The temperature of the saturated steam ( T 2 ) entering the chamber

Question

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Answer 1

Question

Chapter 8.8, Problem 85P

a)

To determine

The temperature of the saturated steam (T2) entering the chamber

a)

Expert Solution

Answer to Problem 85P

The temperature of the saturated steam (T2) entering the chamber is 129.2°C_.

Explanation of Solution

Write expression for the mass balance on the chamber.

m˙1h1+m˙2h2=m˙3h3m˙1h1+m˙2h2=(m˙1+m˙2)h3h2=[(m˙1+m˙2)h3]−m˙1h1m˙2 (I)

Here, mass rate of liquid water at state 1 is m˙1, mass flow rate of saturated steam at state 2 is m˙2, mixture mass flow rate of liquid water and saturated steam at state 3 is m˙3, enthalpy at state 1,2 and 3 is h1,h2 and h3 respectively.

Conclusion:

Refer to Table A-4, “Saturated water-temperature table”, obtain the following properties at the temperature of 20°C.

h1,hf@20°C=h0=83.91 kJ/kgs1,sf@20°C=s0=0.29649 kJ/kg⋅K

Here, enthalpy of saturated liquid is hf, entropy of saturated liquid is sf, dead-state enthalpy is h0 and dead-state entropy is s0.

Refer to Table A-4, “Saturated water-temperature table”, obtain the following properties at the temperature of 45°C and state 1 quality (x1) of 0.

h3,hf@45°C=188.44 kJ/kgs3,sf@45°C=0.63862 kJ/kg⋅K

Substitute 4.6 kg/s for m˙1, 83.91 kJ/kg for h1, 0.19 kg/s for m˙2, 188.44 kJ/kg for h3 in Equation (I).

h2=[(4.6 kg/s+0.19 kg/s)188.44 kJ/kg]−(4.6 kg/s)(83.91 kJ/kg)0.19 kg/s=[(4.79 kg/s)188.44 kJ/kg]−385.986 kJ/s0.19 kg/s=902.627 kJ/s−385.986 kJ/s0.19 kg/s=2,719 kJ/kg

Refer to Table A-4, “saturated water–temperature table”, for the enthalpy of 2719 kJ/kg obtain the following properties using interpolation method.

Enthalpy kJ/kg	Temperature, °C
2713.1	125
2719	?
2720.1	130

Here, temperature of the saturated steam at state 2 is T2 and enthalpy of saturated vapor is hg.

Substitute 2713.1 kJ/kg for x1, 2719 kJ/kg for x2, 2720.1 kJ/kg for x3, 125°C for y1, and 130°C for y3 in Equation (IV).

T2=(2719−2713.1)(130−125)(2720.1−2713.1)+125=4.2+125=129.2°C

Thus, the temperature of the saturated steam (T2) entering the chamber is 129.2°C_.

b)

To determine

The exergy destruction (X˙dest) during the mixing process.

b)

Expert Solution

Answer to Problem 85P

The exergy destruction (X˙dest) during the mixing process is 105.05 kW_.

Explanation of Solution

Calculate the specific exergy of state 2 (ψ2).

ψ2=h2−h0−T0(s2−s0) (II)

Here, dead state temperature is T0.

Calculate the specific exergy of state 3 (ψ3).

ψ3=h3−h0−T0(s3−s0) (III)

Calculate the specific exergy of state 1 (ψ1).

ψ1=h1−h0−T0(s1−s0) (IV)

Calculate the exergy destruction (X˙dest) during the mixing process.

X˙dest=m˙1ψ1+m˙2ψ2−m˙3ψ3=m˙1ψ1+m˙2ψ2−(m˙1+m˙2)ψ3 (V)

Conclusion:

Refer to Table A-4, “saturated water–temperature table”, for the enthalpy of 2719 kJ/kg obtain the following properties using interpolation method.

Temperature, °C	Entropy kJ/kg⋅K
125	7.0771
129.2	?
130	7.0265

Substitute 125°C for x1, 129.2°C for x2, 130°C for x3, 7.0771 kJ/kg⋅K for y1, and 7.0265 kJ/kg⋅K for y3 in Equation (IV).

s2@129.2°C=(129.2−125)(7.0265−7.0771)(130−125)+7.0771=−0.0425+7.0771=7.0348 kJ/kg⋅K

Substitute 2,719 kJ/kg for h2, 83.91 kJ/kg for h0, 20°C for T0, 7.0348 kJ/kg⋅K for s2 and 0.29649 kJ/kg⋅K for s0 in equation (II).

ψ2=[(2,719−83.91) kJ/kg]−(20°C)[(7.0348−0.29649) kJ/kg⋅K]=(2,635.09 kJ/kg)−{[(20+273) K](6.73831 kJ/kg⋅K)}=2,635.09 kJ/kg−[(293 K)(6.73831 kJ/kg⋅K)]=660.76 kJ/kg

Substitute 188.44 kJ/kg for h3, 83.91 kJ/kg for h0, 20°C for T0, 0.63862 kJ/kg⋅K for s3 and 0.29649 kJ/kg⋅K for s0 in equation (III).

ψ2=[(188.44−83.91) kJ/kg]−(20°C)[(0.63862−0.29649) kJ/kg⋅K]=104.53 kJ/kg−[(293 K)(0.34213 kJ/kg⋅K)]=4.276 kJ/kg

Substitute 83.91 kJ/kg for h1, 83.91 kJ/kg for h0, 20°C for T0, 0.29649 kJ/kg⋅K for s1 and 0.29649 kJ/kg⋅K for s0 in equation (IV).

ψ1=[(83.91−83.91) kJ/kg]−(20°C)[(0.29649−0.29649) kJ/kg⋅K]=0

Substitute 4.6 kg/s for m˙1, 0 for ψ1, 0.19 kg/s for m˙2, 660.7 kJ/kg for ψ2 and 4.276 kJ/kg for ψ3 in Equation (V).

X˙dest={[(4.6 kg/s)(0)]+[(0.19 kg/s)(660.7 kJ/kg)]−[(4.6 kg/s+0.19 kg/s)(4.276 kJ/kg)]}=0+125.533 kJ/s−20.482 kJ/s=105.05 kJ/s(kWkJ/s)=105.05 kW

Thus, the exergy destruction (X˙dest) during the mixing process is 105.05 kW_.

c)

To determine

The second law efficiency (nΙΙ) of the mixing chamber

c)

Expert Solution

Answer to Problem 85P

The second law efficiency (nΙΙ) of the mixing chamber is 15.8%_.

Explanation of Solution

Write the expression for the second law efficiency (nΙΙ) of the mixing chamber.

nΙΙ=X˙recoverdX˙expended=m˙1(ψ3−ψ1)m˙2(ψ2−ψ3) (VI)

Here, rate of exergy recovered is X˙recovered and rate of exergy expended is X˙expended.

Conclusion:

Substitute 4.6 kg/s for m˙1, 0 for ψ1, 0.19 kg/s for m˙2, 660.7 kJ/kg for ψ2 and 4.276 kJ/kg for ψ3 in Equation (VI).

nΙΙ=4.6 kg/s(4.276 kJ/kg−0)0.19 kg/s(660.7 kJ/kg−4.276 kJ/kg)=19.66 kJ/s0.19 kg/s(656.424 kJ/kg)=19.66 kJ/s124.720 kJ/s=0.158

=15.8%

Thus, the second law efficiency (nΙΙ) of the mixing chamber is 15.8%_.

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Answer 2

Question

Chapter 8.8, Problem 85P

a)

To determine

The temperature of the saturated steam (T2) entering the chamber

a)

Expert Solution

Answer to Problem 85P

The temperature of the saturated steam (T2) entering the chamber is 129.2°C_.

Explanation of Solution

Write expression for the mass balance on the chamber.

m˙1h1+m˙2h2=m˙3h3m˙1h1+m˙2h2=(m˙1+m˙2)h3h2=[(m˙1+m˙2)h3]−m˙1h1m˙2 (I)

Here, mass rate of liquid water at state 1 is m˙1, mass flow rate of saturated steam at state 2 is m˙2, mixture mass flow rate of liquid water and saturated steam at state 3 is m˙3, enthalpy at state 1,2 and 3 is h1,h2 and h3 respectively.

Conclusion:

Refer to Table A-4, “Saturated water-temperature table”, obtain the following properties at the temperature of 20°C.

h1,hf@20°C=h0=83.91 kJ/kgs1,sf@20°C=s0=0.29649 kJ/kg⋅K

Here, enthalpy of saturated liquid is hf, entropy of saturated liquid is sf, dead-state enthalpy is h0 and dead-state entropy is s0.

Refer to Table A-4, “Saturated water-temperature table”, obtain the following properties at the temperature of 45°C and state 1 quality (x1) of 0.

h3,hf@45°C=188.44 kJ/kgs3,sf@45°C=0.63862 kJ/kg⋅K

Substitute 4.6 kg/s for m˙1, 83.91 kJ/kg for h1, 0.19 kg/s for m˙2, 188.44 kJ/kg for h3 in Equation (I).

h2=[(4.6 kg/s+0.19 kg/s)188.44 kJ/kg]−(4.6 kg/s)(83.91 kJ/kg)0.19 kg/s=[(4.79 kg/s)188.44 kJ/kg]−385.986 kJ/s0.19 kg/s=902.627 kJ/s−385.986 kJ/s0.19 kg/s=2,719 kJ/kg

Refer to Table A-4, “saturated water–temperature table”, for the enthalpy of 2719 kJ/kg obtain the following properties using interpolation method.

Enthalpy kJ/kg	Temperature, °C
2713.1	125
2719	?
2720.1	130

Here, temperature of the saturated steam at state 2 is T2 and enthalpy of saturated vapor is hg.

Substitute 2713.1 kJ/kg for x1, 2719 kJ/kg for x2, 2720.1 kJ/kg for x3, 125°C for y1, and 130°C for y3 in Equation (IV).

T2=(2719−2713.1)(130−125)(2720.1−2713.1)+125=4.2+125=129.2°C

Thus, the temperature of the saturated steam (T2) entering the chamber is 129.2°C_.

b)

To determine

The exergy destruction (X˙dest) during the mixing process.

b)

Expert Solution

Answer to Problem 85P

The exergy destruction (X˙dest) during the mixing process is 105.05 kW_.

Explanation of Solution

Calculate the specific exergy of state 2 (ψ2).

ψ2=h2−h0−T0(s2−s0) (II)

Here, dead state temperature is T0.

Calculate the specific exergy of state 3 (ψ3).

ψ3=h3−h0−T0(s3−s0) (III)

Calculate the specific exergy of state 1 (ψ1).

ψ1=h1−h0−T0(s1−s0) (IV)

Calculate the exergy destruction (X˙dest) during the mixing process.

X˙dest=m˙1ψ1+m˙2ψ2−m˙3ψ3=m˙1ψ1+m˙2ψ2−(m˙1+m˙2)ψ3 (V)

Conclusion:

Refer to Table A-4, “saturated water–temperature table”, for the enthalpy of 2719 kJ/kg obtain the following properties using interpolation method.

Temperature, °C	Entropy kJ/kg⋅K
125	7.0771
129.2	?
130	7.0265