# Example Problem with Complete Solution

5C-5 : Open Feedwater Heater 6 pts
Steam at 350oC and 650 kPa is mixed with subcooled water at 30oC and 650 kPa in an open feedwater heater (FWH) as a way to produce saturated liquid water at the same pressure.
Assuming the open FWH is adiabatic, determine the mass flow rate of steam required per kilogram of subcooled liquid water fed to the open FWH.

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 30 oC P3 650 kPa
P1 650 kPa x3 0 kg vap/kg
T2 350 oC Q 0
P2 650 kPa
Find: mdot1 / mdot,2 =  ???
Diagram:
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.
Eqn 1
The assumptions in the list above allow us to simplify the 1st Law considerably:
Eqn 2
Conservation of mass on the feedwater heater operating at steady-state tells us that :
Eqn 3
We can solve Eqn 3 for mdot,3 and use the result to eliminate mdot,3 from Eqn 2.  The result is:
Eqn 4
The easiest way to determine mdot,1 / mdot,2 is to divide Eqn 4 by mdot,2.
Eqn 5
Now, we can solve Eqn 5 for mdot,1 / mdot,2 : 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(650kPa) = 161.98 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 126.32 kJ/kg H2 3165.1 kJ/kg
H3 684.1 kJ/kg
Now, plug these values into Eqn 6 to obtain : mdot1 / mdot,2 =  4.448
Verify: None of the assumptions made in this problem solution can be verified.
Answers : mdot1 / mdot,2 =  4.45