The heat transferred and the work produced during the process.

Question

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Answer 1

Question

Chapter 5, Problem 38P

To determine

The heat transferred and the work produced during the process.

Expert Solution & Answer

Explanation of Solution

Given:

Initial pressure of steam is 75 kPa.

Initial quality is 8%.

Initial volume is 2 m3.

Final volume is 5 m3.

Final pressure of the steam is 225 kPa.

Calculation:

From the Table A-5, to obtain the value of the specific volume of saturated liquid is vf, the specific volume of saturated vapour is vg, the specific internal energy of saturated liquid is uf, the specific internal energy change upon vaporization is vfg at initial pressure of 250 kPa.

vf1=0.001037 m3/kgvg1=2.2172 m3/kguf1=384.36 kJ/kgufg1=2111.8 kJ/kg

The specific volume of the spring-loaded piston cylinder device is,

v1=vf1+xvfg1

v1=(0.001037 m3/kg)+(0.08)×(2.2172 m3/kg−0.001037 m3/kg)=(0.001037 m3/kg)+(0.08)×(2.216163 m3/kg)=0.1783 m3/kg

The initial specific internal energy of the spring-loaded piston cylinder device is,

u1=uf1+xufg1

u1=(384.36 kJ/kg)+(0.08)×(2111.8 kJ/kg)=(384.36 kJ/kg)+(168.94 kJ/kg)=553.30 kJ/kg

The mass of the system is,

m=ν1v1

m=2 m30.1783 m3/kg=11.217 kg

Determine the final specific volume of the piston cylinder device.

v2=V2m

v2=(5 m3)(11.217 kg)=0.44575 m3/kg≅0.4458 m3/kg

From the Table A-5, to obtain the value of the specific volume of saturated liquid is vf, the specific volume of saturated vapour is vg, the specific internal energy of saturated liquid is uf, the specific internal energy change upon vaporization is vfg at final pressure of 225 kPa.

vf2=0.001064 m3/kgvg2=0.79329 m3/kguf2=520.47 kJ/kgufg2=2012.7 kJ/kg

Determine the quality of final state for the spring-loaded piston-cylinder device.

x2=v2−vf2(vg2−vf2)

x2=(0.4458 m3/kg)−(0.001064 m3/kg)(0.79329 m3/kg−0.001064 m3/kg)=(0.444736 m3/kg)(0.792226 m3/kg)=0.561375

The final specific internal energy of the spring-loaded piston cylinder device is,

u2=uf2+x2ufg2

u2=(520.47 kJ/kg)+(0.561375)×(2012.7 kJ/kg)=(520.47 kJ/kg)+(1129.75 kJ/kg)=1650.35 kJ/kg≅1650.4 kJ/kg

The work done during the constant pressure process is,

Wb,out=∫12P dν=P1+P22(ν2−ν1)

Wb,out=(75+225)kPa2(5−2)m3=300 kPa2×3m3=150 kPa×3m3=450 kJ

Thus, the work produced during the process is 450 kJ.

The energy balance equation is,

Ein−Eout=ΔEsystemQin−Wb,out=ΔUQin=m(u2−u1)+Wb,out

Qin=(11.217 kg)(1650.4 kJ/kg−553.30 kJ/kg)+450 kJ=(11.217 kg)(1097.1 kJ/kg)+450 kJ=12756 kJ

Thus, the heat transferred during the process is 12,756 kJ.

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Answer 2

Question

Chapter 5, Problem 38P

To determine

The heat transferred and the work produced during the process.

Expert Solution & Answer

Explanation of Solution

Given:

Initial pressure of steam is 75 kPa.

Initial quality is 8%.

Initial volume is 2 m3.

Final volume is 5 m3.

Final pressure of the steam is 225 kPa.

Calculation:

From the Table A-5, to obtain the value of the specific volume of saturated liquid is vf, the specific volume of saturated vapour is vg, the specific internal energy of saturated liquid is uf, the specific internal energy change upon vaporization is vfg at initial pressure of 250 kPa.

vf1=0.001037 m3/kgvg1=2.2172 m3/kguf1=384.36 kJ/kgufg1=2111.8 kJ/kg

The specific volume of the spring-loaded piston cylinder device is,

v1=vf1+xvfg1

v1=(0.001037 m3/kg)+(0.08)×(2.2172 m3/kg−0.001037 m3/kg)=(0.001037 m3/kg)+(0.08)×(2.216163 m3/kg)=0.1783 m3/kg

The initial specific internal energy of the spring-loaded piston cylinder device is,

u1=uf1+xufg1

u1=(384.36 kJ/kg)+(0.08)×(2111.8 kJ/kg)=(384.36 kJ/kg)+(168.94 kJ/kg)=553.30 kJ/kg

The mass of the system is,

m=ν1v1

m=2 m30.1783 m3/kg=11.217 kg

Determine the final specific volume of the piston cylinder device.

v2=V2m

v2=(5 m3)(11.217 kg)=0.44575 m3/kg≅0.4458 m3/kg

From the Table A-5, to obtain the value of the specific volume of saturated liquid is vf, the specific volume of saturated vapour is vg, the specific internal energy of saturated liquid is uf, the specific internal energy change upon vaporization is vfg at final pressure of 225 kPa.

vf2=0.001064 m3/kgvg2=0.79329 m3/kguf2=520.47 kJ/kgufg2=2012.7 kJ/kg

Determine the quality of final state for the spring-loaded piston-cylinder device.

x2=v2−vf2(vg2−vf2)

x2=(0.4458 m3/kg)−(0.001064 m3/kg)(0.79329 m3/kg−0.001064 m3/kg)=(0.444736 m3/kg)(0.792226 m3/kg)=0.561375

The final specific internal energy of the spring-loaded piston cylinder device is,

u2=uf2+x2ufg2

u2=(520.47 kJ/kg)+(0.561375)×(2012.7 kJ/kg)=(520.47 kJ/kg)+(1129.75 kJ/kg)=1650.35 kJ/kg≅1650.4 kJ/kg

The work done during the constant pressure process is,

Wb,out=∫12P dν=P1+P22(ν2−ν1)

Wb,out=(75+225)kPa2(5−2)m3=300 kPa2×3m3=150 kPa×3m3=450 kJ

Thus, the work produced during the process is 450 kJ.

The energy balance equation is,

Ein−Eout=ΔEsystemQin−Wb,out=ΔUQin=m(u2−u1)+Wb,out

Qin=(11.217 kg)(1650.4 kJ/kg−553.30 kJ/kg)+450 kJ=(11.217 kg)(1097.1 kJ/kg)+450 kJ=12756 kJ

Thus, the heat transferred during the process is 12,756 kJ.

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Answer 3

Question

Chapter 5, Problem 38P

To determine

The heat transferred and the work produced during the process.

Expert Solution & Answer

Explanation of Solution

Given:

Initial pressure of steam is 75 kPa.

Initial quality is 8%.

Initial volume is 2 m3.

Final volume is 5 m3.

Final pressure of the steam is 225 kPa.

Calculation:

From the Table A-5, to obtain the value of the specific volume of saturated liquid is vf, the specific volume of saturated vapour is vg, the specific internal energy of saturated liquid is uf, the specific internal energy change upon vaporization is vfg at initial pressure of 250 kPa.

vf1=0.001037 m3/kgvg1=2.2172 m3/kguf1=384.36 kJ/kgufg1=2111.8 kJ/kg

The specific volume of the spring-loaded piston cylinder device is,

v1=vf1+xvfg1

v1=(0.001037 m3/kg)+(0.08)×(2.2172 m3/kg−0.001037 m3/kg)=(0.001037 m3/kg)+(0.08)×(2.216163 m3/kg)=0.1783 m3/kg

The initial specific internal energy of the spring-loaded piston cylinder device is,

u1=uf1+xufg1

u1=(384.36 kJ/kg)+(0.08)×(2111.8 kJ/kg)=(384.36 kJ/kg)+(168.94 kJ/kg)=553.30 kJ/kg

The mass of the system is,

m=ν1v1

m=2 m30.1783 m3/kg=11.217 kg

Determine the final specific volume of the piston cylinder device.

v2=V2m

v2=(5 m3)(11.217 kg)=0.44575 m3/kg≅0.4458 m3/kg

From the Table A-5, to obtain the value of the specific volume of saturated liquid is vf, the specific volume of saturated vapour is vg, the specific internal energy of saturated liquid is uf, the specific internal energy change upon vaporization is vfg at final pressure of 225 kPa.

vf2=0.001064 m3/kgvg2=0.79329 m3/kguf2=520.47 kJ/kgufg2=2012.7 kJ/kg

Determine the quality of final state for the spring-loaded piston-cylinder device.

x2=v2−vf2(vg2−vf2)

x2=(0.4458 m3/kg)−(0.001064 m3/kg)(0.79329 m3/kg−0.001064 m3/kg)=(0.444736 m3/kg)(0.792226 m3/kg)=0.561375

The final specific internal energy of the spring-loaded piston cylinder device is,

u2=uf2+x2ufg2

u2=(520.47 kJ/kg)+(0.561375)×(2012.7 kJ/kg)=(520.47 kJ/kg)+(1129.75 kJ/kg)=1650.35 kJ/kg≅1650.4 kJ/kg

The work done during the constant pressure process is,

Wb,out=∫12P dν=P1+P22(ν2−ν1)

Wb,out=(75+225)kPa2(5−2)m3=300 kPa2×3m3=150 kPa×3m3=450 kJ

Thus, the work produced during the process is 450 kJ.

The energy balance equation is,

Ein−Eout=ΔEsystemQin−Wb,out=ΔUQin=m(u2−u1)+Wb,out

Qin=(11.217 kg)(1650.4 kJ/kg−553.30 kJ/kg)+450 kJ=(11.217 kg)(1097.1 kJ/kg)+450 kJ=12756 kJ

Thus, the heat transferred during the process is 12,756 kJ.

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Answer 4

Question

Chapter 5, Problem 38P

To determine

The heat transferred and the work produced during the process.

Expert Solution & Answer

Explanation of Solution

Given:

Initial pressure of steam is 75 kPa.

Initial quality is 8%.

Initial volume is 2 m3.

Final volume is 5 m3.

Final pressure of the steam is 225 kPa.

Calculation:

From the Table A-5, to obtain the value of the specific volume of saturated liquid is vf, the specific volume of saturated vapour is vg, the specific internal energy of saturated liquid is uf, the specific internal energy change upon vaporization is vfg at initial pressure of 250 kPa.

vf1=0.001037 m3/kgvg1=2.2172 m3/kguf1=384.36 kJ/kgufg1=2111.8 kJ/kg

The specific volume of the spring-loaded piston cylinder device is,

v1=vf1+xvfg1

v1=(0.001037 m3/kg)+(0.08)×(2.2172 m3/kg−0.001037 m3/kg)=(0.001037 m3/kg)+(0.08)×(2.216163 m3/kg)=0.1783 m3/kg

The initial specific internal energy of the spring-loaded piston cylinder device is,

u1=uf1+xufg1

u1=(384.36 kJ/kg)+(0.08)×(2111.8 kJ/kg)=(384.36 kJ/kg)+(168.94 kJ/kg)=553.30 kJ/kg

The mass of the system is,

m=ν1v1

m=2 m30.1783 m3/kg=11.217 kg

Determine the final specific volume of the piston cylinder device.

v2=V2m

v2=(5 m3)(11.217 kg)=0.44575 m3/kg≅0.4458 m3/kg

From the Table A-5, to obtain the value of the specific volume of saturated liquid is vf, the specific volume of saturated vapour is vg, the specific internal energy of saturated liquid is uf, the specific internal energy change upon vaporization is vfg at final pressure of 225 kPa.

vf2=0.001064 m3/kgvg2=0.79329 m3/kguf2=520.47 kJ/kgufg2=2012.7 kJ/kg

Determine the quality of final state for the spring-loaded piston-cylinder device.

x2=v2−vf2(vg2−vf2)

x2=(0.4458 m3/kg)−(0.001064 m3/kg)(0.79329 m3/kg−0.001064 m3/kg)=(0.444736 m3/kg)(0.792226 m3/kg)=0.561375

The final specific internal energy of the spring-loaded piston cylinder device is,

u2=uf2+x2ufg2

u2=(520.47 kJ/kg)+(0.561375)×(2012.7 kJ/kg)=(520.47 kJ/kg)+(1129.75 kJ/kg)=1650.35 kJ/kg≅1650.4 kJ/kg

The work done during the constant pressure process is,

Wb,out=∫12P dν=P1+P22(ν2−ν1)

Wb,out=(75+225)kPa2(5−2)m3=300 kPa2×3m3=150 kPa×3m3=450 kJ

Thus, the work produced during the process is 450 kJ.

The energy balance equation is,

Ein−Eout=ΔEsystemQin−Wb,out=ΔUQin=m(u2−u1)+Wb,out

Qin=(11.217 kg)(1650.4 kJ/kg−553.30 kJ/kg)+450 kJ=(11.217 kg)(1097.1 kJ/kg)+450 kJ=12756 kJ

Thus, the heat transferred during the process is 12,756 kJ.

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Answer 5

Chapter 5, Problem 38P

To determine

The heat transferred and the work produced during the process.

Expert Solution & Answer

Explanation of Solution

Given:

Initial pressure of steam is 75 kPa.

Initial quality is 8%.

Initial volume is 2 m3.

Final volume is 5 m3.

Final pressure of the steam is 225 kPa.

Calculation:

From the Table A-5, to obtain the value of the specific volume of saturated liquid is vf, the specific volume of saturated vapour is vg, the specific internal energy of saturated liquid is uf, the specific internal energy change upon vaporization is vfg at initial pressure of 250 kPa.

vf1=0.001037 m3/kgvg1=2.2172 m3/kguf1=384.36 kJ/kgufg1=2111.8 kJ/kg

The specific volume of the spring-loaded piston cylinder device is,

v1=vf1+xvfg1

v1=(0.001037 m3/kg)+(0.08)×(2.2172 m3/kg−0.001037 m3/kg)=(0.001037 m3/kg)+(0.08)×(2.216163 m3/kg)=0.1783 m3/kg

The initial specific internal energy of the spring-loaded piston cylinder device is,

u1=uf1+xufg1

u1=(384.36 kJ/kg)+(0.08)×(2111.8 kJ/kg)=(384.36 kJ/kg)+(168.94 kJ/kg)=553.30 kJ/kg

The mass of the system is,

m=ν1v1

m=2 m30.1783 m3/kg=11.217 kg

Determine the final specific volume of the piston cylinder device.

v2=V2m

v2=(5 m3)(11.217 kg)=0.44575 m3/kg≅0.4458 m3/kg

From the Table A-5, to obtain the value of the specific volume of saturated liquid is vf, the specific volume of saturated vapour is vg, the specific internal energy of saturated liquid is uf, the specific internal energy change upon vaporization is vfg at final pressure of 225 kPa.

vf2=0.001064 m3/kgvg2=0.79329 m3/kguf2=520.47 kJ/kgufg2=2012.7 kJ/kg

Determine the quality of final state for the spring-loaded piston-cylinder device.

x2=v2−vf2(vg2−vf2)

x2=(0.4458 m3/kg)−(0.001064 m3/kg)(0.79329 m3/kg−0.001064 m3/kg)=(0.444736 m3/kg)(0.792226 m3/kg)=0.561375

The final specific internal energy of the spring-loaded piston cylinder device is,

u2=uf2+x2ufg2

u2=(520.47 kJ/kg)+(0.561375)×(2012.7 kJ/kg)=(520.47 kJ/kg)+(1129.75 kJ/kg)=1650.35 kJ/kg≅1650.4 kJ/kg

The work done during the constant pressure process is,

Wb,out=∫12P dν=P1+P22(ν2−ν1)

Wb,out=(75+225)kPa2(5−2)m3=300 kPa2×3m3=150 kPa×3m3=450 kJ

Thus, the work produced during the process is 450 kJ.

The energy balance equation is,

Ein−Eout=ΔEsystemQin−Wb,out=ΔUQin=m(u2−u1)+Wb,out

Qin=(11.217 kg)(1650.4 kJ/kg−553.30 kJ/kg)+450 kJ=(11.217 kg)(1097.1 kJ/kg)+450 kJ=12756 kJ

Thus, the heat transferred during the process is 12,756 kJ.

Answer 6

Chapter 5, Problem 38P

To determine

The heat transferred and the work produced during the process.

Expert Solution & Answer

Explanation of Solution

Given:

Initial pressure of steam is 75 kPa.

Initial quality is 8%.

Initial volume is 2 m3.

Final volume is 5 m3.

Final pressure of the steam is 225 kPa.

Calculation:

From the Table A-5, to obtain the value of the specific volume of saturated liquid is vf, the specific volume of saturated vapour is vg, the specific internal energy of saturated liquid is uf, the specific internal energy change upon vaporization is vfg at initial pressure of 250 kPa.

vf1=0.001037 m3/kgvg1=2.2172 m3/kguf1=384.36 kJ/kgufg1=2111.8 kJ/kg

The specific volume of the spring-loaded piston cylinder device is,

v1=vf1+xvfg1

v1=(0.001037 m3/kg)+(0.08)×(2.2172 m3/kg−0.001037 m3/kg)=(0.001037 m3/kg)+(0.08)×(2.216163 m3/kg)=0.1783 m3/kg

The initial specific internal energy of the spring-loaded piston cylinder device is,

u1=uf1+xufg1

u1=(384.36 kJ/kg)+(0.08)×(2111.8 kJ/kg)=(384.36 kJ/kg)+(168.94 kJ/kg)=553.30 kJ/kg

The mass of the system is,

m=ν1v1

m=2 m30.1783 m3/kg=11.217 kg

Determine the final specific volume of the piston cylinder device.

v2=V2m

v2=(5 m3)(11.217 kg)=0.44575 m3/kg≅0.4458 m3/kg

From the Table A-5, to obtain the value of the specific volume of saturated liquid is vf, the specific volume of saturated vapour is vg, the specific internal energy of saturated liquid is uf, the specific internal energy change upon vaporization is vfg at final pressure of 225 kPa.

vf2=0.001064 m3/kgvg2=0.79329 m3/kguf2=520.47 kJ/kgufg2=2012.7 kJ/kg

Determine the quality of final state for the spring-loaded piston-cylinder device.

x2=v2−vf2(vg2−vf2)

x2=(0.4458 m3/kg)−(0.001064 m3/kg)(0.79329 m3/kg−0.001064 m3/kg)=(0.444736 m3/kg)(0.792226 m3/kg)=0.561375

The final specific internal energy of the spring-loaded piston cylinder device is,

u2=uf2+x2ufg2

u2=(520.47 kJ/kg)+(0.561375)×(2012.7 kJ/kg)=(520.47 kJ/kg)+(1129.75 kJ/kg)=1650.35 kJ/kg≅1650.4 kJ/kg

The work done during the constant pressure process is,

Wb,out=∫12P dν=P1+P22(ν2−ν1)

Wb,out=(75+225)kPa2(5−2)m3=300 kPa2×3m3=150 kPa×3m3=450 kJ

Thus, the work produced during the process is 450 kJ.

The energy balance equation is,

Ein−Eout=ΔEsystemQin−Wb,out=ΔUQin=m(u2−u1)+Wb,out

Qin=(11.217 kg)(1650.4 kJ/kg−553.30 kJ/kg)+450 kJ=(11.217 kg)(1097.1 kJ/kg)+450 kJ=12756 kJ

Thus, the heat transferred during the process is 12,756 kJ.

Answer 7

Chapter 5, Problem 38P

To determine

The heat transferred and the work produced during the process.

Answer 8

Expert Solution & Answer

Explanation of Solution

Given:

Initial pressure of steam is 75 kPa.

Initial quality is 8%.

Initial volume is 2 m3.

Final volume is 5 m3.

Final pressure of the steam is 225 kPa.

Calculation:

From the Table A-5, to obtain the value of the specific volume of saturated liquid is vf, the specific volume of saturated vapour is vg, the specific internal energy of saturated liquid is uf, the specific internal energy change upon vaporization is vfg at initial pressure of 250 kPa.

vf1=0.001037 m3/kgvg1=2.2172 m3/kguf1=384.36 kJ/kgufg1=2111.8 kJ/kg

The specific volume of the spring-loaded piston cylinder device is,

v1=vf1+xvfg1

v1=(0.001037 m3/kg)+(0.08)×(2.2172 m3/kg−0.001037 m3/kg)=(0.001037 m3/kg)+(0.08)×(2.216163 m3/kg)=0.1783 m3/kg

The initial specific internal energy of the spring-loaded piston cylinder device is,

u1=uf1+xufg1

u1=(384.36 kJ/kg)+(0.08)×(2111.8 kJ/kg)=(384.36 kJ/kg)+(168.94 kJ/kg)=553.30 kJ/kg

The mass of the system is,

m=ν1v1

m=2 m30.1783 m3/kg=11.217 kg

Determine the final specific volume of the piston cylinder device.

v2=V2m

v2=(5 m3)(11.217 kg)=0.44575 m3/kg≅0.4458 m3/kg

From the Table A-5, to obtain the value of the specific volume of saturated liquid is vf, the specific volume of saturated vapour is vg, the specific internal energy of saturated liquid is uf, the specific internal energy change upon vaporization is vfg at final pressure of 225 kPa.