# Thermodynamics problem involving engine efficiency

• Stephen Bulking
In summary: I have not been able to do. :(In summary, the engine has power of 26.5 kW, needs 9 kg coal during an hour for energy. The heat capacity of coal is 7800 cal/g. The engine's efficiency is 33.3%.
Stephen Bulking
Homework Statement
I need help with question 2,5,6,7,9,10. I have posted a word file with all the questions below.
Relevant Equations
Entropy
1st and 2nd law of Thermodynamics
Heat engine, Carnot engine, C.O.P, efficiency
Q2:
An engine has power of 26.5 kW, needs 9 kg coal during an hour for energy. The heat capacity of coal is 7800 cal/g. Define the engine’s efficiency.
Qh= mcΔT = 9000x7800xΔT ( stuck)
P=26.5kW(is this the power output of the engine in an hour?)
If only I could find ΔT, thenI would be able to calculate Efficiency e=W/Qh; I might get W from the given P, but still the time interval on which the engine operates is not given...
Q5: ( using the given information from Q4)
e(carnot)= 33.3%
Qc = 1000J (I think so)
e=1-Qc/Qh => Qh=1.499J
e= w/Q =>W=Qhxe=0.499; there is an answer very close ti this number but still I can't accept the 0.003 difference.
Q6:
e(carnot)= 1 - Tc/Th=24/25
P =150kW, again I'm not sure on which time interval does the engine has this power but let's just assume it's a second then ==> P(hour)=125/3
e=24/25=P/W ( I think) ==> then W would be 43.4 J ?
Q7:
W=-P(V1-V2) = -0.8(2-9)x10^-3 =5.6x10^-3
Q9 and 10: I think I got this one but I'm not sure about the units of measurements like whether it should be Pa or atm, L or m3 or cm3
W=nRTxln(Vi/Vf)=3000=8,314Txln(Vi/25)
We can find T using PfVf=nRT. This would leave us with one equation and one unknown Vi which could be easily found through a few simple mathematical operations.

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Hello Stephen, !
Stephen Bulking said:
interval on which the engine operate
It says
Stephen Bulking said:
9 kg coal during an hour

In problem 6, where does the 24/25 come from?

Welcome to the PF.
Stephen Bulking said:
I have posted a word file with all the questions below.
In the future, please post a PDF copy of your work instead of a DOC file, or even better learn to post using LaTeX (see the tutorial at the top of the page under INFO, Help). Word or Excel file present security issues that are avoided when you use a free PDF writer like PrimoPDF. Thanks.

Stephen Bulking
Chestermiller said:
In problem 6, where does the 24/25 come from?
Oh it's from the this equation
e(carnot)=1-Tc/Th
with T(cold) being 20 and T(hot) being 500

Fahrenheit, Kelvin, Reaumur ?

Stephen Bulking said:
Q2:
An engine has power of 26.5 kW, needs 9 kg coal during an hour for energy.The heat capacity heat of combustion of coal is 7800 cal/g. (Define) then calculate the engine’s efficiency.

A little better.

The question is written to trap a student missing some of the basics. (Either that or your teacher is just mean).

Actively review/figure out :

- the difference between energy and power (aka kWh vs kW)
- why Carnot efficiency is only a theoretical maximum
- how an engine's efficiency could be determined, and what its relationship to Carnot efficiency is.
- for fun, what specific heat ("heat capacity") really is, how it's related to the definition of a calorie, and what metric would be more useful in the question than "cal/g" (which you'll have to figure out anyways).

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Stephen Bulking
Stephen Bulking said:
Oh it's from the this equation
e(carnot)=1-Tc/Th
with T(cold) being 20 and T(hot) being 500
That equation is based on absolute temperature, not centigrade.

Sorry I'm still lost. I don't know how to make use of the given Power.

Which question are you stuck on ? and why are you stuck.

Question 2 and 6. I do not know what to do with the given Power. I tried a lot of guesses but they all failed.
Also question 8 is confusing me: from question 7 I SUCCESSFULLY calculated the work to be -567 J, that has to be the right answer but then since the work is done ON the system, it's negative which means the 1st law should go like

delta E(int)=Q-W=400-(-567)=967 (J) but there is no answer. Is this wrong?

Okay, let's tackle #6.
A Carnot engine has a power output of 150kW . The engine operates between two reservoirs at 150°C and 20°C . How much energy does it take in per hour ?
Do you understand what the problem is about ? and the elements it contains ?

What steps do you think should be accomplished to solve ?
How far have you gotten, so far ?

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well for #6, I don't think I completely understand the problem as in I can't imagine a specific engine that matches the description, is this like a refrigerator, a heat pump or a heat engine? They stated that there is a power output which I believe to be the work done by the system when there's heat transfer from the hot to cold reservoir. But I don't remember any heat engine that takes in work or energy; heat pumps or refrigerator would take in work though.
Confused as I am, the only step I could take is to calculate the Carnot efficiency given the two temperatures.

Stephen Bulking said:
Confused as I am, the only step I could take is to calculate the Carnot efficiency given the two temperatures.

If that's a useful step, then ...

And how'rwe coming along on the other questions I asked - those are the sentences with the "?" at the end.

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Yeah thanks I just got 869 Mega Joules but it was a wild guess, I thought of the efficiency as the amount of useful power out of the input power and through a few calculations got the result I wanted. The thing is I used this method before and failed and I only succeeded this time cause I assumed the given power to be 150 KWh, is this right?

But still I don't think the same goes for #2

Stephen Bulking said:
Yeah thanks I just got 869 Mega Joules but it was a wild guess,

Well,there's 4 answers to choose from, so I'd say you've a 1 in 4 chance of being right. Maybe better, if you've chosen the correct calculations and did them, properly.

The correct abbreviation of "kilowatt-hour" is "kWh".

What is the source of the homework sheet ?

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Stephen Bulking said:
But still I don't think the same goes for #2

Question #2 is complete nonsense. It must read:
"An engine has power of 26.5 kW, needs 9 kg coal during an hour for energy. The heating value of coal is 7800 cal/g. Define the engine’s efficiency. "

The heating value of a substance is the amount of heat released during the combustion of a specified amount of it. https://en.wikipedia.org/wiki/Heat_of_combustion

Stephen Bulking
hmmm27 said:
Well,there's 4 answers to choose from, so I'd say you've a 1 in 4 chance of being right. Maybe better, if you've chosen the correct calculations and did them, properly.

The correct abbreviation of "kilowatt-hour" is "kWh".

What is the source of the homework sheet ?
It's a file my teachers/professors in VIETNAM made and/or copied ( from Serway or other sources online) so there might be misinterpretations.

Lord Jestocost said:
Question #2 is complete nonsense. It must read:
"An engine has power of 26.5 kW, needs 9 kg coal during an hour for energy. The heating value of coal is 7800 cal/g. Define the engine’s efficiency. "

The heating value of a substance is the amount of heat released during the combustion of a specified amount of it. https://en.wikipedia.org/wiki/Heat_of_combustion
THANK YOU. I am convinced that you are right, I checked the data on Wikipedia and converted coal's "heating value" to 7763 cal/g which of course would be rounded up to 7800. I wasn't clear on this before, all the wrong terminology lead me to the wrong calculations, man I was even wondering how to get delta(T) to find Q.
Anyway since the heating value is Q itself, I did a few unit conversion and found out the efficiency to be 0.32.

And if you're also wondering why these questions are poorly made, please take into consideration that I am a Vietnamese University freshman and the professors here might have made misinterpretations and/or misconceptions, even grammar and spelling mistakes on the files they made (or copied online and from the Serway textbook)

## 1. What is engine efficiency and why is it important?

Engine efficiency is a measure of how well an engine converts energy into work. It is important because a higher efficiency means the engine can do more work while using less fuel, resulting in cost savings and reduced environmental impact.

## 2. How is engine efficiency calculated?

The efficiency of an engine can be calculated by dividing the work output by the energy input. The work output is the useful work produced by the engine, while the energy input is the amount of energy supplied to the engine.

## 3. What factors affect engine efficiency?

There are several factors that can affect engine efficiency, including the type of fuel used, the design and size of the engine, and the operating conditions. In general, engines with higher compression ratios and lower friction tend to have higher efficiency.

## 4. What is the Carnot cycle and how does it relate to engine efficiency?

The Carnot cycle is a theoretical thermodynamic cycle that describes the most efficient way to convert heat into work. It serves as a benchmark for the maximum possible efficiency of any engine. Real engines may not achieve the same efficiency as the Carnot cycle due to factors such as friction and heat loss.

## 5. How can engine efficiency be improved?

There are several ways to improve engine efficiency, such as using higher quality fuels, optimizing the design and size of the engine, and implementing technologies like turbocharging and hybrid systems. Regular maintenance and proper driving habits can also help improve engine efficiency.

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