# Thermodynamic Systems - Power rating and thermal efficiency

• Lewishio
In summary: KJQ - W + Σmi x Hi - Σme x He = 020,740,500 KJ°c p/h - 93,900,000 KJ p/h + 4,468,350KJ - 47,644,200KJ = - 116,335,350KJ- 116,335,350KJ + 116,335,350KJ = 0Power (watts) 116,335,350 / 3600 = 32,315.375 KWIn summary, a coal-fired steam plant produces 15 tonnes of steam per hour at a pressure of 60 bar and temperature
Lewishio
Homework Statement
Hi,

I have had a go at this question but have gone wrong somewhere, any ideas?

Thanks.

A coal fired steam plant takes in feed water at a temperature of 70°C and produces 15 tonnes of steam per hour at a pressure of 60 bar and temperature 400°C.
The fuel consumption rate is 1.5 tonnes per hour and the calorific value of the fuel is 40MJkgˉ¹.
Determine the power rating of the boiler and its thermal efficiency.
Relevant Equations
Q in = E x V
Q out = McΔT
Efficiency = Q out / Q in x 100
A coal fired steam plant takes in feed water at a temperature of 70°C and produces 15 tonnes of steam per hour at a pressure of 60 bar and temperature 400°C.
The fuel consumption rate is 1.5 tonnes per hour and the calorific value of the fuel is 40MJkgˉ¹.
Determine the power rating of the boiler and its thermal efficiency.Power rating:
Q in = E x V
40,000,000Jkg^-1 x 1500kg p/h / 3600 = 16,666,666,67 J/s

Q out = McΔT
15,000kg x 4.19 KJ/kg°c x (400°c – 70°c) / 3600 = 5,761.25 J/s

Thermal efficiency:
Efficiency = Q out / Q in x 100
5,761.25 J/s / 16,666,666.67 x 100 = 0.034%

From your steam tables, what is the enthalpy of steam at 400 C and 60 Bar. What is the enthalpy of liquid water at 70 C and 60 bar?

Lewishio
Steam Enthalpy at 400°c at 60 bar = 3176.28KJ/kg
Water Enthalpy at 70°c at 60 bar = 297.89KJ/kg

Lewishio said:
Steam Enthalpy at 400°c at 60 bar = 3176.28KJ/kg
Water Enthalpy at 70°c at 60 bar = 297.89KJ/kg
How is the heat addition rate related to the steam mass flow rate and the enthalpy change per unit mass?

Lewishio
I'm not sure

Lewishio said:
I'm not sure
Have you learned about the open system (control volume) version of the first law of thermodynamics?

Incidentally, in your original post, you omitted the heat of vaporization. You are aware of that now, right?

Lewishio
I will do some research on open systems for the first law of thermodynamics and I can now see I haven't included the heat of vaporization which I believe is 2,260KJ/kg. I am not too sure which equations are to be used to figure this out.

Lewishio said:
I will do some research on open systems for the first law of thermodynamics and I can now see I haven't included the heat of vaporization which I believe is 2,260KJ/kg. I am not too sure which equations are to be used to figure this out.
Once you understand the application of the open system version of the first law to this, you will. I can help answer your questions.

Lewishio
Hi Chester,

Do I need to find the volume of the mass entering the system?
If so, there are 15 tonnes entering, so this would be 15,000L.
Would I then multiply this by the Water Enthalpy at 70°c at 60 bar which is 297.89KJ/kg?
This would be 4,468,350 KJ.
Would I then multiply the mass by the heat of vaporization which I believe is 2,260KJ/kg?
This would be 33,900,000 KJ.

would the sum of these be the energy entering the system?
4,468,350 + 33,900,000 = 38,368,350 KJ

Is this the correct formula:

Q - W + Σm in (Hi + V1^2/2 + gzi) - Σm exit (He + V1^2/2 + gze)

Lewishio said:
Is this the correct formula:

Q - W + Σm in (Hi + V1^2/2 + gzi) - Σm exit (He + V1^2/2 + gze)
I don't see an equal sign in the above "equation." But this is definitely the correct starting point. For the case of the single water stream passing through the boiler, what does this equation reduce to (assuming negligible change in kinetic and potential energy)?

Equation assuming negligible change in kinetic and potential energy:
Q - W + Σmi x Hi - Σme x He = 0

I’m not sure if this is correct but thought I’d have a go:

Q - W + Σmi x Hi - Σme x He = 0

Q (Heat transfer):

McΔT
15,000kg x 4.19KJ/kg°c x (400°c - 70°c) = 20,740,500 KJ°c p/h

W (Work done):
Consumption rate x calorific value
1,500kg x 40,000KJkg = 60,000,000 KJ p/h

Heat of vaporization:
Q = mL
15,000kg x 2,260KJ/kg = 33,900,000 KJ

Total work done: 60,000,000KJ p/h + 33,900,000 KJ = 93,900,000 KJ p/h

H (Enthalpy):
MH

Steam Enthalpy at 400°c at 60 bar = 3176.28KJ/kg
Water Enthalpy at 70°c at 60 bar = 297.89KJ/kg

15,000kg x 297.89KJ/kg = 4,468,350KJ
15,000kg x 3176.28KJ/kg = 47,644,200KJ

Q - W + Σmi x Hi - Σme x He = 0
20,740,500 KJ°c p/h - 93,900,000 KJ p/h + 4,468,350KJ - 47,644,200KJ = - 116,335,350KJ

- 116,335,350KJ + 116,335,350KJ = 0

Power (watts)
116,335,350 / 3600 = 32,315.375 KW

Lewishio said:
I’m not sure if this is correct but thought I’d have a go:

Q - W + Σmi x Hi - Σme x He = 0

Q (Heat transfer):

McΔT
15,000kg x 4.19KJ/kg°c x (400°c - 70°c) = 20,740,500 KJ°c p/h

W (Work done):
Consumption rate x calorific value
1,500kg x 40,000KJkg = 60,000,000 KJ p/h

Heat of vaporization:
Q = mL
15,000kg x 2,260KJ/kg = 33,900,000 KJ

Total work done: 60,000,000KJ p/h + 33,900,000 KJ = 93,900,000 KJ p/h

H (Enthalpy):
MH

Steam Enthalpy at 400°c at 60 bar = 3176.28KJ/kg
Water Enthalpy at 70°c at 60 bar = 297.89KJ/kg

15,000kg x 297.89KJ/kg = 4,468,350KJ
15,000kg x 3176.28KJ/kg = 47,644,200KJ

Q - W + Σmi x Hi - Σme x He = 0
20,740,500 KJ°c p/h - 93,900,000 KJ p/h + 4,468,350KJ - 47,644,200KJ = - 116,335,350KJ

- 116,335,350KJ + 116,335,350KJ = 0

Power (watts)
116,335,350 / 3600 = 32,315.375 KW
This is totally screwed up. See my next post.

Lewishio said:
Equation assuming negligible change in kinetic and potential energy:
Q - W + Σmi x Hi - Σme x He = 0
I'm going to write this equation in a little different way:
$$\dot{Q}-\dot{W}+\dot{m}_{in}h_{in}-\dot{m}_{out}h_{out}=0$$
where ##\dot{Q}## = rate of heat transfer to water inside boiler (J/sec)
##\dot{W}## = rate at which water inside boiler does shaft work (J/sec)
##\dot{m}_{in}## = rate at which water enters the boiler (kg/sec)
##\dot{m}_{out}## = rate at which water (steam) exits the boiler (kg/sec)
##h_{in}## = enthalpy per unit mass of water entering the boiler (J/kg)
##h_{out}## = enthalpy per unit mass of water (steam) exiting the boiler (J/kg)

Do you really feel that the water and steam inside the boiler does shaft work? If so, what is the force and displacement? If not, then ##\dot{W}=0##

In this problem what is the rate at which liquid water enters the boiler ##\dot{m}_{in}##?

In this problem, what is the rate at which water (steam) exits the boiler ##\dot{m}_{out}##?

From your steam tables, what is the enthalpy per unit mass of liquid water entering the boiler ##h_{in}##?

From your steam tables, what is the enthalpy per unit mass of the water (steam) exiting the boiler ##h_{out}##?

Lewishio
Thanks for your help, how does this look:

In this problem what is the rate at which liquid water enters the boiler ˙minm˙in?
4.16kg/sec
In this problem, what is the rate at which water (steam) exits the boiler ˙moutm˙out?
4.16kg/sec
From your steam tables, what is the enthalpy per unit mass of liquid water entering the boiler hinhin?
Water Enthalpy at 70°c at 60 bar = 297,890J/kg
From your steam tables, what is the enthalpy per unit mass of the water (steam) exiting the boiler houthout?
Steam Enthalpy at 400°c at 60 bar = 3,176,280J/kg

Lewishio said:
Thanks for your help, how does this look:

In this problem what is the rate at which liquid water enters the boiler ˙minm˙in?
4.16kg/sec
In this problem, what is the rate at which water (steam) exits the boiler ˙moutm˙out?
4.16kg/sec
From your steam tables, what is the enthalpy per unit mass of liquid water entering the boiler hinhin?
Water Enthalpy at 70°c at 60 bar = 297,890J/kg
From your steam tables, what is the enthalpy per unit mass of the water (steam) exiting the boiler houthout?
Steam Enthalpy at 400°c at 60 bar = 3,176,280J/kg
These numbers look about right. So, what is the heat load ##\dot{Q}## in J/sec?

Lewishio
Would this be:
4.16kg/sec x 297,890J/kg = 1,239,222.4J/sec

Lewishio said:
Would this be:
4.16kg/sec x 297,890J/kg = 1,239,222.4J/sec
From the equation in post #15, if ##\dot{m}_{in}=\dot{m}_{out}=\dot{m}=4.17\ kg/sec##, then $$\dot{Q}=\dot{m}(h_{out}-h_{in})$$What does this give you?

Lewishio
4.17kg/sec x (3,176,280J/kg - 297,890J/kg) = 12,002,886.3J/sec

Lewishio said:
4.17kg/sec x (3,176,280J/kg - 297,890J/kg) = 12,002,886.3J/sec
Good. So what’s the efficiency?

Lewishio
Thanks Chester, I can't seem to get the right answer for this:

Efficiency = Q out / Q in x 100

(4.17 x 3,176,280) ÷ (4.17 x 297,890) x 100 = 1066.26%

Lewishio said:
Thanks Chester, I can't seem to get the right answer for this:

Efficiency = Q out / Q in x 100

(4.17 x 3,176,280) ÷ (4.17 x 297,890) x 100 = 1066.26%
This is incorrect. The heating efficiency is defined as the heat that goes into generating steam divided by the energy available from burning the fuel (times 100)

Lewishio
1,500kg/hour x 40,000,000Jkg / 3600 = 16,666,666.67 J/sec

12,002,886,3J/sec / 16,666,666.67J/sec x 100 = 72.01%

Chestermiller
Thanks for your help, am I right in saying the power rating is 12MW?

Lewishio said:
Thanks for your help, am I right in saying the power rating is 12MW?
I'm not sure it should be called that. Is there a second part to this problem?

Lewishio
No there isn't another part to this question.

Lewishio said:
No there isn't another part to this question.
If there was information on the turbine after the boiler, in my judgment, that is would I would use to determine the power rating. But, to do that, I would need to know the outlet pressure from the turbine (unless the steam emerging from the turbine were dry saturated).

Lewishio
Thanks for your help on this, really appreciated.

## 1. What is the power rating of a thermodynamic system?

The power rating of a thermodynamic system refers to the maximum amount of work that the system can produce or absorb in a given amount of time. This can be calculated by dividing the amount of work done by the time it takes to do that work.

## 2. How is thermal efficiency defined for a thermodynamic system?

Thermal efficiency is a measure of how well a thermodynamic system converts heat energy into work. It is calculated by dividing the amount of work done by the amount of heat energy put into the system.

## 3. What factors affect the power rating of a thermodynamic system?

The power rating of a thermodynamic system is affected by several factors, including the type of system, the temperature difference between the hot and cold reservoirs, the amount of heat energy input, and the efficiency of the system.

## 4. How can the thermal efficiency of a thermodynamic system be improved?

The thermal efficiency of a thermodynamic system can be improved by reducing energy losses, increasing the temperature difference between the hot and cold reservoirs, and using more efficient components such as heat exchangers and turbines.

## 5. What are some real-world applications of thermodynamic systems with high power ratings and thermal efficiencies?

Thermodynamic systems with high power ratings and thermal efficiencies are commonly used in power plants, such as steam turbines and gas turbines, to generate electricity. They are also used in refrigeration and air conditioning systems, as well as in engines for vehicles and aircrafts.

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