Saturated water vapor at 300°C is expanded while its pressure is kept constant until its temperature is 700°C. Calculate the change in the specific enthalpy and entropy using ( a ) the departure charts and ( b ) the property tables.

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Answer 1

Textbook Question

Chapter 12.6, Problem 72P

Saturated water vapor at 300°C is expanded while its pressure is kept constant until its temperature is 700°C. Calculate the change in the specific enthalpy and entropy using (a) the departure charts and (b) the property tables.

(a)

Expert Solution

To determine

The change in specific enthalpy and entropy of saturated water per unit mass using departure charts.

Answer to Problem 72P

The change in specific enthalpy and entropy of saturated water per unit mass using departure charts is $973 kJ/kg$ and $1.2954 kJ/kg⋅K$ respectively.

Explanation of Solution

At ideal gas state, the enthalpy is the function of temperature only.

Write the formula for difference in molar specific enthalpy of water vapor at ideal gas state.

$(h2¯−h1¯)ideal=h2¯ideal(T2)−h1¯ideal(T1)$ (I)

Here, the molar enthalpy at ideal gas state corresponding to the temperature is $h¯(T)$ and the subscripts 1 and 2 indicates initial and final states.

Write the formula for change in molar specific entropy.

$(s2¯−s1¯)ideal=s2∘¯−s1∘¯−RulnP2P1$ (II)

Here, the molar specific entropy at reference sate is $s°¯$ , the universal gas constant is $Ru$ , the pressure is $P$ , and the subscripts 1 and 2 indicates initial and final states.

Write formula for enthalpy departure factor $(Zh)$ on molar basis.

$Zh=(h¯ideal−h¯)T,PRuTcr$ (III)

Here, the molar enthalpy at ideal gas state is $h¯ideal$ , the molar enthalpy and normal state is $h¯$ , the universal gas constant is $Ru$ , and the critical temperature is $Tcr$ ; The subscripts $T,P$ indicates the correspondence of given temperature and pressure.

Rearrange the Equation (III) to obtain $h¯$ .

$h¯=h¯ideal−ZhRuTcr$ (IV)

Refer Equation (IV) express as two states of enthalpy difference (final – initial).

$h¯2−h¯1=(h¯2−h¯1)ideal−(Zh2−Zh1)RuTcr$ (V)

Write formula for entropy departure factor $(Zs)$ on molar basis.

$Zs=(s¯ideal−s¯)T,PRu$ (VI)

Here, the molar entropy at ideal gas state is $s¯ideal$ , the molar entropy and normal state is $s¯$ , the universal gas constant is $Ru$ , and the critical temperature is $Tcr$ ; The subscripts $T,P$ indicates the correspondence of given temperature and pressure.

Rearrange the Equation (VI) to obtain $s¯$ .

$s¯=s¯ideal−ZsRu$ (VII)

Refer Equation (VII) express as two states of entropy difference (final – initial).

$s¯2−s¯1=(s¯2−s¯1)ideal−(Zs2−Zs1)Ru$ (VIII)

Write the formula for enthalpy $(h)$ and entropy $(s)$ changes per unit mass.

$h2−h1=h2¯−h1¯M$ (IX)

$s2−s1=s2¯−s1¯M$ (X)

Here, the molar enthalpy is $h¯$ , the molar entropy $s¯$ , and the molar mass of water is $M$ .

Refer Table A-1, “Molar mass, gas constant, and critical-point properties”.

The critical temperature and pressure of water vapor gas is as follows.

$Tcr=647.1 KPcr=22.06 MPa$

Refer Table A-1, “Molar mass, gas constant, and critical-point properties”.

The molar mass $(M)$ of water is $18.015 kg/kmol$ .

The pressure is kept constant until its temperature reaches to $700 °C$ .

$P1=P2=Psat @ 300 °C$

Refer Table A-4, “Saturated water-Temperature table”.

The saturation pressure corresponding to the temperature of $700 °C$ is $8587.9 kPa≃8588 kPa(8.588 MPa)$ .

The reduced pressure $(PR1)$ and temperature $(TR1)$ at initial state is expressed as follows.

$TR1=T1Tcr=(300+273) K647.1 K=0.885$

$PR1=P1Pcr=8.588 MPa22.06 MPa=0.389$

The reduced pressure $(PR2)$ and temperature $(TR2)$ at final state is expressed as follows.

$TR2=T2Tcr=(700+273) K647.1 K=1.504$

$PR2=P2Pcr=8.588 MPa22.06 MPa=0.389$

At initial:

Refer Figure A-29, “Generalized enthalpy departure chart”.

The enthalpy departure factor $(Zh1)$ corresponding the reduced pressure $(PR1)$ and reduced temperature $(TR1)$ is $0.609$ .

Refer Figure A-30, “Generalized entropy departure chart”.

The entropy departure factor $(Zs1)$ corresponding the reduced pressure $(PR1)$ and reduced temperature $(TR1)$ is $0.481$ .

At final:

Refer Figure A-29, “Generalized enthalpy departure chart”.

The enthalpy departure factor $(Zh2)$ corresponding the reduced pressure $(PR2)$ and reduced temperature $(TR2)$ is $0.204$ .

Refer Figure A-30, “Generalized entropy departure chart”.

The entropy departure factor $(Zs2)$ corresponding the reduced pressure $(PR2)$ and reduced temperature $(TR2)$ is $0.105$ .

Refer Table A-19, “Ideal-gas properties of water vapor, $H2O$ ”.

Obtain the initial properties corresponding to the temperature of $573 K$ .

$h1¯=19426 kJ/kmols1∘¯=211.263 kJ/kmol⋅K$

Obtain the final properties corresponding to the temperature of $973 K$ .

$h2¯=34775 kJ/kmols2∘¯=231.473 kJ/kmol⋅K$

The universal gas constant $(Ru)$ is $8.314 kJ/kmol⋅K$ .

Conclusion:

Substitute $34775 kJ/kmol$ for $h2¯ideal(T2)$ and $19426 kJ/kmol$ for $h1¯ideal(T1)$ in

Equation (I).

$(h2¯−h1¯)ideal=34775 kJ/kmol−19426 kJ/kmol=15349 kJ/kmol$

Substitute $231.473 kJ/kmol⋅K$ for $s2∘¯$ , $211.263 kJ/kmol⋅K$ for $s1∘¯$ , $8.314 kJ/kmol⋅K$ for $Ru$ , and $8.588 MPa$ for $P2$ , $P1$ in Equation (II).

$(s2¯−s1¯)ideal=[231.473 kJ/kmol⋅K−211.263 kJ/kmol⋅K−(8.314 kJ/kmol⋅K)ln8.588 MPa8.588 MPa]=20.21 kJ/kmol⋅K−0=20.21 kJ/kmol⋅K$

Substitute $15349 kJ/kmol$ for $(h¯2−h¯1)ideal$ , $0.204$ for $Zh2$ , $0.609$ for $Zh1$ , $8.314 kJ/kmol⋅K$ for $Ru$ , and $647.1 K$ for $Tcr$ in Equation (V).

$h¯2−h¯1={15349 kJ/kmol−[(0.204−0.609)(8.314 kJ/kmol⋅K)(647.1 K)]}=15349 kJ/kmol+2178.8957 kJ/kmol=17527.8957 kJ/kmol≃17528 kJ/kmol$

Substitute $20.21 kJ/kmol⋅K$ for $(s¯2−s¯1)ideal$ , $0.105$ for $Zs2$ , $0.481$ for $Zs1$ , and $8.314 kJ/kmol⋅K$ for $Ru$ , in Equation (VIII).

$s¯2−s¯1={20.21 kJ/kmol⋅K−[(0.105−0.481)(8.314 kJ/kmol⋅K)]}=20.21 kJ/kmol⋅K+3.1260 kJ/kmol⋅K=23.336 kJ/kmol⋅K$

Substitute $17528 kJ/kmol$ for $h2¯−h1¯$ and $18.015 kg/kmol$ for $M$ in Equation (IX).

$h2−h1=17528 kJ/kmol18.015 kg/kmol=972.9669 kJ/kg=973 kJ/kg$

Substitute $23.336 kJ/kmol⋅K$ for $s2¯−s1¯$ and $18.015 kg/kmol$ for $M$ in Equation (X).

$s2−s1=23.336 kJ/kmol⋅K18.015 kg/kmol=1.2954 kJ/kg⋅K$

Thus, the change in specific enthalpy and entropy of saturated water per unit mass using departure charts is $973 kJ/kg$ and $1.2954 kJ/kg⋅K$ respectively.

(a)

Expert Solution

To determine

The change in specific enthalpy and entropy of water vapor per unit mass using property tables.

Answer to Problem 72P

The change in specific enthalpy and entropy of water vapor per unit mass using property tables is $1129 kJ/kg$ and $1.5405 kJ/kg⋅K$ respectively.

Explanation of Solution

At state 1:

The steam is at state of saturated vapor at the temperature of $300 °C$ . Hence, the enthalpy $(h1)$ and entropy $(s1)$ at state 1 is expressed as follows.

$h1=hg@300 °Cs1=sg@300 °C$

Refer Table A-4, “Saturated water-Temperature table”.

The enthalpy $(h1)$ and entropy $(s1)$ at state 1 corresponding to the temperature of $300 °C$ is $2749.6 kJ/kg$ and $5.7059 kJ/kg⋅K$ respectively.

At state 2:

The water vapor is expanded to the temperature of $700 °C$ by keeping the pressure as constant. This makes the saturated vapor into to superheated vapor.

The pressure is kept constant until its temperature reaches to $700 °C$ .

$P1=P2=Psat @ 300 °C$

Refer Table A-4, “Saturated water-Temperature table”.

The saturation pressure corresponding to the temperature of $700 °C$ is $8587.9 kPa≃8588 kPa(8.588 MPa)$ .

Refer Table A-6, “Superheated water”.

Obtain the enthalpy $(h2)$ and entropy $(s2)$ at state 2 corresponding to the pressure of $8.588 MPa$ and the temperature of $700 °C$ by interpolating the tables.

$h2=3878.6 kJ/kgs2=7.2465 kJ/kg⋅K$

Conclusion:

The enthalpy changes are expressed as follows.

$h2−h1=3878.6 kJ/kg−2749.6 kJ/kg=1129 kJ/kg$

The entropy changes are expressed as follows.

$s2−s1=7.2464 kJ/kg⋅K−5.7059 kJ/kg⋅K=1.5405 kJ/kg⋅K$

Thus, the change in specific enthalpy and entropy of water vapor per unit mass using property tables is $1129 kJ/kg$ and $1.5405 kJ/kg⋅K$ respectively.

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