Answer to Question #281866 in Mechanical Engineering for niloy

Question

One kg of fluid enters the steady flow apparatus at a pressure of P1, velocity 16m/s and

specific volume 0.4 m3/kg. The Inlet is 30 m above the ground level. The fluid leaves the

apparatus at Pressure of P2, velocity of 275 m/s and specific volume 0.6 m3/kg. The outlet is

at the ground level. The total heat loss between the inlet and outlet is 10 KJ/kg of fluid. If 140

KJ/kg of work is done by the system, find the change in specific internal energy and indicate

whether this is a increase or decrease.

P1 = 6 Bar − {

√X5

849 } Pa ; P2 = 1 Bar + {+√99 − X

4

}Pa

x= 51

Answer 1

The steady flow energy equation:

Q − W = mΔH + 1/2m(C₂² − C₁²) + mg(Z₂ − Z₁)

ΔH = (U₂ + p₂$\upsilon$₂) − (U₁ + p₁$\upsilon$₁) = ΔU + p₂$\upsilon$₂ − p₁$\upsilon$₁

$C$ - velocity, $p$ - pressure, $v$ - specific volume, $Z$ - height, $U$ - internal energy.

10 x 10³ J/kg - (-140 x 10³ J/kg) = 1 kg x ΔH + 1/2 x 1 kg x ((16 m/s)² − (275 m/s)²) + 1 kg x 9.81 m/s² x (30 m − 0 m)

150 x 10³ J/kg = ∆H - 37 390 J/kg

∆H = 187 390 J/kg

p₂ = 1 Bar + {+√99 − X4}Pa = 1 Bar + {+√99 − 4 x 51} = 100 000 Pa - 194 Pa = 99 806 Pa

p₁ = 6 Bar − {√X5849} Pa = 600 000 Pa - {√51 x 5849} Pa = 600 000 Pa - 546 Pa = 599 453 Pa

187 390 J/kg = ∆U + 99 806 Pa x 0.6 m³/kg - 599 453 Pa x 0.4 m³/kg

∆U = 367 288 J/kg = 367.3 kJ/kg (positive sign indicates that specific internal energy increased)