A 600 MW power plant has an efficiency of 36 percent with 15
percent of the waste heat being released to the atmosphere as stack
heat and the other 85 percent taken away in the cooling water.
Instead of drawing water from a river, heating it, and returning it
to the river, this plant uses an evaporative cooling tower wherein
heat is released to the atmoshphere as cooling water is
vaporized.
At what rate must 15C makeup water be provided from the river to
offset the water lost in the cooling tower?

The power plant has an efficiency of 36%. This means 64% of the heat form the source (q1) will become waste heat. Of the waste heat, 85% will be taken away by water (qw).

qw = 0.85 * q2

q2 = 0.64 * q1

p = 0.36 * q1

q1 = p /0.36

q2 = 0.64/0.36 * p

qw = 0.85 *0.64/0.36 * p

qw = 0.85 *0.64/0.36 * 600 = 907 MW

In evaporation water becomes vapor absorbing heat without going to the boiling point (similar to how sweating takes heat from the human body)

The latent heat for the vaporization of water is:

SLH = 2.26 MJ/kg

So, to dissipate 907 MW

G = qw * SLH = 907 / 2.26 = 401.3 kg/s

rejkjavik rejkjavik · Answer 2 · 2019-12-02T09:54:42+01:00

Answer:

m = 367.753 kg.s

Explanation:

GIVEN DATA:

Power plant capacity W=600 MW

Plant efficiency is [tex]\eta = 36\%[/tex]

[tex]efficiency = \frac{W}{QH}[/tex]

[tex]0.36 = \frac{600}{QH}[/tex]

QH = 1666.6 MW

from first law of thermodynamics we hvae

QH -QR = W

Amount of heat rejection is QR = 1066.66 MW

As per given information we have 15% heat released to atmosphere

[tex]QR = 0.15 \tiimes 1066.66 = 159.99 MW[/tex]

AND 85% to cooling water

[tex]Q = 0.85 \times 1066.66 = 906.66 MW[/tex]

from saturated water table

at temp 150 degree c we have Hfg = 2465.4 kJ/kg

rate of cooiling water is given as = mhfg

[tex]906.66 \times 1000 KW = m \times 2465.4[/tex]

m = 367.753 kg.s

where m is rate of makeup water that is added to offset