TITLE.PM5

476 ENGINEERING THERMODYNAMICS

dharm

\M-therm\Th10-2.pm5

[ctpdb 2 + W 2 hvapour (2)] + (W 3 – W 2 )h 2 = ctpdb 3 + W 3 hvapour (3)
∴ cp(tdb 3 – tdb 2 ) + W 3 hvapour (3) – W 2 hvapour (2) – (W 3 – W 2 )h 4 = 0
From the steam tables at pv = 0.01452 bar : hg = 2524 kJ/kg
and tdp = tsat = 12.5ºC

∴ 1.005(22 – tdb 2 ) + 0.00904[2524 + 1.88(22 – 12.5)]

• 0.0047 [2524 + 1.88(tdb 2 – 12.5)] – (0.00434 × 4.187 × 10) = 0
  22.11 – 1.005 tdb 2 + 22.97 – 11.86 – 0.0088 tdb 2 + 0.11 – 0.1817 = 0
  1.014tdb 2 = 33.148
  ∴ tdb 2 = 32.7ºC. (Ans.)
  Example 10.14. Cooling tower : A small-size cooling tower is designed to cool 5.5 litres
  of water per second, the inlet temperature of which is 44ºC. The motor-driven fan induces 9 m^3 /s
  of air through the tower and the power absorbed is 4.75 kW. The air entering the tower is at 18ºC,
  and has a relative humidity of 60%. The air leaving the tower can be assumed to be saturated and
  its temperature is 26ºC. Calculate :
  (i)The amount of cooling water (make-up) required per second.
  (ii)The final temperature of the water.
  Assume that the pressure remains constant throughout the tower at 1.013 bar.
  Solution. The cooling tower is shown diagrammatically in Fig. 10.18.
  (i)Make-up water required :
  At 18ºC pvs = 0.0206 bar,
  ∴ pv = φ × pvs = 0.6 × 0.0206 = 0.01236 bar
  ∴ pa 1 = 1.013 – 0.01236 = 1.00064 bar

Hot water in

(m , h )ww 1 1

Cold water out

(m , h )ww2 2

Air out

Wi

26 C, = 1

m , h , m , h

º φ

aa v v 2 2 2

18 C, = 60%

m , h , m , h

º φ

aa v v 1 1 1

Fig. 10.18