The entropy change of the refrigerant.

Question

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Answer 1

Question

Chapter 8, Problem 42P

a)

To determine

The entropy change of the refrigerant.

a)

Expert Solution

Explanation of Solution

Given:

The initial pressure (P1) is 200 kPa.

The initial quality of the steam is 40%.

The final pressure (P2) is 400 kPa.

The volume of the tank (ν) is 0.5 m3.

The source temperature (Tsource) is 35°C.

Calculation:

Refer the Table A-12, “Saturated refrigerant 134a– pressure table”, obtain the following properties at saturated pressure of 200 kPa.

uf=38.26 kJ/kgufg=186.25 kJ/kg

sf=0.15449 kJ/kg⋅Ksfg=0.78339 kJ/kg⋅K

vf=0.0007532m3/kgvg=0.099951 m3/kg

Calculate the initial internal energy of the refrigerant (u1).

u1=uf+x1ufg

u1=38.26 kJ/kg+(0.4)(186.25 kJ/kg)=112.76 kJ/kg

Calculate the initial entropy of the refrigerant (s1) .

s1=sf+x1sfg

s1=0.15449 kJ/kg⋅K+(0.4)(0.78339 kJ/kg⋅K)=0.4678 kJ/kg⋅K

Calculate the initial specific volume of the refrigerant (v1).

v1=vf+x1(vg−vf)

v1=0.0007532m3/kg+(0.4)(0.099951 m3/kg−0.0007532m3/kg)=0.04043m3/kg

Specific volume remains constant for a rigid tank (v2=v1).

Refer the Table A-12, “Saturated refrigerant 134a– pressure table”, obtain the following properties at saturated pressure of 400 kPa.

vf=0.0007905 m3/kgvg=0.051266 m3/kg

uf=63.61 kJ/kgufg=171.49kJ/kg

sf=0.24757 kJ/kg⋅Ksfg=0.67954 kJ/kg⋅K

Calculate the final quality of the refrigerant using the specific volume.

v2=vf+x2(vg−vf)

0.04043m3/kg=[0.0007905 m3/kg+x2(0.051266 m3/kg−0.0007905 m3/kg)]x2=0.04043−0.00079050.051266−0.0007905 =0.7853

Calculate the final internal energy of the refrigerant (u2) .

u2=uf+x2ufgu2=63.61 kJ/kg+(0.7853)(171.49 kJ/kg)=198.29 kJ/kg

Calculate the final entropy of the refrigerant (s2).

s2=sf+x2sfgs2=0.24757 kJ/kg⋅K+(0.7853)(0.67954 kJ/kg⋅K)=0.7813 kJ/kg⋅K

Calculate the mass of the refrigerant (m).

m=νv1m=0.5 m30.04043m3/kg=12.37 kg

Calculate the expression for the entropy change of the refrigerant (ΔSsystem).

ΔSsystem=m(s2−s1)ΔSsystem=12.37 kg(0.7813 kJ/kg⋅K−0.4678 kJ/kg⋅K)=3.878 kJ/K

Thus, the entropy change of the refrigerant is 3.878 kJ/K.

b)

To determine

The entropy change of the heat source

b)

Expert Solution

Explanation of Solution

Write the expression for the energy balance equation for closed system.

Ein−Eout=ΔEsystem (I)

Here, energy transfer into the control volume is Ein, energy transfer exit from the control volume is Eout and change in internal energy of system is ΔEsystem.

Substitute Ein=Qin, Eout=0 and ΔEsystem=m(u2−u1) in Equation (I).

Qin−0=m(u2−u1)Qin=m(u2−u1)

Qin=(12.37 kg)(198.29 kJ/kg−112.76 kJ/kg)=1058 kJ

Heat transfer for the source (Qsource) is equal to heat transfer input (Qin). The heat transfer for the source (Qsource) is equal in magnitude but opposite in direction.

Qsource=−Qin=−1058 kJ

Calculate the entropy change of the heat source.

ΔSsource=QsourceTsource (X)

ΔSsource=−1058 kJ35°C=−1058 kJ(35+273)K=−3.435 kJ/K

The entropy change of the heat source is −3.435 kJ/K.

c)

To determine

The total entropy change during the process.

c)

Expert Solution

Explanation of Solution

Calculate the total entropy change during the process.

ΔStotal=ΔSsystem+ΔSsourceΔStotal=3.876 kJ/K−3.434 kJ/K=0.442 kJ/K

Thus, the total entropy change during the process is 0.442 kJ/K.

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Students have asked these similar questions

1.1 m 1.3 m B 60-mm diameter Brass 40-mm diameter Aluminum PROBLEM 2.52 - A rod consisting of two cylindrical portions AB and BC is restrained at both ends. Portion AB is made of brass (E₁ = 105 GPa, α = 20.9×10°/°C) and portion BC is made of aluminum (Ę₁ =72 GPa, α = 23.9×10/°C). Knowing that the rod is initially unstressed, determine (a) the normal stresses induced in portions AB and BC by a temperature rise of 42°C, (b) the corresponding deflection of point B.

30 mm D = 40 MPa -30 mm B C 80 MPa PROBLEM 2.69 A 30-mm square was scribed on the side of a large steel pressure vessel. After pressurization, the biaxial stress condition at the square is as shown. For E = 200 GPa and v=0.30, determine the change in length of (a) side AB, (b) side BC, (c) diagnonal AC.

Please solve in detail this problem thank you

Answer 2

Question

Chapter 8, Problem 42P

a)

To determine

The entropy change of the refrigerant.

a)

Expert Solution

Explanation of Solution

Given:

The initial pressure (P1) is 200 kPa.

The initial quality of the steam is 40%.

The final pressure (P2) is 400 kPa.

The volume of the tank (ν) is 0.5 m3.

The source temperature (Tsource) is 35°C.

Calculation:

Refer the Table A-12, “Saturated refrigerant 134a– pressure table”, obtain the following properties at saturated pressure of 200 kPa.

uf=38.26 kJ/kgufg=186.25 kJ/kg

sf=0.15449 kJ/kg⋅Ksfg=0.78339 kJ/kg⋅K

vf=0.0007532m3/kgvg=0.099951 m3/kg

Calculate the initial internal energy of the refrigerant (u1).

u1=uf+x1ufg

u1=38.26 kJ/kg+(0.4)(186.25 kJ/kg)=112.76 kJ/kg

Calculate the initial entropy of the refrigerant (s1) .

s1=sf+x1sfg

s1=0.15449 kJ/kg⋅K+(0.4)(0.78339 kJ/kg⋅K)=0.4678 kJ/kg⋅K

Calculate the initial specific volume of the refrigerant (v1).

v1=vf+x1(vg−vf)

v1=0.0007532m3/kg+(0.4)(0.099951 m3/kg−0.0007532m3/kg)=0.04043m3/kg

Specific volume remains constant for a rigid tank (v2=v1).

Refer the Table A-12, “Saturated refrigerant 134a– pressure table”, obtain the following properties at saturated pressure of 400 kPa.

vf=0.0007905 m3/kgvg=0.051266 m3/kg

uf=63.61 kJ/kgufg=171.49kJ/kg

sf=0.24757 kJ/kg⋅Ksfg=0.67954 kJ/kg⋅K

Calculate the final quality of the refrigerant using the specific volume.

v2=vf+x2(vg−vf)

0.04043m3/kg=[0.0007905 m3/kg+x2(0.051266 m3/kg−0.0007905 m3/kg)]x2=0.04043−0.00079050.051266−0.0007905 =0.7853

Calculate the final internal energy of the refrigerant (u2) .

u2=uf+x2ufgu2=63.61 kJ/kg+(0.7853)(171.49 kJ/kg)=198.29 kJ/kg

Calculate the final entropy of the refrigerant (s2).

s2=sf+x2sfgs2=0.24757 kJ/kg⋅K+(0.7853)(0.67954 kJ/kg⋅K)=0.7813 kJ/kg⋅K

Calculate the mass of the refrigerant (m).

m=νv1m=0.5 m30.04043m3/kg=12.37 kg

Calculate the expression for the entropy change of the refrigerant (ΔSsystem).

ΔSsystem=m(s2−s1)ΔSsystem=12.37 kg(0.7813 kJ/kg⋅K−0.4678 kJ/kg⋅K)=3.878 kJ/K

Thus, the entropy change of the refrigerant is 3.878 kJ/K.

b)

To determine

The entropy change of the heat source

b)

Expert Solution

Explanation of Solution

Write the expression for the energy balance equation for closed system.

Ein−Eout=ΔEsystem (I)

Here, energy transfer into the control volume is Ein, energy transfer exit from the control volume is Eout and change in internal energy of system is ΔEsystem.

Substitute Ein=Qin, Eout=0 and ΔEsystem=m(u2−u1) in Equation (I).

Qin−0=m(u2−u1)Qin=m(u2−u1)

Qin=(12.37 kg)(198.29 kJ/kg−112.76 kJ/kg)=1058 kJ

Heat transfer for the source (Qsource) is equal to heat transfer input (Qin). The heat transfer for the source (Qsource) is equal in magnitude but opposite in direction.

Qsource=−Qin=−1058 kJ

Calculate the entropy change of the heat source.

ΔSsource=QsourceTsource (X)

ΔSsource=−1058 kJ35°C=−1058 kJ(35+273)K=−3.435 kJ/K

The entropy change of the heat source is −3.435 kJ/K.

c)

To determine

The total entropy change during the process.

c)

Expert Solution

Explanation of Solution

Calculate the total entropy change during the process.

ΔStotal=ΔSsystem+ΔSsourceΔStotal=3.876 kJ/K−3.434 kJ/K=0.442 kJ/K

Thus, the total entropy change during the process is 0.442 kJ/K.

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Answer 3

Question

Chapter 8, Problem 42P

a)

To determine

The entropy change of the refrigerant.

a)

Expert Solution

Explanation of Solution

Given:

The initial pressure (P1) is 200 kPa.

The initial quality of the steam is 40%.

The final pressure (P2) is 400 kPa.

The volume of the tank (ν) is 0.5 m3.

The source temperature (Tsource) is 35°C.

Calculation:

Refer the Table A-12, “Saturated refrigerant 134a– pressure table”, obtain the following properties at saturated pressure of 200 kPa.

uf=38.26 kJ/kgufg=186.25 kJ/kg

sf=0.15449 kJ/kg⋅Ksfg=0.78339 kJ/kg⋅K

vf=0.0007532m3/kgvg=0.099951 m3/kg

Calculate the initial internal energy of the refrigerant (u1).

u1=uf+x1ufg

u1=38.26 kJ/kg+(0.4)(186.25 kJ/kg)=112.76 kJ/kg

Calculate the initial entropy of the refrigerant (s1) .

s1=sf+x1sfg

s1=0.15449 kJ/kg⋅K+(0.4)(0.78339 kJ/kg⋅K)=0.4678 kJ/kg⋅K

Calculate the initial specific volume of the refrigerant (v1).

v1=vf+x1(vg−vf)

v1=0.0007532m3/kg+(0.4)(0.099951 m3/kg−0.0007532m3/kg)=0.04043m3/kg

Specific volume remains constant for a rigid tank (v2=v1).

Refer the Table A-12, “Saturated refrigerant 134a– pressure table”, obtain the following properties at saturated pressure of 400 kPa.

vf=0.0007905 m3/kgvg=0.051266 m3/kg

uf=63.61 kJ/kgufg=171.49kJ/kg

sf=0.24757 kJ/kg⋅Ksfg=0.67954 kJ/kg⋅K

Calculate the final quality of the refrigerant using the specific volume.

v2=vf+x2(vg−vf)

0.04043m3/kg=[0.0007905 m3/kg+x2(0.051266 m3/kg−0.0007905 m3/kg)]x2=0.04043−0.00079050.051266−0.0007905 =0.7853

Calculate the final internal energy of the refrigerant (u2) .

u2=uf+x2ufgu2=63.61 kJ/kg+(0.7853)(171.49 kJ/kg)=198.29 kJ/kg

Calculate the final entropy of the refrigerant (s2).

s2=sf+x2sfgs2=0.24757 kJ/kg⋅K+(0.7853)(0.67954 kJ/kg⋅K)=0.7813 kJ/kg⋅K

Calculate the mass of the refrigerant (m).

m=νv1m=0.5 m30.04043m3/kg=12.37 kg

Calculate the expression for the entropy change of the refrigerant (ΔSsystem).

ΔSsystem=m(s2−s1)ΔSsystem=12.37 kg(0.7813 kJ/kg⋅K−0.4678 kJ/kg⋅K)=3.878 kJ/K

Thus, the entropy change of the refrigerant is 3.878 kJ/K.

b)

To determine

The entropy change of the heat source

b)

Expert Solution

Explanation of Solution

Write the expression for the energy balance equation for closed system.

Ein−Eout=ΔEsystem (I)

Here, energy transfer into the control volume is Ein, energy transfer exit from the control volume is Eout and change in internal energy of system is ΔEsystem.

Substitute Ein=Qin, Eout=0 and ΔEsystem=m(u2−u1) in Equation (I).

Qin−0=m(u2−u1)Qin=m(u2−u1)

Qin=(12.37 kg)(198.29 kJ/kg−112.76 kJ/kg)=1058 kJ

Heat transfer for the source (Qsource) is equal to heat transfer input (Qin). The heat transfer for the source (Qsource) is equal in magnitude but opposite in direction.

Qsource=−Qin=−1058 kJ

Calculate the entropy change of the heat source.

ΔSsource=QsourceTsource (X)

ΔSsource=−1058 kJ35°C=−1058 kJ(35+273)K=−3.435 kJ/K

The entropy change of the heat source is −3.435 kJ/K.

c)

To determine

The total entropy change during the process.

c)

Expert Solution

Explanation of Solution

Calculate the total entropy change during the process.

ΔStotal=ΔSsystem+ΔSsourceΔStotal=3.876 kJ/K−3.434 kJ/K=0.442 kJ/K

Thus, the total entropy change during the process is 0.442 kJ/K.

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Answer 4

Question

Chapter 8, Problem 42P

a)

To determine

The entropy change of the refrigerant.

a)

Expert Solution

Explanation of Solution

Given:

The initial pressure (P1) is 200 kPa.

The initial quality of the steam is 40%.

The final pressure (P2) is 400 kPa.

The volume of the tank (ν) is 0.5 m3.

The source temperature (Tsource) is 35°C.

Calculation:

Refer the Table A-12, “Saturated refrigerant 134a– pressure table”, obtain the following properties at saturated pressure of 200 kPa.

uf=38.26 kJ/kgufg=186.25 kJ/kg

sf=0.15449 kJ/kg⋅Ksfg=0.78339 kJ/kg⋅K

vf=0.0007532m3/kgvg=0.099951 m3/kg

Calculate the initial internal energy of the refrigerant (u1).

u1=uf+x1ufg

u1=38.26 kJ/kg+(0.4)(186.25 kJ/kg)=112.76 kJ/kg

Calculate the initial entropy of the refrigerant (s1) .

s1=sf+x1sfg

s1=0.15449 kJ/kg⋅K+(0.4)(0.78339 kJ/kg⋅K)=0.4678 kJ/kg⋅K

Calculate the initial specific volume of the refrigerant (v1).

v1=vf+x1(vg−vf)

v1=0.0007532m3/kg+(0.4)(0.099951 m3/kg−0.0007532m3/kg)=0.04043m3/kg

Specific volume remains constant for a rigid tank (v2=v1).

Refer the Table A-12, “Saturated refrigerant 134a– pressure table”, obtain the following properties at saturated pressure of 400 kPa.

vf=0.0007905 m3/kgvg=0.051266 m3/kg

uf=63.61 kJ/kgufg=171.49kJ/kg

sf=0.24757 kJ/kg⋅Ksfg=0.67954 kJ/kg⋅K

Calculate the final quality of the refrigerant using the specific volume.

v2=vf+x2(vg−vf)

0.04043m3/kg=[0.0007905 m3/kg+x2(0.051266 m3/kg−0.0007905 m3/kg)]x2=0.04043−0.00079050.051266−0.0007905 =0.7853

Calculate the final internal energy of the refrigerant (u2) .

u2=uf+x2ufgu2=63.61 kJ/kg+(0.7853)(171.49 kJ/kg)=198.29 kJ/kg

Calculate the final entropy of the refrigerant (s2).

s2=sf+x2sfgs2=0.24757 kJ/kg⋅K+(0.7853)(0.67954 kJ/kg⋅K)=0.7813 kJ/kg⋅K

Calculate the mass of the refrigerant (m).

m=νv1m=0.5 m30.04043m3/kg=12.37 kg

Calculate the expression for the entropy change of the refrigerant (ΔSsystem).

ΔSsystem=m(s2−s1)ΔSsystem=12.37 kg(0.7813 kJ/kg⋅K−0.4678 kJ/kg⋅K)=3.878 kJ/K

Thus, the entropy change of the refrigerant is 3.878 kJ/K.

b)

To determine

The entropy change of the heat source

b)

Expert Solution

Explanation of Solution

Write the expression for the energy balance equation for closed system.

Ein−Eout=ΔEsystem (I)

Here, energy transfer into the control volume is Ein, energy transfer exit from the control volume is Eout and change in internal energy of system is ΔEsystem.

Substitute Ein=Qin, Eout=0 and ΔEsystem=m(u2−u1) in Equation (I).

Qin−0=m(u2−u1)Qin=m(u2−u1)

Qin=(12.37 kg)(198.29 kJ/kg−112.76 kJ/kg)=1058 kJ

Heat transfer for the source (Qsource) is equal to heat transfer input (Qin). The heat transfer for the source (Qsource) is equal in magnitude but opposite in direction.

Qsource=−Qin=−1058 kJ

Calculate the entropy change of the heat source.

ΔSsource=QsourceTsource (X)

ΔSsource=−1058 kJ35°C=−1058 kJ(35+273)K=−3.435 kJ/K

The entropy change of the heat source is −3.435 kJ/K.

c)

To determine

The total entropy change during the process.

c)

Expert Solution

Explanation of Solution

Calculate the total entropy change during the process.

ΔStotal=ΔSsystem+ΔSsourceΔStotal=3.876 kJ/K−3.434 kJ/K=0.442 kJ/K

Thus, the total entropy change during the process is 0.442 kJ/K.

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Answer 5

Chapter 8, Problem 42P

a)

To determine

The entropy change of the refrigerant.

a)

Expert Solution

Explanation of Solution

Given:

The initial pressure (P1) is 200 kPa.

The initial quality of the steam is 40%.

The final pressure (P2) is 400 kPa.

The volume of the tank (ν) is 0.5 m3.

The source temperature (Tsource) is 35°C.

Calculation:

Refer the Table A-12, “Saturated refrigerant 134a– pressure table”, obtain the following properties at saturated pressure of 200 kPa.

uf=38.26 kJ/kgufg=186.25 kJ/kg

sf=0.15449 kJ/kg⋅Ksfg=0.78339 kJ/kg⋅K

vf=0.0007532m3/kgvg=0.099951 m3/kg

Calculate the initial internal energy of the refrigerant (u1).

u1=uf+x1ufg

u1=38.26 kJ/kg+(0.4)(186.25 kJ/kg)=112.76 kJ/kg

Calculate the initial entropy of the refrigerant (s1) .

s1=sf+x1sfg

s1=0.15449 kJ/kg⋅K+(0.4)(0.78339 kJ/kg⋅K)=0.4678 kJ/kg⋅K

Calculate the initial specific volume of the refrigerant (v1).

v1=vf+x1(vg−vf)

v1=0.0007532m3/kg+(0.4)(0.099951 m3/kg−0.0007532m3/kg)=0.04043m3/kg

Specific volume remains constant for a rigid tank (v2=v1).

Refer the Table A-12, “Saturated refrigerant 134a– pressure table”, obtain the following properties at saturated pressure of 400 kPa.

vf=0.0007905 m3/kgvg=0.051266 m3/kg

uf=63.61 kJ/kgufg=171.49kJ/kg

sf=0.24757 kJ/kg⋅Ksfg=0.67954 kJ/kg⋅K

Calculate the final quality of the refrigerant using the specific volume.

v2=vf+x2(vg−vf)

0.04043m3/kg=[0.0007905 m3/kg+x2(0.051266 m3/kg−0.0007905 m3/kg)]x2=0.04043−0.00079050.051266−0.0007905 =0.7853

Calculate the final internal energy of the refrigerant (u2) .

u2=uf+x2ufgu2=63.61 kJ/kg+(0.7853)(171.49 kJ/kg)=198.29 kJ/kg

Calculate the final entropy of the refrigerant (s2).

s2=sf+x2sfgs2=0.24757 kJ/kg⋅K+(0.7853)(0.67954 kJ/kg⋅K)=0.7813 kJ/kg⋅K

Calculate the mass of the refrigerant (m).

m=νv1m=0.5 m30.04043m3/kg=12.37 kg

Calculate the expression for the entropy change of the refrigerant (ΔSsystem).

ΔSsystem=m(s2−s1)ΔSsystem=12.37 kg(0.7813 kJ/kg⋅K−0.4678 kJ/kg⋅K)=3.878 kJ/K

Thus, the entropy change of the refrigerant is 3.878 kJ/K.

b)

To determine

The entropy change of the heat source

b)

Expert Solution

Explanation of Solution

Write the expression for the energy balance equation for closed system.

Ein−Eout=ΔEsystem (I)

Here, energy transfer into the control volume is Ein, energy transfer exit from the control volume is Eout and change in internal energy of system is ΔEsystem.

Substitute Ein=Qin, Eout=0 and ΔEsystem=m(u2−u1) in Equation (I).

Qin−0=m(u2−u1)Qin=m(u2−u1)

Qin=(12.37 kg)(198.29 kJ/kg−112.76 kJ/kg)=1058 kJ

Heat transfer for the source (Qsource) is equal to heat transfer input (Qin). The heat transfer for the source (Qsource) is equal in magnitude but opposite in direction.

Qsource=−Qin=−1058 kJ

Calculate the entropy change of the heat source.

ΔSsource=QsourceTsource (X)

ΔSsource=−1058 kJ35°C=−1058 kJ(35+273)K=−3.435 kJ/K

The entropy change of the heat source is −3.435 kJ/K.

c)

To determine

The total entropy change during the process.

c)

Expert Solution

Explanation of Solution

Calculate the total entropy change during the process.

ΔStotal=ΔSsystem+ΔSsourceΔStotal=3.876 kJ/K−3.434 kJ/K=0.442 kJ/K

Thus, the total entropy change during the process is 0.442 kJ/K.

Answer 6

Chapter 8, Problem 42P

a)

To determine

The entropy change of the refrigerant.

a)

Expert Solution

Explanation of Solution

Given:

The initial pressure (P1) is 200 kPa.

The initial quality of the steam is 40%.

The final pressure (P2) is 400 kPa.

The volume of the tank (ν) is 0.5 m3.

The source temperature (Tsource) is 35°C.

Calculation:

Refer the Table A-12, “Saturated refrigerant 134a– pressure table”, obtain the following properties at saturated pressure of 200 kPa.

uf=38.26 kJ/kgufg=186.25 kJ/kg

sf=0.15449 kJ/kg⋅Ksfg=0.78339 kJ/kg⋅K

vf=0.0007532m3/kgvg=0.099951 m3/kg

Calculate the initial internal energy of the refrigerant (u1).

u1=uf+x1ufg

u1=38.26 kJ/kg+(0.4)(186.25 kJ/kg)=112.76 kJ/kg

Calculate the initial entropy of the refrigerant (s1) .

s1=sf+x1sfg

s1=0.15449 kJ/kg⋅K+(0.4)(0.78339 kJ/kg⋅K)=0.4678 kJ/kg⋅K

Calculate the initial specific volume of the refrigerant (v1).

v1=vf+x1(vg−vf)

v1=0.0007532m3/kg+(0.4)(0.099951 m3/kg−0.0007532m3/kg)=0.04043m3/kg

Specific volume remains constant for a rigid tank (v2=v1).

Refer the Table A-12, “Saturated refrigerant 134a– pressure table”, obtain the following properties at saturated pressure of 400 kPa.

vf=0.0007905 m3/kgvg=0.051266 m3/kg

uf=63.61 kJ/kgufg=171.49kJ/kg

sf=0.24757 kJ/kg⋅Ksfg=0.67954 kJ/kg⋅K

Calculate the final quality of the refrigerant using the specific volume.

v2=vf+x2(vg−vf)

0.04043m3/kg=[0.0007905 m3/kg+x2(0.051266 m3/kg−0.0007905 m3/kg)]x2=0.04043−0.00079050.051266−0.0007905 =0.7853

Calculate the final internal energy of the refrigerant (u2) .

u2=uf+x2ufgu2=63.61 kJ/kg+(0.7853)(171.49 kJ/kg)=198.29 kJ/kg

Calculate the final entropy of the refrigerant (s2).

s2=sf+x2sfgs2=0.24757 kJ/kg⋅K+(0.7853)(0.67954 kJ/kg⋅K)=0.7813 kJ/kg⋅K

Calculate the mass of the refrigerant (m).

m=νv1m=0.5 m30.04043m3/kg=12.37 kg

Calculate the expression for the entropy change of the refrigerant (ΔSsystem).

ΔSsystem=m(s2−s1)ΔSsystem=12.37 kg(0.7813 kJ/kg⋅K−0.4678 kJ/kg⋅K)=3.878 kJ/K

Thus, the entropy change of the refrigerant is 3.878 kJ/K.

Answer 7

Chapter 8, Problem 42P

a)

To determine

The entropy change of the refrigerant.

Answer 8

a)

Expert Solution

Explanation of Solution

Given:

The initial pressure (P1) is 200 kPa.

The initial quality of the steam is 40%.

The final pressure (P2) is 400 kPa.

The volume of the tank (ν) is 0.5 m3.

The source temperature (Tsource) is 35°C.

Calculation:

Refer the Table A-12, “Saturated refrigerant 134a– pressure table”, obtain the following properties at saturated pressure of 200 kPa.

uf=38.26 kJ/kgufg=186.25 kJ/kg

sf=0.15449 kJ/kg⋅Ksfg=0.78339 kJ/kg⋅K

vf=0.0007532m3/kgvg=0.099951 m3/kg

Calculate the initial internal energy of the refrigerant (u1).

u1=uf+x1ufg

u1=38.26 kJ/kg+(0.4)(186.25 kJ/kg)=112.76 kJ/kg

Calculate the initial entropy of the refrigerant (s1) .

s1=sf+x1sfg

s1=0.15449 kJ/kg⋅K+(0.4)(0.78339 kJ/kg⋅K)=0.4678 kJ/kg⋅K

Calculate the initial specific volume of the refrigerant (v1).

v1=vf+x1(vg−vf)

v1=0.0007532m3/kg+(0.4)(0.099951 m3/kg−0.0007532m3/kg)=0.04043m3/kg

Specific volume remains constant for a rigid tank (v2=v1).

Refer the Table A-12, “Saturated refrigerant 134a– pressure table”, obtain the following properties at saturated pressure of 400 kPa.

vf=0.0007905 m3/kgvg=0.051266 m3/kg

uf=63.61 kJ/kgufg=171.49kJ/kg

sf=0.24757 kJ/kg⋅Ksfg=0.67954 kJ/kg⋅K

Calculate the final quality of the refrigerant using the specific volume.

v2=vf+x2(vg−vf)

0.04043m3/kg=[0.0007905 m3/kg+x2(0.051266 m3/kg−0.0007905 m3/kg)]x2=0.04043−0.00079050.051266−0.0007905 =0.7853

Calculate the final internal energy of the refrigerant (u2) .

u2=uf+x2ufgu2=63.61 kJ/kg+(0.7853)(171.49 kJ/kg)=198.29 kJ/kg

Calculate the final entropy of the refrigerant (s2).

s2=sf+x2sfgs2=0.24757 kJ/kg⋅K+(0.7853)(0.67954 kJ/kg⋅K)=0.7813 kJ/kg⋅K

Calculate the mass of the refrigerant (m).

m=νv1m=0.5 m30.04043m3/kg=12.37 kg

Calculate the expression for the entropy change of the refrigerant (ΔSsystem).

ΔSsystem=m(s2−s1)ΔSsystem=12.37 kg(0.7813 kJ/kg⋅K−0.4678 kJ/kg⋅K)=3.878 kJ/K

Thus, the entropy change of the refrigerant is 3.878 kJ/K.

Answer 9

a)

Expert Solution

Answer 10

a)

Expert Solution

Answer 11

Expert Solution

Answer 12

b)

To determine

The entropy change of the heat source

b)

Expert Solution

Explanation of Solution

Write the expression for the energy balance equation for closed system.

Ein−Eout=ΔEsystem (I)

Here, energy transfer into the control volume is Ein, energy transfer exit from the control volume is Eout and change in internal energy of system is ΔEsystem.

Substitute Ein=Qin, Eout=0 and ΔEsystem=m(u2−u1) in Equation (I).

Qin−0=m(u2−u1)Qin=m(u2−u1)

Qin=(12.37 kg)(198.29 kJ/kg−112.76 kJ/kg)=1058 kJ

Heat transfer for the source (Qsource) is equal to heat transfer input (Qin). The heat transfer for the source (Qsource) is equal in magnitude but opposite in direction.

Qsource=−Qin=−1058 kJ

Calculate the entropy change of the heat source.

ΔSsource=QsourceTsource (X)

ΔSsource=−1058 kJ35°C=−1058 kJ(35+273)K=−3.435 kJ/K

The entropy change of the heat source is −3.435 kJ/K.

Answer 13

b)

To determine

The entropy change of the heat source

Answer 14

b)

Expert Solution

Explanation of Solution

Write the expression for the energy balance equation for closed system.

Ein−Eout=ΔEsystem (I)

Here, energy transfer into the control volume is Ein, energy transfer exit from the control volume is Eout and change in internal energy of system is ΔEsystem.

Substitute Ein=Qin, Eout=0 and ΔEsystem=m(u2−u1) in Equation (I).

Qin−0=m(u2−u1)Qin=m(u2−u1)

Qin=(12.37 kg)(198.29 kJ/kg−112.76 kJ/kg)=1058 kJ

Heat transfer for the source (Qsource) is equal to heat transfer input (Qin). The heat transfer for the source (Qsource) is equal in magnitude but opposite in direction.

Qsource=−Qin=−1058 kJ

Calculate the entropy change of the heat source.

ΔSsource=QsourceTsource (X)

ΔSsource=−1058 kJ35°C=−1058 kJ(35+273)K=−3.435 kJ/K

The entropy change of the heat source is −3.435 kJ/K.

Answer 15

b)

Expert Solution

Answer 16

b)

Expert Solution

Answer 17

Expert Solution

Answer 18

c)

To determine

The total entropy change during the process.

c)

Expert Solution

Explanation of Solution

Calculate the total entropy change during the process.

ΔStotal=ΔSsystem+ΔSsourceΔStotal=3.876 kJ/K−3.434 kJ/K=0.442 kJ/K

Thus, the total entropy change during the process is 0.442 kJ/K.

Answer 19

c)

To determine

The total entropy change during the process.

Answer 20

c)

Expert Solution

Explanation of Solution

Calculate the total entropy change during the process.

ΔStotal=ΔSsystem+ΔSsourceΔStotal=3.876 kJ/K−3.434 kJ/K=0.442 kJ/K

Thus, the total entropy change during the process is 0.442 kJ/K.

Answer 21

c)

Expert Solution

Answer 22

c)

Expert Solution

Answer 23

Expert Solution

Answer 24

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The entropy change of the refrigerant.

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The entropy change of the refrigerant.

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The entropy change of the heat source

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The total entropy change during the process.

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