Lesson 2, Topic 5
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Worked Out Problems for Heat Transfer

Abdulaziz July 5, 2020
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Problem 1:
A freezer compartment consists of a cubical cavity that is 2 m on a side. Assume the bottom to be perfectly insulated. What is the minimum thickness of Styrofoam insulation (k=0.030W/m.K) which must be applied to the top and side walls to ensure a heat load less than 500 W, when the inner and outer surfaces are -10 ºC and 350C?
Known: Dimensions of freezer component, inner and outer surfaces temperatures.
Find: Thickness of Styrofoam insulation needed to maintain heat load below prescribed value.


Assumptions: (1) perfectly insulted bottom, (2) one-dimensional conduction through five walls of areas A=4m2, (3) steady-state conditions

Analysis: Using Fourier’s law, the heat rate is given by

q=\dot { q\times A=k\frac { \Delta T }{ L } { A }_{ total } }

{ A }_{ total }\quad =\quad 5\times { W }^{ 2 }

L=\frac { 5k\Delta T{ W }^{ 2 } }{ q }

L=\frac { 5\times 0.03\times 45\times { 4 } }{ 500 } =0.054m

Comments: The corners will cause local departures from one–dimensional conduction and, for a prescribed value of L, a slightly larger heat loss.

Problem 2:
A square silicon chip (k=150W/m.k) is of width W=5mm on a side and of thickness t=1mm.
the chip is mounted in a substrate such that its side and back surfaces are insulated, while the front surface is exposed to a coolant. If 4W are being dissipated in circuits mounted to the back surface of the chip, what is the steady-state temperature difference between back and front surfaces?
Known: Dimensions and thermal conductivity of a chip. Power dissipated on one surface.
Find: temperature drop across the chip


Assumptions: (1) steady-state conditions, (2) constant properties, (3) uniform 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, Fourier’s law,

P=q=kA\frac { \Delta T }{ t } \

Delta T=\frac { t\times P }{ k{ W }^{ 2 } } =\frac { 0.001m\times 4W }{ 150W/m.K(0.005{ m }^{ 2 }) }

Delta T={ 1.1 }^{ o }C

Comments: for fixed P, the temperature drop across the chip decreases with increasing k and W, as well as with decreasing t.

Problem 3:
Air at 300 °C flows over a plate of dimensions 0.50 m, by 0.25 m. if the convection heat transfer coefficient is 250 W/m2.K; determine the heat transfer rate from the air to one side of the plate when the plate is maintained at 400 °C.

Known: air flow over a plate with prescribed air and surface temperature and convection heat transfer coefficient.

Find: heat transfer rate from the air to the plate


Assumptions: (1) temperature is uniform over plate area, (2) heat transfer coefficient is uniform over plate area

Analysis: the heat transfer coefficient rate by convection from the airstreams to the plate can be determined from Newton’s law of cooling written in the form,

q = q” .A hA(T∞ – Ts )

where A is the area of the plate. Substituting numerical values,

q = 250 W/m2 * K(0.25 – 0.50)m2 (300 – 40)°C

q = 8125 W

Comments: recognize that Newtown’s law of cooling implies a direction for the convection heat transfer rate. Written in the form above, the heat rate is from the air to plate.

Problem 4 :
A water cooled spherical object of diameter 10 mm and emissivity 0.9 is maintained at 400°C. What is the net transfer rate from the oven walls to the object?
Known: spherical object maintained at a prescribed temperature within a oven.
Find: heat transfer rate from the oven walls to the object


Assumptions: (1) oven walls completely surround spherical object, (2) steady-state condition, (3) uniform temperature for areas of sphere and oven walls, (4) oven enclosure is evacuated and large compared to sphere.

Analysis: heat transfer rate will be only due to the radiation mode. The rate equation is
qrad = ε Asσ (Tsur4 − Ts4)
Where As=πD2, the area of the sphere, substituting numerical values,
qrad = 0.9* π(10*10-3) m2 *5.67 *10-8 W/m .K[(400+273)4 – (80 + 273)4]K
qrad = 3.04W