TITLE.PM5

PSYCHROMETRICS 477

dharm

\M-therm\Th10-2.pm5

Then, &

.

(. ) ( )

ma= ××

××+

10 1 00064 9

0 287 10 18 273

5

3 = 10.78 kg/s

and &

.

(. ) ( )

mv 1 10 0 01236 9

0 4618 10 18 273

5

= 3

××

××+ = 0.0828 kg/s

(Script ‘v’ denotes vapour and the script ‘a’ denotes the air).

At exit at 26ºC, pvs = 0.0336 bar and φ = 1

∴ pv = pvs = 0.0336 bar

W 2 =

0 622 0 622 0 0336

1 013 0 0336

...

..

p

pp

v

tv−

= ×

− = 0.02133 kg.

But W =

m

m

v
a
∴ m&v 2 = 0.02133 × 10.78 = 0.23 kg/s
Hence, make-up water required
= 0.23 – 0.0828 = 0.1472 kg/s. (Ans.)
(ii)Final temperature of the water :
Also, m&w 1 = 5.5 × 1 = 5.5 kg/s
and mm&&ww 21 = – (make-up water)
= 5.5 – 0.1472 = 5.353 kg/s
Applying the steady flow energy equation and neglecting changes in kinetic energy and
potential energy, we have

Wmh mh mhiww aa vv+++&&& 11 11 11 =++mh mh mh&&&aa 22 vv 22 ww2 2

Now, Wi (i.e., work input) = 4.75 kW = 4.75 kJ/s.

Evaluating the enthalpies from a datum of 0ºC, we have :

hw 1 = hf at 44ºC = 184.3 kJ/kg,

ha 1 = 1.005 (18 – 0) = 18.09 kJ/kg,

hv 1 = 2519.7 + 1.88 (18 – 10) = 2534.74 kJ/kg.

[Corresponding to pv = 0.01236 bar, ts = tdp (^) ~− 10ºC i.e., the vapour is superheated]
hv 1 = hg at 26ºC = 2549 kJ/kg
ha 2 =1.005 (26 – 0) = 26.13 kJ/kg.
Then, substituting, we get
4.75 + 5.5 × 184.3 + 10.78 × 18.09 + 0.0828 × 2534.74
= 10.78 × 26.13 + 0.23 × 2549 + 5.353 × hw 2
or 5.353 hw 2 = 1423.28 – 867.95 = 555.33
or hw 2 = 103.74 kJ/kg.
By interpolation, hf = 103.74 kJ/kg at 26.7ºC.
Hence, final temperature of water = 26.7ºC. (Ans.)