5/20/2018 Homework 1 Solution MEEN3650-CHEN3453 S2014

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PROBLEM 1.16

KNOWN: Dimensions and thermal conductivity of a chip. Power dissipated on one surface.

FIND: Temperature drop across the chip.

SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) Constant properties, (3) Uniform heat

dissipation, (4) Negligible heat loss from back and sides, (5) One-dimensional conduction in

chip.

ANALYSIS: All of the electrical power dissipated at the back surface of the chip is

transferred by conduction through the chip. Hence, from Fouriers law,

P = q = kAT

t

or

( )

t P 0.001 m 4 WT =

kW 150 W/m K 0.005 m2 2

=

T = 1.1 C.o

5/20/2018 Homework 1 Solution MEEN3650-CHEN3453 S2014

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PROBLEM 1.26

KNOWN: Chip width and maximum allowable temperature. Coolant conditions.

FIND: Maximum allowable chip power for air and liquid coolants.

SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) Negligible heat transfer from sides and

bottom, (3) Chip is at a uniform temperature (isothermal), (4) Negligible heat transfer by

radiation in air.

ANALYSIS: All of the electrical power dissipated in the chip is transferred by convection to

the coolant. Hence,

P = q

and from Newtons law of cooling,

P = hA(T - T) = h W2(T - T).

In air,

Pmax= 200 W/m2K(0.005 m)

2(85 - 15) C = 0.35 W.

5/20/2018 Homework 1 Solution MEEN3650-CHEN3453 S2014

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PROBLEM 1.27

KNOWN: Upper temperature set point, Tset, of a bimetallic switch and convection heat

transfer coefficient between clothes dryer air and exposed surface of switch.

FIND: Electrical power for heater to maintain Tsetwhen air temperature is T

= 50

C.

SCHEMATIC:

ASSUMPTIONS:

(1) Steady-state conditions, (2) Electrical heater is perfectly insulatedfrom dryer wall, (3) Heater and switch are isothermal at Tset, (4) Negligible heat transfer from

sides of heater or switch, (5) Switch surface, As, loses heat only by convection.

ANALYSIS: Define a control volume around the bimetallic switch which experiences heat

input from the heater and convection heat transfer to the dryer air. That is,

( )E - E = 0

q - hA T T 0.outin

s setelec =

& &

The electrical power required is,

( )q = hA T Ts setelec

( )q = 25 W/m K 30 10 m 70 50 K=15 mW.2 -6 2elec

5/20/2018 Homework 1 Solution MEEN3650-CHEN3453 S2014

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PROBLEM 1.30

KNOWN: Diameter and emissivity of spherical interplanetary probe. Power dissipation

within probe.

FIND: Probe surface temperature.

SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) Negligible radiation incident on the probe.

ANALYSIS: Conservation of energy dictates a balance between energy generation within

the probe and radiation emission from the probe surface. Hence, at any instant

-E + E = 0out g& &

A T Es s4

g= &

ETD

1/ 4g

s 2

=

&

( )

150WT

2 40.8 0.5 m 5.67 10

1/ 4

s 2 8 W/m K

=

T K.s= 254 7.

5/20/2018 Homework 1 Solution MEEN3650-CHEN3453 S2014

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PROBLEM 1.69

KNOWN: Average heat sink temperature when total dissipation is 20 W with prescribed air and

surroundings temperature, sink surface area and emissivity.

FIND: Sink temperature when dissipation is 30 W.

SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) All dissipated power in devices is transferred

to the sink, (3) Sink is isothermal, (4) Surroundings and air temperature remain the same for both

power levels, (5) Convection coefficient is the same for both power levels, (6) Heat sink is a small

surface within a large enclosure, the surroundings.

ANALYSIS: Define a control volume around the heat sink. Power dissipated within the devicesis transferred into the sink, while the sink loses heat to the ambient air and surroundings by

convection and radiation exchange, respectively.

( ) ( )E E 0

P hA T T A T T 0.

outin4 4

e s s s s sur

=

=

& &

(1)

Consider the situation when Pe= 20 W for which Ts= 42C; find the value of h.

( ) ( )h= P / A T T / T T4 4e s s sur s

( ) ( )2

h= 20 W/0.045 m 0.8 5.67 10 W/m K 315 300 K / 315 300 K 8 2 4 4 4 4

h = 24.4 W / m K.2

For the situation when Pe= 30 W, using this value for h with Eq. (1), obtain

( )30 W - 24.4 W/m K 0.045 m T 300 K 2 2 s

( )0.045 m 0.8 5.67 10 W/m K T 300 K 02 8 2 4 4 4 4s = ( ) ( )30 1.098 T 300 2.041 10 T 300 .9 4 4s s= +

By trial-and-error, find

T K = 49 C.s 322 o