Question

A FIRE ENGINE LIFT 103KG OF WATER PER-MINUTE THROUGH VERTICAL HEIGHT OF 30M AND DISCHARGE IT THROUGH THE NOZZLE WITH A SPEED 055M/S-1 CALCULATE 1. THE WORK DONE PER MINUTE INLIFTING THE WATER 2. WORK DOME PER-MINUTE IN GIVING THE WATER NOZZLE VELOCITY AND 3. THE MINIMUM POWER OF THE ENGINE IN KILO WATTS REQUIRED TO WORK THE PUMP TAKE G=10MS-2.

Answer 1

Answer:

Explanation:

First let's find the power using 2 formulas

P = W/t

  = F.s/t ----(1) (power to lift water)

  = F.v  -----(2) (power in nozzle)

using first formula

P1 = mg*h/t = 103*10*30/60 = 1030/2 = 515 Watt

so using W = P.t

we get work done to lift water = W1 = 515 * 60 = 30900 J = 30.9 kJ

using second formula

P2 = F.v = mg/v = 1030/0.55 = 1872.7  Watt

using W = P.t

we get work  done by nozzle = W2 = 1872.7 * 60 = 112356 J = 112.3 KJ

Minimim power of engine = P1 = 515 Watt = 0.5 KW