Steam enters a steady flow turbine with a velocity of 120 m/s at 600C and 6000 K
ID: 2995375 • Letter: S
Question
Steam enters a steady flow turbine with a velocity of 120 m/s at 600C and 6000 Kpa. The steam leaves the turbine as a saturated vapor at 200 kPa with a velocity of 180 m/s. Heat is transferred from the turbine to the surroundings in the amount of 30 kJ for every kilogram of steam flowing through the turbine. The mass flow rate of the steam through the turbines is 20 kg/s.
a. Determine the rate of work done by the steam turbine in KW
b. Determine the flow area at the entrance to the turbine in m^2.
Explanation / Answer
Given
m = 20 kg/s
qin = -30 kJ/kg
Qin = m*qin = -600 kW
INLET
P1 = 6000 kPa
T1 = 600 C
V1 = 120 m/s
At P1 and T1 From Steam Tables
h1 = 3658 kJ/kg
v1 = 0.06525 m^3/kg
EXIT
Saturated vapor at P2 = 200 kPa
V2 = 180 m/s
Saturated vapor properties at P2 from steam tables
h2 = hg = 2707 kJ/kg
Applying first law b/w inlet and exit
Qin + m*(h1+V1^2/2000) = Wout + m*(h2+V2^2/2000) (units in kW)
-600 + 20*(3658 + 120^2/2000) = Wout + 20*(2707 + 180^2/2000)
Wout = 18,240 kW
Worj done by the steam is 18,240 kW
b)
From
m = A1*V1/v1
A1 = m*v1/V1 = 20*0.06525/120
A1 = 0.010875 m^2