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### Topic: cylindical heat exchanger  (Read 11550 times)

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#### Bibinou

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##### cylindical heat exchanger
« on: June 10, 2006, 02:15:50 PM »
Hello,
I have some problems in resolving this exercise.
Consider a cylindrical heat exchanger, whose you know all caracteristics (such as : inside diameter, outside diameter and its length). This exchanger is used to cool an oil phase with water.

We know the flow of water, its Cp and its temperature, moreover we also know the flow of oil, its Cp and its temperature.
We also know the global transfert coefficient K expressed in kW.m-².K-1

How can we determine the temperature of oil at the exit of the exchanger?

Thank you very much.

#### Donaldson Tan

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##### Re: cylindical heat exchanger
« Reply #1 on: June 10, 2006, 06:41:42 PM »
I am trying to understand your question.

1. Heat Exchanger Configuration
- Counter-Current, Co-Current?
- What is flowing inside the inner pipe - oil or water?
- U should be defined for the diameter of the internal pipe.

2. Inlet and Outlet Temperature
- Are both known for water?

Typically, if the outlet temperature of oil is unknown, I would use the NTU method.

NTU = U.S/(M.Cp)min
? = (M.Cp)min / (M.Cp)max
? = [ 1 - exp{ -NTU(1-?) } ] / [ 1 - ?.exp{ -NTU(1-?) } ] (for counter-current)
Q = ?(M.Cp)min(Tinlet,oil-Tinlet,water)

(M.Cp)oil = Foil * Cpoil
(M.Cp)water = Fwater * Cpwater
where F is the mass flowrate and Cp is the specific heat capacity.

(M.Cp)min refers to the lower of the 2 bulk heat capacities above.
(M.Cp)max refers to the higher of the 2 bulk heat capacities above.

S refers to the contact surface area for heat exchange. Since this is a cylindrical heat exchanger, then S = Pi.D.L where D is the diameter of the internal pipe, and L is the length of internal pipe.

Once you had worked out the heat transfer rate Q, you can derive the actual outlet oil temperature from a simple heat balance, ie. Q = rate of heat gain by water = rate of heat loss by oil

"Say you're in a [chemical] plant and there's a snake on the floor. What are you going to do? Call a consultant? Get a meeting together to talk about which color is the snake? Employees should do one thing: walk over there and you step on the friggin� snake." - Jean-Pierre Garnier, CEO of Glaxosmithkline, June 2006

#### Bibinou

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##### Re: cylindical heat exchanger
« Reply #2 on: June 11, 2006, 01:48:00 AM »
Hello,

I'm sorry I wasn't accurate.

This exchanger is a co-current cylindrical exchanger
The flowing in the inner pipe is oil.

The different data are :
- oil flow : 50g/s
- water flow : 60 g/s
- Cp oil = 2 kJ / ( kg . K)
- Cp water = 4.18  kJ / ( kg . K)
- lengt of the both pipe : 1m
- diameter of oil pipe : 25mm
- diameter of water pipe : 55mm
- temperature of oil AT THE ENTRANCE : 420 K
- temperature of water AT THE ENTRANCE : 290 K
- global coefficient tranfert K = 0.8729 kW/(m². K)

Thank you very much for your help

#### mbeychok

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##### Re: cylindical heat exchanger
« Reply #3 on: June 11, 2006, 03:50:11 AM »
Bibinou:

Why have you chosen to use a concurrent exchanger? Is there some special reason for doing so?  If not, it is better to use a countercurrent flow exchanger.  If you do use a countercurrent exchanger, the general design approach is:

(1) Select the coolant exit temperature by assuming how closely the coolant exit temperature will "approach" the hot fluid inlet temperature.  A good first estimate is that the approach will be 5 to 8 °C. Keep in mind that the closer the "approach", the longer your exchanger will be.

(2) The "approach" sets the coolant exit temperature.  Now, calculate the heat absorbed by the coolant:
-----  kJ/s = (kg/s of water)  (cp of water in kJ/kg/°C) (?T of water in °C).

(3) The heat lost by the hot fluid must equal the heat absorbed by the coolant. Now you can calculate the ?T of the oil:
-----  ?T of oil in °C = (kJ/s) ÷ [(kg/s of oil)  (cp of oil in kJ/kg/°C)]

(4) Now you have the inlet and outlet temperature of both the coolant and the hot fluid. You could then use your global heat transfer coefficient (if that is valid) to calculate the required m2 of your inner pipe (the oil pipe).  Knowing the diameter of the oil pipe, you can then calculate the piping length required.

I don't know if your overall global heat transfer coefficient is valid or not for the water and oil velocities involved.  If you are not sure of the validity, then you must calculate your water and oil heat transfer coefficients ... which is much more complex.

One other question. The cooling water inlet temperature of 290 K (16.9 °C) is quite cool.  Unless the annual mean air temperature in your locality is quite cold, that water temperature would not be available say from a cooling tower.  Are you sure of that water temperature?

Or is this whole exercise simply a homework problem?
Milton Beychok
(Visit me at www.air-dispersion.com)

#### Donaldson Tan

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##### Re: cylindical heat exchanger
« Reply #4 on: June 11, 2006, 06:27:27 AM »
NTU = U.S/(M.Cp)min
? = (M.Cp)min / (M.Cp)max
? = [ 1 - exp{ -NTU(1-?) } ] / [ 1 - ?.exp{ -NTU(1-?) } ] (for counter-current)
Q = ?(M.Cp)min(Tinlet,oil-Tinlet,water)

(M.Cp)oil = Foil * Cpoil
(M.Cp)water = Fwater * Cpwater
where F is the mass flowrate and Cp is the specific heat capacity.

Once you had worked out the heat transfer rate Q, you can derive the actual outlet oil temperature from a simple heat balance, ie. Q = rate of heat gain by water = rate of heat loss by oil

You can still use the NTU method that I had shown you.

However the ? function is different for a co-current heat exchanger.

For a co-current heat exchanger,
? = [ 1 - exp{ -NTU(1 + ?) } ] / (1 + ?)
"Say you're in a [chemical] plant and there's a snake on the floor. What are you going to do? Call a consultant? Get a meeting together to talk about which color is the snake? Employees should do one thing: walk over there and you step on the friggin� snake." - Jean-Pierre Garnier, CEO of Glaxosmithkline, June 2006

#### Donaldson Tan

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##### Re: cylindical heat exchanger
« Reply #5 on: June 11, 2006, 07:07:26 AM »
? is the effectiveness of the heat exchanger, define as the ratio of the actual heat transfer rate to the maximum heat transfer rate.

U represents the heat transfer coefficient, and it also refers to k.
« Last Edit: June 11, 2006, 01:10:37 PM by geodome »
"Say you're in a [chemical] plant and there's a snake on the floor. What are you going to do? Call a consultant? Get a meeting together to talk about which color is the snake? Employees should do one thing: walk over there and you step on the friggin� snake." - Jean-Pierre Garnier, CEO of Glaxosmithkline, June 2006

#### Bibinou

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##### Re: cylindical heat exchanger
« Reply #6 on: June 11, 2006, 12:41:34 PM »
Hello

thank you both for your explanations.
This was in fact a question of a previous exam I wasn't able to solve.

thx

#### technologist

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##### Re: cylindical heat exchanger
« Reply #7 on: June 14, 2006, 12:30:04 AM »
Your figures give a value of ~87°C of Oil temperature & ~41°C water Temp at the outlet by using simple thermal simulation for the exchanger once u know overall HTC.

However, you are using chilled water at 17°C whereas if u increase the length of pipe by only 250 mm u can use cooling water at 34°C with same flow rate of 60 g/s. (I've assumed it to be Gm/sec & not Gallon/sec, which is too high for this size)