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 5, Problem 83P
To determine

The temperature of the device at the end of the 5-min operating period with and without the heat sink.

Expert Solution & Answer
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Explanation of Solution

Given:

The rate of electrical work to be done into the system (W˙e,in) is 25W.

The mass of the electronic device (mdevice) is 20 g.

The specific heat of constant pressure for electronic device (cp) is 850J/kg°C.

The change in time (Δt) is 5 min.

The initial temperature of the system (T1) is 25°C.

Calculation:

Write the expression for the energy balance equation.

  EinEout=ΔEsystem        (I)

Here, the total energy entering the system is Ein, the total energy leaving the system is Eout, and the change in the total energy of the system is ΔEsystem.

Simplify Equation (I) and write energy balance electrical device without heat sink.

  We,in=ΔUdevice=m(u2u1)        (II)

Here, the electrical work to be done into the system is We,in, the change in the internal energy of device is ΔUdevice, the mass of the electronic device is m, the initial specific internal energy of device is u1, and the final specific internal energy of device is u2.

Rewrite the Equation (II) into per unit time.

  W˙e,inΔt=mcp(T2T1)

  (25W)(5min)=(20g)(850J/kg°C)(T225°C)(25W×(1J/s1W))(5min×(60sec1min))=[(20g×(103kg1g))(850J/kg°C)(T225°C)](7500J)=(17J/°C)(T225°C)T2=466°C

Thus, the temperature of the device at the end of the 5-min operating period without the heat sink is 466°C_.

Simplify Equation (I) and write energy balance electrical device with heat sink.

  We,in=ΔUdevice+ΔUheatsink=m(u2u1)device+m(u2u1)heatsink        (III)

Here, the electrical work to be done into the system is We,in, the change in the internal energy of device is ΔUdevice, the change in the internal energy of heat sink is ΔUheat sink, the mass of the electronic device is m, the initial specific internal energy of device is u1, the final specific internal energy of device is u2, the initial specific internal energy of heat sink is u1, the final specific internal energy of heat sink is u2.

Rewrite the Equation (II) into per unit time.

  W˙e,inΔt=mcp(T2T1)device+mcp(T2T1)heat sink

From the Table A-3(b), “Properties of common liquids, solids, and foods”, obtain the value of specific heat of constant pressure for aluminium at 300 K temperature is 0.902kJ/kg°C.

Substitute W˙e,in=25W, Δt= 5 min, mdevice=20 g, cpdevice=850J/kg°C, T1=25 C, cpdevice=0.902kJ/kg°C, and mheatsink=0.5 kg in the above Equation

  (25W)(5min)=(20g)(850J/kg°C)(T225°C)+(0.5kg)(0.902kJ/kg°C)(T225°C)(25W×(1J/s1W))(5min×(60sec1min))=[[(20g×(103kg1g))(850J/kg°C)(T225°C)]+[(0.5kg)(0.902kJ/kg°C×(1000J/kg°C1kJ/kg°C))(T225°C)]](7500J)=(17J/°C)(T225°C)+(451J/°C)(T225°C)T2=41.0°C

Thus, the temperature of the device at the end of the 5-min operating period with the heat sink is 41.0°C_.

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Chapter 5 Solutions

Fundamentals of Thermal-Fluid Sciences

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