TITLE.PM5

HEAT TRANSFER 813

dharm

\M-therm\Th15-2.pm5

( ) Physical configurationa

Fluid

flow

tf

ts

t<tsf

Q Surface

ts tf

()

1

hA

Q

( ) Equivalent circuitb

Pr = Prandtl number

c

k

F pμ

HG

I

KJ

,

D

L = Diameter to length ratio,

Z = A constant to be determined

experimentally,

ρ = Density,

μ = Dynamic viscosity, and

C = Velocity.

The mechanisms of convection in which

phase changes are involved lead to the

important fields of boiling and condensation.

Refer Fig. 15.29 (b). The quantity

1

1

28 44)

hA

Q

tt

hA

L = s− f

N

M

O

Q

(/ )... Eqn (. P is called convection thermal

resistance [(Rth)conv.] to heat flow.

l Dimensionless numbers :

Reynolds numbers, Re = VL

ν

Prandtl number, Pr =

μ ν

α

c

k

p=

Nusselt number, Nu = hL

k

Stanton number, St =

h

Vc

Nu

ρ p Re Pr

=

×

Peclet number, Pe =

LV Re Pr

α

()=.

Graetz number, G = Pe

πD

4

F

HG

I

KJ

Grashoff number, Gr =

ρβ

μ

23

2

gtL∆
.
Example 15.14. A hot plate 1 m × 1.5 m is maintained at 300°C. Air at 25°C blows over the
plate. If the convective heat transfer coefficient is 20 W/m^2 °C, calculate the rate of heat transfer.
Solution. Area of the plate exposed to heat transfer, A = 1 × 1.5 = 1.5 m^2
Plate surface temperature, ts = 300°C
Temperature of air (fluid), tf = 20°C
Convective heat-transfer coeffficient, h = 20 W/m^2 °C
Rate of heat transfer, Q :
From Netwon’s law of cooling,
Q = hA (ts – tf)
= 20 × 1.5(300 – 20) = 8400 W or 8.4 kW. (Ans.)
Example 15.15. A wire 1.5 mm in diameter and 150 mm long is submerged in water at
atmospheric pressure. An electric current is passed through the wire and is increased until the
water boils at 100°C. Under the condition if convective heat transfer coefficient is 4500 W/m^2 °C
find how much electric power must be supplied to the wire to maintain the wire surface at 120°C?

Fig. 15.29. Convective heat-transfer.