Solution Manual

For
Heat Convection 2nd
edition
By
Latif M. Jiji

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PROBLEM 1.1
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Heat is removed from a rectangular surface by L

convection to an ambient fluid at Tf . The heat transfer
coefficient is h. Surface temperature is given by
A 0 x W
Ts = 1 / 2
x

where A is constant. Determine the steady state heat

transfer rate from the plate.
L
(1) Observations. (i) Heat is removed from the surface
by convection. Therefore, Newton's law of cooling is dq s
applicable. (ii) Ambient temperature and heat transfer 0 x W
coefficient are uniform. (iii) Surface temperature varies
along the rectangle. dx
(2) Problem Definition. Find the total heat transfer rate by convection from the surface of a
plate with a variable surface area and heat transfer coefficient.

(3) Solution Plan. Newton's law of cooling gives the rate of heat transfer by convection.
However, in this problem surface temperature is not uniform. This means that the rate of heat
transfer varies along the surface. Thus, Newton’s law should be applied to an infinitesimal area
dAs and integrated over the entire surface to obtain the total heat transfer.

(4) Plan Execution.

(i) Assumptions. (1) Steady state, (2) negligible radiation, (3) uniform heat transfer
coefficient and (4) uniform ambient fluid temperature.

(ii) Analysis. Newton's law of cooling states that

q s = h As (Ts - Tf) (a)
where
As = surface area, m2
h = heat transfer coefficient, W/m2-oC
q s = rate of surface heat transfer by convection, W
Ts = surface temperature, oC
Tf = ambient temperature, oC
Applying (a) to an infinitesimal area dAs
d q s = h (Ts - Tf) dAs (b)
The next step is to express Ts ( x) in terms of distance x along the triangle. Ts ( x) is specified as
A
Ts = 1 / 2 (c)
x
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PROBLEM 1.1 (continued)

The infinitesimal area dAs is given by

dAs = W dx (d)
where
x = axial distance, m
W = width, m
Substituting (c) and into (b)
A
d q s = h( 1/ 2
- Tf) Wdx (e)
x
Integration of (f) gives q s
L

³ ³
q s = dqs = hW ( Ax 1/ 2  Tf )dx
0
(f)

Evaluating the integral in (f)

qs >
hW 2 AL1/ 2  LTf @
Rewrite the above
qs >
hWL 2 AL1/ 2  Tf @ (g)
Note that at x = L surface temperature Ts (L) is given by (c) as

Ts ( L) AL1/ 2 (h)
(h) into (g)
qs hWL >2Ts ( L)  Tf @ (i)

(iii) Checking. Dimensional check: According to (c) units of C are o C/m1/ 2 . Therefore units
q s in (g) are W.
Limiting checks: If h = 0 then q s = 0. Similarly, if W = 0 or L = 0 then q s = 0. Equation (i)
satisfies these limiting cases.

(5) Comments. Integration is necessary because surface temperature is variable.. The same
procedure can be followed if the ambient temperature or heat transfer coefficient is non-uniform.
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PROBLEM 1.2

A right angle triangle is at a uniform surface temperature Ts. Heat is removed by convection to
an ambient fluid at Tf . The heat transfer coefficient h varies along the surface according to

C
h=
x1 / 2

where C is constant and x is distance along the base measured from the apex. Determine the
total heat transfer rate from the triangle.

(1) Observations. (i) Heat is removed from the surface by convection. Therefore, Newton's
law of cooling may be helpful. (ii) Ambient temperature and surface temperature are uniform.
(iii) Surface area and heat transfer coefficient vary along the triangle.

(2) Problem Definition. Find the total heat transfer rate by convection from the surface of a
plate with a variable surface area and heat transfer coefficient.

(3) Solution Plan. Newton's law of cooling gives the rate of

heat transfer by convection. However, in this problem surface dqs W
area and heat transfer coefficient are not uniform. This means
that the rate of heat transfer varies along the surface. Thus,
Newton’s law should be applied to an infinitesimal area dAs x
and integrated over the entire surface to obtain the total heat dx
transfer.
L
(4) Plan Execution.

(i) Assumptions. (1) Steady state, (2) negligible radiation and (3) uniform ambient fluid
temperature.

(ii) Analysis. Newton's law of cooling states that

q s = h As (Ts - Tf) (a)
where
As = surface area, m2
h = heat transfer coefficient, W/m2-oC
q s = rate of surface heat transfer by convection, W
Ts = surface temperature, oC
Tf = ambient temperature, oC
Applying (a) to an infinitesimal area dAs
d q s = h (Ts - Tf) dAs (b)
The next step is to express h and dAs in terms of distance x along the triangle. The heat transfer
coefficient h is given by
C
h= (c)
x1 / 2
The infinitesimal area dAs is given by
 PROBLEM 1.2 (continued)

dAs = y(x) dx (d)

where
x = distance along base of triangle, m
y(x) = height of the element dAs, m
Similarity of triangles give
W
y(x) = x (e)
L
where
L = base of triangle, m
W = height of triangle, m
Substituting (c), (d) and (e) into (b)
C W
d qs = 1/ 2
(Ts - Tf) x dx (f)
x L
Integration of (f) gives qs. Keeping in mind that C, L, W, Ts and Tf are constants, (f) gives
L
CW x
³
q s = dqs =
L
(Ts  Tf )
³0 x1 / 2
dx (g)

Evaluating the integral in (g)

2
qs = C W L1/2 (Ts - Tf) (h)
3

(iii) Checking. Dimensional check: According to (c) units of C are W/m3/2-oC. Therefore
units of q s in (h) are

q s = C(W/m3/2-oC) W(m) L1/2(m1/2) (Ts - Tf)(oC) = W

Limiting checks: If h = 0 (that is C = 0) then q s = 0. Similarly, if W = 0 or L = 0 or Ts = Tf

then q s = 0. Equation (h) satisfies these limiting cases.

(5) Comments. Integration was necessary because both area and heat transfer coefficient vary
with distance along the triangle. The same procedure can be followed if the ambient temperature
or surface temperature is non-uniform.
 PROBLEM 1.3
A high intensity light bulb with surface heat flux (q / A) s is cooled by a fluid at Tf . Sketch the
fluid temperature profiles for three values of the heat transfer coefficients: h1, h2, and h3, where
h1 < h2 < h3.

(1) Observations. (i) Heat flux leaving the surface is specified (fixed). (ii) Heat loss from the
surface is by convection and radiation. (iii) Convection is described by Newton's law of cooling.
(iv). Changing the heat transfer coefficient affects temperature distribution. (v). Surface
temperature decreases as the heat transfer coefficient is increased. (vi) Surface temperature
gradient is described by Fourier’s law.(vii) Ambient temperature is constant.

(2) Problem Definition. Determine effect of heat transfer coefficient on surface temperature and
surface gradient..

(3) Solution Plan. (i) Apply Newton's law of cooling to examine surface temperature. (ii) Apply
Fourier’s law to determine temperature gradient at the surface.

(4) Plan Execution.

(i) Assumptions. (1) Steady state, (2) no radiation ,(3) uniform ambient fluid temperature
and (4) constant properties.
(ii) Analysis. Newton’s law of cooling
q / As h(Ts  Tf ) (a)
Solve for Ts
(q / A) s
Ts Tf  (b)
h
This result shows that for constant (q / A) s surface temperature decreases as h is increased.
Apply Fourier’s law
y
§ wT ·
q / As k ¨¨ ¸¸ (c)
© wy ¹ y 0
where y is the distance normal to the h1
surface. Rewrite (c)
Tf h2 ( q / A) s
§ wT · q / Aw h3
¨¨ ¸¸  (d)
© wy ¹ y 0 k
This shows that temperature gradient at T
Ts
the surface remains constant independent
of h. Based on (b) and (d) the temperature
profiles corresponding to three values of
h are shown in the sketch.

(iii) Checking. Dimensional check: (1) Each term in (b) has units of temperature
(q / A) s ( w/m 2 )
Ts ( o C) Tf ( o C)  2 o
o
C
h( w/m  C)
 PROBLEM 1.3 (continued)
o
(2) Each term in (d) has units of C/m

§ wT · q / Aw ( o C/m 2 )
¨¨ ¸¸ ( o C/m)  o
o
C/m
© wy ¹ y 0 k ( W/m- C)

Limiting check: (i) for h = 0 (no heat leaves the surface), surface temperature is infinite. Set h = 0
in (b) gives Ts f.
(5) Comments. Temperature gradient at the surface is the same for all values of h as long as the
thermal conductivity of the fluid is constant and radiation is neglected.

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 PROBLEM 1.4

Explain why fanning gives a cool sensation. y

Tf
(1) Observations. (i) Metabolic heat leaves no fan
body at the skin by convection and radiation.
(ii) Convection is described by Newton's law
of cooling. (iii). Fanning increases the heat fan
transfer coefficient and affects temperature
distribution, including surface temperature.
Ts
skin T
(iv). Surface temperature decreases as the qcsc
heat transfer coefficient is increased. (v)
Surface temperature is described by Newton’s
law of cooling. (vi) Ambient temperature is
constant.

(2) Problem Definition. Determine effect of heat transfer coefficient on surface temperature.

(3) Solution Plan. Apply Newton's law of cooling to examine surface temperature.

(4) Plan Execution.

(i) Assumptions. (1) Steady state, (2) no radiation ,(3) uniform ambient fluid temperature,
(4) constant surface heat flux and (5) constant properties.
(ii) Analysis. Newton’s law of cooling
q csc h(Ts  Tf ) (a)
where
h = heat transfer coefficient, W/m 2  o C
q csc surface heat flux, W/m 2
Ts = surface temperature, o C
Tf =ambient temperature, o C
Solve (a) for Ts
q csc
Ts Tf  (b)
h
This result shows that for constant q csc , surface temperature decreases as h is increased. Since
fanning increases h it follows that it lowers surface temperature and gives a cooling sensation.

(iii) Checking. Dimensional check: Each term in (b) has units of temperature
q csc ( w/m 2 )
Ts ( o C) Tf ( o C)  2 o
o
C
h( w/m  C)
 PROBLEM 1.4 (continued)

Limiting check: for h = 0 (no heat leaves the surface), surface temperature is infinite. Set h = 0 in
(b) gives Ts f.
(5) Comments. (i) The analysis is based on the assumption that surface heat flux remains
constant. (ii) Although surface temperature decreases with fanning, temperature gradient at the
surface remains constant. This follows from the application of Fourier’s law at the surface
§ wT ·
q csc  k ¨¨ ¸¸
© wy ¹ s
Solving for (wT / wy ) s
§ wT · q csc
¨¨ ¸¸  constant
© wy ¹ s k

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 PROBLEM 1.5

A block of ice is submerged in water at the melting temperature. Explain why stirring the water
accelerates the melting rate.
y
(1) Observations. (i) Melting rate of ice depends on
the rate of heat added at the surface. (ii) Heat is added no stirring
to the ice from the water by convection. (iii) Newton's stirring
law of cooling is applicable. (iv). Stirring increases
surface temperature gradient and the heat transfer water ice

coefficient. An increase in gradient or h increases the

rate of heat transfer. (v) Surface temperature remains
qcsc T
constant equal to the melting temperature of ice. (vi) Ts 0
water temperature is constant.
ice
(2) Problem Definition. Determine effect of stirring
on surface heat flux.

(3) Solution Plan. Apply Newton's law of cooling to examine surface heat flux.

(4) Plan Execution.

(i) Assumptions. (1) no radiation ,(2) uniform water temperature, (3) constant melting
(surface) temperature.
(ii) Analysis. Newton’s law of cooling
q csc h(Ts  Tf ) (a)
where
h = heat transfer coefficient, W/m 2  o C
q csc surface heat flux, W/m 2
Ts = surface temperature, o C
Tf =ambient water temperature, o C

Stirring increases h . Thus, according to (a) surface heat flux increases with stirring. This will
accelerate melting.

(iii) Checking. Dimensional check: Each term in (a) has units of heat flux.
Limiting check: For Tf Ts (water and ice are at the same temperature), no heat will be added to
the ice. Set Tf Ts in (a) gives q csc 0.
(5) Comments. An increase in h is a consequence of an increase in surface temperature gradient.
Application of Fourier’s law at the surface gives
§ wT ·
q csc  k ¨¨ ¸¸ (b)
© wy ¹ s
 PROBLEM 1.5 (continued)
Combining (a) and (b)
§ wT ·
 k ¨¨ ¸¸
© wy ¹ s
h (c)
Ts  Tf
According to (c), for constant Ts and Tf , increasing surface temperature gradient increases h.
 PROBLEM 1.6
Consider steady state, incompressible, axisymmetric parallel flow in a tube of radius ro . The
axial velocity distribution for this flow is given by
r2
u u (1  2 )
ro
where u is the mean or average axial velocity. Determine the three components of the total
acceleration for this flow.

(1) Observations. (i) This problem is described by cylindrical coordinates. (ii) For parallel
streamlines v r v T 0 . (iii) Axial velocity is independent of axial and angular distance.

(2) Problem Definition. Determine the total acceleration in the r, T and z directions.

(3) Solution Plan. Apply total derivative in cylindrical coordinates.

(4) Plan Execution.

(ii) Assumptions. (1) Constant radius tube, (2) constant density and (3) streamlines are
parallel to surface.
(ii) Analysis. Total acceleration in cylindrical coordinates is given by
dv r Dv r wv r v T wv T v T2 wv wv
vr    vz r  r (1.23a)
dt Dt wr r wT r wz wt
dv T Dv T wv v wv T v r v T wv wv
vr T  T   vz T  T (1.23b)
dt Dt wr r wT r wz wt
dv z Dv z wv z v T wv z wv z wv z
vr   vz  (1.23c)
dt Dt wr r wT wz wt

For streamlines parallel to surface

vr vT 0 (a)
The axial velocity u v z is given by
r2
vz u u (1  ) (b)
ro2
From (b) it follows that
wv z wv z
0 (c)
wz wt
Substituting into (1.23a), (1.23b) and (1.23c)

dv r Dv r
Radial acceleration: 0
dt Dt

dv T Dv T
Angular acceleration; 0
dt Dt
 PROBLEM 1.6 (continued)

dv z Dv z
Axial acceleration: 0
dt Dt

(5) Comments. All three acceleration components vanish for this flow.
 PROBLEM 1.7

Consider transient flow in the neighborhood of a vortex line where the

velocity is in the tangential direction given by
V
*o ª § r 2 ·º
V (r , t ) «1  exp¨¨  ¸»
¸
r
2 S r ¬« © 4ǎ t ¹¼»
Here r is the radial coordinate, t is time, * o is circulation
(constant) ǎ is kinematic viscosity. Determine the three components
of total acceleration.

(1) Observations. (i) This problem is described by cylindrical coordinates. (ii) streamlines are
concentric circles. Thus the velocity component in the radial direction vanishes ( v r 0 ). (iii)
For one-dimensional flow there is no motion in the z-direction ( v z 0 ). (iv) The T -velocity
component, v T , depends on distance r and time t.

(2) Problem Definition. Determine the total acceleration in the r, T and z directions.

(3) Solution Plan. Apply total derivative in cylindrical coordinates.

(4) Plan Execution.

(ii) Assumptions. (1) streamlines are concentric circles (2) no motion in the z-direction.
(ii) Analysis. Total acceleration in cylindrical coordinates is given by

The three components of the total acceleration in the cylindrical coordinates r ,T , z are

dv r Dv r wv r v T wv r v T2 wv wv
vr    vz r  r (1.23a)
dt Dt wr r wT r wz wt
dv T Dv T wv T v T wv T v r v T wv wv
vr    vz T  T (1.23b)
dt Dt wr r wT r wz wt
dv z Dv z wv z v T wv z wv wv
vr   vz z  z (1.23c)
dt Dt wr r wT wz wt
For the flow under consideration the three velocity component, v r , v T and v z are
vr 0 (a)

*o ª § r 2 ·º
vT (r , t ) «1  exp¨ ¸ (b)
2 S r «¬ ¨ 4ǎt ¸»
© ¹»¼
vz 0 (c)
Radial acceleration: (a) and (c) into (1.23a)