A well-insulated tank in a vapor power plant operates at steady state. Water enters at inlet 1 at a rate of 125 lb/s at 14.7 lbf/in.2. Make-up water to replenish steam losses from the plant enters at inlet 2 at a rate of 10 lb/s at 14.7 lbf/in.2 and 60°F. Water exits the tank at 14.7 lbf/in.2 Neglecting kinetic and potential energy effects, determine for the water exiting the tank.

a. the mass flow rate, in lb/s.
b. the specific enthalpy, in Btu/lb
c. the temperature, in F

Explanation:

KNOWN:  Two water lines enter a well-insulated tank with one exit.

FIND:For the exiting water (a) the mass flow rate, in lb/s, and (b) the specific enthalpy, in Btu/lb.

ENGINEERING MODEL:

1. The control volume shown with the schematic is at steady state.

2. For the control volume, and kinetic and potential energy effects can be  ignored.

ANALYSIS:

(a) The steady state form of the mass rate balance

∑=∑

reduced to

by substitution values we get

(b) The steady state form of the energy rate balance

∑∑

simplifies to

solving for exit enthalpy gives

Specific enthalpy of saturated liquid water at inlet 1 is obtained from Table A-3E at  p1 = 14.7 lbf/in^2

Make-up water at inlet 2 is a liquid. Thus, specific enthalpy, h2, is  h2 ≈ hf (60o F) = 28.08 Btu/lb

Substituting values into the energy rate balance and solving for the exit specific enthalpy yield

(c) Since, h3 < hf4, state 3 is a compressed liquid. The corresponding temperature can be

determined from the approximation h3 ≈ hf3 at T3. Interpolating in Table A-2E, T3 = 200.8oC.