16 replies to this topic

Helloooo Cn u plz solve this q's for me :S i have to many try with no result it's urgent for my exam ... thanx alot

Steam in the condenser of a power plant is to be condensed at a temperature of 30 C with cooling water from a nearby lake . which enters the tubes of the condenser at 14 C and leaves at 22C the surface area of the tubes is 45 m2 and the overall heat transfer cofficient is 2100 W/m2.C Determine :

1- The mass flow rate of the cooling water
2- the rate of condensation of the stream in the condenser

the latent of vaporization of water at 30C = 2431 KJ/Kg and the specific heat of cold water at the average temperature of 18C is Cp = 4184 j/Kg.c

regards

It's a heat exchanger problem .. plz anyone ?!

[Art Montemayor]

What do you plan to do with the answers?

Turn them in for credit towards your degree for work done by others?

If you can't do the work yourself, how do you know you are being given the CORRECT answers to turn in?

i have tried to solve it so many type with trial and error assumption for Thi but nothing came up with me . also i have tried to solve using NTU method with no result . i dont know i think there is some thing is missing . i am having my exam on sat this is not a solution for an exam problem . this came up while i am studying

[joerd]

i have tried to solve it so many type with trial and error assumption for Thi but nothing came up with me . also i have tried to solve using NTU method with no result . i dont know i think there is some thing is missing . i am having my exam on sat this is not a solution for an exam problem . this came up while i am studying

All the info is there. Your hot side temperature is 30 C, you can safely ignore any superheat of the steam. Assume steam comes in at 30 C and condensate leaves at 30 C.

[katmar]

This really is very basic stuff and if you are struggling with this you have got a lot of preparation to do if you want to pass your exams. I agree totally with Art that it will not help you at all if someone simply hands you the answers, so I will just give you some tips to get you started.

Heat lost by steam = heat picked up by cold water.

Look at your 3 basic heat transfer equations ie
1. what is the formula for relating the sensible heat picked up by a flowing fluid in terms of its flow rate, temperature change and the specific heat?
2. what is the formula that relates heat trasferred across a surface to the area of the surface and the temperature difference?
3. what is the formula for relating the latent heat lost by a condensing vapor in terms of the flow rate?

It really is that basic. No trial and error required.

thank U a buuunch .. i wasnnt trying to get the answer i just want a hint about this problem . thank u

[latexman]

So, you played with us?

[Qalander (Chem)]

So, you played with us?

Dear engzooz,

This is definitely highly unfair,disgusting and even agonising to 'pinch -away valuable manhours' of our forum's learned friends/colleagues like this.

May I just request that this not only shatters image of Upcoming Engineer's in general,
But also creates a breach of trust thus discourages future responses/helps to usher for queries of people's in dire need.
Only becuse you used the word 'Urgent'

Please abstain in future and do offer clear-cut unconditional apology to all forum colleagues came forward to help you.

Hope you do submit with positive attitude indeed!
Best Regards
Qalander

[Allen]

This really is the most trivial of problems which can be easily solved without recourse to trial and error. if you are having difficulty with this problem then your first actions should be to discuss your problem with your lecturer or tutor or failing that read your text books. This forum is not the place to deal with elementary heat transfer problems.

I don't mean these comments unkindly - all engineers need to develop the ability to think through and solve their own problems otherwise who is going to help you when you get a job?

[djack77494]

Guys please. You're being too harsh on engzooz. We don't know his background, and maybe he's a second year student taking his first heat and material balances course. I don't think it's at all unethical to provide a few very general suggestions such as to look into the energy balance or even to suggest using a Q=UAdT type equation. He's not asking for an answer or anything close - just a "hint".

[katmar]

Well said djack - we need to remember that we were all in that position once. And we all still have areas in which we are ignorant.

I know I have been guilty in the past of simply jumping in and solving the problem for the student, and in truth that does not help them at all. The process of giving hints until the student can run with the problem themselves is a lot more work, but is more helpfull in the long run.

[Qalander (Chem)]

Well said djack - we need to remember that we were all in that position once. And we all still have areas in which we are ignorant.

I know I have been guilty in the past of simply jumping in and solving the problem for the student, and in truth that does not help them at all. The process of giving hints until the student can run with the problem themselves is a lot more work, but is more helpfull in the long run.

Dear Doug/Katmar Esq. and Our Young engzooz

I tender my apology if I had been too harsh in any way.

It was the OP(engzooz)'s 22nd January's post& comment from our always helping friend Latexman; which sparked my reaction.

Although I should have applied better self restraint.
Regards
Qalander

[Allen]

Sorry - I don't think I was being too harsh. If you're going to refer back to when we were students - I didn't have the benefit of internet forums to seek help - there was no Internet. I solved my problems by reading the text books, speaking to my tutor, speaking to my lecturers, speaking to fellow students and using the library. This taught me a discipline which has served me well. The internet is an excellent source of information and I would have used it to help me had it been around. But to be frank, the use of forums for the "easy answer" doesn't do the student any favours.

[katmar]

I am sure that all of the non-students here (and probably most of the students as well) agree that students should not be spoon-fed answers to their problems. In this particular thread the OP set off on the wrong foot by asking "Cn u plz solve this q's for me?", but was quickly put on track by Art Montemayor and indeed I warned that this was very basic stuff and the student needed to pull their socks up.

But Allen, I ask you to seriously consider what is the difference between the way we used to discuss problems with our profs, tutors and fellow students, and modern students being able to add the internet into that mix? Provided there is no cheating, and the answers are not spoon fed, I see no difference in obtaining the answer from the library or from the internet. It's just new technology achieving the same result more efficiently.

When I was a young engineer and I needed to look up some data it meant either a walk to the departmental book shelf or a further walk to the company library. On a bad day it meant driving to the university and using that library. Now I sit at my desk and find it all online. Should I forego this advantage because it makes me fat and lazy? I don't think so because it gets the job done more quickly. Early in my career I was fortunate to be in a situation where there were several senior engineers who could mentor us newbies. With all the downsizing over the last while many youngsters no longer have that advantage and my presence here is in a large measure me paying back to the engineering fraternity what it gave me all those years ago.

[pjadams]

What's the world coming to? Gizzzzzz, he just asked a simple question, you guys really showed your rudeness and brutalness. Enough is enough, if a college freshman can not get a little help from this board, what good is this board created for????????????? Complaint after complaint, sometimes this makes me think if you guys know anything at all.

I guess I am a rookie, it's okay for me to answer this question. His exam is already done any way, this is a little mental exercise for me and you engzooz

The heat transfer is a simple process, if you have a heat source in contact with a cold source, the heat will go across, observing Newton's law of cooling (the same as an apple falling from the tree):

dQ/dt = h * A * (To - Ti)

where A is the area, To - Ti is the temperature difference, h is the coefficient.

In Che Engineering, especially for an heat exchanger, however, this equation became

Q = A*U*dT

A is the same, but U is called the overall heat transfer coefficient. dT is the log mean temp diff (LMTD) defined as:

dT = {(Ti - ti) - (To - to)}/ln{(Ti - ti) / (To - to)}

So, the heat exchanger is more complicated, you have Hot stream (Ti, To) and cold stream (ti, to); i for in and o for out. If you used the equation correctly you will end up with 11.54 deg. C.

Without further arithmetics, based on this equation, the Q, in this case it is really heat transfer rate is calculated as

Q = 1,090530 watts

Remember, W is equivalent to Joul per second!

Now, you just solved one piece of puzzle, this transferred heat is obsorbed by heating up water and you know the heat capacity, this is how you calc the cooling water flow rate (F)

Q = F * Cp * (delt-T)

In this case delt-T is simply = to - ti = 36 deg. C

So if everything works out nicely, you will get F = Q/(Cp * Del-T) = 7.2 Kg/s.

Now for the steam gets cool down by releasing the latent heat (DeltH) which can be evaluated by

Q = DelH * F so the flow is roughly 0.4486 Kg/s

I am in a hurry and I may have made some mistakes, I will leave all the smart people to correct them for me.

Nothing to it.

[Allen]

But Allen, I ask you to seriously consider what is the difference between the way we used to discuss problems with our profs, tutors and fellow students, and modern students being able to add the internet into that mix? Provided there is no cheating, and the answers are not spoon fed, I see no difference in obtaining the answer from the library or from the internet. It's just new technology achieving the same result more efficiently.

I agree - that was the point I was making. The key point is "no cheating or spoon feeding"