200c and 1 atm, with the velocity 0.2m/s. The jet of water exits at 230c, 1 atm with a velocity 50m/s at an elevation of 5m. At steady state the magnitude of the heal transfer rate from power unit to the surroundings is 10% of the power input. Determine the power input to the motor in kW. - Mumbai university > MECH > SEM 3 > THERMO

Marks: 5M

Year: Dec 2013

Given:

m = 0.1kg/s; T1 = 293K; $P1 = 10^5$ pa; v1 = 0.2 m/s

$P2 = 10^5$ pa; T2 = 296K ; v2= 50m/s;

Z2 - Z1 = 5m;

Q= -0.1W

Formula:

Q-W = ∆Potential energy + ∆Kinetic energy + ∆Pressure energy

$-0.1 W - W = mg(z2 - z1) + 1/2(V2^2 - V1^2) + (H2 - H1)$

= $0.1 × 9.8 × (5-0) + 1/2 × 0.1 × (2500 - 0.004) + mCp(T2 - T1)$

= $4.9 + 125 + 0.1 × 4200 × (296 - 293)$

$-1.1W = 1386$

$W = -1260 J/s = -1.26kW$ (-ve sign means work is extracted from the fluid)