Answer:
The maximum power that can be generated is 127.788 kW
Explanation:
Using the steam table
Enthalpy at 20 bar = 2799 kJ/kg
Enthalpy at 2 bar = 2707 kJ/kg
Change in enthalpy = 2799 - 2707 = 92 kJ/kg
Mass flow rate of steam = 5000 kg/hr = 5000 kJ/hr × 1 hr/3600 s = 1.389 kg/s
Maximum power generated = change in enthalpy × mass flow rate = 92 kJ/kg × 1.389 kg/s = 127.788 kJ/s = 127.788 kW
Answer:
V1=5<u>ft3</u>
<u>V2=2ft3</u>
n=1.377
Explanation:
PART A:
the volume of each state is obtained by multiplying the mass by the specific volume in each state
V=volume
v=especific volume
m=mass
V=mv
state 1
V1=m.v1
V1=4lb*1.25ft3/lb=5<u>ft3</u>
state 2
V2=m.v2
V2=4lb*0.5ft3/lb= <u> 2ft3</u>
PART B:
since the PV ^ n is constant we can equal the equations of state 1 and state 2
P1V1^n=P2V2^n
P1/P2=(V2/V1)^n
ln(P1/P2)=n . ln (V2/V1)
n=ln(P1/P2)/ ln (V2/V1)
n=ln(15/53)/ ln (2/5)
n=1.377