4

To solve the system of equations above we need to specify initial and boundary
conditions.

3 Steady State Laminar Boundary Layer on a Flat

Plate.
We consider a flat plate at y = 0 with a stream with constant speed U parallel to the
plate. We are interested in the steady state solution. We are not interested in how
the flow outside the boundary layer reached the speed U . In this case, we need only to
∂u
consider boundary conditions and the equation (2.18) simplifies since ∂t
= 0. At the
plate surface there is no flow across it, which implies that

v = 0 at y = 0. (3.24)

Due to the viscosity we have the no slip condition at the plate. In other words,

u = 0 at y = 0. (3.25)

At infinity (outside the boundary layer), away from the plate, we have that

u → U as y → ∞. (3.26)

For the flow along a flat plate parallel to the stream velocity U , we assume no pressure
gradient, so the momentum equation in the x direction for steady motion in the boundary
layer is

∂u ∂u ∂2u
u +v =ν 2 (3.27)
∂x ∂y ∂y

and the appropriate boundary conditions are:

 5

At y = 0, x > 0;u = v = 0, (3.28)

At y → ∞, all x;u = U, (3.29)
At x = 0;u = U. (3.30)

These conditions demand an infinite gradient in speed at the leading edge x = y = 0,

which implies a singularity in the mathematical solution there. However, the assump-
tions implicit in the boundary layer approximation break down for the region of slow
flow around the leading edge. The solution given by the boundary layer approximation
is not valid at the leading edge. It is valid downstream of the point x = 0. We would
like to reduce the boundary layer equation (3.27) to an equation with a single dependent
variable. We consider the stream function ψ related to the velocities u and v according
to the equations

∂ψ
u= , (3.31)
∂y
∂ψ
v =− . (3.32)
∂x
If we substitute equations (3.31) and (3.32) into the equation (3.27), we obtain a partial
differential equation for ψ, which is given by

∂ψ ∂ 2 ψ ∂ψ ∂ 2 ψ ∂3ψ
− =ν 3, (3.33)
∂y ∂x∂y ∂x ∂y 2 ∂y
and the boundary conditions (3.28) and (3.29) assume the form

∂ψ ∂ψ
At y = 0, x > 0; = = 0, (3.34)
∂y ∂x
∂ψ
At y → ∞, all x; = U. (3.35)
∂y
The boundary value problem admits a similarity solution. We would like to reduce the
partial differential equation (3.33) to an ordinary differential equation. We would like
to find a change of variables which allows us to perform the reduction mentioned above.
 Flat Plate Boundary Layer Theory
Laminar Flow Analysis
For steady, incompressible flow over a flat plate, the laminar boundary layer
equations are:
∂u + ∂v =
Conservation of mass: 0
∂x ∂ y

u ∂ u + v ∂u = −
1 d p 1 ∂  ∂ u 
'X' momentum: + µ
∂x ∂y ρ dx ρ ∂ y  ∂ y

'Y' momentum: −∂p = 0

∂y

The solution to these equations was obtained in 1908 by Blasius, a student of

Prandtl's. He showed that the solution to the velocity profile, shown in the table
below, could be obtained as a function of a single, non-dimensional variable η
defined as
Table 7.1 the Blasius Velocity Profile
1/ 2

η= y  U∞x 
υ 

with the resulting ordinary

differential equation:
1
f ′′ ′ + f f ′′ = 0
2
u
and f ′(η ) =
U∞

Boundary conditions for the differential equation are expressed as follows:

at y = 0, v = 0 → f (0) = 0 ; y component of velocity is zero at y = 0

at y = 0 , u = 0 → f ′(0 ) = 0 ; x component of velocity is zero at y = 0

VII-3
The key result of this solution is written as follows:

∂2 f  τw
 = 0.332 =
∂η2  y= 0 µ U∞ U∞ / υ x

With this result and the definition of the boundary layer thickness, the following
key results are obtained for the laminar flat plate boundary layer:

5x
Local boundary layer thickness δ (x) = Re x
Local skin friction coefficient: 0.664
C fx =
(defined below) Re x
Total drag coefficient for length L ( integration 1.328
of τw dA over the length of the plate, per unit CD =
2 Re x
area, divided by 0.5 ρ U∞ )
τ (x ) FD / A
where by definition C fx = 1 w 2 and CD = 1 2
2
ρ U∞ 2
ρ U ∞

With these results, we can determine local boundary layer thickness, local wall
shear stress, and total drag force for laminar flow over a flat plate.

Example:
Air flows over a sharp edged flat plate with L = 1 m, a width of 3 m and
U∞ = 2 m/s . For one side of the plate, find: δ(L), Cf (L), τw(L), CD, and FD.
3 2
Air: ρ = 1.23 kg/m ν = 1.46 E-5 m /s
U∞ L 2 m / s* 2.15 m
First check Re: Re L = = = 294,520 < 500,000
υ 1.46E − 5m / s
2

Key Point: Therefore, the flow is laminar over the entire length of the plate and
calculations made for any x position from 0 - 1 m must be made using laminar
flow equations.

VII-4
 Hot fluid (outlet)
Baffle plates
Shell Tubes

Hot fluid
(inlet)

Cold fluid (outlet) Cold fluid (inlet)

Fig. 5 Schematic of Shell & Tube Heat Exchangera

Effectiveness of a heat exchanger

The effectiveness (ϵ) of a heat exchanger is defined as the ratio of the actual heat transfer to the
maximum possible heat transfer.

actual heat transfer

∈= … (7)
maximum possible heat transfer

Actual heat transfer = Q = mh Cph t h1 − t h2 = mc Cpc t c2 − t c1 … (8)

where

mh . Cph = Ch = hot fluid capacity rate

mc . Cpc = Cc = Cold fluid capacity rate

Maximum possible heat transfer = Q max = Ch t h1 − t c1 or

= Cc t h1 − t c1

Q max is the minimum of these two values i. e.

Q max = Cmin t h1 − t c1 … (9)

a
Rajput, R.K., 2007,Engineering Thermodynamics, Laxmi Publications, New Delhi.
 C h t h 1 −t h 2
∈ = or …(10)
C min t h 1 −t c1

C c t c 2 −t c 1
=C t h 1 −t c 1
…(11)
min

Determination of overall heat transfer coefficient

To determine the overall heat transfer coefficient U for a given heat exchanger, we use the following
relation:

UA
NTU = … (12)
C min

where,
NTU = Number of Transfer Units Dimensionless
W
U = Overall heat transfer coefficient
m2 K
2
A = Heat transfer surface area (m )

Cmin = Minimum of Ch or Cc (kJ/K)

In the present study, steam is condensing while passing through the heat exchanger. Hence, Ch → ∞.
Thus, the capacity ratio, Cr = Cmin / Cmax = 0. For such a case, NTU can be calculated using the
following relationship between ϵ and NTUb:

NTU = − ln(1 − ϵ) … (13)

Hence, from equations (12) and (13), we have

U = − Cmin × A × ln(1 − ϵ) … (14)

b
Kays, W. M., and London, A.L., 1984,Compact Heat Exchangers, McGraw-Hill, New York.
 Heat Transfer
Lecturer : Dr. Rafel Hekmat Hameed University of Babylon
Subject : Heat Transfer College of Engineering
Year : Third B.Sc. Mechanical Engineering Dep.

TRANSIENT HEAT FLOW IN A SEMI-INFINITE SOLID

Consider the semi-infinite solid shown in Figure below maintained at some initial
temperature Ti. The surface temperature is suddenly lowered and maintained at a temperature T0,
and we seek an expression for the temperature distribution in the solid as a function of time.

This temperature distribution may subsequently be used to calculate heat flow at any x position in
the solid as a function of time. For constant properties, the differential equation for the temperature
distribution T(x, τ) is

(1)

The boundary and initial conditions are

This is a problem that may be solved by the Laplace-transform technique. The solution is given as

(2)

1
where the Gauss error function is defined as
(3)

It will be noted that in this definition η is a dummy variable and the integral is a function of its
upper limit. When the definition of the error function is inserted in Equation (2), the expression
for the temperature distribution becomes

(4)

The heat flow at any x position may be obtained from

Performing the partial differentiation of Equation (4) gives

(5)

At the surface (x=0) the heat flow is

(6)

The surface heat flux is determined by evaluating the temperature gradient at x=0 from Equation
(5). A plot of the temperature distribution for the semi-infinite solid is given in Figure below.

2
Values of the error function are tabulated and an abbreviated tabulation is given in Appendix A of
Holman book.
Figure (a) represents a sudden change in surface temperature and (b) instantaneous surface pulse
of Q0/A J/m2.

Constant Heat Flux on Semi-Infinite Solid

For the same uniform initial temperature distribution, we could suddenly expose the surface to a
constant surface heat flux q0/A. The initial and boundary conditions on Equation (1) would then
become

The solution for this case is

(7-a)

Energy Pulse at Surface

Equation (7-a) presents the temperature response that results from a surface heat flux that remains
constant with time. A related boundary condition is that of a short, instantaneous pulse of energy
at the surface having a magnitude of Q0/A. The resulting temperature response is given by
(7-b)

In contrast to the constant-heat-flux case where the temperature increases indefinitely for all x and
times, the temperature response to the instantaneous surface pulse will die out with time, or
T −Ti→0 for all x as τ→∞
This rapid exponential decay behavior is illustrated in Figure (b).

SOLVED EXAMPLES
- See EX. 4-2, 4-3, 4-4 in Holman book
A very large slab of copper is initially at a temperature of 300◦C. The surface temperature is
suddenly lowered to 35◦C. What is the temperature at a depth of 7.5 cm 4 min after the surface
temperature is changed?