6,702 ÷ .85 = 7,885 brake horse power at the engine
This, with mechanical efficiency of 90% is equivalent to
7,885 ÷ .9 = 8,761 indicated horse power
Since 15 lbs. of feed water are required for the engines and auxiliaries per indicated horse power per hour, the total feed water or evaporation required to generate 5,000 kw. is
15 × 8,761 = 131,415 lbs. per hour.
that is to say, the boilers must be of sufficient capacity to generate 131,415 lbs. of steam per hour from water at a temperature of 150° Fahr. This must be multiplied by the factor of evaporation for steam at 175 lbs. pressure from feed water at a temperature of 150°, in order to get the equivalent evaporation "from and at 212°."
The formula for the factor of evaporation is
| H - h | ||
| factor of evaporation = | ——— | (1) |
| 965.7 |
in which
H = total heat of steam at the observed pressure;
h = total heat of feed water of the observed temperature;
965.7 = latent heat, of steam at atmospheric pressure.
Substituting in (1) values for the observed pressure and temperature as obtained from the steam table