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 163RQ

(a)

To determine

The exit temperature of the air.

(a)

Expert Solution
Check Mark

Explanation of Solution

Given:

The inlet pressure (P1) is 1 MPa.

The inlet temperature (T1) is 550 K.

The mass flow rate (m˙) is 800kg/min.

The heat transfer rate (Q˙1) is 2700kJ/s.

The inlet temperature of regenerator (Tin)reg is 800 K.

The exit temperature of regenerator (Tout)reg is 600 K.

The inlet pressure of regenerator (Pin)reg is 140 kPa.

The exit pressure of regenerator (Pout)reg is 130 kPa.

Consider the system is in steady state. Hence, the inlet and exit mass flow rates are equal.

The mass flow rate of air (m˙air) is as follows.

  m˙1=m˙2=m˙air

The mass flow rate of exhaust gas (m˙eg) is as follows.

  m˙3=m˙4=m˙eg

Write the energy rate balance equation for one inlet and one outlet system.

  [Q˙1+W˙1+m˙(h1+V122+gz1)][Q˙2+W˙2+m˙(h2+V222+gz2)]=ΔE˙system        (I)

Here, the rate of heat transfer is Q˙, the rate of work transfer is W˙, the enthalpy is h and the velocity is V, the gravitational acceleration is g, the elevation from the datum is z and the rate of change in net energy of the system is ΔE˙system; the suffixes 1 and 2 indicates the inlet and outlet of the system.

The system is at steady state. Hence, the rate of change in net energy of the system becomes zero.

  ΔE˙system=0

Neglect the work transfer, potential and kinetic energies. The heat transfer occurs from the exhaust gas to the air.

Consider the air inlet and outlet alone.

The Equations (I) reduced as follows for air.

  [Q˙1+0+m˙air(h1+0+0)][0+0+m˙air(h2+0+0)]=0Q˙1+m˙airh1m˙airh2=0Q˙1=m˙air(h2h1)        (II)

Rearrange the Equation (II) to obtain exit enthalpy (h2).

  h2=Q˙1m˙air+h1        (III)

For air:

At inlet:

Refer Table A-17, “Ideal-gas properties of air”.

Obtain the inlet enthalpy (h1) corresponding to the temperature of 550K.

  h1=555.74kJ/kg

Calculation:

Substitute  Q˙1=2700kJ/s, m˙air=800kg/min and h1=555.74kJ/kg in

Equation (III)

  h2=2700kJ/s800kg/min+555.74kJ/kg=2700kJ/s800kg/min×1min60s+555.74kJ/kg=202.5kJ/kg+555.74kJ/kg=758.24kJ/kg

Refer Table A-17, “Ideal-gas properties of air”.

Obtain the exit temperature of air (T2) corresponding to the enthalpy of 758.24kJ/kg-using interpolation method.

Write the formula of interpolation method of two variables.

  y2=(x2x1)(y3y1)(x3x1)+y1        (IV)

Show the temperature and enthalpy values from the Table A-17 as in below table.

S.No.xy
Enthalpy (h), in kJ/kgTemperature (T),in K
1756.44740
2758.24?
3767.29750

Substitute x1=756.44, x2=758.24, x3=767.29, y1=740, and y3=750 in Equation (IV).

  y2=(758.24756.44)(750740)(767.29756.44)+740=741.6589K741K

Thus, the exit temperature of the air is 741K.

(b)

To determine

The mass flow rate of exhaust gas.

(b)

Expert Solution
Check Mark

Explanation of Solution

Consider the exhaust gas inlet and outlet alone.

The heat transfer occurs from the exhaust gas to the air. The heat losses from the exhaust gas.

The Equations (I) reduced as follows for exhaust gas.

  [0+0+m˙eg(h3+0+0)][Q˙2+0+m˙eg(h4+0+0)]=0m˙egh3(Q˙2+m˙egh4)=0m˙egh3Q˙2m˙egh4=0m˙eg(h3h4)=Q˙2

  m˙eg=Q˙2h3h4        (V)

Heat loss by the exhaust gas is equal to the heat gained by the air.

  Q˙eg=Q˙air=Q˙2=2700kJ/s

For exhaust gas:

Consider the exhaust gas as air and refer the property tables of air.

At inlet:

Refer Table A-17, “Ideal-gas properties of air”.

Obtain the inlet enthalpy (h3) corresponding to the temperature of 800K.

  h3=821.95kJ/kg

Obtain the exit enthalpy (h4) corresponding to the temperature of 600K.

  h4=607.02kJ/kg

Calculation:

Substitute Q˙2=2700kJ/s, h3=821.95kJ/kg and h4=607.02kJ/kg in

Equation (V).

  m˙eg=2700kJ/s821.95kJ/kg607.02kJ/kg=2700kJ/s214.93kJ/kg=12.5622kg/s12.6kg/s

Thus, the mass flow rate of exhaust gas is 12.6kg/s.

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Chapter 6 Solutions

Fundamentals of Thermal-Fluid Sciences

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