. Traditionally, cast iron, cast steel, and wrought steel have been used for crank shafts. The use of selectively carburized and hardened bearing surfaces is also every day. 

Crankshaft Design Procedure

The subsequent procedure has to be followed for design.

1. Calculate the magnitude of the different loads acts upon the crank shaft.
2. According to the loads, calculate the distance between the support structures and positions.
3. For simplistic and safe design, the shaft has to be supported at the bearings’ center and all the forces and reactions has to be acts upon at those points. The distance between the supports be subject to on the length of the bearing, which usually depend on the shaft’s dia as of the tolerable bearing pressures.
4. The thickness of the webs is expected to be from 0.4ds to 0.6ds, wherever “ds” is the shaft’s diameter. It usually considers as 0.22*D to 0.32*D, where D is the cylinder’s bore diameter in mm.
5. Here and now estimate the distance between the support structures.
6. Assuming the acceptable bending and shear stresses for Crank shaft material, find the dimension of the crankshaft.

Crankshaft Deflection

The crankshaft consists of the main shaft segments, individually reinforced by the main bearing, and then several web-shafts on which the specific piston connecting rod will rotate. The throw crank that is the crank pins and the connecting arms must be square with no deflection. If this is not the case, it causes unusual wear on the main bearings. A dial gauge detects the misalignment of the crank shaft between the crank arms. It is the uneven wear that occurs between the several segments of the crankshaft’s central axis.

Crankshafts Deflection Curve Plotting

• From the centerline of the crank shaft, A straight line is drawn parallel to it, and then perpendicular lines from each unit are drawn towards this parallel line.
• After taking the crank shaft deflection of each unit, the values derived are noted above every unit of the crank web in the above graph.
• Plot the distance -5.0 mm, which is the first deflection reading, downwards (for negative value and upwards for positive value) from the reference line on the center line of the unit and have the line “a-b” that is at an angle proportionate to the deflection at ‘a’.
• This line is extended to intersect the center line of the next unit. The subsequent step is to calculate the deflection from this point of joint and join the point from the preceding point, which will escalate to the line “b-c”. The steps have to be repeated again till completion.
• Plot a smooth curve between these points and compare this curve’s position with respect to the baseline XY. In the above graph, the curve drawn from the readings of unit 1 and 2 is being too far away from the baseline compared to the rest of the curve and hence need attention.

Marine Crankshaft failure Case Study

The case study done is about the tragic failure of a web marine crankshaft.  The crank shaft is subjected to high bending and torsion, and its combined effect on failure of the crank shaft is analyzed.  The microscopic observation suggested that the crack initiation began on the crankpin’s filet due to rotary bending, and the propagation was a combination of cyclic bending and steady torsion.  The number of cycles from crack initiation to the crank shaft’s final failure was found by readings of the main engine operation on board. Benchmarks left on the fatigue crack surface are taken into consideration.

By using the linear elastic fracture mechanics, the cycles calculated depicted that the propagation was quick. It also shows that the level of bending stress was quite high compared with total cycles of the main engine in service. Microstructure defects or inclusions were not observed; thus, it indicates that the failure was due to external cause and not the internal intrinsic defect.

The crank shaft material had configuration (42CrMo4 + Ni + V) (chemical composition, %: C = 0.39; Si = 0.27; Mn = 0.79; P = 0.015; S = .014; Cr = 1.14; Mo = 0.21; Ni = 0.45; V = 0.10). The crank shaft of the main engine has damaged. The crank-web no. 4 has broken. Material near the crack initiation region was analyzed, and it showed bainitic microstructure. The material had hardness vickers285.

The fatigue looks as if in two different surfaces, one vertical to the crank shaft and the other in the horizontal plane with the crank shaft with changeover zones among two planes. Thus, the tragic failure of the above marine crank shaft was by fatigue and a combined with the rotating-bending with the steady torsion. The research and observation and development of new crank shafts are in progress to avoid this type of failure.

Reference:

Fonte MA, Freitas MM. Marine main engine crank shaft failure analysis: A case study, Engineering Failure Analysis 16 (2009) 1940–1947

Failure Analysis of Boxer Diesel Crankshaft: Case Study

The report is about the failure mode analysis of boxer diesel engine crank shaft. Crank shaft is the component that experience a higher complex dynamic loading because of rotating bending supplemented with torsion and bending on crankpin. Crank shafts are subjected to multi-axial loading. Bending-stress and shear-stress due to twisting and torsional-loading because of power-transmissions. Crank shafts are manufactured from forged steel, nodular cast iron and aus-tempered ductile-iron.

They should possess adequate strength, toughness, hardness, and high fatigue strength. They must be easy to machine and heat treat and shaped. Heat treatment increases wear resistance; thus, all diesel crank shafts are heat treated. They are surface hardened to enhance fatigue strength. High-level stresses are observed on critical zones like web fillets and the effects of centrifugal force due to power transmission and vibrations. The fatigue fracture near the web fillet region is the major cause of crank shaft failure since the crack generation, and propagation occurs through this zone. 

The specifications of the crankshaft of a box motor are: displacement = 2000 cu. cm, diameter cylinder = 100 mm, max power = 150 HP, max torque = 350 N m. It has been observed that after 95,000 km in service, the failure of crank shaft takes place. Fatigue failure has occurred at nearly 2000 manufactured engines. After analysis, it has been noted that the weakness of two central steel shells and the yielding of bedplate bridges due to cracking were the main culprits of failure of crank shaft.

The crank shaft’s bending amplitude increases from cracked steel shells’ weakness and the bridges of the bedplate, which are beneath them. There was certainly no evidence of material defects or misalignment of main journal bearings. The devastating failure of the crank shaft was due to flawed design of steel support shells and bedplate bridges. The improved design from the manufacturer will solve this problem.

Reference:

M. Fonte et al., Crankshaft failure analysis of a boxer diesel motor, Engineering Failure Analysis 56 (2015) 109–115.

Crankshaft Fatigue Failure Analysis: A Review

In this paper, the root cause of fracture of the air compressor’s crank shaft is being analyzed using various methods and parameters like chemical composition, mechanical property, macroscopic, microscopic characteristics, and theoretic calculations. This paper also aims in improving the design, fatigue strength and work reliability of the crank shaft. The crank shaft used in this study is 42CrMo steel which is forged and heat-treated and nitridated to increase the fatigue strength of crank shaft.  The analyzing procedure for the cause of crank shaft fracture is carried out in three parts:

• Experimental analysis of crank shaft
• Macroscopic features and microstructure analysis
• Theoretical calculations

The chemical element analysis is being done to accurately determine the crank shaft material’s chemical composition and check if they are under the standard permissible values. It is done with the help of spectrometer. The fractured surfaces are classified into three regions: (1) fatigue crack initiation region, (2) fatigue expansion region and (3) static fracture region.

During the analysis, its w found that the fatigue crack growth rate is high due to high bending. The misalignment of main journals and small fillet to lubrication hole are the leading causes of high bending. The fatigue crack was initiated on the edge of the lubrication hole and thus led to the fracture. The beach marks produced due to small overloads because of starting and stopping the compressor were not visible. In a particular rotating cycle after a period of standard work, micro-cracks due to high bending stress concentration appeared on the lubrication hole’s fillet. However, the crank shaft can still close to normal working condition.

As the operating time went on increasing, the fluctuation also increased, leading the cracks to propagate to the static fracture region, leading to complete failure. The microscopic observation of the fracture surface measured utilizing Scanning Electron Microscopy (SEM), which showed that crack at the edge of the lubricating hole was the reason to fracture the crank shaft. According to the theoretical calculation, the curve for safety for the lubrication hole and fillet region is obtained, which helps identify the weakest sections.

By improving the surface quality and reducing surface roughness reliability of the crank shaft can be increased.  Proper alignment of main journals will reduce induced bending stress and increase the fatigue life of crank shaft.

Reference:

W.Li et al., Analysis of Crankshaft fatigue failure, Engineering Failure Analysis 55 (2015) 139–147.

Failure of Diesel Engine Crankshaft: A Case Study

In this paper, the failure analysis, modal, and stress analysis of a diesel engine’s crankshaft is conducted. To evaluate the fracture of crankshaft material, both the visual inspection and investigation were done. The engine used was S-4003, and its crank shaft was ruptured near the crankpin four after 5500 hours of operation.  The crankshaft was broken after about 30h to 700h of engine operation. The additional analysis showed the presence of micro-cracks near the 2nd crankpin and 2nd journal. The study showed that the primary reason behind the failure was a faulty grinding process.

For further experimental analysis, the specimen was cut from the damaged part. Non-linear Finite element analysis was used to identify the reasons for the abrupt failure of crankshaft. The analysis was performed for determining the stresses induced in the shaft due to cyclic loading conditions when the engine runs at maximum power.

Numerical analysis is used to find the relation between the connecting rod and the crankshaft by applying complex boundary conditions. For the determination of modes and frequency of free vibration, numerical modal analysis of the crankshaft was performed.

After the analysis, it was observed that the stress value in the fillet of the crankpin no.4 was about 6% of the yield stress of crankshaft material. The modal analysis gave the result that during the second mode of free vibration, the high-stress area was found in the area where the crack generation took place (critical zone).

On further observation, it was discovered that crankshaft failure occurred by resonant vibration generated due to unbalanced masses on the shaft, which induced high cyclic stress conditions, causing it to decrease crankshaft’s fatigue life.

Reference:

Lucjan Witek et al., Failure investigation of a crankshaft of diesel engine, Procedia Structural Integrity 5 (2017) 369–376

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In real life, we may encounter many cases in which material is applied to tension or compression in 2 perpendicular direction at that time. The stress applied in such a case is known as biaxial stress. A balloon is a perfect example of it.

These tensile/compressive stresses also produce shear stress in the material. To calculate the net tensile and shear stress produced in the material, a graphical method is used known as Mohr’s Circle for biaxial stress.

Mohr’s circle is an advantageous and easy way to solve stress equations. It gives the information about the stresses on various planes.

Topic of Discussion: Mohr’s Circle

How to Draw Mohr’s Circle | How Do You Plot Mohr’s Circle?

Let us consider a thin sheet subjected to biaxial tension, as shown in the following figure. The normal and shear stress on a plane whose normal n have an angle of ϕ with the x-axis are specified as follows:

From the above equations, it can be said that these equations can be plotted as a circle in a normal stress-shear stress plane where angle ϕ acts as a parameter.

As we know:

So, normal stress and shear stress can be represented in more compact form as follow:

By solving the above equations and eliminating parameter ϕ.

Substitute of this in the first of

The above equation denotes the standard form of the equation of a circle.

On solving the above equation, we get that radius of the circle formed is

The Center of the circle formed is on σ-axis denoted as

Circle formed on the σ-τ plane with the above parameters is known as Mohr ‘s Circle.

If the applied principal stress applied is of compression kind, it must be taken with the negative sign.

Thus, the origin of Mohr ‘s circle always lies on the σ-axis.

Mohr’s Circle Equations

Following are the standard equations formed of Mohr circle.

Where,

How to Use Mohr’s Circle

Mohr’s circle is the circle drawn in the plane of σ-τ. σ is on the x-axis, which is the total of the normal force acting on the material. τ is on the y-axis, which is the total shear stress acting on the same plane, hence if we take any point on the Mohr’s circle, its x-coordinate gives the value of total normal stress acting on the material, and y-coordinate gives the value of total shear stress acting on the material.

For figure 2, let’s take a point D on it. The x-coordinate gives the value of total normal stress acting on it, and the y-coordinate gives the value of total shear stress acting on it.

From the geometry, it can be seen that the coordinates of point D are

Where OE is an x-coordinate, and DE is a y-coordinate.

For each condition of the material in figure 1 defined by ϕ, there is a corresponding point denoting it on the Mohr’s circle in figure 2.

Let’s say, when ϕ = 0, and normal n coincides the x-axis, and it gives σ_n = σ_x

And τ = 0.

When ϕ = 90⁰, the normal n coincides with the x-axis, and it gives σ_n = σ_y

And τ = 0.

When ϕ = 45⁰, the normal n coincides with the x-axis, and it gives

And

The Mohr’s failure envelope

Failure is the particular value of normal stress or shear stress at which material breaks or develops a crack.

Mohr ’s circle can be used to know the normal and shear stress values at the point of failure.

A material has multiple failure values of shear stresses and normal stresses. Thus, the Mohr Failure Envelope is a locus of all failure such failure points.

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Mohr’s Circle Pressure Vessel

The stress experienced by any pressure vessel is the biaxial type of stress. It gives the impression due to pressure experienced by the wall of the pressure vessel can also have stresses generated by the weight of the under pressure fluid inside, its weight, and externally applied load and by an functional torque.

Mohr’s circle is used to denote the stresses developed in the vessel.

Questions and Answers

What is Mohr’s circle used for?

In real life, we may come across many cases in which material is subjected to tension or compression in two perpendicular direction at the identical time. The stress applied in such a case is known as biaxial stress. A balloon is a perfect example of it.

These tensile/compressive stresses also produce shear stress in the material. To calculate the net tensile and shear stress produced in the material, a graphical method is used known as Mohr ‘s Circle for biaxial stress.

What are the principal stresses?

Principal stresses are maximum and minimum stresses at a point on the material. These stresses include only normal stresses and do not include shear stresses.

What are the three principal stresses?

There are mainly three principal stresses as follows:

1) σ1= maximum (most tensile) principal stress

2) σ3= minimum (most compressive) principal stress,

3) σ2= intermediate principal stress.

What is Mohr’s circle of stress?

In real life, we may see numerous cases in which material is subjected to tension or compression in two perpendicular direction at that time. The stress applied in such a case is known as biaxial stress. A balloon is a perfect example of it.

These tensile/compressive stresses also produce shear stress in the material. To calculate the net tensile and shear stress produced in the material, a graphical method is used known as Mohr ‘s Circle for biaxial stress.

What is the radius of Mohr’s circle?

The radius of Mohr’s Circle formed is as follows:

τ_max=1/2(σ_x-σ_y)

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