EBK THERMODYNAMICS: AN ENGINEERING APPR
EBK THERMODYNAMICS: AN ENGINEERING APPR
8th Edition
ISBN: 8220102809444
Author: CENGEL
Publisher: YUZU
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Chapter 5.5, Problem 99P

(a)

To determine

The power rating of the electric heater.

(a)

Expert Solution
Check Mark

Answer to Problem 99P

The power rating of the electric heater is 3.151kW.

Explanation of Solution

Write the formula to calculate the volume of room (V).

V=lbh (I)

Here, length of room is l, width of the room is b and height of the room is h.

Write the ideal gas equation to calculate the total mass of air in the room (m).

m=P1VRT1 (II)

Here, initial pressure of air is P1, gas constant of air is R and initial temperature of air is T1.

Assume the entire room as the steady-flow system that is a control volume as mass traverses the boundary.

Write the energy balance for system in the rate form

E˙inE˙out=ΔE˙system

Here, rate of net energy transfer into the control volume is E˙in, rate of net energy transfer’s exit from the control volume is E˙out and rate of change in internal energy of system is ΔE˙system.

At steady state, rate of change in internal energy of the system is zero. Thus rewrite the energy balance equation for the system.

E˙in=E˙outW˙e,in+W˙fan,inQ˙out=ΔUΔt(W˙e,in+W˙fan,inQ˙out)=mcv,avg(T2T1)

W˙e,in=Q˙outW˙fan,in+mcv,avg(T2T1)Δt (III)

Here, power rating of the electric heater or electrical work input is W˙e,in, electrical work input of fan is W˙fan,in, heat dissipated out of the room is Q˙out, internal energy change is ΔU, time required by air to evacuate the room is Δt, constant volume specific heat of air at room temperature is cv and exit temperature of air is T2.

Conclusion:

Substitute 4m for l, 5m for b, and 6m for h in Equation (I).

V=4m×5m×6m=120m3

Refer Table A-1, “Gas constant of common gases”, obtain the gas constant of air as 0.287kPam3/kgK.

Substitute 98kPa for P1, 120m3 for V, 0.287kPam3/kgK for R and 15°C for T1 in Equation (II).

m=(98kPa)(120m3)(0.287kPam3/kgK)(15°C)=11,760kPam3(0.287kPam3/kgK)(15+273)K=142.3kg

Refer Table A-2, “Ideal – gas specific heats of common gases”, obtain the constant volume specific heat of air as 0.718kJ/kg°C.

Substitute 150kJ/min for Q˙out, 0.2kJ/s for W˙fan,in, 142.3kg for m, 0.718kJ/kg°C for cv, 25°C for T2, 15°C for T1 and 20min for Δt in Equation (III).

W˙e,in=(150kJ/min)0.2kJ/s+(142.3kg)(0.718kJ/kg°C)(25°C15°C)(20min)=[(150kJ/min)(1kJ/min60kJ/s)0.2kJ/s+(142.3kg)(0.718kJ/kg°C)(25°C15°C)(20min)(60s1min)]=3.151kJ/s

=3.151kJ/s(1kW1kJ/s)=3.151kW

Thus, the power rating of the electric heater is 3.151kW.

(b)

To determine

The temperature rise of air as it passes through the heating duct.

(b)

Expert Solution
Check Mark

Answer to Problem 99P

The temperature rise of air as it passes through the heating duct is 5°C.

Explanation of Solution

Write the mass balance equation for the flow of air.

m˙2=m˙1=m˙

Here, mass flow rate of air at the inlet is m˙1, and mass flow rate of air at the outlet is m˙2.

Assume the heating duct as the steady-flow system that controls the volume as mass traverses the boundary.

Write the energy balance for system in the rate form as follows:

E˙inE˙out=ΔE˙system

Re-write the energy balance equation for the system as follows:

E˙in=E˙outW˙e,in+W˙fan,in+m˙h1=m˙h2W˙e,in+W˙fan,in=m˙(h2h1)

W˙e,in+W˙fan,in=m˙cp(T2T1)T2T1=W˙e,in+W˙fan,inm˙cp (IV)

Here, initial specific enthalpy of air is h1 and final specific enthalpy of air h2 and specific heat at constant pressure for air at room temperature is cp.

Conclusion:

Refer Table A-2,“Ideal – gas specific heats of common gases”, obtain the constant pressure specific heat of air as 1.005kJ/kg°C.

Substitute 3.151kW for W˙e,in, 0.2kJ/s for W˙fan,in, 1.005kJ/kg°C for cp and 40kJ/min for m˙ in Equation (IV).

T2T1=3.151kW+(0.2kJ/s)(40kJ/min)(1.005kJ/kg°C)=3.151kW(1kJ/s1kW)+(0.2kJ/s)[(40kJ/min60kJ/s)(1kg/s)](1.005kJ/kg°C)=3.151kJ/s+0.2kJ/s0.666kg/s(1.005kJ/kg°C)

=3.351kJ/s0.666kg/s(1.005kJ/kg°C)=5°C

Thus, the temperature rise of air as it passes through the heating duct is 5°C.

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Chapter 5 Solutions

EBK THERMODYNAMICS: AN ENGINEERING APPR

Ch. 5.5 - 5–11 A spherical hot-air balloon is initially...Ch. 5.5 - A desktop computer is to be cooled by a fan whose...Ch. 5.5 - 5–13 A pump increases the water pressure from 100...Ch. 5.5 - Refrigerant-134a enters a 28-cm-diameter pipe...Ch. 5.5 - Prob. 15PCh. 5.5 - Prob. 16PCh. 5.5 - 5–17C What is flow energy? Do fluids at rest...Ch. 5.5 - How do the energies of a flowing fluid and a fluid...Ch. 5.5 - Prob. 19PCh. 5.5 - Prob. 20PCh. 5.5 - Refrigerant-134a enters the compressor of a...Ch. 5.5 - Steam is leaving a pressure cooker whose operating...Ch. 5.5 - A diffuser is an adiabatic device that decreases...Ch. 5.5 - The kinetic energy of a fluid increases as it is...Ch. 5.5 - Prob. 25PCh. 5.5 - Air enters a nozzle steadily at 50 psia, 140F, and...Ch. 5.5 - The stators in a gas turbine are designed to...Ch. 5.5 - The diffuser in a jet engine is designed to...Ch. 5.5 - Air at 600 kPa and 500 K enters an adiabatic...Ch. 5.5 - Prob. 30PCh. 5.5 - Prob. 31PCh. 5.5 - Air at 13 psia and 65F enters an adiabatic...Ch. 5.5 - Carbon dioxide enters an adiabatic nozzle steadily...Ch. 5.5 - Refrigerant-134a at 700 kPa and 120C enters an...Ch. 5.5 - Prob. 35PCh. 5.5 - Refrigerant-134a enters a diffuser steadily as...Ch. 5.5 - Prob. 38PCh. 5.5 - Air at 80 kPa, 27C, and 220 m/s enters a diffuser...Ch. 5.5 - 5–40C Consider an air compressor operating...Ch. 5.5 - Prob. 41PCh. 5.5 - Somebody proposes the following system to cool a...Ch. 5.5 - 5–43E Air flows steadily through an adiabatic...Ch. 5.5 - Prob. 44PCh. 5.5 - Prob. 45PCh. 5.5 - Steam flows steadily through an adiabatic turbine....Ch. 5.5 - Prob. 48PCh. 5.5 - Steam flows steadily through a turbine at a rate...Ch. 5.5 - Prob. 50PCh. 5.5 - Carbon dioxide enters an adiabatic compressor at...Ch. 5.5 - Prob. 52PCh. 5.5 - 5–54 An adiabatic gas turbine expands air at 1300...Ch. 5.5 - Prob. 55PCh. 5.5 - Prob. 56PCh. 5.5 - Air enters the compressor of a gas-turbine plant...Ch. 5.5 - Why are throttling devices commonly used in...Ch. 5.5 - Would you expect the temperature of air to drop as...Ch. 5.5 - Prob. 60PCh. 5.5 - During a throttling process, the temperature of a...Ch. 5.5 - Refrigerant-134a is throttled from the saturated...Ch. 5.5 - A saturated liquidvapor mixture of water, called...Ch. 5.5 - Prob. 64PCh. 5.5 - A well-insulated valve is used to throttle steam...Ch. 5.5 - Refrigerant-134a enters the expansion valve of a...Ch. 5.5 - Prob. 68PCh. 5.5 - Consider a steady-flow heat exchanger involving...Ch. 5.5 - Prob. 70PCh. 5.5 - Prob. 71PCh. 5.5 - Prob. 72PCh. 5.5 - Prob. 73PCh. 5.5 - Prob. 74PCh. 5.5 - Prob. 76PCh. 5.5 - Steam is to be condensed on the shell side of a...Ch. 5.5 - Prob. 78PCh. 5.5 - Air (cp = 1.005 kJ/kgC) is to be preheated by hot...Ch. 5.5 - Prob. 80PCh. 5.5 - Refrigerant-134a at 1 MPa and 90C is to be cooled...Ch. 5.5 - Prob. 82PCh. 5.5 - An air-conditioning system involves the mixing of...Ch. 5.5 - The evaporator of a refrigeration cycle is...Ch. 5.5 - Steam is to be condensed in the condenser of a...Ch. 5.5 - Steam is to be condensed in the condenser of a...Ch. 5.5 - Two mass streams of the same ideal gas are mixed...Ch. 5.5 - Prob. 89PCh. 5.5 - A 110-volt electrical heater is used to warm 0.3...Ch. 5.5 - The fan on a personal computer draws 0.3 ft3/s of...Ch. 5.5 - Prob. 92PCh. 5.5 - 5–93 A scaled electronic box is to be cooled by...Ch. 5.5 - Prob. 94PCh. 5.5 - Prob. 95PCh. 5.5 - Prob. 96PCh. 5.5 - Prob. 97PCh. 5.5 - A computer cooled by a fan contains eight PCBs,...Ch. 5.5 - Prob. 99PCh. 5.5 - A long roll of 2-m-wide and 0.5-cm-thick 1-Mn...Ch. 5.5 - Prob. 101PCh. 5.5 - Prob. 102PCh. 5.5 - A house has an electric heating system that...Ch. 5.5 - Steam enters a long, horizontal pipe with an inlet...Ch. 5.5 - Refrigerant-134a enters the condenser of a...Ch. 5.5 - Prob. 106PCh. 5.5 - Water is heated in an insulated, constant-diameter...Ch. 5.5 - Prob. 108PCh. 5.5 - Air enters the duct of an air-conditioning system...Ch. 5.5 - A rigid, insulated tank that is initially...Ch. 5.5 - 5–113 A rigid, insulated tank that is initially...Ch. 5.5 - Prob. 114PCh. 5.5 - A 0.2-m3 rigid tank equipped with a pressure...Ch. 5.5 - Prob. 116PCh. 5.5 - Prob. 117PCh. 5.5 - Prob. 118PCh. 5.5 - Prob. 119PCh. 5.5 - An air-conditioning system is to be filled from a...Ch. 5.5 - Oxygen is supplied to a medical facility from ten...Ch. 5.5 - Prob. 122PCh. 5.5 - A 0.3-m3 rigid tank is filled with saturated...Ch. 5.5 - Prob. 124PCh. 5.5 - Prob. 125PCh. 5.5 - Prob. 126PCh. 5.5 - The air-release flap on a hot-air balloon is used...Ch. 5.5 - An insulated 0.15-m3 tank contains helium at 3 MPa...Ch. 5.5 - An insulated 40-ft3 rigid tank contains air at 50...Ch. 5.5 - A vertical pistoncylinder device initially...Ch. 5.5 - A vertical piston-cylinder device initially...Ch. 5.5 - Prob. 135RPCh. 5.5 - Prob. 136RPCh. 5.5 - Air at 4.18 kg/m3 enters a nozzle that has an...Ch. 5.5 - An air compressor compresses 15 L/s of air at 120...Ch. 5.5 - 5–139 Saturated refrigerant-134a vapor at 34°C is...Ch. 5.5 - A steam turbine operates with 1.6 MPa and 350C...Ch. 5.5 - Prob. 141RPCh. 5.5 - Prob. 142RPCh. 5.5 - Prob. 143RPCh. 5.5 - Steam enters a nozzle with a low velocity at 150C...Ch. 5.5 - Prob. 146RPCh. 5.5 - Prob. 147RPCh. 5.5 - Prob. 148RPCh. 5.5 - Prob. 149RPCh. 5.5 - Cold water enters a steam generator at 20C and...Ch. 5.5 - Prob. 151RPCh. 5.5 - An ideal gas expands in an adiabatic turbine from...Ch. 5.5 - Prob. 153RPCh. 5.5 - Prob. 154RPCh. 5.5 - Prob. 155RPCh. 5.5 - Prob. 156RPCh. 5.5 - Prob. 157RPCh. 5.5 - Prob. 158RPCh. 5.5 - Prob. 159RPCh. 5.5 - Prob. 160RPCh. 5.5 - Prob. 161RPCh. 5.5 - Prob. 162RPCh. 5.5 - Prob. 163RPCh. 5.5 - The ventilating fan of the bathroom of a building...Ch. 5.5 - Determine the rate of sensible heat loss from a...Ch. 5.5 - An air-conditioning system requires airflow at the...Ch. 5.5 - The maximum flow rate of standard shower heads is...Ch. 5.5 - An adiabatic air compressor is to be powered by a...Ch. 5.5 - Prob. 171RPCh. 5.5 - Prob. 172RPCh. 5.5 - Prob. 173RPCh. 5.5 - Prob. 174RPCh. 5.5 - Prob. 175RPCh. 5.5 - A tank with an internal volume of 1 m3 contains...Ch. 5.5 - A liquid R-134a bottle has an internal volume of...Ch. 5.5 - Prob. 179RPCh. 5.5 - Prob. 181RPCh. 5.5 - Prob. 182RPCh. 5.5 - Prob. 184RPCh. 5.5 - A pistoncylinder device initially contains 1.2 kg...Ch. 5.5 - In a single-flash geothermal power plant,...Ch. 5.5 - The turbocharger of an internal combustion engine...Ch. 5.5 - A building with an internal volume of 400 m3 is to...Ch. 5.5 - Prob. 189RPCh. 5.5 - Prob. 190RPCh. 5.5 - Prob. 191RPCh. 5.5 - Prob. 192FEPCh. 5.5 - Prob. 193FEPCh. 5.5 - An adiabatic heat exchanger is used to heat cold...Ch. 5.5 - A heat exchanger is used to heat cold water at 15C...Ch. 5.5 - An adiabatic heat exchanger is used to heat cold...Ch. 5.5 - In a shower, cold water at 10C flowing at a rate...Ch. 5.5 - Prob. 198FEPCh. 5.5 - Hot combustion gases (assumed to have the...Ch. 5.5 - Steam expands in a turbine from 4 MPa and 500C to...Ch. 5.5 - Steam is compressed by an adiabatic compressor...Ch. 5.5 - Refrigerant-134a is compressed by a compressor...Ch. 5.5 - Prob. 203FEPCh. 5.5 - Prob. 204FEPCh. 5.5 - Air at 27C and 5 atm is throttled by a valve to 1...Ch. 5.5 - Steam at 1 MPa and 300C is throttled adiabatically...Ch. 5.5 - Air is to be heated steadily by an 8-kW electric...
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