FIRST LAW OF THERMODYNAMICS 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? Solution: m(h_in pe_in ke_in) Q m(h_out pe_out ke_out) W 1) Assume and neglect changes in pe and ke Q 0 m(h_in) m(h_out) W 2) (h_in  - h_out) where is the rate of doing work by the pump W m W h_in u_in nPin h_out u_out    nP_out (h_in  - h_out) (u_in  - u_out) (nP_in  -  nP_out) = u (nP_in  -  nP_out) D 3) Since no heat is transferred, u = 0 and the specific volume out of the pump is D the same as the specific volume entering since water is incompressible. (h_in  - h_out) =  n(P_in  - P_out) 4) Substituting the expression for work, (h_in-h_out) we have: W m . W m  n(P_in P_out) 5) Using 0.01246 for specific volume. = 3.0 x 10⁷  lbm/hr (0.01246 ft³/lbm) (-90psia) (144 in²/ft²)/778 ft-lbf/Btu W = -6.23 x 10⁶  Btu/hr or -2446 hp W Note:   The  minus  sign  indicating  work  put  into  the  fluid  by  the  pump.    1  hp  =  2545 Btu/hr. HT-01 Page 64 Rev. 0