Thermodynamics anyone? need help

[jabran]

:salams

I have a question I need to answer for assignmnet and i need a little help.
It's about a non-flow system and how to calculate the heat transfer.
i can type out the entire question if need be.

[_mike_]

Well, the hitchikers guide has the answer to anything ;) it's 42

Not the answer you were looking for... maybe that your answer is on howstuffworks.com or wikipedia, have you looked there yet?

[jabran]

yeh i have, but they don't have something on this for some reason or it's not explained enough.

[Mace]

I feel bad, I want to help, but the last thermodynamic problem I did was literally in 1989. 18 years ago! :o


But, if you want to post the question, maybe we can figure out a way to work through it?

If I'm not mistaken, Dante is much closer to college age and probably has this fresher in his mind. Or the engineers on this forum could probably help, like THH.

[BubbleGum]

I feel bad, I want to help, but the last thermodynamic problem I did was literally in 1989. 18 years ago! :o


What, You didn't save your textbooks, Mace? :D

I have both me heat transfer and thermodynamics textbooks in the bookshelf behind me. What's the question, Jabran? And what is the course level, ie Uni first year?

~Bub

[Mace]

What, You didn't save your textbooks, Mace? :D

I have both me heat transfer and thermodynamics textbooks in the bookshelf behind me. What's the question, Jabran? And what is the course level, ie Uni first year?

~Bub

Oh good, I hadn't seen you around lately. You'd be perfect to help with this. :up:

You know, most of my textbooks are gone now. Various nieces, nephews, cousins came by to borrow things and somehow they don't get back to me. I have all my econ books mostly because nobody else really studied that (at least quantitative econ at that level).

My library has gradually been filled more with political science, investments, cars, religion/philosophy, and of course software. More current interests. :)

[jabran]

Thanks for the replys guys, much appreciated.

It's for my BTEC National Diploma course in College.
Here is the full question:
Q4. In a non-flow process carried out on 4.8kg of a fluid substance, there is a specific energy decrease of 50kJ/kg and a work transfer from the substance of 80kJ/kg. Determine the heat transfer.

I tried to do it using the "E2-E1=Q-W" formula but not sure if that was correct(not even sure if I have remebered it correctly off hand).

thanks

[BubbleGum]

Thanks for the replys guys, much appreciated.

It's for my BTEC National Diploma course in College.
Here is the full question:
Q4. In a non-flow process carried out on 4.8kg of a fluid substance, there is a specific energy decrease of 50kJ/kg and a work transfer from the substance of 80kJ/kg. Determine the heat transfer.

I tried to do it using the "E2-E1=Q-W" formula but not sure if that was correct(not even sure if I have remebered it correctly off hand).

thanks

Sounds very simple but I must be missing something. The substance looses 50J/g energy yet that creates 80J/g of work. That's an efficiency of 160% which is impossible. But the answer would be 4.8kg*50Kj/Kg, assuming specific energy loss refers to heat.

However, if it is 50J/g 'energy' decrease and a loss of 80J/g of work removed, assuming all energy is equal, then answer would be 4.8kg(50kJ/kg + 80kJ/kg).

I don't know what specific energy is. There is a property called specific heat that has the units J/g*Temperature.

Just checked my Thermodynamics tome. No mention of specific energy.

Sorry, I don't think that's gonna help. It's either a badly written questions or some terminology differences. BTEC is in the UK, right?


~Bub

[jabran]

Yeah it's in the u.k, the teacher says you should learn to translate the words etc and thats where i usually get lost.

well tehe book i have says

heat input to the system is positive
heat output from the system is neagtive
work input to the system is negative
work output from the systme is positive

Q-W=E2-E1

Now i am not sure if this is right or not but 80-50 = -30kj/kg and as this is work from the above statemnet it would mean it is work input, what do you think.

Thanks for the input.

[BubbleGum]

That makes more since. Now just multiply the value by the mass of the fluid. Just make sure you get the the +/- correct or its -130 and not -30.

:)


~Bub

[jabran]

Thanks for the bubblegum. I ahve another one if you don't mind as this one is really getting to me. I'll type the question out to save time.

In a steady-flow open system a fluid substance flows at the rate of 4kg/s. it enters the system at a pressure of 600kN/m^2, a velocity of 220m/, internal energy 2200 Kj/kg and specific volume 0.42 m^3/kg. during its passage through the system, the substance has a loss by heat transfer of 40 kj/kg to the surroundings. determine the power of the system, stating whether it is from or to the sysemt. neglect any change of gravitational potential energy.

Would be good if you could help me with this one if you have the time.

By the way what do you do? I mean like work or study?


Thank you :)

[BubbleGum]

I graduated in '88 with in mechanical engineering.

Again, I am not quite clear. I think this one is deceptive. The power removed from the system is 40kJ/Kg and the flow rate is 4Kg/s. So the power output is 40kJ/Kg*4Kg/s = 160Kj/s or 160 kilowatts. Pressure, velocity, density, etc. are irrellavent.

Do other questions sometimes offer information that is not needed?

~Bub