TITLE.PM5

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\M-therm\Th15-3.pm5

826 ENGINEERING THERMODYNAMICS

Solution. Given : m&h = 0.2 kg/s ; m&c = 0.5 kg/s ; th 1 = 75°C ;

th 2 = 45°C ; tc 1 = 20°C ; hi = h 0 = 650 W/m^2 °C.

The area of heat exchanger, A :

The heat exchanger is shown diagrammatically in Fig. 15.40.

The heat transfer rate, Q = mh × cph × (–)tthh 12

= 0.2 × 4.187 × (75 – 45) = 25.122 kJ/s

Cold water

Cold water

Hot water

( ) Flow arrangement.a

t = 20 Cc 1 º

t = 75 Ch 1 º 45 C (t )º h 2

t = 20 Cc 1 º tc 2

tc 2

Temperature

t = 75 Ch 1 º

t = 20 Cc 1 º

q 1

Hot water

Cold water

q 2

Area / Length

( ) Temperature distribution.b

t = 45 Ch 2 º

t = 32 Cc 2 º

Fig. 15.40. Parallel-flow heat exchanger.

Heat lost by hot water = Heat gained by cold water

m&h × cph × (–)tthh 12 = m&c × cpc × (–)ttcc 21

0.2 × 4.187 × (75 – 45) = 0.5 × 4.187 × (–)tc 2 20

∴ tc 2 = 32°C

Logarithmic mean temperature difference (LMTD) is given by,

θm =

θθ

θθ

12

12

−

ln ( / ) ...[Eqn. (15.54)]

or θm =

()( )

ln [( ) / ( )]

tt tt

tttt

hc hc

hc hc

11 2 2

11 22

−− −

−−

=

()()

ln [( ) / ( )] ln ( / )

75 20 45 32

75 20 45 32

55 33

55 13

−−−

−−

= − = 29.12°C

Overall heat transfer coefficient U is calculated from the relation

111 1

650

1

650

1

Uh hio 325

=+=+=

∴ U = 325 W/m^2 °C

Also, Q = U A θm