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, [tex]W_{CV} =0;Q_{CV}=0[/tex]and kinetic and potential energy effects can be ignored.
ANALYSIS:
(a) The steady state form of the mass rate balance
∑[tex]m_{i}[/tex]=∑[tex]m_{e}[/tex]
reduced to
[tex]m_{1}+ m_{2}= m_{3}[/tex]
by substitution values we get
[tex]m_{3} =125lb/s+10lb/s=135lb/s[/tex]
(b) The steady state form of the energy rate balance
[tex]0=Q_{CV}- W_{CV}+[/tex]∑[tex]m_{i} (h_{i}+ \frac{V_{i}^2 }{2}+ g_{Zi} )-[/tex]∑[tex]m_{e}(h_{e}+ \frac{V_{e}^2 }{2}+g_{Ze})[/tex]
simplifies to
[tex]0=m_{1} h_{1}+m_{2} h_{2}-m_{3} h_{3}[/tex]
solving for exit enthalpy gives
[tex]h_{3}=m_{1} h_{1} +m_{2} h_{2}/m_{3}[/tex]
Specific enthalpy of saturated liquid water at inlet 1 is obtained from Table A-3E at p1 = 14.7 lbf/in^2
[tex]h_{1}=h_{f1}=180.15Btu/lb[/tex]
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
[tex]h_{3}= (125lb/s)(180.15Btu/lb)+(10lb/s)(20.08Btu/lb)/(135lb/s)\\h_{3} =168.89Btu/lb[/tex]
(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.
