PROBLEM 3 Liquid water at 1 bar and 25 Cis pressurized in a pump to 3 bar under

Question

PROBLEM 3 Liquid water at 1 bar and 25 Cis pressurized in a pump to 3 bar under steady state operation. You may assume that the water is incompressible. Under these conditions, the density and heat capacity of water are 1000 kg m3 and 4.18 J g1K-1, respectively. Assuming the process is reversible and adiabatic, determine the temperature of the water leaving the pump. (a) (b) Of course, the real process is irreversible. It is found experimentally that, for a water flow rate of 1 mol/s, 64 W is consumed by the pump to pressurize the water to 3 bar. Assuming the process is still adiabatic, determine the actual temperature of the water leaving the pump. Also determine the rate of total entropy generation.

Explanation / Answer

(a) Work (per unit mass) for pump can be calculated as:

W = V(P2-P1) or (P2-P1)/?    ... Eq (1)

where V is specific volume = 1/? = 1/ density.

The steady state energy balance for pump is:

h1 + W + Q = h2 ... Eq (2)

where h1 and h2 are enthalpy at inlet and exit respectively. h2 - h1 = Cp (T2 - T1) ... Eq (3)

Q is heat transfer = 0 for an adiabatic process

Substitutine Eq (1) and Eq (3) in Eq (2) gives:

(3-1)*100/1000 = 4.18 (T2 - 25) Eq (4)

T2 = 25.05 oC.

(b) Pump work required to increase the pressure = 64 W per1 (mol/s) of liquid

Specific work = 64 x 1000/18 = 3555 J/kg = 3.555 kJ/kg

Now the energy balance will also have this pump work in addition to LHS of Eq (4).

0.2 + 3.555 = 4.18 (T2 -25)

T2= 25.9 oC

Entropy generation Sgen = S2 - S1 +Q/T

Here Q =0 (adiabatic) and S2 - S1= Cp ln(T2/T1) T in K

Sgen = Cp ln(T2/T1) = 4.18 ln (25.9+273.15/25+273.15)

Entropy generation: Sgen= 13.75 kJ/kg.K

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