FIRST LAW OF THERMODYNAMICS
The pumps used for returning the fluid to the heat source can be analyzed as a thermodynamic
system also. One such example is illustrated in Example 3.
Example 3: Open System - Coolant
A power pump is used to return the fluid from the heat exchanger back to the core. The
flow rate through the pump is about 3.0 x 107 lbm/hr with the fluid entering the pump as
saturated liquid at 540°F. The pressure rise across the pump is 90 psia. What is the
work of the pump, neglecting heat losses and changes in potential and kinetic energy?
and neglect changes in pe and ke
(hin - hout) where
is the rate of doing work by the pump
(hin - hout)
(uin - uout)
(nPin - nPout) =
(nPin - nPout)
Since no heat is transferred,
u = 0 and the specific volume out of the pump is
the same as the specific volume entering since water is incompressible.
(hin - hout) = n(Pin - Pout)
Substituting the expression for work,
(hin-hout) we have:
Using 0.01246 for specific volume.
= 3.0 x 107 lbm/hr (0.01246 ft3/lbm) (-90psia) (144 in2/ft2)/778 ft-lbf/Btu
= -6.23 x 106 Btu/hr or -2446 hp
Note: The minus sign indicating work put into the fluid by the pump. 1 hp = 2545