Example Problem with Complete Solution

5C-5 : Open Feedwater Heater 6 pts
An open feedwater heater in a steam power plant operates at steady-state with liquid entering at T1 = 40oC and P1 = 7 bar.  Water vapor at T2 = 200oC and P2 = 7 bar enters in a second feed stream.  The effluent is saturated liquid water at P3 = 7 bar.  If heat exchange with the surroundings and changes in potential and kinetic energies are negligible, determine the ratio of the mass flow rates of the two feed streams, mdot1 / mdot,2.
               
Read : The feedwater heater is just a fancy mixer.  When we write the MIMO form of the 1st Law at steady-state, there are three unknowns: the three mass flow rates.  The states of all three streams are fixed, so we can determine the specific enthalpy of each of them.
Mass conservation tells us that m3 = m1 + m2.  We can use this to eliminate m3 from the 1st Law.  Then we can solve the 1st Law for m1 / m2 !
Given : T1 40 oC Find :
P1 700 kPa mdot1 / mdot,2 =  ???
T2 200 oC
P2 700 kPa
P3 700 kPa
Q 0 kW
Diagram: Figure 1
Assumptions:
1 - The feedwater heater operates at steady-state.
2 - Changes in potential and kinetic energies are negligible.
3 - Heat transfer is negligible.
4 - No shaft work crosses the system boundary in this process.
Equations / Data / Solve :
An open feedwater heater is essentially a mixer in which superheated vapor is used to raise the temperature of a subcooled liquid.  We can begin our analysis with the steady-state form of the 1st Law.
Equation 1
Eqn 1
The assumptions in the list above allow us to simplify the 1st Law considerably:
Equation 2 Eqn 2
Conservation of mass on the feedwater heater operating at steady-state tells us that :
Equation 3 Eqn 3
We can solve Eqn 3 for mdot,3 and use the result to eliminate mdot,3 from Eqn 2.  The result is:
Equation 4 Eqn 4
The easiest way to determine mdot,1 / mdot,2 is to divide Eqn 4 by mdot,2.
Equation 5 Eqn 5
Now, we can solve Eqn 5 for mdot,1 / mdot,2 :
Equation 6 Eqn 6
Now, all we need to do is to determine the specific enthalpy of all three streams and plug these values into Eqn 6 to complete the problem.
First we must determine the phase(s) present in each stream.
Tsat(700kPa)= 164.95 oC
Therefore: Stream 1 is a subcooled liquid because T1 < Tsat
Stream 2 is a superheated vapor because T2 > Tsat
T3 = Tsat because it is a saturated liquid.
Data from the Steam Tables of the NIST Webbook (using the default reference state) :
H1 168.15 kJ/kg H2 2845.3 kJ/kg
H3 697 kJ/kg
Now, plug these values into Eqn 6 to obtain : mdot1 / mdot,2 =  4.062
Verify : None of the assumptions made in this problem solution can be verified.
Answers : mdot1 / mdot,2 =  4.06