Thermodynamics problem involving engine efficiency

[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.

Please Help.

 

[BvU]

Hello Stephen, !

interval on which the engine operate

It says

9 kg coal during an hour

 

[Lord Jestocost]

The heat capacity of coal is 7800 cal/g.

Question 2 is poorly formulated. Seems that somebody has mistaken the calorific value of coal with "heat capacity" (see units).
https://en.wikipedia.org/wiki/Heat_of_combustionhttp://www.coalmarketinginfo.com/coal-basics/

 

[Chestermiller]

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

 

[berkeman]

Welcome to the PF.

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]

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

 

[BvU]

Fahrenheit, Kelvin, Reaumur ?

 

[hmmm27]

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).

 

[Chestermiller]

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.

 

[Stephen Bulking]

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

 

[hmmm27]

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

 

[Stephen Bulking]

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?

 

[hmmm27]

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 ?

 

[Stephen Bulking]

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.

 

[hmmm27]

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.

 

[Stephen Bulking]

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?

 

[Stephen Bulking]

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

 

[hmmm27]

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 ?

 

[Lord Jestocost]

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]

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.

 

[Stephen Bulking]

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)