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Anonymous Poster

Heat Transfer Process - URGENT!!

06/06/2009 8:57 AM

Hi! I need urgent help with the following task! Could someone have a look at it and tell me if the solutions are correct? Many thanks!

A coil is wrapped around a pipe that carries a fluid which needs to be heated.

Pipe
(U) change in thermal energy = ?
(m) mass flow rate = 0.1kg/s
(C) specific heat capacity = 1970J/kgK
(Tpin) Initial fluid temperature = 50 (Converted to Kelvin 323K)
(Tpout) Final fluid temperature = 140 (Converted to Kelvin 413)
(T) Change in temperature

Coil
(U) change in thermal energy = ?
(m) mass flow rate = 0.15kg/s
(C) specific heat capacity = 500J/kgK
(Tcin) Inlet temperature = 500 (Converted to Kelvin 773)
(Tcout) Outlet temperature = 140 (Converted to Kelvin 413)
(T) Change in temperature

(Q2) Heat energy rate of coil:

U = mCT
U = 0.15kg/s * 500J/kgK * (773 - 413)
U = 27000J/s or 27x10^3 J/s

(Q2) Efficiency of heat transfer process:

Efficiency = Tpin-Tpout/(Tpout-Tcin)
Efficiency = (323-413)/(413-773)
Efficiency = 0.25
or 0.25x100 = 25%

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#1

Re: Heat Transfer Process - URGENT!!

06/06/2009 9:12 AM

You are correct. The below information is from:

http://fordhamprep.org/gcurran/sho/sho/lessons/lesson210.htm

The formula for heat transfer calculations is:

amount of heat transferred = mass x change in temperature x specific heat

I will remind you that the symbols that are used for the formula will vary from textbook to textbook, but the values that they represent never change. One way to write the heat transfer formula is shown below:

q = m(DT)Cp

Where q = heat transferred, DT = the change in temperature and Cp = the specific heat.

The SI units for heat transferred are joules, however calories are still often used for problems involving water. You should memorize the conversion factor; 4.18 J = 1 cal. The units for specific heat are joules/grams x degrees Celsius (J/g x oC) or Calories/grams x degrees Celsius (cal/g x oC). Temperature is usually given in degrees Celsius.

You will solve these problems logically and algebraically. Logically, meaning you will strive to understand the logic of performing each step, and that you will check to make sure that your answer makes sense. As in any algebra problem, you will only have one unknown. The rest of the information will be provided for you. We will work through one example of each of the possible types of heat transfer problems that you will be responsible for. Then you can move on to try the worksheets and the online quiz programs.


Type 1. Heat Transferred (q) is the unknown:

Ex. Aluminum has a specific heat of 0.902 J/g x oC. How much heat is lost when a piece of aluminum with a mass of 23.984 g cools from a temperature of 415.0 oC to a temperature of 22.0 oC?

Step 1: First read the question and try to understand what they are asking you. Can you picture a piece of aluminum foil that is taken out of an oven. Imagine the aluminum losing heat to its surroundings until the temperature goes from 415.0 oC to 22.0 oC.

Step 2: Write the original formula.

q = m(DT)Cp

Step 3: List the known and unknown factors. Looking at the units in the word problem will help you determine which is which.

q = ?
m = 23.984 g
DT = (415.0 oC - 22.0 oC) = 393.0 oC (remember, they asked for the change in temperature)
Cp = 0.902 J/g x oC

Step 4. Substitute your values into the formula

q = ?
m = 23.984 g
DT = (415.0 oC - 22.0 oC) = 393.0 oC
Cp = 0.902 J/g x oC

q = m(DT)Cp

q = 23.984 g x 393.0 oC x 0.902 J/g x oC

Step 5. Cross out units where possible, and solve for unknown.

q = 23.984 g x 393.0 oC x 0.902 J/g x oC

q = 8501.992224 J

Step 6. Round to the correct number of significant digits and check to see that you answer makes sense.

q = 8.50 x 103 J

Our answer makes sense because joules (J) are acceptable units for q, and the value should be positive based on the wording of the question.


Type 2. mass (m) is the unknown:

Ex. The temperature of a sample of water increases by 69.5 oC when 24 500 J are applied. The specific heat of liquid water is 4.18 J/g x oC. What is the mass of the sample of water?

Step 1: First read the question and try to understand what they are asking you. Energy is being used to change the temperature of a sample of water by 69.5 oC. What size sample of water would require 24 500 J to make that change?

Step 2: Write the original formula, and then modify it isolate the unknown.

q = m(DT)Cp

q = m(DT)Cp
--- -------------
(DT)Cp (DT)Cp

m = q/(DT)Cp

Step 3: List the known and unknown factors. Looking at the units in the word problem will help you determine which is which.

q = 24 500 J
m = ?
DT = 69.5 oC
Cp = 4.18 J/g x oC

Step 4. Substitute your values into the formula.

q = 24 500 J
m = ?
DT = 69.5 oC
Cp = 4.18 J/g x oC

m = q/(DT)Cp

m = 24 500 J/69.5 oC x 4.18 J/g x oC

Step 5. Cross out units where possible, and solve for unknown.

m = 24 500 J/69.5 oC x 4.18 J/g x oC

m = 84.3344463184 g

Step 6. Round to the correct number of significant digits and check to see that you answer makes sense.

m = 84.3 g

Our answer makes sense because grams are the correct units for mass, and the value should be positive.


Type 3. change in temperature (DT) is the unknown:

Ex. 850 calories of heat are applied to a 250 g sample of liquid water with an initial temperature of 13.0 oC. Find a) the change in temperature and b) the final temperature. (remember, the specific heat of liquid water, in calories, is 1.00 cal/g x oC.)

Step 1: First read the question and try to understand what they are asking you. Here they are heating up a sample of water. They want to know how many degrees increase will result from 850 calories of heat. Further, they want to know the final temperature of the water, which will simply be equal to the initial temperature + the change in temperature.

Step 2: Write the original formula, and then modify it isolate the unknown.

q = m(DT)Cp

q = m(DT)Cp
---- ---------------
m Cp m Cp

DT = q/m x Cp

Step 3: List the known and unknown factors. Looking at the units in the word problem will help you determine which is which.

q = 850 cal
m = 250 g
DT = ?
Cp = 1.00 cal/g x oC

Step 4. Substitute your values into the formula

q = 850 cal
m = 250 g
DT = ?
Cp = 1.00 cal/g x oC

DT = q/m x Cp

DT = 850 cal/250 g x 1.00 cal/g x oC

Step 5. Cross out units where possible, and solve for unknown.

DT = 850 cal/250 g x 1.00 cal/g x oC

Answer to step a) DT = 3.4 oC

Answer to step b) final temperature = 13.0 oC + 3.4 oC = 16.4 oC

Step 6. Round to the correct number of significant digits and check to see that you answer makes sense.

Answers are already rounded correctly. They make sense because they show the correct units for temperature and because the final temperature is higher than the initial temperature, as it should be.


Type 4. Specific Heat (Cp) is the unknown:

Ex. When 34 700 J of heat are applied to a 350 g sample of an unknown material the temperature rises from 22.0 oC to 173.0 oC. What must be the specific heat of this material?

Step 1: First read the question and try to understand what they are asking you. Specific heat is a concept that some students struggle with. The question is about finding the specific heat by seeing how much the temperature changes when a certain amount of heat is applied. Metal heats up faster than water because it has a low specific heat. If a material has a low specific heat, the temperature change will be greater for a given amount of heat, when all other things are equal.

Step 2: Write the original formula, and then modify it to isolate the unknown.

q = m(DT)Cp

q = m(DT)Cp
--- -------------
m(DT) m(DT)

Cp = q/m(DT)

Step 3: List the known and unknown factors. Looking at the units in the word problem will help you determine which is which.

q = 34 700 J
m = 350 g
DT = (173.0oC - 22.0oC) = 151.0 oC
Cp = ?

Step 4. Substitute your values into the formula

q = 34 700 J
m = 350 g
DT = (173.0oC - 22.0oC) = 151.0 oC
Cp = ?

Cp = q/m(DT)

Cp = 34 700 J/350 g x 151.0 oC

Step 5. Cross out units where possible, and solve for unknown.

Cp = 34 700 J/350 g x 151.0 oC

Cp = 0.65657521286 J/g x oC

Step 6. Round to the correct number of significant digits and check to see that you answer makes sense.

Cp = 0.66 J/g x oC

Our answer is logical, and the units are correct.

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Anonymous Poster
#2
In reply to #1

Re: Heat Transfer Process - URGENT!!

06/06/2009 9:28 AM

I did actually read this just before I posted my question....

The problem is:

1. I don't have the mass but mass flow rate. (but I think it can be applied in place of mass)

2. If I apply mass flow rate in place of mass what will q represent? heat transferred(energy rate)/sec?

3. As my data is in J/kgK and kg/s I have to use Kelvin's rather than Celsius.

4. I don't know if I'm doing it correctly as someone on another forum gave me an answer of 50J/s and I can't reproduce the result.

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#11
In reply to #2

Re: Heat Transfer Process - URGENT!!

06/07/2009 6:49 PM

If you use mass flow rate instead of mass, it means that you divided your mass value over time. The algebric expression in the other side will be heat over time as well, or, if you prefer, power, in this case expressed by J/s, or W.

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Anonymous Poster
#12
In reply to #11

Re: Heat Transfer Process - URGENT!!

06/08/2009 6:45 AM

Thanks for the explonation, but i'm still wondering how to get this post deleted. That was hard work for me and I dont want people from my course to just google it and get the answers just like that...

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Anonymous Poster
#3
In reply to #1

Re: Heat Transfer Process - URGENT!!

06/06/2009 9:34 AM

Sorry, just noticed "You are correct." J So you think the first bit is correct, what about the efficiency of the process?

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#4
In reply to #1

Re: Heat Transfer Process - URGENT!!

06/06/2009 9:43 AM

There is a general policy of CR4 NOT to make the home works of students but only to correct what they present.

It is in their interest to make the work, meet the difficulties and correct them.

It was clear from the presentation that it is a home work.

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Anonymous Poster
#6
In reply to #4

Re: Heat Transfer Process - URGENT!!

06/06/2009 10:32 AM

I don't expect anyone to do my homework, simply push me in the right direction as the question (and subject itself) can be interpreted in many different ways (if not explained correctly). I did make the work, now I simply need a yay o nay.

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Anonymous Poster
#10
In reply to #4

Re: Heat Transfer Process - URGENT!!

06/06/2009 11:36 AM

Going back to the topic of homework... Is it possible to delete this post now, all I needed is conformation, and unlike me there are people on my course that wont do the work, just come over here and change the numbers (it's easy enugh to find this post on google).

Many Thanks

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#5

Re: Heat Transfer Process - URGENT!!

06/06/2009 9:53 AM

Didn't know this was a homework problem!

Efficiency = (Output - Losses) X 100 / Input

I guess you have to do the math!

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Anonymous Poster
#7
In reply to #5

Re: Heat Transfer Process - URGENT!!

06/06/2009 10:37 AM

What about efficiency = heat energy of fluid (output) / heat energy of coil (input) ?

Efficiency 17730 / 27000 = 0.66

0.66 x 100 = 66%

Many thanks.

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Anonymous Poster
#8
In reply to #5

Re: Heat Transfer Process - URGENT!!

06/06/2009 10:40 AM

unless by (Output - Losses) you mean (Output eg. Losses) / Input X 100. Then it's all god

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Anonymous Poster
#9
In reply to #8

Re: Heat Transfer Process - URGENT!!

06/06/2009 10:54 AM

I mean good

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#13

Re: Heat Transfer Process - URGENT!!

06/08/2009 7:39 AM

Yes, with a clarification:

X 100 can either be in the dividend or times the quotient of the calculation.

Efficiency = ((Output *includes process losses*)-(Losses *includes ambient losses*)/input) X 100

*Process losses are driven by boiler efficiencies, heat exchanger efficiencies, etc.

*Ambient losses are driven by uninsulated piping or equipment adding to the thermal loss of a process. No one, with exception of a design engineer, remembers the additional losses of a complete system. Those of us in the energy field find a significant savings in energy use by correcting the additional loss as well as increasing the efficiency of the process.

In your problem, above, I assume that your coil is in direct contact with the process heating pipe and the thermal transfer is from fluid "A" to copper pipe "A" to copper coil "B" to fluid "B". If you are solving for the efficiency of the entire system, efficiencies must be calculated starting with the energy input for process "A" through your output to process "B".

Now show me what you can do! Work the math!

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Anonymous Poster
#14
In reply to #13

Re: Heat Transfer Process - URGENT!!

06/08/2009 8:19 AM

Doing it this way my brain tells me that:

Input = 27000J/s

Output = 17730J/s

Losses = (Input - Output) = 9270

17730 - 9270 / 27000 * 100 = 31.3%

But that doesn't seem right....

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Anonymous Poster
#15
In reply to #13

Re: Heat Transfer Process - URGENT!!

06/08/2009 8:25 AM

But I don't understand what is wrong with

Efficiency = 17730 / 27000 = 0.66

0.66 x 100 = 66%

Where u have (Output *after losses*)/(input) * 100 = Efficiency?

As it takes loss into account already

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Anonymous Poster
#16
In reply to #13

Re: Heat Transfer Process - URGENT!!

06/08/2009 8:32 AM

And it's the other way around:

A coil is wrapped around a pipe that carries a fluid which needs to be heated.

The process is gas to coil - coil to pipe - pipe - to fluid in pipe

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Anonymous Poster
#17
In reply to #13

Re: Heat Transfer Process - URGENT!!

06/08/2009 2:20 PM

Also I think the other assumption is that no other losses in energy are taken into account (like ambient losses)?

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#18

Re: Heat Transfer Process - URGENT!!

06/08/2009 6:44 PM

Well, after all that, I guess you need to look at the problem and simplify:

1. The coil fluid is your heat source, your input. Its heat content assumed 100% at the beginning of the process.

2. The pipe fluid is the result of your process. It absorbs something less than the 100% of the coil heat content.

2. Calculate the heat transfer based on the flow rates given, specific heat and delta T for each fluid.

3. Calculate the efficiency

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Anonymous Poster
#19
In reply to #18

Re: Heat Transfer Process - URGENT!!

06/09/2009 6:42 AM

I will stay with my initial calculations:

COIL (INPUT)

U = mCT
U = 0.15kg/s * 500J/kgK * (773 - 413)
U = 27000J/s or 27x10^3 J/s

PIPE FLUID (OUTPUT)

U = mCT
U = 0.1kg/s * 1970J/kgK * (413-323)
U = 17730J/s or 17.73x10^3 J/s

Efficiency = (output heat rate J/s, what you get) / (input heat rate J/s, what you pay for) * 100

Efficiency 17730 / 27000 * 100 = 65.66%

Thats how I understand it. I think I understand it to a level that is required from me now(or I hope so). the problem is that there are so many different terms in heat transfer for the same thing that it gets confusing after a while.

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#20

Re: Heat Transfer Process - URGENT!!

06/09/2009 8:53 AM

Welcome to Physics & Thermodynamics!!!

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