Fundamentals of Thermal-Fluid Sciences
Fundamentals of Thermal-Fluid Sciences
5th Edition
ISBN: 9780078027680
Author: Yunus A. Cengel Dr., Robert H. Turner, John M. Cimbala
Publisher: McGraw-Hill Education
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Chapter 6, Problem 90P

(a)

To determine

The power rating of the electric heater.

(a)

Expert Solution
Check Mark

Explanation of Solution

Given:

The length of room (l) is 4m.

The width of the room (b) is 5m.

The height of the room (h) is 6m.

The initial pressure of air (P1) is 98kPa.

The initial temperature of air (T1) is 15°C.

The electrical work input of fan (W˙fan,in) is 200 W (or) 0.2kJ/s.

The heat dissipated out of the room (Q˙out) is 150kJ/min.

The mass flow rate of water (m˙) is 40 kg/min.

The time required by air to evacuate the room (Δt) is 20min.

The exit temperature of air (T2) is 25°C.

Calculation:

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

Calculate the volume of room (V).

  V=lbh

  V=4m×5m×6m=120m3

Calculate the total mass of air in the room (m) using the ideal gas equation.

  m=P1VRT1

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

Consider 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)

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

Calculate the power rating of the electric heater (W˙e,in) using the 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

Explanation of Solution

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

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

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.

  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 6 Solutions

Fundamentals of Thermal-Fluid Sciences

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