Page 240 - Standard Handbook Of Petroleum & Natural Gas Engineering

Thermodynamics 2 13

Example 2-23. Hydroelectric Power System
A hydroelectric power plant proposes to use 1,500 ft’/s of river water to generate
electricity. The water enters the system at 1 atm and 50°F and is discharged at 1 atm
and 50.4”F after passing through a turbine generator. The discharge point is 600 ft
below the inlet. The increase in enthalpy of the water is known to be 0.36 Btu/lb,,,.
Assuming 70% efficiency for the conversion, what power output can be expected
from the power plant?

Solution
The following assumptions pertain to this open system:
1. Steady-state flow.
2. No heat transferred between system and surroundings.
3. Change in kinetic energy of the flow streams is negligible.
With these assumptions, we take as a reference point the discharge level of the water,
and apply Equation 2-110. Thus Z, = 0, Z, = 600 ft and the energy balance becomes
g
AH+-AZ =-W,,,
g<
fti j(1j
1,500 ft‘ 62.4 lb, 3,600 s
whereAH =[0.36E)(y)(
= 1.213~ 10’ Btu/hr

= 35,553 kW

ft
32.2, 62.4 lb, )(%)
and -AZ g =I ft lb, ](0-600ft)[7)( 1,500 ft’ ft3
gc 32.2
Ib, s2
~
= -2.599~ lo8 Btu/hr

= -76,163 kW
Therefore,

W,,eL -35,553 kW + 76,163 kW = 40,610 kW
At 70% efficiency, this would yield
= (0.70)(40,610) = 28,427 kW = 28.427 mega-Watts.

Entropy and the Second Law

The second law of thermodynamics provides a basis for determining whether or
not a process is possible. It is concerned with availability of the energy of a given