EBK THERMODYNAMICS: AN ENGINEERING APPR
EBK THERMODYNAMICS: AN ENGINEERING APPR
8th Edition
ISBN: 8220100257056
Author: CENGEL
Publisher: YUZU
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Chapter 8.8, Problem 105RP

a)

To determine

The exit temperature

a)

Expert Solution
Check Mark

Answer to Problem 105RP

The exit temperature is 12.3°C_.

Explanation of Solution

Write energy balance equation for a closed system of steam.

EinEout=ΔEsystemQout=ΔUQout=m(u1u2) (I)

Here, temperature at inlet and outlet condition is T1andT2 respectively, change in internal energy of a system is ΔU, change in energy transfer of a system is ΔEsystem, internal energy at state 1 and 2 is u1andu2, energy transfer at inlet and outlet condition is EinandEout respectively, heat transfer leaving from the steam radiator is Qout, and total mass of air is m.

Write the expression for the dryness fraction at state 2(x2).

x2=v2vfvgvf (II)

Here, v2=v1.

Write the expression of the internal energy at state 2(u2).

u2=uf+x2ufg (III)

Write the expression of the entropy at state 2(s2).

s2=sf+x2sfg (IV)

Write the expression of the mass of steam (m).

m=V1v1 (V)

Here, initial volume of steam is V1.

Write the expression of the mass of air (mair).

mair=P1V1RT1 (VI)

Here, initial temperature is T1 and volume of air is V1.

Write the expression of the amount of fan work done in 24 min.

Wfan,in=W˙fan,inΔt (VII)

Here, change in time is Δt.

Write the expression of energy balance equation for a closed system of air.

EinEout=ΔEsystemQin+Wfan,inWb,out=ΔUQin+Wfan,in=ΔHQin+Wfan,in=maircp(T2T1) (VIII)

Here, amount of heat transfer injected to the steam radiator is Qin, boundary work output is Wb,out, and work input of fan is Wfan,in.

Conclusion:

From Table A-1, “molar mass, gas constant, and critical point properties”, Obtain the gas constant (R) of air as 0.287kPam3/kgK.

From Table A-3, “properties of common liquids, solids, and foods”, Obtain the specific heat (cp) of the air as 1.005kJ/kgK.

From Table A-6E, “Superheated water”, at the pressure of 200kPa and temperature of 200°C obtain the following properties.

v1=1.0805m3/kgu1=2654.6kJ/kgs1=7.5081kJ/kgK

From to Table A-5, “saturated water – pressure table”, obtain the following properties at the pressure of 100kPa.

vf=0.001043m3/kgvg=1.6941m3/kguf=417.40kJ/kgufg=2088.2kJ/kg

sf=1.3028kJ/kgKsfg=6.0562kJ/kgK

Substitute 1.0805m3/kg for v2, 0.001043m3/kg for vf, and 1.6941m3/kg for vg.in equation (II).

x2=1.0805m3/kg0.001043m3/kg1.6941m3/kg0.001043m3/kg=0.6376

Substitute 0.6376 for x2, 417.40kJ/kg for uf, and 2088.2kJ/kg for ufg in equation (III).

u2=417.40kJ/kg+0.6376(2088.2kJ/kg)=1748.7kJ/kg

Substitute 0.6376 for x2, 1.3028kJ/kgK for sf, and 6.0562kJ/kgK for sfg in equation (IV).

s2=1.3028kJ/kgK+0.6376(6.0562kJ/kgK)=5.1639kJ/kgK

Substitute 0.015m3 for V1 and 1.0805m3/kg for v1 in equation (V).

m=0.015m31.0805m3/kg=0.01388kg

0.01388 kg for m, 2654.6kJ/kg for u1, and 1748.7kJ/kg for u2 in equation (I) calculate Qout.

Qout=0.01388kg(2654.6kJ/kg1748.7kJ/kg)=12.58kJ

Calculate the volume of air.

V1=4m×4m×5m=80m3

Substitute 283 K for T1, 100 kPa for P1, 0.2870kPam3/kgK for R, and 80m3 for V1 in equation (VI).

mair=(100kPa)(80m3)0.2870kPam3/kgK(10°C)=(100kPa)(80m3)0.2870kPam3/kgK(10+273K)=(100kPa)(80m3)0.2870kPam3/kgK(283K)=98.5kg

Substitute 0.150kJ/s for W˙fan,in and 24 min for Δt in equation (VII).

Wfan,in=0.150kJ/s(24min)=0.150kJ/s(24min×60s1min)=216kJ

Substitute 12.58 kJ for Qin, 216 kJ for Wfan,in, 98.5 kg for mair, 1.005kJ/kgK for cp, and 10°C for T1 in Equation (VIII).

12.58kJ+216kJ=(98.5kg)(1.005kJ/kgK)(T210°C)T2=12.3°C

Thus, the exit temperature is 12.3°C_.

b)

To determine

The entropy change of the steam.

b)

Expert Solution
Check Mark

Answer to Problem 105RP

The entropy change of the steam is 0.0325kJ/K_.

Explanation of Solution

Write the expression the entropy change of the steam.

ΔSsteam=m(s2s1) (IX)

Conclusion:

Substitute 0.01388 kg for m, 5.1639kJ/kgK for s2, and 7.5081kJ/kgK for s1 in equation (IX)

ΔSsteam=0.01388kg(5.1639kJ/kgK7.5081kJ/kgK)=0.0325kJ/K

Thus, the entropy change of the steam is 0.0325kJ/K_.

c)

To determine

The entropy change of the air

c)

Expert Solution
Check Mark

Answer to Problem 105RP

The entropy change of the air is 0.8012kJ/K_.

Explanation of Solution

Write the expression for the entropy change of the air.

ΔSair=mcplnT2T1mRlnP2P10=mcplnT2T1 (X)

Conclusion:

Substitute 98.5 kg for mair, 1.005kJ/kgK for cp, 12.3°C for T2, and 10°C for T1 in Equation (X).

ΔSair=(98.5kg)(1.005kJ/kgK)ln12.3°C10°C=(98.5kg)(1.005kJ/kgK)ln(12.3+273)K(10+273)K=(98.5kg)(1.005kJ/kgK)ln285.3K283K=0.8012kJ/K

Thus, the entropy change of the air is 0.8012kJ/K_.

d)

To determine

The energy destroyed during the process

d)

Expert Solution
Check Mark

Answer to Problem 105RP

The energy destroyed during the process is 218kJ_.

Explanation of Solution

For a closed system, write the simplification rate form of the entropy balance for the room.

SinSout+Sgen=ΔSsystem0+Sgen=ΔSsteam+ΔSairSgen=ΔSsteam+ΔSair (XI)

Here, entropy generation is Sgen, change of entropy is ΔSsystem, entropy at inlet condition is Sin, entropy at outlet condition is Sout.

Calculate the energy destroyed during the process (Xdestroyed).

Xdestroyed=(T0)(ΔSsteam+ΔSair) (XII)

Here, dead state temperature is T0.

Conclusion:

Substitute 0.8012kJ/K for ΔSair and 0.0325kJ/K for ΔSsteam in Equation (XI).

Sgen=0.0325kJ/K+0.8012kJ/K=0.7687kJ/K

Substitute 283 K for T0, 0.8012kJ/K for ΔSair and 0.0325kJ/K for ΔSsteam

Xdestroyed=283K(0.0325kJ/K+0.8012kJ/K)=218kJ

Thus, the energy destroyed during the process is 218kJ_.

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Chapter 8 Solutions

EBK THERMODYNAMICS: AN ENGINEERING APPR

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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