What are the thermodynamic properties and efficiency of a steam plant?

In summary, the steam enters the turbine at 8 MPa and 500°C. The steam extends in an isentropic process and leaves the turbine at 200 kPa, then enters and leaves the condenser at constant pressure. The work done by the turbine is 7 kj/kg. The net power of the plant is 19783.2 kj/s. The efficiency of the cycle is 29.63%. The two processes (3-4, 4-1) plotted on a T-s diagram are shown.
  • #1
JasonHathaway
115
0

Homework Statement


The steam enter the turbine (3) of the steam plant at 8 MPa and 500°C. The steam extends in an isentropic process and leaves the turbine (4) at 200 kPa, then enters and leaves the condenser at constant pressure (You may treat the steam as a saturated liquid when it leaves the condenser). If the work of the pump is 7 kj/kg and both the kinetic and the potential energies are negligible. Compute:
1- The states of the steam when it enters and leaves the turbine.
2- The work done by the turbine in kj/kg.
3- The net power of the plant if the steam flows with 30 kg/s.
4- The amount of heat transfer of both the boiler and the condenser.
5- The efficiency of the cycle.
6- The two processes (3-4, 4-1) plotted on T-s diagram.

t14iMdO.jpg


Homework Equations


Q-W=m(Δh)
efficiency=Wnet/Qin * 100
∑Qnet = ∑ Wnet

The Attempt at a Solution


1-[/B] P3=8 MPa, T3=500°C
Using Tables: T3>TS ----> The inlet state is superheated vapor.
P4=200 kPa, s3=s4=6.7240 kg/kgK (Isentropic Process)
Using Tables: sf < s4 < sg ---> The exit state is mixture of saturated liquid and vapor.

2- Since q3-4=0 in the turbine, and both KE and PE are negligible, we get:
W3-4=h3-h4 (The first law of thermodynamics, h is the enthalpy)
h3= 3398.3 kj/kg, h4=2545.86 kj/kg
W3-4=652.44 kj/kg

3- Wnet=W3-4+Wpump=652.44+7=659.44 kj/kg
W`net=Wnet * 30 kg/s = 19783.2 kj/s

4- Since W4-1=0 in the condenser, then q4-1=h1-h4
h1=504.7 kj/kg, h4=2545.86 kj/kg
q4-1=-2041.6 kj/kg.

Since qnet=Wnet --> The law of closed cycle
q4-1+qboiler=Wnet
qboiler=2701.04 kj/kg

5- eff. = Wnet/qboiler * 100 = 24.41%

6- http://i.imgur.com/xXeDpu8.jpg
 
Last edited:
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  • #2
I'm not sure whether I'm applying the first law for closed cycle right or not (∑Qnet = ∑ Wnet). Should I take the signs in the account or not?
 
  • #3
Check the arithmetic on your calculation of W3-4.

Your overall thermal efficiency looked on the low side, and this may be part of the problem.
 
  • #4
I can't what's wrong with W3-4. You know the first law of thermodynamics (for open systems) states:
Q`-W`=m`(Δh+ΔKE+ΔPE) [kj/s]
Dividing both sides by m`
q-w=(Δh+ΔKE+ΔPE) [kj/kg]
Since q=ΔKE=ΔPE=0
-w=Δh --> w=-Δh --> w=h1-h2
Isn't that right?
 
  • #5
JasonHathaway said:
I can't what's wrong with W3-4. You know the first law of thermodynamics (for open systems) states:
Q`-W`=m`(Δh+ΔKE+ΔPE) [kj/s]
Dividing both sides by m`
q-w=(Δh+ΔKE+ΔPE) [kj/kg]
Since q=ΔKE=ΔPE=0
-w=Δh --> w=-Δh --> w=h1-h2
Isn't that right?
It isn't the thermodynamics that's the problem, it's the arithmetic (you know, addition, subtraction, etc.) Check you calculation...
JasonHathaway said:
h3= 3398.3 kj/kg, h4=2545.86 kj/kg
W3-4=652.44 kj/kg

Is 3398.3 - 2545.86 = 652.44? That's what needs checking.
 
  • #6
Okay, my bad ^^", but the approach is correct, right?
 
  • #7
JasonHathaway said:
Okay, my bad ^^", but the approach is correct, right?
Yes, but you still must be careful of errors in calculation. Correct your mistake and re-post your work.
 
  • #8
W3-4=852.44 kJ/kg
Wnet=859.44 kj/kg
W`net=2.5783e04 kj/s
qboiler=2.901e03
eff. = Wnet/qboiler * 100 = 29.63%
 
  • #9
JasonHathaway said:
W3-4=852.44 kJ/kg
Wnet=859.44 kj/kg
W`net=2.5783e04 kj/s
qboiler=2.901e03
eff. = Wnet/qboiler * 100 = 29.63%
The net work is what comes out of the turbine, W3-4, not the turbine + feed pump. You must supply the energy to run the feed pump from some external source, thus this is work which must be input to the system, and not work which is coming out of the system.

The thermal efficiency of the cycle is based on this value of net work, W3-4.

The energy which is used to raise the pressure of the feed water to return it to the boiler is added to the enthalpy of the condensate coming out of the condenser, and thus reduces slightly the amount of heat input which must be furnished by the boiler.
 
  • #10
Also, remember that work is measured in a special derived unit called the watt.
 

1. What is thermodynamics and how does it relate to steam plants?

Thermodynamics is the branch of physics that deals with the study of energy and its transformation from one form to another. In the context of steam plants, thermodynamics is used to understand the behavior of steam as it is heated, expands, and does work to power turbines.

2. What is a steam plant and how does it work?

A steam plant is a type of power plant that uses steam to generate electricity. It works by heating water in a boiler to produce steam, which then flows through a turbine to turn a generator and produce electricity. The steam is then condensed back into water and the process repeats.

3. What is the role of pressure and temperature in a steam plant?

Pressure and temperature are crucial factors in the operation of a steam plant. The pressure of the steam determines how much work it can do in the turbine, while the temperature affects the efficiency of the process. Higher pressure and temperature result in more efficient and powerful steam plants.

4. How does the second law of thermodynamics apply to steam plants?

The second law of thermodynamics states that energy cannot be converted from one form to another without some loss of usable energy. In the case of steam plants, this means that not all of the heat energy from the boiler can be converted into mechanical energy to power the turbine. Some of the energy is lost as heat.

5. What are the main challenges in designing and operating a steam plant?

One of the main challenges in designing and operating a steam plant is achieving and maintaining high efficiency. This involves careful control of pressure and temperature, as well as minimizing heat losses. Another challenge is ensuring the safety and reliability of the plant, as high temperatures and pressures can be dangerous if not properly managed.

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