Question 6.169: Reconsider Prob. 6–168. Using EES (or other) software, inves......

Problem 6-168 is reconsidered. The effects of turbine exit area and turbine exit pressure on the exit velocity and power output of the turbine as the exit pressure varies from 10 \mathrm{kPa} to 50 \mathrm{kPa} (with the same quality), and the exit area to varies from 1000 \mathrm{~cm}^{2} to 3000 \mathrm{~cm}^{2} is to be investigated. The exit velocity and the power output are to be plotted against the exit pressure for the exit areas of 1000, 2000, and 3000 \mathrm{cm} 2.

Analysis The problem is solved using EES, and the results are tabulated and plotted below.

Fluid$=’Steam_IAPWS’

A[1]=150 [cm^2]
T[1]=550 [C]
P[1]=10000 [kPa]
Vel[1]= 60 [m/s]
A[2]=1400 [cm^2]
P[2]=25 [kPa]
q_out = 30 [kJ/kg]
m_dot = A[1]*Vel[1]/v[1]*convert(cm^2,m^2)
v[1]=volume(Fluid$, T=T[1], P=P[1]) “specific volume of steam at state 1”
Vel[2]=m_dot*v[2]/(A[2]*convert(cm^2,m^2))
v[2]=volume(Fluid$, x=0.95, P=P[2]) “specific volume of steam at state 2”
T[2]=temperature(Fluid$, P=P[2], v=v[2]) “[C]” “not required, but good to know”

“[conservation of Energy for steady-flow:”
“Ein_dot – Eout_dot = DeltaE_dot” “For steady-flow, DeltaE_dot = 0”
DELTAE_dot=0 “[kW]”
“For the turbine as the control volume, neglecting the PE of each flow steam:”
E_dot_in=E_dot_out
h[1]=enthalpy(Fluid$,T=T[1], P=P[1])
E_dot_in=m_dot*(h[1]+ Vel[1]^2/2*Convert(m^2/s^2, kJ/kg))
h[2]=enthalpy(Fluid$,x=0.95, P=P[2])
E_dot_out=m_dot*(h[2]+ Vel[2]^2/2*Convert(m^2/s^2, kJ/kg))+ m_dot *q_out+ W_dot_out
Power=W_dot_out
Q_dot_out=m_dot*q_out