Example of change of enthalpy:
Q 0.5kg/s of a fluid flows in steady state process. The properties of fluid at the entrance are measured as p_{1} = 1.4bar, density = 2.5kg/m^{3}, u_{1} = 920Kj/kg while at exit properties are p_{2 }= 5.6 bar, density = 5 kg/m^{3}, u_{2} = 720Kj/kg. The velocity at entrance is 200m/sec, while at exit it is 180m/sec. It rejects 60kw of heat and rises through 60m during the flow. Find change of enthalpy and rate of work done.
Sol: Given that:
m_{f} = 0.5kg/s
P_{1} = 1.4bar, density = 2.5kg/m^{3},
u_{1} = 920Kj/kg
P_{2 } = 5.6 bar, density = 5 kg/m3,
u_{2} = 720Kj/kg. V_{1} = 200m/sec
V_{2} = 180m/sec Q = - 60kw
Z_{2} - Z_{1} = 60m h = ?
W_{S} = ?
Since h_{2} - h_{1} = U + P
h_{2} - h_{1} = [U_{2} - U_{1} + (P_{2}/ 2 - P_{1}/ 1)]
= [(720 -920) × 103 + (5.6/5 - 1.4/2.5) × 105]
= [-200 × 103 + 0.56 × 105] = - 144KJ/kg
H = mf × (h_{2} - h_{1}) = 0.5 × (-144) Kj/kg = -72KJ/sec .......ANS
By Applying SFEE
60 × 103 - W_{S} = 0.5[ - 144 × 103 + (180^{2} - 100^{2})/2 + 9.81 × 60]
WS = 13605.7 W = 136.1KW .......ANS