TITLE.PM5

814 ENGINEERING THERMODYNAMICS

dharm

\M-therm\Th15-2.pm5

Solution. Diameter of the wire, d = 1.5 mm = 0.0015 m

Length of the wire, l = 150 mm = 0.15 m

∴ Surface area of the wire (exposed to heat transfer),

A = π d l = π × 0.0015 × 0.15 = 7.068 × 10–4 m^2

Wire surface temperature, ts = 120°C

Water temperature, tf = 100°C

Convective heat transfer coefficient, h = 4500 W/m^2 °C

Electric power to be supplied :

Electric power which must be supplied = Total convection loss (Q)

∴ Q = hA(ts – tf) = 4500 × 7.068 × 10–4 (120 – 100) = 63.6 W. (Ans.)

Example 15.16. Water flows inside a tube 45 mm in diameter and 3.2 m long at a velocity

of 0.78 m/s. Determine the heat transfer co-efficient and the rate of heat transfer if the mean

water temperature is 50°C and the wall is isothermal at 70°C. For water at 50°C take k = 0.66 W/

mK, ν = 0.478 × 10–6 m^2 /s and Prandtl number = 2.98.

Solution. Diameter of the tube, D = 45 mm = 0.045 m

Length of the tube, l = 3.2 m

Velocity of water, u = 0.78 m/s

For water at 60°C, k = 0.66 W/mK

Kinematic viscosity, ν = 0.478 × 10–6 m^2 /s

Pr = 2.98

Reynolds number is given by

Re =

Du

ν

= ×

× −

0 045 0 78

0 478 10^6

..

.

= 73431

From Dittus and Boelter equation, Nusselt number,

Nu = 0.023 (Re)0.8 (Pr)0.4

hD

k

= 0.023 (73431)0.8 (2.98)0.4

h×0 045

066

.

.

= 0.023 × 7810.9 × 1.547
∴ h = 4076 W/m^2 K
i.e., Heat transfer co-efficient = 4076 W/m^2 K (Ans).
Q = hA (tw – tf)
= 4076 × πDL (70 – 50)
= 4076 × π × 0.045 × 3.2 × 20 = 36878 or 36.878 kW
i.e., Rate of heat transfer = 36.878 kW. (Ans).
Example 15.17. When 0.5 kg of water per minute is passed through a tube of 20 mm
diameter, it is found to be heated from 20°C to 50°C. The heating is accomplished by condensing
steam on the surface of the tube and subsequently the surface temperature of the tube is main-
tained at 85°C. Determine the length of the tube required for developed flow.
Take the thermo-physical properties of water at 60°C as :
ρ = 983.2 kg/m^2 , cp = 4.178 kJ/kg K, k = 0.659 W/m°C, ν = 0.478 × 10–6 m^2 /s.
Solution. Given : m = 0.5 kg/min, D = 20 mm = 0.02 m, ti = 20°C, to = 50°C
Length of the tube required for fully developed flow, L :

The mean film temperature, tf =

1

2

85 20 50

2

F + + 60

HG

I

KJ

=°C