The changes in pressure and entropy.

Question

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Answer 1

Question

Chapter 13, Problem 1P

To determine

The changes in pressure and entropy.

Expert Solution & Answer

Explanation of Solution

Given:

Initial pressure (p1) is 120 psi.

Initial temperature (T1) is 40 °F.

Final temperature (T2) is 120°F.

Calculation:

Consider oxygen as compressible.

Write an equation for pressure ratio (p1p2) and calculate final pressure (p2).

p1p2=ρRT1ρRT2p1p2=T1T2

120 psip2=460+40°F460+120°Fp2=139.2psi

Calculate the change in pressure (Δp).

Δp=p2−p1

Δp=(139.2−120)psi=19.2 psi

Thus, the change is pressure is 19.2 psi_.

From Appendix A table of “Physical properties of Gases”, for Oxygen, obtain the values for the gas constant (R) as 1,554ft⋅lb/slug⋅°R and specific heat ratio (k) as 1.4.

Write an equation for change in enthalpy (Δh) and calculate specific heat at constant pressure (cp).

Δh=kRk−1(T2−T1)cpΔT=kRk−1ΔTcp=kRk−1

cp=1,554ft⋅lb/slug⋅°R ×1.41.4−1=5439 ft⋅lb/slug⋅°R

Calculate the change in entropy (Δs).

(Δs)=cplnT2T1−Rlnp2p1={(5439ft⋅lb/slug⋅°R)[ln(580°R500°R)]−(1554ft⋅lb/slug⋅°R)[ln(139.2psi120psi)]}=576.61ft⋅lb/slug⋅°R

Thus, entropy change (Δs) is 576.61ft⋅lb/slug⋅°R_.

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Required information Water initially at 200 kPa and 300°C is contained in a piston-cylinder device fitted with stops. The water is allowed to cool at constant pressure until it exists as a saturated vapor and the piston rests on the stops. Then the water continues to cool until the pressure is 100 kPa. NOTE: This is a multi-part question. Once an answer is submitted, you will be unable to return to this part. Water 200 kPa 300°C On the T-V diagram, sketch, with respect to the saturation lines, the process curves passing through the initial, intermediate, and final states of the water. Label the T, P, and V values for end states on the process curves. Please upload your response/solution by using the controls provided below.

A piston-cylinder device contains 0.87 kg of refrigerant-134a at -10°C. The piston that is free to move has a mass of 12 kg and a diameter of 25 cm. The local atmospheric pressure is 88 kPa. Now, heat is transferred to refrigerant-134a until the temperature is 15°C. Use data from the tables. R-134a -10°C Determine the change in the volume of the cylinder of the refrigerant-134a if the specific volume and enthalpy of R-134a at the initial state of 90.4 kPa and -10°C and at the final state of 90.4 kPa and 15°C are as follows: = 0.2418 m³/kg, h₁ = 247.77 kJ/kg 3 v2 = 0.2670 m³/kg, and h₂ = 268.18 kJ/kg The change in the volume of the cylinder is m

Answer 2

Question

Chapter 13, Problem 1P

To determine

The changes in pressure and entropy.

Expert Solution & Answer

Explanation of Solution

Given:

Initial pressure (p1) is 120 psi.

Initial temperature (T1) is 40 °F.

Final temperature (T2) is 120°F.

Calculation:

Consider oxygen as compressible.

Write an equation for pressure ratio (p1p2) and calculate final pressure (p2).

p1p2=ρRT1ρRT2p1p2=T1T2

120 psip2=460+40°F460+120°Fp2=139.2psi

Calculate the change in pressure (Δp).

Δp=p2−p1

Δp=(139.2−120)psi=19.2 psi

Thus, the change is pressure is 19.2 psi_.

From Appendix A table of “Physical properties of Gases”, for Oxygen, obtain the values for the gas constant (R) as 1,554ft⋅lb/slug⋅°R and specific heat ratio (k) as 1.4.

Write an equation for change in enthalpy (Δh) and calculate specific heat at constant pressure (cp).

Δh=kRk−1(T2−T1)cpΔT=kRk−1ΔTcp=kRk−1

cp=1,554ft⋅lb/slug⋅°R ×1.41.4−1=5439 ft⋅lb/slug⋅°R

Calculate the change in entropy (Δs).

(Δs)=cplnT2T1−Rlnp2p1={(5439ft⋅lb/slug⋅°R)[ln(580°R500°R)]−(1554ft⋅lb/slug⋅°R)[ln(139.2psi120psi)]}=576.61ft⋅lb/slug⋅°R

Thus, entropy change (Δs) is 576.61ft⋅lb/slug⋅°R_.

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Answer 3

Question

Chapter 13, Problem 1P

To determine

The changes in pressure and entropy.

Expert Solution & Answer

Explanation of Solution

Given:

Initial pressure (p1) is 120 psi.

Initial temperature (T1) is 40 °F.

Final temperature (T2) is 120°F.

Calculation:

Consider oxygen as compressible.

Write an equation for pressure ratio (p1p2) and calculate final pressure (p2).

p1p2=ρRT1ρRT2p1p2=T1T2

120 psip2=460+40°F460+120°Fp2=139.2psi

Calculate the change in pressure (Δp).

Δp=p2−p1

Δp=(139.2−120)psi=19.2 psi

Thus, the change is pressure is 19.2 psi_.

From Appendix A table of “Physical properties of Gases”, for Oxygen, obtain the values for the gas constant (R) as 1,554ft⋅lb/slug⋅°R and specific heat ratio (k) as 1.4.

Write an equation for change in enthalpy (Δh) and calculate specific heat at constant pressure (cp).

Δh=kRk−1(T2−T1)cpΔT=kRk−1ΔTcp=kRk−1

cp=1,554ft⋅lb/slug⋅°R ×1.41.4−1=5439 ft⋅lb/slug⋅°R

Calculate the change in entropy (Δs).

(Δs)=cplnT2T1−Rlnp2p1={(5439ft⋅lb/slug⋅°R)[ln(580°R500°R)]−(1554ft⋅lb/slug⋅°R)[ln(139.2psi120psi)]}=576.61ft⋅lb/slug⋅°R

Thus, entropy change (Δs) is 576.61ft⋅lb/slug⋅°R_.

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Answer 4

Question

Chapter 13, Problem 1P

To determine

The changes in pressure and entropy.

Expert Solution & Answer

Explanation of Solution

Given:

Initial pressure (p1) is 120 psi.

Initial temperature (T1) is 40 °F.

Final temperature (T2) is 120°F.

Calculation:

Consider oxygen as compressible.

Write an equation for pressure ratio (p1p2) and calculate final pressure (p2).

p1p2=ρRT1ρRT2p1p2=T1T2

120 psip2=460+40°F460+120°Fp2=139.2psi

Calculate the change in pressure (Δp).

Δp=p2−p1

Δp=(139.2−120)psi=19.2 psi

Thus, the change is pressure is 19.2 psi_.

From Appendix A table of “Physical properties of Gases”, for Oxygen, obtain the values for the gas constant (R) as 1,554ft⋅lb/slug⋅°R and specific heat ratio (k) as 1.4.

Write an equation for change in enthalpy (Δh) and calculate specific heat at constant pressure (cp).

Δh=kRk−1(T2−T1)cpΔT=kRk−1ΔTcp=kRk−1

cp=1,554ft⋅lb/slug⋅°R ×1.41.4−1=5439 ft⋅lb/slug⋅°R

Calculate the change in entropy (Δs).

(Δs)=cplnT2T1−Rlnp2p1={(5439ft⋅lb/slug⋅°R)[ln(580°R500°R)]−(1554ft⋅lb/slug⋅°R)[ln(139.2psi120psi)]}=576.61ft⋅lb/slug⋅°R

Thus, entropy change (Δs) is 576.61ft⋅lb/slug⋅°R_.

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Answer 5

Chapter 13, Problem 1P

To determine

The changes in pressure and entropy.

Expert Solution & Answer

Explanation of Solution

Given:

Initial pressure (p1) is 120 psi.

Initial temperature (T1) is 40 °F.

Final temperature (T2) is 120°F.

Calculation:

Consider oxygen as compressible.

Write an equation for pressure ratio (p1p2) and calculate final pressure (p2).

p1p2=ρRT1ρRT2p1p2=T1T2

120 psip2=460+40°F460+120°Fp2=139.2psi

Calculate the change in pressure (Δp).

Δp=p2−p1

Δp=(139.2−120)psi=19.2 psi

Thus, the change is pressure is 19.2 psi_.

From Appendix A table of “Physical properties of Gases”, for Oxygen, obtain the values for the gas constant (R) as 1,554ft⋅lb/slug⋅°R and specific heat ratio (k) as 1.4.

Write an equation for change in enthalpy (Δh) and calculate specific heat at constant pressure (cp).

Δh=kRk−1(T2−T1)cpΔT=kRk−1ΔTcp=kRk−1

cp=1,554ft⋅lb/slug⋅°R ×1.41.4−1=5439 ft⋅lb/slug⋅°R

Calculate the change in entropy (Δs).

(Δs)=cplnT2T1−Rlnp2p1={(5439ft⋅lb/slug⋅°R)[ln(580°R500°R)]−(1554ft⋅lb/slug⋅°R)[ln(139.2psi120psi)]}=576.61ft⋅lb/slug⋅°R

Thus, entropy change (Δs) is 576.61ft⋅lb/slug⋅°R_.

Answer 6

Chapter 13, Problem 1P

To determine

The changes in pressure and entropy.

Expert Solution & Answer

Explanation of Solution

Given:

Initial pressure (p1) is 120 psi.

Initial temperature (T1) is 40 °F.

Final temperature (T2) is 120°F.

Calculation:

Consider oxygen as compressible.

Write an equation for pressure ratio (p1p2) and calculate final pressure (p2).

p1p2=ρRT1ρRT2p1p2=T1T2

120 psip2=460+40°F460+120°Fp2=139.2psi

Calculate the change in pressure (Δp).

Δp=p2−p1

Δp=(139.2−120)psi=19.2 psi

Thus, the change is pressure is 19.2 psi_.

From Appendix A table of “Physical properties of Gases”, for Oxygen, obtain the values for the gas constant (R) as 1,554ft⋅lb/slug⋅°R and specific heat ratio (k) as 1.4.

Write an equation for change in enthalpy (Δh) and calculate specific heat at constant pressure (cp).

Δh=kRk−1(T2−T1)cpΔT=kRk−1ΔTcp=kRk−1

cp=1,554ft⋅lb/slug⋅°R ×1.41.4−1=5439 ft⋅lb/slug⋅°R

Calculate the change in entropy (Δs).

(Δs)=cplnT2T1−Rlnp2p1={(5439ft⋅lb/slug⋅°R)[ln(580°R500°R)]−(1554ft⋅lb/slug⋅°R)[ln(139.2psi120psi)]}=576.61ft⋅lb/slug⋅°R

Thus, entropy change (Δs) is 576.61ft⋅lb/slug⋅°R_.

Answer 7

Chapter 13, Problem 1P

To determine

The changes in pressure and entropy.

Answer 8

Expert Solution & Answer

Explanation of Solution

Given:

Initial pressure (p1) is 120 psi.

Initial temperature (T1) is 40 °F.

Final temperature (T2) is 120°F.

Calculation:

Consider oxygen as compressible.

Write an equation for pressure ratio (p1p2) and calculate final pressure (p2).

p1p2=ρRT1ρRT2p1p2=T1T2

120 psip2=460+40°F460+120°Fp2=139.2psi

Calculate the change in pressure (Δp).

Δp=p2−p1

Δp=(139.2−120)psi=19.2 psi

Thus, the change is pressure is 19.2 psi_.

From Appendix A table of “Physical properties of Gases”, for Oxygen, obtain the values for the gas constant (R) as 1,554ft⋅lb/slug⋅°R and specific heat ratio (k) as 1.4.