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\M-therm\Th15-3.pm5

HEAT TRANSFER 827

or A =

Q

Uθm

= ×

×

25 122 1000

325 29 12

.

. = 2.65 m

(^2). (Ans.)
Example 15.19. In a counter-flow double pipe heat exchanger, water is heated from 25°C to
65 °C by an oil with a specific heat of 1.45 kJ/kg K and mass flow rate of 0.9 kg/s. The oil is cooled
from 230°C to 160°C. If the overall heat transfer coefficient is 420 W/m^2 °C, calculate the following :
(i)The rate of heat transfer,
(ii)The mass flow rate of water, and
(iii)The surface area of the heat exchanger.
Solution. Given : tc 1 = 25°C ; tc 2 = 65°C, cph = 1.45 kJ/kg K ; m&h= 0.9 kg/s ;
th 1 = 230°C ; th 2 = 160°C, U = 420 W/m^2 °C.
(i) The rate of heat transfer, Q :
Q = m&h × cph × (–)tthh 12
or Q = 0.9 × (1.45) × (230 – 160) = 91.35 kJ/s
(ii) The mass flow rate of water, m&c :
Heat lost by oil (hot fluid) = Heat gained by water (cold fluid)
m&h × chp × (–)tthh 12 = m&c × cpc × (–)ttcc 21
91.35 = m&c × 4.187 (65 – 25)
∴ m&c =
91.35
4.187 (65 25)×− = 0.545 kg/s
Water
Water
Oil
( ) Flow arrangement.a
t=65Cc 2 º
t = 230 Ch 1 º 160 C (t )º h 2
t=65Cc 2 º
25 C (t )º c 1
25 C (t )º c 1
Temp.
t = 230 Ch 1 º
t=65Cc 2 º
Area / Length
( ) Temperature distribution.b
t = 160 Ch 2 º
t=25Cc 1 º
Q 2
Oil (Hot fluid)
Water (Cold fluid)
q 1
Fig. 15.41. Counter-flow heat exchanger.
(iii) The surface area of heat exchanger, A :
Logarithmic mean temperature difference (LMTD) is given by
θm =
θθ
θθ
12
12
−
ln ( / )