Engineering

Topic of discussion: Isentropic process

Isentropic Definition

A typical case of an adiabatic process that has no transfer of heat or matter through the process while the entropy of the system remains constant is known as an isentropic process.

The thermodynamic process where the entropy of the gas or fluid remains constant can also be coined as the reversible adiabatic process. This type of process that is both adiabatic in nature and internally reversible while considering that it is frictionless enables the engineering sector to view this as an idealized process and a model for comparing actual processes.

Ideally, enthalpy of the system is used in the particular isentropic process as the only variables changing are internal energy dU and system volume ΔV while the entropy remains unchanged.

 The T-s diagram for an isentropic process is plotted based on the known traits varying from different states such as the pressure and temperature quantity. Since,

 ΔS = 0 or s1 = s2

And,

H = U + PV

They are intrinsically related to the first law of thermodynamics in terms of enthalpy measure. Since it both reversible and adiabatic, the equations formed would be as follows:

In enthalpy terms,

Or,

The water, refrigerants, and ideal gas can be derived using the equations in the molar form to deal with the enthalpy and temperature relation. At the same time, the specific entropy of the system remains unchanged.

From the enthalpy equation abiding by the first law of thermodynamics, VdP is considered a flow process work where a mass flow is involved as work is required to transfer the fluid in or out of the boundaries of the control volume. This flow energy (work) is generally utilized for systems with the difference in pressure dP, like an open flow system found in turbines or pumps. By simplifying the energy transfer description, it is derived that enthalpy change is equivalent to flow energy or process work done on or by the system at constant entropy.

For,

Isentropic process for an ideal gas

Now, for an ideal gas, the isentropic process where entropy changes are involved can be represented as:

Integrating and rearranging,

(this is by assuming constant specific heats)

Where k is the specific heat ratio

Now, setting

Consolidated expressions of the three relations of the equations in compact form can be projected as:

If the specific heat constant assumptions are invalid, the entropy change would be:

If the numerator of the above equation is construed as the relative pressure, then:

Pressure vs temperature values are tabulated against each other. Hence, the ideal gas relation produces:

Defining the relative specific volume,

Isentropic process derivation

The total energy change in a system:

A reversible condition involving work with pressure is,

As established earlier,

For isentropic,

And,

Now,

Capacity ratio:

Where, r=density

As dS=0,

After substitution of PV=rRT equation in the above equation,

Differentiating,

Substituting the gamma equation,

Simplifying the equation:

Integrating,

For the flow brought to rest isentropically, the total pressure and density occurring can be evaluated as a constant.

pt being the total pressure and rt being the total density of the system.

Now, by combining the equations:

Isentropic work equation

While satisfying the isentropic equations respectively under enthalpy and entropy values.

Isentropic turbine and isentropic expansion

For the purpose of calculations, the adiabatic process for the steady flow devices such as turbines, compressors or pumps is ideally generated as an isentropic process. Specific ratios are evaluated for calculating the efficiency of steady flow machines by including parameters that intrinsically affect the overall system of the process.

Typically, the particular device’s efficiency ranges from 0.7-0.9, which is about 70-90%.

While,

Summary and conclusion

The Isentropic process, ideally known as a reversible adiabatic process, is exclusively used in the various thermodynamic cycles such as Carnot, Otto, Diesel, Rankine, Brayton cycle and so on. The numerous mathematical equations and tables plotted utilizing the isentropic process parameters are basically used to determine the efficiency of gases and flows of the systems that are steady in nature such as turbines, compressors, nozzles, etc.

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Contents: Deflection of Beam

• Deflection Curve Definition
• Deflection Angle Definition
• Deflection Definition
• Beam deflection boundary conditions
• Relationship between Loading forces, shear force, bending moment, slope, and deflection
• Beam Bending equations and relations
• Beam deflection table and Formulas for standard load cases
• Beam Deflection and slope with examples Case I: Overhanging Beam
• Case II: Determine the maximum deflection of simply supported beam with point load at the center
• Case III: Determine the maximum deflection of simply supported beam with a concentrated point load at a distance ‘a’ from support A
• Double Integration Method
• Procedure for Double Integration Method
• Double integration method for finding beam deflection using Example of a cantilever beam with Uniformly distributed load
• Double integration method for Triangular Loading

In engineering, deflection is the degree to which a structural element is displaced under a load (due to its deformation). It may refer to an angle or a distance. The deflection distance of a member under a load can be calculated by integrating the function that mathematically describes the slope of the deflected shape of the member under that load. Standard formulas exist for the deflection of common beam configurations and load cases at discrete locations. Otherwise methods such as virtual work, direct integration, Castigliano’s method, Macaulay’s method or the direct stiffness method are used.

Deflection Curve

When beams are loaded by lateral or longitudinal loads, the initial straight longitudinal axis is deformed into a curve known as the beam’s elastic curve or deflection curve. The deflection curve is the deformed axis of the selected beam.

Deflection Angle

The slope can be defined as the angle between the beam’s longitudinal axis and the tangent constructed to the beam’s deformation curve at any desired location. It is the angle of rotation of the neutral axis of the beam. It is measured in Radians.

Deflection

Deflection is the translation or displacement of any point on the axis of the beam, measured in the y-direction from the initial straight longitudinal axis to the point on the deflection curve of the beam. It is measured in mm. Deflection represents the deviation of the straight longitudinal axis due to transverse loading. In contrast, buckling of the beam represents the deviation of the initial straight longitudinal axis due to axial compressive load. It is usually represented by ‘y’

If the beam bends like the arc of a circle, it is called circular bending; otherwise, it is called non-circular bending. Suppose a Prismatic beam is subjected to a variable bending moment. In that case, it results in a non-circular type bending, and if it is subjected to constant Bending moment results in circular bending of the beam.

Beam deflection boundary conditions

1. y is zero at a pin or roller support.
2. y is zero at a built-in or cantilever support.
3. Suppose the bending moment and flexural rigidity are discontinuous functions of the x. In that case, a single differential equation cannot be written for the entire beam; the equations of the curve for two adjacent segments should satisfy the given two conditions at the junction between segments:

• 1. The y for the left-hand section must be equal to the y for the right-hand section.
• 2. The slope for the left-hand section must be equal to the slope for the right-hand section.

Relationship between Loading forces, shear force, bending moment, slope, and deflection

Consider a Horizontal Beam AB in unloaded condition. If AB deflects under the load, the new position will be A’B’. The slope at any point C will be

i=\\frac{dy}{dx}

Usually, the deflection is minimal, and for a small radius of curvature,

ds=dx=Rdi
\\\\\\frac{di}{dx}=1/R

But\\;i=\\frac{dy}{dx}

Thus,

\\frac{d^2 y}{ dx^2}=1/R  

According to the simple bending moment theory

\\frac{M}{I}=\\frac{E}{R}

\\frac{1}{R}=\\frac{M}{EI}

Thus,

\\frac{d^2 y}{dx^2}=\\frac{1}{R}=\\frac{M}{EI}

Where,

E = Young’s Modulus of the material

I = Area moment of inertia

M = Maximum Moment

R = Radius of curvature of the beam

This is the Basic differential equation for the deflection of the beam.

Beam Bending equations and relations

Deflection = y

Slope = \\frac{dy}{dx}

Bending\\;moment =EI\\frac{d^2y}{dx^2}

Shear\\; Force = EI\\frac{d^3y}{dx^3}

Load \\;distribution =EI\\frac{d^4y}{dx^4}

Beam deflection table and Formulas for standard load cases:

• Maximum slope and deflection in a cantilever beam occur at the free end of the beam, while no slope or deflection is observed on the clamped end of a cantilever beam.
•  For a simply supported beam with symmetric loading conditions, the maximum deflection can be found at the midspan. The maximum slope can be observed at the supports of the beam. Maximum deflection occurs where the slope is zero.

Beam Deflection and slope with examples

Case I: Overhanging Beam

Consider an overhanging steel beam carrying a concentrated load P = 50 kN at end C.

For The overhanging beam, (a) determine the slope and maximum deflection, (b) evaluate slope at 7m from A and maximum deflection from given data I = 722 cm² , E = 210 GPa.

Solution: The Free body diagram for the given beam is

The value of the reaction at A and B can be calculated by applying Equilibrium conditions

\\sum F_y=0\\;\\sum M_A=0

For vertical Equilibrium, F_y = 0

R_A+R_B=P

Taking a moment about A, Clockwise moment positive and Counter Clockwise moment is taken negative.

P(L+a)-R_B*L=0
\\\\R_B=P(1+a/L)

Thus,

R_A+P(1+\\frac{a}{L})=P

R_A= \\frac{-Pa}{L}

Consider any section AD at a distance x from support A

The moment at point D is

M= \\frac{-Pa}{L x}

Using the differential equation of the curve,

EI \\frac{d^2 y}{dx^2}= \\frac{-Pa}{L x}

Integrating twice, we get

EI \\frac{dy}{dx}= \\frac{-1}{2}  \\frac{Pa}{L }x^2+C_1……………..[1]

EIy=\\frac{-1}{6}  \\frac{Pa}{L }x^3+C_1x+C_2……………..[2]

We find the constants of integration by using the boundary conditions available to us

At x = 0, y = 0; from equation [2] we get,

C_2=0

At x = L, y = 0; from equation [2] we get,

0=\\frac{-1}{6}  \\frac{Pa}{L }*L^3+C_1*L+0

C_1= \\frac{PaL}{6}

Thus, the equation of slope so obtained by substituting the values of C₁ and C₂ in [1]

EI \\frac{dy}{dx}= \\frac{-1}{2}  \\frac{Pa}{L }x^2+\\frac{PaL}{6}……………..[3]

Thus, the equation of deflection so obtained by substituting the values of C₁ and C₂ in [2]

EIy=\\frac{-1}{6}  \\frac{Pa}{L }x^3+\\frac{PaL}{6}x……………..[4]

Maximum deflection takes place when the slope is zero. Thus, the location of the point of maximum deflection can be found from [3]:

0= \\frac{-1}{2}  \\frac{Pa}{L }x^2+\\frac{PaL}{6}

 \\frac{1}{2}  \\frac{Pa}{L }x^2=\\frac{PaL}{6}

x_m=\\frac{L}{\\sqrt 3}

x_m=0.577 L

Putting the value of x in equation [4]

EIy_{max}=\\frac{-1}{6}  \\frac{Pa}{L }x_m^3+\\frac{PaL}{6}x_m

EIy_{max}=\\frac{-1}{6}  \\frac{Pa}{L }*0.577 L^3+\\frac{PaL}{6}*0.577 L

y_{max}=0.064\\frac{Pal^2}{EI}

Evaluate slope at 7m from A from given data:

 I = 722 \\;cm^4=72210^{-8}\\; m^4 , E = 210\\; GPa = 210*10^9\\; Pa

Using equation [3]

EI \\frac{dy}{dx}= \\frac{-1}{2}  \\frac{Pa}{L }x^2+\\frac{PaL}{6}

210*10^9*722*10^{-8}* \\frac{dy}{dx}= \\frac{-1}{2}  \\frac{50*10^3*4}{15 }*7^2+\\frac{50*10^3*4*15}{6}

\\frac{dy}{dx}=0.5452 \\;radians

maximum deflection in the beam can be given by

y_{max}=0.064\\frac{Pal^2}{EI}

y_{max}=0.064\\frac{50*10^3*4*15^2}{210*10^9*722*10^{-8}}

y_{max}=1.89 \\;m

Case II: Determine the maximum deflection of simply supported beam with point load at the center.

Consider a simply supported steel beam carrying a concentrated load F = 50 kN at Point C. For the Simply supported beam, (a) evaluate slope at A and maximum deflection from given data: I = 722 cm⁴ , E = 210 GPa, L =15 m

The Figure below shows the FBD for a simply supported beam with Point load on it.

According to standard relations and formula

Slope at the end of the beam can be given by

\\frac{dy}{dx}=\\frac{FL^2}{16EI}

\\frac{dy}{dx}=\\frac{50*10^3*15^2}{16*210*10^9*722*10^{-8}}

\\frac{dy}{dx}=0.463

For a simply supported beam with point load acting at the center, Maximum Deflection can be determined by

y_{max}=\\frac{FL^3}{48EI }

y_{max}=\\frac{50*10^3*15^3}{48*210*10^9*722*10^{-8} }

y_{max}=2.31 \\;m

Case III: For Simply supported beam with a concentrated point load at a distance from support A

Consider a simply supported steel beam carrying a concentrated load F = 50 kN at Point C. For the Simply supported beam, (a) evaluate slope at A and B and maximum deflection from given data: I = 722 cm⁴ , E = 210 GPa, L =15 m, a = 7 m, b = 13 m

The Figure below shows the FBD for a simply supported beam with Point load on it.

According to standard relations and formula

Slope at the support A of the beam can be given by

\\theta_1=\\frac{Fb(L^2-b^2)}{6LEI}

\\theta_1=\\frac{50*10^3*13*(20^2-13^2)}{6*20*210*10^9*722*10^{-8}}

\\theta_1=0.825 \\;radians 

The slope at the support B of the beam can be given by

\\theta_2=\\frac{Fab(2L-b)}{6LEI}

\\theta_2=\\frac{50*10^3*7*13*(2*20-13)}{6*20*210*10^9*722*10^{-8}}

\\theta_2=0.675 \\;radians

For a simply supported beam with point load acting at the center, Maximum Deflection can be determined by

y_{max}=\\frac{50*10^3*13}{48*210*10^9*722*10^{-8} }*(3*15^2-4*13^2)

y_{max}=-8.93*10^{-3}\\; m=-8.93\\;mm

Double Integration Method

If Flexural rigidity EI is constant and the moment is the function of distance x, Integration of EI (d² y)/(dx² )=M will yield Slope

EI \\frac{dy}{dx}=\\int M dx+C_1

EIy=\\int \\int Mdxdx+C_1x+C_2

where C₁ and C₂ are constants. They are determined by using the boundary conditions or other conditions on the beam. The above equation gives the deflection y as a function of x; it is called the elastic or deformation curve equation.

The above analysis method of deflection and slope of the beam is known as the double-integration method for calculating beam deflections. If the bending moment and flexural rigidity are continuous functions of the x, a single differential equation can be noted for the entire beam. For a statically determinate Beam, there are two support reactions; each imposes a given set of constraints on the elastic curve’s slope. These constraints are called boundary conditions and are used to determine the two constants of integration.

Double integration method boundary conditions

1. y is zero at a pin or roller support.
2. y is zero at a built-in or cantilever support.
3. Suppose the bending moment and flexural rigidity are discontinuous functions of the x. In that case, a single differential equation cannot be written for the entire beam; the equations of the curve for two adjacent segments should satisfy the given two conditions at the junction between segments:

• 1. The y for the left-hand section must be equal to the y for the right-hand section.
• 2. The slope for the left-hand section must be equal to the slope for the right-hand section.

Procedure for Double Integration Method

• Draw the elastic curve for the beam and consider all the necessary boundary conditions, such as y is zero at a pin or roller support and y is zero at a built-in or cantilever support.
• Determine the bending moment M at an arbitrary distance x from the support using the sections’ method. Use appropriate Bending Moment rules while finding Moment M. for a discontinuous moment, the equations of the curve for two adjacent segments should satisfy the given two conditions at the junction between segments: 1. The y for the left-hand section must be equal to the y for the right-hand section. 2. The slope for the left-hand section must be equal to the slope for the right-hand section.
• Integrate the equation twice to get the slope and deflection, and don’t forget to find the constant integration for every section using boundary conditions.

Examples of double integration method for finding beam deflection

Consider the Cantilever beam of length L shown in the Figure below with Uniformly distributed load. In a Cantilever beam, one end is Fixed while another end is free to move. We will derive the equation for slope and bending moment for this beam using the Double integration method.

The bending moment acting at the distance x from the left end can be obtained as:

M=-wx* \\frac{x}{2}

Using the differential equation of the curve,

\\frac{d^2y}{dx^2}=M = \\frac{-wx^2}{2}

Integrating once we get,

EI \\frac{dy}{dx}= \\frac{-wx^3}{6}+C_1………..[1]

Integrating equation [1] we get,

EIy= \\frac{-wx^4}{24}+C_1 x+C_2……..[2]

The constants of integrations can be obtained by using the boundary conditions,

At x = L, dy/dx = 0; since support at A resists motions. Thus, from equation [1], we get,

C_1=\\frac{wL^3}{6}

At x = L, y = 0, No deflection at the support or fixed end A Thus, from equation [2], we get,

0= \\frac{-wL^4}{24}+\\frac{wL^3}{6} *L+C_2

C_2= \\frac{-wL^4}{8}

 Substituting the constant’s value in [1] and [2] we get new sets of equation as

EI \\frac{dy}{dx}= \\frac{-wx^3}{6}+\\frac{wL^3}{6}………..[3]

EIy= \\frac{-wx^4}{24}+\\frac{wL^3}{6} -\\frac{wL^4}{8}……..[4]

evaluate slope at x = 12 m and maximum deflection from given data: I = 722 cm⁴ , E = 210 GPa, L =20 m, w = 20 N-m

From the above equations: at x = 12 m,

EI \\frac{dy}{dx}= \\frac{-wx^3}{6}+\\frac{wL^3}{6}

210*10^9*722*10^{-8}* \\frac{dy}{dx}= \\frac{-20*12^3}{6}+\\frac{20*20^3}{6}

\\frac{dy}{dx}=0.01378 \\;radians

From equation [4]

EIy= \\frac{-wx^4}{24}+\\frac{wL^3}{6} -\\frac{wL^4}{8}

210*10^9*722*10^{-8}*y= \\frac{-20*12^4}{24}+\\frac{20*20^3}{6} -\\frac{20*20^4}{8}

y=-0.064 \\;m

Double integration method for Triangular Loading

Consider the Simply supported beam of length L shown in the Figure below with Triangular Loading. We will derive the equation for slope and bending moment for this beam using the Double integration method.

Since the loading is symmetric, each support reaction will bear half of the total loading. The reaction at A and B are found to be wL/4.

Moment at any point at a distance x from R_A is

M=\\frac{wL}{4} x- \\frac{wx^2}{L}\\frac{x}{3}=\\frac{w}{12L} (3L^2 x-4x^3 ) 

 \\frac{d^2 y}{dx^2}=M=\\frac{w}{12L} (3L^2 x-4x^3 ) 

Integrating twice will get us the equations,

EI \\frac{dy}{dx}=\\frac{w}{12L}(\\frac{3L^2x^2}{2}-x^4)+C_1...........................[1]

EIy=\\frac{w}{12L} (\\frac{L^2x^3}{2}-\\frac{x^5}{5})+C_1 x+C_2……..[2]

At x = 0, y = 0; from equation [2] we get,

C_2=0

Due to symmetry of Load, the slope at midspan is zero. Thus, dy/dx = 0 at x = L/2

0=\\frac{w}{12L}(\\frac{3L^2*L^2}{2*4}-(L^4/16))+C_1

C_1=\\frac{-5wL^3}{192}

Substituting the constants value in [1] and [2] we get,

EI \\frac{dy}{dx}=\\frac{w}{12L}(\\frac{3L^2x^2}{2}-x^4)+\\frac{-5wL^3}{192}...........................[3]

EIy=\\frac{w}{12L} (\\frac{L^2x^3}{2}-\\frac{x^5}{5})+\\frac{-5wL^3}{192} x……..[4]

The Maximum deflection will be observed at the center of the beam. i.e., at L/2

EIy=\\frac{w}{12L} (\\frac{L^2(L/2)^3}{2}-\\frac{(L/2)^5}{5})+\\frac{-5wL^3}{192}(L/2)

EIy_{max}=\\frac{w}{12L} (\\frac{L^5}{16}-\\frac{L^5}{160})+\\frac{-5wL^4}{384}

EIy_{max}=\\frac{-wL^4}{120}

evaluate slope at x = 12 m and maximum value of y from given data: I = 722 cm⁴ , E = 210 GPa, L =20 m, w = 20 N-m

From the above equations: at x = 12 m,

EI \\frac{dy}{dx}=\\frac{w}{12L}(\\frac{3L^2x^2}{2}-x^4)+\\frac{-5wL^3}{192}

210*10^9*722*10^{-8}* \\frac{dy}{dx}=\\frac{20}{12*20}(\\frac{3*20^2*12^2}{2}-12^4)+\\frac{-5*20*20^3}{192}

\\frac{dy}{dx}=8.60*10^{-4 } \\;radians

From equation [4]

EIy_{max}=\\frac{-wL^4}{120}

210*10^9*722*10^{-8}*y=\\frac{-20*20^4}{120}

y=-0.01758\\;m

To know about Strength of material(click here)and Moment Area method Click here.

Topic of Discussion: Adiabatic Process

• Adiabatic process definition
• Adiabatic process examples
• Adiabatic process formula
• Adiabatic process derivation
• Adiabatic process work done
• Reversible adiabatic process and Irreversible adiabatic process
• Adiabatic graph

Adiabatic process definition

Abiding the first law of thermodynamics, the process occurring during expansion or compression where there is no heat exchanged from the system to the surroundings can be known as an adiabatic process. Differing from the isothermal process, adiabatic process transfers energy to the surrounding in the form of work. It can be either reversible or an irreversible process.

In reality, a perfectly adiabatic process can never be obtained since no physical process can happen spontaneously nor a system can be perfectly insulated.

Following the first law of thermodynamics that says when energy (as work, heat, or matter) passes into or out of a system, the system’s internal energy changes accordingly with the law of conservation of energy, where ​E​ can be denoted as the internal energy, while ​Q​ is the heat added to the system and ​W​ is the work done.

ΔE=Q−W

For an adiabatic process where there is no heat exchanged,

ΔE=−W

Conditions required for an adiabatic process to take place are:

• The system must be completely insulated from its surroundings.
• For the heat transfer to occur in a sufficient amount of time, the process must be performed quickly.

Adiabatic process Example

1. Expansion process in an internal combustion engine found among hot gases.
2. The quantum-mechanic analogue of an oscillator classically known as the quantum harmonic oscillator.
3. Gases liquified in a cooling system.
4. Air released from a pneumatic tire is the most significant and common instance of an adiabatic process.
5. Ice stored in an icebox follows the principles of heat not being transferred in and out to the surroundings.
6. Turbines, using heat as a medium to generate work, is considered an excellent example as it reduces the efficiency of the system as the heat is lost to the surroundings.

Adiabatic process formula

The expression of an adiabatic process in mathematical terms can be given by:

ΔQ=0

Q=0,

ΔU= -W, (since there is no heat flow in the system)

Therefore,

Consider a system where the exclusion of heat and work interactions on a stationary adiabatic process is performed. The only energy interactions are the boundary work by the system in its surroundings.

Ideal gas

The amount of thermal energy per unit temperature unavailable to perform specific work can be defined as the entropy of a system. A speculative gas that comprises the random motion of point particles subject to interparticle molecular interactions is ideal.

The molar form of the ideal gas formula is given by:

Integrating the equations,

Adiabatic process equation can be denoted as:

PVY = constant

Where,

• P= pressure
• V= volume
• Y= adiabatic index; (C_p/C_v)

For a reversible adiabatic process,

• P^1-YT^Y = constant,
• VT^f/2 = constant,
• TV^Y-1 = constant. (T = absolute temperature)

This process is also known as the isentropic process, an idealized thermodynamic process containing frictionless work transfers and adiabatic. In this reversible process, there is no transfer of heat or work.

Adiabatic process derivation

The alteration in internal energy dU in a system to work done dW plus the heat added dQ to it can be associated as the first law of thermodynamics through which the adiabatic process can be derived.

According to the definition, 

Hence,

Addition of heat escalates the amount of energy U defining the specific heat as the amount of heat added for a unit rise in temperature change for 1 mole of a substance.

(n is the number moles), Therefore:

Derived from the ideal gas law,

Merging equation 1 and 2,

For a constant pressure C_p, heat is added and,

γ is the specific heat

Using the integration and differentiation concepts, it is arrived at:

This equation above becomes real for a given ideal gas that contains the adiabatic process.

Adiabatic process Work done.

For a pressure P and a cross-sectional area A moving through a small distance dx, the force acting would be given by:

And the work done on the system can be written as:

Since,

The net work produced for the expansion of the gas from the volume of the gas V_i to V_f (initial to the final) will be given as

W= area of ABDC from the graph plotted as the adiabatic process takes place. The conditions to be followed are associated with an example of a perfectly non-conducting piston cylinder with a single gram molecule of a perfect gas. The cylinder’s container is to be made of an insulating material, and the curve plotted by the graph should be sharper.

Whereas, in an analytical method to derive the work done on the system would be as follows:

—–(1)

Initially, for an adiabatic change, we can assume:

Which can be,

From (1),

For solving,

Thus,

Which is,

Taking T₁ and T₂ as the initial and final temperatures of the gas respectively,  

 Using this in equation (2),

Or,

—-(3)

The heat required during the expansion process to do the work is:

As R is the universal gas constant and during adiabatic expansion, the work done is directly proportional to the decrease in temperature, while the work done during an adiabatic compression is negative.

Hence,

Or,

This can be given as the work done in an adiabatic process.

And the heat expelled during the process is:

Adiabatic graph

The mathematical representation of the adiabatic expansion curve is represented by:

P,V,T are the pressure, volume, and temperature of the process. Considering the initial stage conditions of the system as P₁, V₁, and T₁, also defining the final stage as P₂, V₂, and T₂ respectively the P-V graph diagram is plotted essentially for a piston cylinder movement heated adiabatically from the initial to final state for a m kg of air.

Adiabatic entropy, adiabatic compression and expansion

A gas allowed to expand freely without the transfer of external energy to it from higher pressure to a lower pressure will essentially cool by the law of adiabatic expansion and compression. Likewise, a gas will heat up if it is compressed from a lower temperature to a more significant temperature without the substance’s transfer of energy.

• Air parcel will expand if the surrounding air pressure is reduced.
• There is a decrease in temperature at higher altitudes due to the diminish in the pressure as they are directly proportional in the case of this process.
• Energy can either be utilized to do work for expansion or to maintain the temperature of the process and not both at the same time.

Reversible adiabatic process

The frictionless process where the system’s entropy remains constant is coined as the term reversible or isentropic process. This means that the change in entropy is constant. The internal energy is equivalent to the work done in the expansion process.

Since there is no heat transfer,

Thus,

Which means that,

Examples of a reversible isentropic process can be found in gas turbines.

Irreversible adiabatic process

As the name suggests, the internal friction dissipation process resulting in the change in entropy of the system during the expansion of gases is an irreversible adiabatic process.

This generally means the entropy increases as the process furthers that cannot be performed at equilibrium and cannot be tracked back to its original state.

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Cohesion and Adhesion

First of all we try to understand some terms useful in surface tension study. Liquid has properties like Cohesion. Cohesion is a property in which one molecule of liquid attracting another nearer molecule. Adhesion is a property in which the fluid molecules are attracted by solid surface contact with it. In short, we can say that the force between similar molecules is Cohesion and the force between dissimilar molecules is adhesion.

Let’s take an example.

If we drop mercury droplet on any surface, it tries to form in droplet because Cohesion is higher than the adhesion force. You will get a notice that Mercury droplet does not stick on the solid surface. The Mercury will try to stay away from the solid surface; it will not wet solid surface.

Now let’s take another example if we consider water particles fall on the surface. It will spread all over the concrete surface. It happens because of the adhesive force is more significant than Cohesive force in that case. The angle of contact between the liquid and the solid surface can describe wetting and non-wetting of the surface.

Refer the above figure the liquid gas and solid surface interface the liquid will where the solid surface when the angle is less than 90 degree (π/2). The wetting of the surface is increasing with decreasing the angle. If the angle is more than 90 degree, the liquid will not wet solid surface. The angle depends on the nature of the surface, types of liquid, solid surface, and cleanliness.

If we consider pure water comes in contact with the clean glass surface. The angle is 0 (zero) degree in that case. If we add impurities in the water: The angle will increase with the addition of impurities. As we have discussed the Mercury is non-wetting liquid, so the angle is lies from 130 to 150 degree.

Surface Tension

In liquid, the molecules are lying below the free surface. Every molecule of liquid is attracting to the molecule nearby. The molecular Cohesion force is the same in all direction. All the forces are the same in magnitude and opposite in direction. So, it will get cancel in liquid. It can be the reason for equilibrium in liquid. There is no resultant force present in the liquid.

Suppose we considered topmost molecules of liquid lying at a free surface as we know that there are no liquid molecules over them. So here, they are getting attracted by liquid molecules lying below them. This free surface liquid molecules will feel pull force interior of the liquid. This force acts like elastic force. Expended per unit area of the surface is called surface tension.

The Surface tension is denoted by Sigma (σ). Surface tension occurs at liquid-gas interface, liquid-liquid interface. The reason behind surface tension is an intermolecular attraction because of Cohesion.

Let’s understand it in depth by considering some practical examples,

• You have seen liquid droplet of a spherical shape. The reason behind its spherical shape is surface tension.
• You might be noticed that if we thoroughly pour water inside the glass. Even if the glass is filled, still we can add some water above the glass limit.
• Suppose we will experiment with a thin glass tube on the water surface. We can quickly notice a capillary rise and depression inside a thin glass tube.
• Birds can drink water from the water body due to surface tension.

Though pressure and the gravity force are higher than surface tension force, the surface tension force plays an important role when there are free surface and small dimensions. The unit of surface tension is N/m. The magnitude of surface tension depends on the following factors:

• Type of liquid
• Type of surrounding state gas, liquid or solid
• The kinetic energy of molecules
• Temperature of molecules

If we increase the temperature of substance like liquid, the intermolecular attraction is decreasing because the distance between molecules increases. The surface tension depends on intermolecular attraction (Cohesion). The value of surface tension for liquid is taken for air as a surrounding medium,

The surface tension for the air-water interface is 0.073 N/m.

The value of surface tension decreases with increasing temperature.

Capillary

If a narrow tube is dipped into the water, the water will rise inside the tube at a certain level. This type of tube is called a capillary tube, and this phenomenon is called the capillary effect. Another name of the capillary effect is the meniscus effect.

The capillary effect is due to surface tension force. The capillary rise and depression are happening because of cohesion and adhesion intermolecular attraction. The adhesion force between tube surface and a water molecule is higher than the Cohesion force between water molecules. Because of this, the water molecules can be observed in concave shape on the tube surface.

Weight of liquid rise or depress in small diameter tube

= ( Area of tube * Rise or fall ) * ( specific weight )

= (π/4 *d²*h) w

Verticle component of surface tension force

= σ cosθ * circumference

= σ cosθ * πd

If we consider equilibrium then upward force balances downward force, so the component of force is given as,

( π/4 * d² *h * w ) = σ cosθ * πd

H = ( 4 σ cosθ/ wd )

It can be observed from an angle that if the angle is between 0 to 90 degree, the value of h is positive, concave shape formation and capillary rise. If the angle is between 90 and 180 degrees, the h is negative, convex shape formation and capillary depression.

If the liquid is Mercury, then the effect is wholly turned opposite. In the case of Mercury, the Cohesion force is more significant than the adhesion force. Because of these, the Mercury molecules form convex shape on the tube surface.

The capillary effect is inversely proportional to the tube diameter. If you want to avoid the capillary effect, then you should not choose a small diameter tube. The minimum tube diameter is recommended for water, and Mercury is 6 mm. The surface inside the tube should be clean.

Evaporation

Evaporation is defined as a change of state from liquid to gaseous. Operation rate is dependent on the pressure and temperature condition of liquid.

Consider one example,

Suppose, the liquid is inside the closed vessel. In this vessel, the vapour molecules possess some pressure. It is called vapour pressure. If the vapour pressure starts decreasing then the molecule starts leaving from liquid surface very fast, this phenomenon is known as boiling.

In boiling, the bubbles are formed inside the liquid. This bubble travels near to higher pressure zone and collapses due to higher pressure. These collapsing bubbles are exerting significantly higher pressure around 100 atmospheric pressure. This pressure causes mechanical erosion on metal. Commonly, this effect is called Cavitation. It is required to study and design hydrodynamic machinery considering Cavitation.

Cavitation has both sides beneficial and non-beneficial. As we know that Cavitation cause erosion in metal, so it is non-beneficial

Some new research areas recently suggest that hydrodynamic Cavitation is useful for some chemical and wastewater treatment. So here, hydrodynamic Cavitation is a beneficial concept.

The vapour pressure of the liquid firmly depends on temperature: It increases with increase in temperature. At the temperature of 20°C, the vapour pressure of water is 0.235 N/cm². The vapour pressure of Mercury is 1.72* 10^-5 N/cm².

If we want to avoid Cavitation in hydraulic machinery: We should not allow liquid pressure to fall below vapour pressure at the local temperature.

You might have thought many times that why the Mercury is used inside the thermometer and manometer. Why not other liquid?

Your answer is here; the Mercury has the lowest value of vapour pressure with high density. Its make Mercury suitable for use in thermometer and manometer. 

Find the capillary effect in a tube of diameter 4mm. When the liquid is water

Questions & Answers

1) What is the difference between Cohesion and adhesion?

Cohesion is an attraction force of molecules between the same matter whereas adhesion is an attraction between molecules of different matter.

2) The Mercury is tried to stay away from the surface, why?

In Mercury, cohesion force is greater than adhesion force. Because of this, Mercury is called non-wetting liquid.

3) What is condition for wetting and non-wetting of liquid with the surface?

The liquid will wet the solid surface is less than 90 degree. If the angle is greater than 90 degree, then the liquid will not wet the solid surface.

4) Explain about surface tension

The liquid molecules on the free surface are getting attracted by liquid molecules lying below them. This free surface liquid molecules will feel pull force interior of the liquid. This force acts like elastic force. Expended per unit area of the surface is called surface tension. The Surface tension is denoted by Sigma (σ). Surface tension occurs at liquid-gas interface, liquid-liquid interface. The reason behind surface tension is an intermolecular attraction because of Cohesion.

5) Give some practical examples of surface tension.

• You might be noticed that if we thoroughly pour water inside the glass. Even if the glass is filled, still we can add some water above the glass limit.
• Suppose we will experiment with a thin glass tube on the water surface. We can easily notice a capillary rise and depression inside a thin glass tube.
• Birds can drink water from the water body due to surface tension.

6) What is the unit of surface tension?

The unit of surface tension is N/m.

7) Give the value of surface tension for air-water and air-Mercury interface at standard pressure and temperature.

The surface tension for the air-water interface is 0.073 N/m.

The surface tension for the air-Mercury interface is 0.480 N/m.

8) What is the capillary effect?

If the narrow tube is dipped into the water, the water will rise inside the tube at a certain level. This type of tube is called a capillary tube, and this phenomenon is called the capillary effect.

9) Is there any relationship between the capillary effect and the surface tension? If yes, what?

Yes. The capillary effect is due to surface tension force. The capillary rise and depression are happening because of cohesion and adhesion intermolecular attraction.

10) Define: Boiling, Cavitation

Boiling: The vapour bubbles form inside the liquid due to temperature and pressure change. The boiling is a change of state) from liquid to vapour.

Cavitation: The formation of a vapour bubble inside machinery due to pressure of the liquid falls below the saturated vapour pressure.

Multiple Choice Questions

1) For wetting liquid, the angle of contact θ should be ________

(a) 0                       (b) θ < π/2                           (c) θ >π/2                            (d) None

2) For non-wetting liquid, the angle of contact θ should be ________

(a) 0                       (b) θ < π/2                           (c) θ >π/2                            (d) None

3) Surface tension value decrease with __________

(a) Constant pressure

(b) Increase in temperature

(c) Increase in pressure

(d) Decrease in temperature

4) If the value angle lies between 0 and 90, then what happens in capillary effect?

 (a) h is positive with concave shape formation

(b) h is negative with concave shape formation

(c) h is negative with convex shape formation

(d) h is positive with convex shape formation

5) Why Mercury is used in thermometer and manometer?

(a) High vapour pressure and low density

(b) High vapour pressure and high density

(c) Low vapour pressure and low density

(d) Low vapour pressure and high density

6) What is approx. collapsing pressure of bubbles in cavitation phenomena?

(a) Around 20 atmospheric pressure

(b) Around 50 atmospheric pressure

(c) Around 75 atmospheric pressure

(d) Around 100 atmospheric pressure

7) What is the value of vapour pressure of water at 20° C temperature?

(a) 0.126 N/cm2

(b) 0.513 N/cm2

(c) 0.235 N/cm2

(d) 0.995 N/cm2

8) What is the value of vapour pressure of Mercury at 20° C temperature?

(a) 1.25* 10-5 N/cm2

(b) 1.72* 10^-5 N/cm2

(c) 1.5* 10-5 N/cm2

(d) 1.25 N/cm2

Conclusion

This article is presented you to understand the concept of surface tension, capillary effect, Cavitation, evaporation and its effects. Some of the practical examples is included in this article to represent it practically. The effort was made to make you correlate fluid mechanics concept with your day to day life.

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