Isentropic Process: 5 Important Factors Related To It

December 31, 2023February 13, 2021 by lambdageeksScience Core SME

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:

$Reversible \\rightarrow dS=\\int_{1}^{2}\\left ( \\frac{\\delta Q}{T} \\right )_{rev}$

$Adiabatic\\rightarrow Q=0 \\Rightarrow dS=0$

In enthalpy terms,

$dH=dQ+VdP$

Or,

$dH=TdS+VdP$

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,

$dQ=0$

$dH=VdP$

$\\rightarrow W=H_{2}-H_{1}$

$\\rightarrow H_{2}-H_{1}=C_{p}\\left ( T_{2}-T_{1} \\right )$

Isentropic process for an ideal gas

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

$\\Delta S=s_{2}-s_{1}$

$=\\int_{1}^{2}C_{v}\\frac{dT}{T}+Rln\\frac{V_{2}}{V_{1}} \\rightarrow \\left ( 1 \\right )$

$=\\int_{1}^{2}C_{p}\\frac{dT}{T}-Rln\\frac{P_{2}}{P_{1}} \\rightarrow \\left ( 2 \\right )$

$\\Delta S\\rightarrow 0$

$equation \\left ( 1 \\right )\\rightarrow 0$

$=\\int_{1}^{2}C_{v}\\frac{dT}{T}-Rln\\frac{V_{2}}{V_{1}} \\rightarrow \\left ( 2 \\right )$

Integrating and rearranging,

$C_{v}ln\\frac{T_{2}}{T_{1}}=-Rln\\frac{V_{2}}{V_{1}}$

(this is by assuming constant specific heats)

$\\frac{T_{2}}{T_{1}}=\\left ( \\frac{V_{2}}{V_{1}} \\right )^{\\frac{R}{C_{v}}}=\\left ( \\frac{V_{2}}{V_{1}} \\right )^{k-1}$

Where k is the specific heat ratio

$k=\\frac{C_{p}}{C_{v}}; R=C_{p}-C_{v}$

Now, setting

$equation \\left ( 2 \\right )\\rightarrow 0$

$\\int_{1}^{2}C_{p}\\frac{dT}{T}=Rln\\frac{P_{2}}{P_{1}}$

$\\Rightarrow C_{p}ln\\frac{T_{2}}{T_{1}}=Rln\\frac{P_{2}}{P_{1}}$

$\\Rightarrow \\frac{T_{2}}{T_{1}}=\\left ( \\frac{P_{2}}{P_{1}} \\right )^{\\frac{R}{C_{p}}}=\\left ( \\frac{P_{2}}{P_{1}} \\right )^{\\frac{k-1}{k}}$

$combining \\left ( 1 \\right ) and \\left ( 2 \\right )relations$

$\\left ( \\frac{P_{2}}{P_{1}} \\right )^{\\frac{k-1}{k}}=\\left ( \\frac{V_{1}}{V_{2}} \\right )^{k}$

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

$TV^{k-1}=constant$

$TP^{\\frac{1-k}{k}}=constant$

$PV^{k}=constant$

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

$\\Delta S=s_{2}-s_{1}$

$s_{2}^{0}-s_{1}^{0}-Rln\\frac{P_{2}}{P_{1}}\\rightarrow \\left ( 1 \\right )$

$equation\\left ( 1 \\right )\\rightarrow 0$

$\\frac{P_{2}}{P_{1}}=\\frac{exp\\left ( \\frac{s_{2}^{0}}{R} \\right )}{exp\\left ( \\frac{s_{1}^{0}}{R} \\right )}$

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

$\\left ( \\frac{P_{2}}{P_{1}} \\right )_{s}=constant=\\frac{P_{r2}}{P_{r1}}$

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

$\\frac{V_{2}}{V_{1}}=\\frac{T_{2}P_{1}}{T_{1}P_{2}}$

$Replacing \\rightarrow \\frac{P_{r2}}{P_{r1}}$

$\\left ( \\frac{V_{2}}{V_{1}} \\right )=\\frac{\\left ( \\frac{T_{2}}{P_{r2}} \\right )}{\\left ( \\frac{T_{1}}{P_{r1}} \\right )}$

Defining the relative specific volume,

$\\left ( \\frac{V_{2}}{V_{1}} \\right )_{s}=constant=\\frac{V_{r2}}{V_{r1}}$

Isentropic process derivation

The total energy change in a system:

$dU=\\delta W+\\delta Q$

A reversible condition involving work with pressure is,

As established earlier,

$dH=dU+pdV+Vdp$

For isentropic,

$\\delta Q_{rev}=0$

And,

$dS=\\frac{\\delta Q_{rev}}{T}=0$

Now,

$dU=\\delta W+\\delta Q=-pdV+0,$

$dH=\\delta W+\\delta Q+pdV+Vdp=-pdV+0+pdV+Vdp=Vdp$

Capacity ratio:

$\\gamma =-\\frac{\\frac{dp}{p}}{\\frac{dV}{V}}$

$cp - cv = R$

$1 - \\frac{1}{\\gamma } = \\frac{R}{C_{p}}$

$\\frac{C_{p}}{R} = \\frac{\\gamma }{\\gamma -1}$

$p = r * R * T$

Where, r=density

$ds = \\frac{C_{p}dT}{T} - R \\frac{dp}{p}$

As dS=0,

$\\frac{C_{p}dT}{T} = R \\frac{dp}{p}$

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

$Cp dT = \\frac{dp}{r}$

$\\Rightarrow (\\frac{C_{p}}{r}) d(\\frac{p}{r}) = \\frac{dp}{r}$

Differentiating,

$(\\frac{C_{p}}{r}) * (\\frac{dp}{r} - \\frac{pdR}{r^{2}}) = \\frac{dP}{r}$

$((\\frac{C_{p}}{r}) - 1) \\frac{dp}{p} = (\\frac{C_{p}}{r}) \\frac{dr}{r}$

Substituting the gamma equation,

$(\\frac{1}{\\gamma -1}) \\frac{dp}{p} = \\left ( \\frac{\\gamma }{\\gamma -1} \\right )\\frac{dr}{r}$

Simplifying the equation:

$\\frac{dp}{p} = \\gamma \\frac{dr}{r}$

Integrating,

$\\frac{p}{r^{\\gamma }} = constant$

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

$\\frac{p}{r^{\\gamma }} = \\frac{pt}{rt^{\\gamma }}$

$\\frac{p}{pt} = \\left ( \\frac{r}{rt} \\right )^{\\gamma }$

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

$\\frac{rt}{(rt * Tt) } = \\left ( \\frac{r}{rt} \\right )^{\\gamma }$

$\\frac{T}{Tt} = \\left ( \\frac{r}{rt} \\right )^{\\gamma -1}$

Now, by combining the equations:

$\\frac{p}{pt} = \\left ( \\frac{T}{Tt} \\right )^{\\frac{\\gamma }{\\gamma -1}}$

Isentropic work equation

$W=\\int_{1}^{2}PdV=\\int_{1}^{2}\\frac{K}{V^{\\gamma }}dV$

$\\Rightarrow W=\\frac{K}{-\\gamma +1}\\left [ \\frac{V_{2}}{V_{2}^{\\gamma }}-\\frac{V_{1}}{V_{1}^{\\gamma }} \\right ]$

$\\Rightarrow W=\\frac{1}{-\\gamma +1}\\left [ \\left ( \\frac{K}{V_{1}^{\\gamma }} \\right )V_{1}-\\left ( \\frac{K}{V_{2}^{\\gamma }} \\right )V_{2} \\right ]$

$\\Rightarrow W=\\left ( \\frac{1}{\\gamma -1} \\right )\\left [ P_{1}V_{1}-P_{2}V_{2} \\right ]$

$\\Rightarrow W=\\left ( \\frac{1}{\\gamma -1} \\right )\\left [ nRT_{2}-nRT_{1} \\right ]$

$\\therefore W=\\frac{nR\\left ( T_{2}-T_{1} \\right )}{\\gamma -1}$

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

Isentropic turbine and isentropic expansion

$\\eta _{T}=\\frac{Actual Turbine work}{Isentropic Turbine work}$

$\\Rightarrow \\frac{W_{real}}{W_{s}}$

$\\Rightarrow \\frac{h_{1}-h_{2r}}{h_{1}-h_{2s}}$

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,

$\\eta _{C}=\\frac{Isentropic Compressor work}{Actual Compressor work}$

$\\Rightarrow \\frac{W_{s}}{W_{real}}$

$\\Rightarrow \\frac{h_{2s}-h_{1}}{h_{2r}-h_{1}}$

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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Deflection of beam | Complete Overview and Important Relations

December 31, 2023February 11, 2021 by Hakimuddin Bawangaonwala

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

y is zero at a pin or roller support.
y is zero at a built-in or cantilever support.
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.

https://www.pinterest.it/pin/711076228644344940/

Slope and deflection for Simply supported beam at various loading conditions

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

y is zero at a pin or roller support.
y is zero at a built-in or cantilever support.
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 .

Adiabatic Process: 7 Interesting Facts To Know

January 3, 2024February 8, 2021 by lambdageeksScience Core SME

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

Expansion process in an internal combustion engine found among hot gases.
The quantum-mechanic analogue of an oscillator classically known as the quantum harmonic oscillator.
Gases liquified in a cooling system.
Air released from a pneumatic tire is the most significant and common instance of an adiabatic process.
Ice stored in an icebox follows the principles of heat not being transferred in and out to the surroundings.
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.

Exaple of adiabetic process — Adiabatic Process Example piston movement. Image credit : Andlaus, Adiabatic-irrevisible-state-change, CC0 1.0

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)

$U= frac{3}{2} nRDelta T= -W$

Therefore,

$W= frac{3}{2} nR(T_{i} - T_{f})$

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.

$delta q=0=dU+delta W,$

$0=dU+PdV$

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:

$P.V=R.T$

$dU = C_{v} . dT$

$C_{v}dT + (frac{R.T}{V})dV = 0$

$rightarrow frac{dT}{T}= -(frac{R}{C_{v}}) frac{dV}{V}$

Integrating the equations,

$ln(frac{T_{2}}{T_{1}}) = (frac{R}{C_{v}})ln(frac{V_{1}}{V_{2}})$

$left ( frac{T_{2}}{T_{1}} right )=left ( frac{V_{1}}{V_{2}} right )frac{R}{C_{v}}$

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.

$dU=dQ-dW$

According to the definition,

$dQ=0$

Hence,

$dQ=0=dU+dW$

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.

$C_{v}=frac{dU}{dT}(frac{1}{n})$

(n is the number moles), Therefore:

$0=PdV+nC_{v}dT$

Derived from the ideal gas law,

$PV=nRT$

$PdV +VdP=nRdT$

Merging equation 1 and 2,

$-PdV =nC_{v}dT = frac{C_{v}}R left ( PdV +VdP right )0 = left ( 1+frac{C_{v}}{R} right )PdV +frac{C_{v}}{R}VdP0=left ( frac{R+C_{v}}{C_{v}} right )frac{dV}{V}+frac{dP}{P}$

For a constant pressure C_p, heat is added and,

$C_{p}=C_{v}+R0 = gamma left ( frac{dV}{V} right )+frac{dP}{P}$

γ is the specific heat

$gamma = frac{C_{p}}{C_{v}}$

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

$dleft ( lnx right )= frac{dx}{x}0=gamma dleft ( lnV right ) + d(lnP)0=d(gamma lnV+lnP) = d(lnPV^{gamma })PV^{gamma }= constant$

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:

$F=PA$

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

$dW=Fdx =PAdx =PdV$

Since,

$dW=PdV$

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:

$W=int_{0}^{W}dW=int_{V_{1}}^{V_{2}}PdV$ —–(1)

Initially, for an adiabatic change, we can assume:

$PV_{gamma }=constant = K$

Which can be,

From (1),

$W=int_{V_{1}}^{V_{2}}frac{K}{V^{gamma }}dV=Kint_{V_{1}}^{V_{2}}V^{-gamma }dV$

$W=kleft | frac{V^{1-gamma }}{1-gamma } right |=frac{K}{1-gamma }left [ V_{2}^{1-gamma }-V_{1}^{1-gamma } right ]$

For solving,

$P_{1}V_{1}^{gamma }=P_{2}V_{2}^{gamma }=K$

Thus,

Which is,

Taking T₁ and T₂as the initial and final temperatures of the gas respectively,

$P_{1}V_{1}^{gamma }=P_{2}V_{2}^{gamma }=K$

Using this in equation (2),

$W=left [ frac{R}{1-gamma } right ]left [ T_{2}-T_{1} right ]$

Or,

$W=left [ frac{R}{gamma-1 } right ]left [ T_{1}-T_{2} right ]$ —-(3)

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

$=left [ frac{R}{J(gamma-1)} right ]left [ T_{1}-T_{2} right ]$

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,

$W=-left [ frac{R}{gamma-1} right ]left [ T_{1}-T_{2} right ]$

Or,

$W=-left [ frac{R}{1-gamma} right ]left [ T_{2}-T_{1} right ] ----left ( 4 right )$

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

And the heat expelled during the process is:

Adiabatic graph

Adiabatic process1 — Various curves in thermodynamic process
image credit : User:Stannered, Adiabatic, CC BY-SA 3.0

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

$PV^{gamma }=C$

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

$dE=frac{dQ}{dT}$

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,

$dQ=0$

Thus,

$frac{dQ}{dT}=0$

Which means that,

$dE=0$

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.

To know about Thermodynamics click here

Surface Tension: 7 Important Factors Related To It

December 31, 2023February 6, 2021 by Dr. Deepakkumar Jani

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.

Surface Tension — Wetting and non-wetting of liquid credit Hisoki

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^-5N/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

(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

(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

(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

(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

(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

(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.

To learn more on fluid mechanics, please click here.

Thermal Insulation: 5 Important Facts You Should Know

December 31, 2023February 3, 2021 by lambdageeksScience Core SME

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.

Conduction
Convention
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

Fiberglass
Polyurethane foam
Cellulose
Polystyrene
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

mineral wools — **Mineral wool**
Image credit: Achim Hering, Rockwool 4lbs per ft3 fibrex5, CC BY 3.0

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

pollystrene — **Polystyrene**
Image credit:Phyrexian, Polistirolo, CC BY-SA 3.0

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.

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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 Insulation: 5 Important Facts You Should Know 31

Thermal transmittance (U):

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

Thermal conductivity (k-value):

AWp75pQ1x5w499fU184t5dlL RsgXVEU v7NmkDEd jUWZscZ 5IABciK9Cd D5hHhUfuzGU15Sufptt X7Dj6wf9gb6TAiMhqItmmI2hCIMCSTlP QfVDVb6vjr lF4JN2obh0

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

T= temperature, (K)

q = heat flow rate, (W/m²)

To learn more about thermodynamics click here!

Macaulay’s Method & Moment Area Method: 11 Important Facts

December 31, 2023February 3, 2021 by Hakimuddin Bawangaonwala

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,

$\\\\M=EI \\frac{d^2 y}{dx^2}=M+Qx……………[1]\\\\\\\\EI \\frac{dy}{dx}=Mx+Q\\frac{x^2}{2} +C_1……………[2]\\\\\\\\EIy=M \\frac{x^2}{2}+Q \\frac{x^3}{6}+C_1 x+C_2……………[3]$

And Beyond Y

$M=EI \\frac{d^2 y}{dx^2}=M+Qx-F(x-a)…………… [4]\\\\\\\\EI \\frac{d y}{dx}=Mx+Q (x^2/2)-F (x^2/2)+Fax+C_3…………… [5]$

$EIy=M (x^2/2)+Q (x^3/6)-F (x^3/6)+Fa (x^2/2) C_3 x+C_4…………… [6]$

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

$Mx+Q (x^2/2)+C_1= Mx+Q (x^2/2)-F (x^2/2)+Fax+C_3$

But at Point Y, x = a

$C_1=-F (a^2/2)+Fa^2+C_3\\\\\\\\C_3=C_1-F (a^2/2)$

Substituting the above equation in [5]

$EI \\frac{dy}{dx}=Mx+Q (x^2/2)-F (x^2/2)+Fax+C_1-F (a^2/2)$

$EI \\frac{dy}{dx}=Mx+Q (x^2/2)-F(x-a)^2/2+C_1………….[7]$

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

$M(a^2/2)+Q(a^3/6)+C_1 a+C_2=M(a^2/2)+Q(a^3/6)-F(a^3/6)+F(a^3/6)+C_3 a+C_4$

On solving these equations and substituting value of C3

$C_4=F(a^3/6)+C_2$

Substituting in equation [6] we get,

$\\large EIy=M x^2/2+Q x^3/6-F x^3/6+Fa (x^2/2)(C_1-F a^2/2)x+F(a^3/6)+C_2$

$\\large EIy=M x^2/2+Q x^3/6-F (x-a)^3/6+C_1 x+C_2…………[8]$

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,

$M=EI \\frac{d^2 y}{dx^2}=M+Qx-F(x-a)$

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

$EI\\frac{d^2 y}{dx^2}=R_A x-w(x^2/2)- W_1 (x-a)-W_2 (x-b)$

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

$EI \\frac{d^2 y}{dx^2}=R_A x-w[(x-a)^2/2]- W_1 [(x-a)]-W_2 [(x-b)]$

Integrating we get,

$EI\\frac{dy}{dx}=R_A(x^2/2)-w[(x-a)^3/6]-W_1 [(x-a)^2/2]-W_2 [(x-b)^2/2]+A$

$EIy=R_A(x^3/6)-w[(x-a)^4/24]-W_1 [(x-a)^3/6]-W_2 [(x-b)^3/6]+Ax+B$

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,

$EI\\frac{d^2 y}{dx^2}=1000x-400(x-1)^2+400(x-4)^2+2600(x-6)$

Integrating it,

$EI\\frac{dy}{dx}=500x^2-400 (x-1)^3/3+400 (x-4)^3/3+1300(x-6)^2+P$

Again, Integrating it,

$EIy=500x^3/3 -100 (x-1)^4/3+100 (x-4)^4/3+1300 (x-6)^3/3+Px+Q….[a]$

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

$EI*0=500*0^3/3-100 (0-1)^4/3+100 (0-4)^4/3+1300 (0-6)^3/3+P*0+Q\\\\\\\\Q=-85100$

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

$EI*0=500*6^3/3-100 *(6-1)^4/3+100 *(6-4)^4/3+1300*(6-6)^3/3+P*6-85100\\\\\\\\0=500*6^3/3-100 *(5)^4/3+100*(2)^4/3+0+P*6-85100\\\\\\\\P= -69400$

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

$EIy=500 x^3/3-100 (x-1)^4/3+100(x-4)^4/3 +1300 (x-6)^3/3-69400x-85100….[b]$

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,

$EIy=500*3^3/3 -100*(3-1)^4/3+100*(3-4)^4/3+1300*(3-6)^3/3-69400*3-85100$

$We\\; have\\; to\\; note\\; that\\; (3-4)^4=0 \\;and \\;(3-6)^3=0$

$EIy=-289333.33 \\;N.m^3$

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]

$EIy=500*8^3/3-100*(8-1)^4/3+100*(8-4)^4/3+1300*(8-6)^3/3-69400*8-85100$

$EIy=-699800 \\;N.m^3$

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.

$\\frac{1}{R}=\\frac{M}{EI} \\;becomes\\;d\\theta=\\frac{M}{EI}dx \\;\\;…………..[a]$

Integrating Eq.(a) over the segment AB yields

$\\int_{B}^{A}d\\theta=\\int_{B}^{A}\\frac{M}{EI}dx\\;\\;……………..[b]$

(a) Elastic curve of the beam (b) B.M.D for the segment.

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

$\\theta_{B/A}=Area\\;of Bending \\;Moment\\; Diagram \\;for\\;section\\;A-B$

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,

$dt=x'd\\theta$

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

$t_{B/A}=\\int_{B}^{A}dt=\\int_{B}^{A}x' d\\theta$

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

$t_{B/A}=\\int_{B}^{A}\\frac{M}{EI}x'dx\\;\\;………………..[c]$

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

$t_{B/A}= Area \\;of \\;M/EI \\;diagram\\; for\\; section\\; A-B* \\bar{x}_B$

$t_{B/A}= Distance \\;of\\; Center\\; of\\; gravity \\;of\\; B.M.D$

$\\bar{x}_B \\; is\\; the \\;Distance \\;of\\; center \\;of \\;gravity \\;of \\;M/EI \\;from \\;point \\;under\\; consideration\\; (B).$

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.]

overhanging Beam with Uniformly distributed Loading 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.

$R_B+R_D=2*3*200$

$R_B+R_D=1200$

$Also\\;\\sum M_B=0$

$(200*3*1.5)+(R_D*10)=200*3*11.5$

$R_D=600 N$

$Thus,\\;R_B=600 N$

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

$above\\;\\theta_c=0$

Thus, tangent at E can be given by,

$\\theta_E=\\theta_c+\\theta_{E/C}=\\theta_{E/C} …………..[1]$

Macaulay 2 — Moment Area Diagram with calculations

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

$A_1= \\frac{-(wa^2)}{2EI}*(L/2)$

$A_1=\\frac{-(200*3^2)}{2*20.18*10^3}*5$

$A_1=-0.2230$

Similarly, for A2

$A_2=(1/3)* \\frac{-(wa^2)}{2EI}*a$

$A_2=(1/3)*\\frac{-(200*3^2)}{2*20.18*10^3}*3$

$A_2=-0.0446$

From equation [1] we get,

$\\theta_E=A_1+A_2$

$\\theta_E=-0.2230-0.0446=-0.2676$

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

$t_{D/C}=A_1*[L/4]$

$t_{D/C}=(-0.2230)*[10/4]$

$t_{D/C}=-0.5575$

Similarly,

$t_{E/C}=A_1*(a+L/4)+A_2 *(3a/4)$

$t_{E/C}=(-0.2230)*(3+10/4)+(-0.0446)*(3*3/4)$

$t_{E/C}=-1.326$

But we know that

$y_E=t_{E/C}-t_{D/C}\\\\y_E=-1.326-(-0.5575)\\\\y_E=-0.7685 m$

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.

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.
Compute the tangential deviation t of support B with respect to A.
Divide the obtained quantity by the span L between the supports A and B.
Since the slope ϴk=0, we must get,

$\\theta_{K/A}= \\theta_K-\\theta_A=-\\theta_A$

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

$Area\\;A=\\theta_{K/A}=-\\theta_A\\;under M/EI\\;Diagram$

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 .

Properties Of Fluids: 13 Things Most Beginner’s Don’t Know

January 1, 2024February 3, 2021 by Dr. Deepakkumar Jani

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.

Properties of fluids — Density Credit Wiki.anton

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.

$Sg=\rho_{fluid}/\rho_{sf}$

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,

$\beta_c=(-1)/V (dV/dp)$

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.

$K=1/\beta_c$

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,

Viscosity — Newton’s law of viscosity Credit Wikipedia

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.

$\tau\; \alpha\; du/dy$

$\tau\; =\; \mu\; du/dy$

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..

$\mu=\tau/((du/dy) )= (N/m^2)/((m/s*1/m) )=(N*s)/m^2 =Pa*s$

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:

$\mu=\mu_{0}/(1+At+Bt^2)$

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:

$\mu_t=\mu_0+\alpha t-\beta t^2$

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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Crankshaft: 9 Important Facts You Should Know

January 3, 2024February 2, 2021 by Hakimuddin Bawangaonwala

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:

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

https://en.wikipedia.org/wiki/Crankshaft

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

Side crankshaft
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.

Calculate the magnitude of the different loads acts upon the crank shaft.
According to the loads, calculate the distance between the support structures and positions.
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.
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.
Here and now estimate the distance between the support structures.
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.

Crankshaft deflection curve — Crankshaft Deflection curve

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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Mohr’s Circle: 9 Important Facts You Should Know

January 2, 2024January 31, 2021 by lambdageeksScience Core SME

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.

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.

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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Thermodynamics Notes: 13 Facts You Should Know

February 10, 2024January 22, 2021 by lambdageeksScience Core SME

Thermodynamics Notes

Thermodynamics: The branch of physics and science that deals with the correlation between heat and other forms of energy that can be transferred from one form and place to another can be defined as thermodynamics. Certain terms to know about when examining thermodynamics can be better understood by following term.

Heat

Heat is a form of energy, the transfer of energy from one body to other happens due to temperature difference and heat-energy flows from a hot body to a cold body, to make it thermal equilibrium and plays a very critical role in the principle of thermodynamics.

Work

An external force applied in the direction of displacement which enables the object to move a particular distance undergoes a certain energy transfer which can be defined as work in the books of physics or science. In mathematical terms, work can be described as the force applied multiplied by the distance covered. If the displacement is involved at an angle Θ when force is exerted, then the equation can be:

W = fs

W = fscosӨ

Where,

f= force applied

s= distance covered

Ө= displacement angle

Thermodynamics is a very vital aspect of our daily life. They follow a set of laws to abide by when applied in terms of physics.

Laws of thermodynamics

The Universe, though it is defined by many laws, only very few are mighty. The laws of thermodynamics as a discipline were formulated and opened ways to numerous other phenomena varying from refrigerators, to chemistry and way beyond life processes.

The four basic laws of thermodynamics consider empirical facts and construe physical quantities, like temperature, heat, thermodynamic work, and entropy, that defines thermodynamic operations and systems in thermodynamic equilibrium. They explain the links between these quantities. Besides their application in thermodynamics, the laws have integrative applications in other branches of science. In thermodynamics, a ‘System’ can be a metal block or a container with water, or even our human body, and everything else is called ‘Surroundings’.

The zero^th law of thermodynamics obeys the transitive property of basic mathematics that if a two systems are in thermal equilibrium with a 3^rd system, then these are in thermal equilibrium state with each other too.

The basic concepts that need to be covered to comprehend the laws of thermodynamics are system and surroundings.

System and Surroundings

The collection of a particular set of items we define or include (something as small as an atom to something as big as the solar system) can be called a system whereas everything that does not fall under the system can be considered as the surroundings and these two concepts are separated by a boundary.

For example, coffee in a flask is considered as a system and surroundings with a boundary.

Essentially, a system consists of three types namely, opened, closed, and isolated.

thermodynamics note — Figure: System and Surroundings in thermodynamics

Thermodynamics equations

The equations formed in thermodynamics are a mathematical representation of the thermodynamic principle subjected to mechanical work in the form of equational expressions.

The various equations that are formed in the thermodynamic laws and functions are as follows:

● ΔU = q + w (first law of TD)

● ΔU = Uf – Ui (internal energy)

● q = m Cs ΔT (heat/g)

● w = -PextΔV (work)

● H = U + PV

ΔH = ΔU + PΔV

ΔU = ΔH – PΔV

ΔU = ΔH – ΔnRT ( enthalpy to internal energy)

● S = k ln Ω (second law in Boltzman formula)

● ΔSrxn° = ΣnS° (products) – ΣnS° (reactants) (third law)

● ΔG = ΔH – TΔS (free energy)

First law of thermodynamics

The 1^st law of thermodynamics elaborate that when energy (as work, heat, or matter) carries in or out of a system, the system’s internal energy will change according to the law of conservation of energy (which means that energy can neither be created nor destroyed and can only be transferred or converted from one form to another), i.e., perpetual motion machine of the 1^st kind ( a machine which actually works without energy i/p) are un-attainable.

For example, lighting a bulb is a law of electrical energy being converted light energy which actually illuminates and some part will be lost as heat energy.

ΔU= q + w

ΔU is the total internal energy change of a system.
q is the heat transfer between a system and its surroundings.
w is the work done by the system.

Thermodynamics notes : First law of Thermodynamics

Second law of thermodynamics

The second law of thermodynamics defines an important property of a system called entropy. The entropy of the universe is always increasing and mathematically represented as ΔSuniv > 0 where ΔSuniv is the change in the entropy of the universe.

Entropy

Entropy is the measurement of the system’s randomness or it is the measure of energy or chaos with in an isolate system, this can be contemplated as a quantitative index that described the classification of energy.

The second law also gives the upper limit of efficiency of systems and the direction of the process. It is a basic concept that heat does not flow from an object of lesser temperature to an object of greater temperature. For that to happen, and external work input is to be supplied to the system. This is an explanation for one of the fundamentals of the second law of thermodynamics called “Clausius statement of second law “. It states that “It is impossible to transfer heat in a cyclic process from low temperature to high temperature without work from an external source”.

2nd law 1 — Figure: Second law of thermodynamics Image source : NASA

A real-life example of this statement is refrigerators and heat pumps. It is also known that a machine that can’t convert all of the energy supplied to a system cannot be converted to work with an efficiency of 100 percent. This then guides us to the following statement called the “Kelvin-Planck statement of second law”. The statement is as follows “It is impossible to construct a device (engine) operating in a cycle that will produce no effect other than extraction of heat from a single reservoir and convert all of it into work”.

Mathematically, the Kelvin-Planck statement can be written as: Wcycle ≤ 0 (for a single reservoir) A machine that can produce work continuously by taking heat from a single heat reservoir and converting all of it into work is called a perpetual motion machine of the second kind. This machine directly violates the Kelvin-Planck statement. So, to put it in simple terms, for a system to produce to work in a cycle it has to interact with two thermal reservoirs at different temperatures.

Thus, in layman’s term the 2nd law of thermodynamics elaborates, when energy conversion happens from one to other state, entropy will not decreases but always increases regardless within a closed-system.

Third law of thermodynamics

In layman’s terms, the third law states that the entropy of an object approaches zero as the absolute temperature approaches zero (0K). This law assists to find an absolute credential point to obtain the entropy. The 3^rd law of thermodynamics has 2 significant characteristics as follows.

The sign of the entropy of any particular substance at any temperature above 0K is recognized as positive sign, and it gives a fixed reference-point to identify the absolute-entropy of any specific substance at any temp.

Figure: TS diagram Image source: Wikipedia commons

Different measures of energy

ENERGY

Energy is defined as the capacity to do work. It is a scalar quantity. It is measured in KJ in SI units and Kcal in MKS units. Energy can have many forms.

FORMS OF ENERGY:

Energy can exist in numerous forms such as

1. Internal energy
2. Thermal energy
3. Electrical energy
4. Mechanical energy
5. Kinetic energy
6. Potential energy
7. Wind energy and
8. Nuclear energy

This further categorized in

(a) Stored energy and (b) Transit energy.

Stored Energy

The stored form of energy can be either of the following two types.

Macroscopic forms of energy: Potential energy and kinetic energy.
Microscopic forms of energy: Internal energy.

Transit Energy

Transit energy means energy in transition, basically represented by the energy possessed by a system that is capable of crossing the boundaries

Heat:

It is a transfer form of energy that flows between two systems under the temperature difference between them.

(a) Calorie (cal) It is the heat needed to raise the temperature of 1 g of H2O by 1 deg C

(b) British thermal unit (BTU) It is the heat needed to raise the temperature of 1 lb of H2O by 1 deg F

Work:

An energy interaction between a system and its surroundings during a process can be regarded as work transfer.

Enthalpy:

Enthalpy (H) defined as the summation of the system’s internal energies and the product of it’s pressure and volume and enthalpy is a state function used in the field of, physical, mechanical, and chemical systems at a constant pressure, represented in Joules (J) in SI units.

Relationship between the units of measurement of energy (with respect to Joules, J)

Unit	Equivalent to
1eV	1.1602 x 10-19 J
1 cal	4.184 J
1 BTU	1.055 kJ
1 W	1 J/sec

Table: Relation table

Maxwell’s Relations

The four most traditional Maxwell relations are the equalities of the second derivatives of every one of the four thermodynamic perspectives, concerning their mechanical variables such as Pressure (P) and Volume (V) plus their thermal variables such as Temperature (T) and Entropy (S).

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Equation: common Maxwell’s Relations

Conclusion

This article on Thermodynamics gives you a glimpse of the fundamental laws, definitions, equations relations, and its few applications, although the content is short, it can be used to quantify many unknowns. Thermodynamics finds its use in various fields as some quantities are easier to measure than others, though this topic is profound by itself, thermodynamics is fundamental, and its fascinating phenomena gives us a deep understanding of the role of energy in this universe

Some questions related to the field of Thermodynamics

What are the applications of thermodynamics in engineering?

There are several applications of thermodynamics in our daily lives as well as in the domain of engineering. The laws of thermodynamics are intrinsically used in the automobile and the aeronautical sector of engineering such as in IC engines and gas turbines in the respective departments. It is also applied in heat engines, heat pumps, refrigerators, power plants, air conditioning, and more following the principles of thermodynamics.

Why is thermodynamics important?

There are various contributions of thermodynamics in our daily life as well as in the engineering sector. The processes that occur naturally in our daily life fall under the guidance of thermodynamic laws. The concepts of heat transfer and the thermal systems in the environment are explained by the thermodynamic fundamental which is why the subject is very important to us.

How long does it take a bottle of water to freeze while at a temperature of 32˚F?

In terms of a conceptual solution to the given question, the amount of time taken to freeze a bottle of water at a temperature of 32F will be depending upon the nucleation point of the water which can be defined as the point where the molecules in the liquid are gathered to turn into a crystal structure of solid where pure water will freeze at -39C.

Other factors into consideration are the latent heat of fusion of water which is the amount of energy required to change its state, essentially liquid to solid or solid to liquid. The latent heat of water at 0C for fusion is 334 joules per gram.

What is cut-off ratio and how does it affect the thermal efficiency of a diesel engine?

Cut off ratio is inversely proportional to the diesel cycle as there is an increase in efficiency of the cut-off ratio, there is a decrease or reduction in the efficiency of a diesel engine. The cut-off ratio is based on its equation where the correspondence of the cylinder volume before and after combustion is in proportion to each other.

It goes as follows:

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What is a steady-state in thermodynamics?

The current state of a system that contains a flow through it over time and the variables of that particular process remains constant, then that state can be defined as a steady-state system in the subject of thermodynamics.

What are the examples of fixed boundary and movable boundary in the case of thermodynamics?

A moveable boundary or in other terms, control mass is a certain class of system where matter cannot move across the boundary of the system while the boundary itself acts as a flexible character that can expand or contract without allowing any mass to flow in or out of it. A simple example of a moveable boundary system in basic thermodynamics would be a piston in an IC engine where the boundary expands as the piston is displaced while the mass of the gas in the cylinder remains constant allowing work to be done.

5ormIhuE6JPsN Og1pk2xsKt3x3WtRRbpVzne2HBNSTyDGjszLIgEXkLr3BhdWzARk1f2lTSwv XjZHGsH2A79In7X8qqL1csLigCTBzqv3inA NMtg91TiUb0KPO XYD6OUM2qI — Figure: Piston movement

Whereas in the case of a fixed boundary, there is no work being permitted as they keep volume constant while the mass is allowed to flow in and out freely in the system. It can also be called a control volume process. Example: gas flowing out of a household cylinder connected to a stove while the volume is fixed.

What are the similarities and dissimilarities of heat and work in thermodynamics?

Similarities:

● Both these energies are considered as path functions or process quantities.
● They are also inexact differentials.
● Both the form of energies are not stored and can be transferred in and out of the system following the transient phenomenon.

Dissimilarities:

● Heat flow in a system is always associated with the entropy function whereas there is no entropy transfer along with the work system.
● Heat cannot be converted a hundred percent into work, while work can be converted into heat a 100%.
● Heat is considered as a low-grade energy meaning, it is easy to convert heat into other forms while work is high-grade energy.


	*Equation 1: Cut-Off Ratio*