A steam turbine is fed with steam having an enthalpy of 3100 kJ/kg. It
moves out of the turbine with an enthalpy of 2100 kJ/kg. Feed heating is done at a pressure of
3.2 bar with steam enthalpy of 2500 kJ/kg. The condensate from a condenser with an enthalpy of
125 kJ/kg enters into the feed heater. The quantity of bled steam is 11200 kg/h. Find the power
developed by the turbine. Assume that the water leaving the feed heater is saturated liquid at
3.2 bar and the heater is direct mixing type. Neglect pump work.
Answers
Explanation:
answer for this question
To solve this problem, we need to use the First Law of Thermodynamics, which states that the change in internal energy of a system is equal to the heat added to the system minus the work done by the system:
ΔU = Q - W
where ΔU is the change in internal energy, Q is the heat added to the system, and W is the work done by the system.
First, let's draw a schematic of the system:
Feed Steam
Water Inlet
| |
| Q1 = m1(
h1
-
h4
)
|_______________|
/ Heater |
/ |
Condensate / |
Inlet / |
| / \
| / \
| / Turbine \ Steam
| / \ Outlet
| / \
| / Q2 = m2(
h2
-
h3
) \
| /_________________________________\
| / | |
|/ | W |
Condenser Pump Condenser
To solve this problem, we need to use the First Law of Thermodynamics, which states that the change in internal energy of a system is equal to the heat added to the system minus the work done by the system:
ΔU = Q - W
where ΔU is the change in internal energy, Q is the heat added to the system, and W is the work done by the system.
First, let's draw a schematic of the system:
css
Copy code
Feed Steam
Water Inlet
| |
| Q1 = m1(
h1
-
h4
)
|_______________|
/ Heater |
/ |
Condensate / |
Inlet / |
| / \
| / \
| / Turbine \ Steam
| / \ Outlet
| / \
| / Q2 = m2(
h2
-
h3
) \
| /_________________________________\
| / | |
|/ | W |
Condenser Pump Condenser
where:
h1 = 2500 kJ/kg is the enthalpy of the steam entering the feed heater
h2 = 3100 kJ/kg is the enthalpy of the steam entering the turbine
h3 = 2100 kJ/kg is the enthalpy of the steam leaving the turbine
h4 = 125 kJ/kg is the enthalpy of the condensate entering the feed heater
m1 is the mass flow rate of the steam entering the feed heater
m2 is the mass flow rate of the steam leaving the turbine
Q1 is the heat added to the feed water in the feed heater
Q2 is the heat removed from the steam in the turbine
W is the work done by the turbine
Now, let's apply the First Law of Thermodynamics to the feed heater and the turbine separately:
Feed Heater:
ΔU = 0 (steady state)
Q1 = m1(h1 - h4)
Turbine:
ΔU = 0 (steady state)
Q2 = m2(h3 - h2)
W = Q2 - Q1 = m2(h3 - h2) - m1(h1 - h4)
We also need to use the mass balance equation for the steam:
m1 = m2 + 11200 kg/h
Now, we can solve for the power developed by the turbine:
W = m2(h3 - h2) - m1(h1 - h4)
W = m2(2100 - 3100) - (m2 + 11200)(2500 - 125)
W = -m2(1000) + 11200(2375)
W = 26,500,000 W
Therefore, the power developed by the turbine is 26,500,000 W or 26.5 MW.
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