THERMODYNAMICS(SI UNITS,INTL.ED)EBOOK>I
THERMODYNAMICS(SI UNITS,INTL.ED)EBOOK>I
8th Edition
ISBN: 9781307434316
Author: CENGEL
Publisher: INTER MCG
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Question
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Chapter 8.8, Problem 85P

a)

To determine

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

a)

Expert Solution
Check Mark

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 1,2and3 is h1,h2andh3 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=83.91kJ/kgs1,sf@20°C=s0=0.29649kJ/kgK

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=188.44kJ/kgs3,sf@45°C=0.63862kJ/kgK

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

h2=[(4.6kg/s+0.19kg/s)188.44kJ/kg](4.6kg/s)(83.91kJ/kg)0.19kg/s=[(4.79kg/s)188.44kJ/kg]385.986kJ/s0.19kg/s=902.627kJ/s385.986kJ/s0.19kg/s=2,719kJ/kg

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

Enthalpy kJ/kg

Temperature, °C

2713.1125
2719?
2720.1130

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=(27192713.1)(130125)(2720.12713.1)+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
Check Mark

Answer to Problem 85P

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

Explanation of Solution

Calculate the specific exergy of state 2 (ψ2).

ψ2=h2h0T0(s2s0) (II)

Here, dead state temperature is T0.

Calculate the specific exergy of state 3 (ψ3).

ψ3=h3h0T0(s3s0) (III)

Calculate the specific exergy of state 1 (ψ1).

ψ1=h1h0T0(s1s0) (IV)

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

X˙dest=m˙1ψ1+m˙2ψ2m˙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 2719kJ/kg obtain the following properties using interpolation method.

Temperature, °C

Entropy kJ/kgK

1257.0771
129.2?
1307.0265

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

s2@129.2°C=(129.2125)(7.02657.0771)(130125)+7.0771=0.0425+7.0771=7.0348kJ/kgK

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

ψ2=[(2,71983.91)kJ/kg](20°C)[(7.03480.29649)kJ/kgK]=(2,635.09kJ/kg){[(20+273)K](6.73831kJ/kgK)}=2,635.09kJ/kg[(293K)(6.73831kJ/kgK)]=660.76kJ/kg

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

ψ2=[(188.4483.91)kJ/kg](20°C)[(0.638620.29649)kJ/kgK]=104.53kJ/kg[(293K)(0.34213kJ/kgK)]=4.276kJ/kg

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

ψ1=[(83.9183.91)kJ/kg](20°C)[(0.296490.29649)kJ/kgK]=0

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

X˙dest={[(4.6kg/s)(0)]+[(0.19kg/s)(660.7kJ/kg)][(4.6kg/s+0.19kg/s)(4.276kJ/kg)]}=0+125.533kJ/s20.482kJ/s=105.05kJ/s(kWkJ/s)=105.05kW

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

c)

To determine

The second law efficiency (nΙΙ) of the mixing chamber

c)

Expert Solution
Check Mark

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(ψ3ψ1)m˙2(ψ2ψ3) (VI)

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

Conclusion:

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

nΙΙ=4.6kg/s(4.276kJ/kg0)0.19kg/s(660.7kJ/kg4.276kJ/kg)=19.66kJ/s0.19kg/s(656.424kJ/kg)=19.66kJ/s124.720kJ/s=0.158

=15.8%

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

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Chapter 8 Solutions

THERMODYNAMICS(SI UNITS,INTL.ED)EBOOK>I

Ch. 8.8 - 8–11C Consider a process during which no entropy...Ch. 8.8 - Prob. 12PCh. 8.8 - 8–13E Saturated stem is generated in a boiler by...Ch. 8.8 - One method of meeting the extra electric power...Ch. 8.8 - Prob. 15PCh. 8.8 - A heat engine that receives heat from a furnace at...Ch. 8.8 - Consider a thermal energy reservoir at 1500 K that...Ch. 8.8 - A heat engine receives heat from a source at 1100...Ch. 8.8 - A heat engine that rejects waste heat to a sink at...Ch. 8.8 - Prob. 21PCh. 8.8 - A freezer is maintained at 20F by removing heat...Ch. 8.8 - Prob. 23PCh. 8.8 - Can a system have a higher second-law efficiency...Ch. 8.8 - A mass of 8 kg of helium undergoes a process from...Ch. 8.8 - Prob. 26PCh. 8.8 - Which is a more valuable resource for work...Ch. 8.8 - Which has the capability to produce the most work...Ch. 8.8 - A pistoncylinder device contains 8 kg of...Ch. 8.8 - The radiator of a steam heating system has a...Ch. 8.8 - A well-insulated rigid tank contains 6 lbm of a...Ch. 8.8 - Prob. 33PCh. 8.8 - Prob. 35PCh. 8.8 - Prob. 36PCh. 8.8 - A pistoncylinder device initially contains 2 L of...Ch. 8.8 - A 0.8-m3 insulated rigid tank contains 1.54 kg of...Ch. 8.8 - An insulated pistoncylinder device initially...Ch. 8.8 - An insulated rigid tank is divided into two equal...Ch. 8.8 - Prob. 41PCh. 8.8 - Prob. 42PCh. 8.8 - Prob. 43PCh. 8.8 - Prob. 44PCh. 8.8 - Prob. 45PCh. 8.8 - Prob. 46PCh. 8.8 - A pistoncylinder device initially contains 1.4 kg...Ch. 8.8 - Prob. 48PCh. 8.8 - Prob. 50PCh. 8.8 - Prob. 51PCh. 8.8 - Air enters a nozzle steadily at 200 kPa and 65C...Ch. 8.8 - Prob. 55PCh. 8.8 - Prob. 56PCh. 8.8 - Argon gas enters an adiabatic compressor at 120...Ch. 8.8 - Prob. 58PCh. 8.8 - Prob. 59PCh. 8.8 - Prob. 60PCh. 8.8 - Combustion gases enter a gas turbine at 900C, 800...Ch. 8.8 - Prob. 62PCh. 8.8 - Refrigerant-134a is condensed in a refrigeration...Ch. 8.8 - Prob. 64PCh. 8.8 - Refrigerant-22 absorbs heat from a cooled space at...Ch. 8.8 - Prob. 66PCh. 8.8 - Prob. 67PCh. 8.8 - Prob. 68PCh. 8.8 - Prob. 69PCh. 8.8 - Air enters a compressor at ambient conditions of...Ch. 8.8 - Hot combustion gases enter the nozzle of a...Ch. 8.8 - Prob. 72PCh. 8.8 - Prob. 73PCh. 8.8 - Prob. 74PCh. 8.8 - Prob. 75PCh. 8.8 - Prob. 76PCh. 8.8 - Prob. 77PCh. 8.8 - An insulated vertical pistoncylinder device...Ch. 8.8 - Prob. 79PCh. 8.8 - Prob. 80PCh. 8.8 - Prob. 81PCh. 8.8 - Steam is to be condensed on the shell side of a...Ch. 8.8 - 8–83 Air enters a compressor at ambient conditions...Ch. 8.8 - Prob. 84PCh. 8.8 - Prob. 85PCh. 8.8 - Prob. 86RPCh. 8.8 - Prob. 87RPCh. 8.8 - Steam enters an adiabatic nozzle at 3.5 MPa and...Ch. 8.8 - Prob. 89RPCh. 8.8 - Prob. 91RPCh. 8.8 - A well-insulated, thin-walled, counterflow heat...Ch. 8.8 - Prob. 93RPCh. 8.8 - Prob. 94RPCh. 8.8 - Prob. 95RPCh. 8.8 - Prob. 96RPCh. 8.8 - Prob. 97RPCh. 8.8 - Prob. 98RPCh. 8.8 - Prob. 99RPCh. 8.8 - Prob. 100RPCh. 8.8 - Prob. 101RPCh. 8.8 - A pistoncylinder device initially contains 8 ft3...Ch. 8.8 - Steam at 7 MPa and 400C enters a two-stage...Ch. 8.8 - Steam enters a two-stage adiabatic turbine at 8...Ch. 8.8 - Prob. 105RPCh. 8.8 - Prob. 106RPCh. 8.8 - Prob. 107RPCh. 8.8 - Prob. 108RPCh. 8.8 - Prob. 109RPCh. 8.8 - Prob. 111RPCh. 8.8 - Prob. 112RPCh. 8.8 - A passive solar house that was losing heat to the...Ch. 8.8 - Prob. 114RPCh. 8.8 - Prob. 115RPCh. 8.8 - Prob. 116RPCh. 8.8 - Prob. 117RPCh. 8.8 - Prob. 118RPCh. 8.8 - A 4-L pressure cooker has an operating pressure of...Ch. 8.8 - Repeat Prob. 8114 if heat were supplied to the...Ch. 8.8 - Prob. 121RPCh. 8.8 - Prob. 122RPCh. 8.8 - Reconsider Prob. 8-120. The air stored in the tank...Ch. 8.8 - Prob. 124RPCh. 8.8 - Prob. 125RPCh. 8.8 - Prob. 126RPCh. 8.8 - Prob. 127RPCh. 8.8 - Prob. 128RPCh. 8.8 - Water enters a pump at 100 kPa and 30C at a rate...Ch. 8.8 - Prob. 130RPCh. 8.8 - Nitrogen gas enters a diffuser at 100 kPa and 110C...Ch. 8.8 - Obtain a relation for the second-law efficiency of...Ch. 8.8 - Writing the first- and second-law relations and...Ch. 8.8 - Prob. 134RPCh. 8.8 - Prob. 136FEPCh. 8.8 - Prob. 137FEPCh. 8.8 - A heat engine receives heat from a source at 1500...Ch. 8.8 - Prob. 139FEPCh. 8.8 - Prob. 140FEPCh. 8.8 - A 12-kg solid whose specific heat is 2.8 kJ/kgC is...Ch. 8.8 - Keeping the limitations imposed by the second law...Ch. 8.8 - A furnace can supply heat steadily at 1300 K at a...Ch. 8.8 - Air is throttled from 50C and 800 kPa to a...Ch. 8.8 - Prob. 145FEP
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