Steam enters the adiabatic system shown below at 20 MPa and 450degreeC at a mass
ID: 1717353 • Letter: S
Question
Steam enters the adiabatic system shown below at 20 MPa and 450degreeC at a mass flow rate of 3.00 kg/s. The steam flows through a heat exchanger, and expands through a throttling valve. The steam leaving the throttling valve is two-phase, and the vapor and liquid are separated in the liquid receiver. Saturated liquid water at 120degreeC leaves the liquid receiver. The steam vapor is returned through the heat exchanger and leaves the system at 200 kPa and 400degreeC. The flow through the device is steady, with negligible changes in kinetic and potential energies. Determine the mass flow rate {kg/s} of liquid water from te liquid receiver.Explanation / Answer
At P1 = 20Mpa and Ti =500C from super heated steam tables
h1 = 3375.1KJ /mol
At P2 = 200Kpa
hf =191.81KJ / kg
hfg = 2392.1KJ / kg
calculate the enthalpy at state 2
h2 = hf + x2hfg
h2 =191.8 + 1(2392.1)
h2 = 2583.9 KJ / kg
Write the energy balance equation for a steady state process
Q - W = m((h2 -h1 +( V22 - Vi2 ) / 2 + g(z2 -z1)
Since the process is adibatic q = 0
And neglecting the kinetic and potential energies Ke = 0 , pe = 0
0-dW / dt =( dm / dt) ( h2 - h1) +0+0
dm / dt = -dW / dt (h2-h1)
By knowing power out put mass flow rate would be found. For power out put we need the system efficiency.