Practice English Speaking&Listening with: Lecture - 27 Heat Exchangers - 3

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We defined the term called mean temperature difference and we derived expressions for

the mean temperature difference in parallel flow and in counter flow. You would recall

that the logarithm of delta Ti delta Te occurs in the denominator of the expressions, both

the expressions. For this reason the mean temperature difference in a heat exchanger

is also referred to as the logarithmic mean temperature difference and the symbol capital

LMTD - all letters in capital - LMTD is used instead of delta Tm by many authors and many

writers.

Then, we moved on to discussing how to obtain the mean temperature difference in cross flow.

Now, you will recall that the, what emerged after discussing the temperature profiles

on the hot side of the hot side fluid and the cold side was that in cross flow - unlike

the parallel flow or counter flow - the temperature is not a function of one variable, the temperature

is a function of 2 variables x and y. If both fluids are unmixed - temperature Th is a function

of x and y, temperature Tc is function of x and y. If both fluids are mixed, one fluid

is a temperature, one fluid's temperature is a function of x, the other fluid's temperature

is a function of y. And if one is mixed and the other is unmixed, the unmixed is a function

of x and y whereas the mixed one is a function only of one variable. But regardless of whichever

case it is, the fact is the temperature on the two sides are not function of one variable

x as is the case in counter flow or cross flow but are a function of x and y. And therefore,

in order to obtain the mean temperature difference it becomes necessary to perform numerical

integrations to obtain expressions for the mean temperature difference. And you will

also recall that towards the end, I said the mean temperature difference is given, the

values are given in terms of a correction factor F and F is defined as the mean temperature

difference in cross flow divided by the mean temperature difference if the arrangement

had been counter flow; this is where we stopped. So, let us take off from that again.

We were deriving getting results for the mean temperature difference in cross flow; we were deriving the results

for this case and, towards the end, we said F - we are going to express the results in

the form of correction factor F. And F is equal to the delta Tm which we were looking

for - The delta Tm in cross flow, this is the quantity we are looking for, divided by

the delta Tm. If the arrangement had been counter flow, delta Tm if the arrangement

was counter flow and we know how to calculate the denominator and in case the fluid is unmixed

on one side, you have to take the mean temperature of the fluid. The bulk mean temperature of

the fluid leaving the heat exchanger and substitute that in the expression for delta Tm for obtaining

the counter flow delta Tm.

Now the results for given, before I go to the actual presentation of the values of F,

how they are presented, let me just make a statement. For given values of Thi, The, Tci

and Tce that is given values of the inlet and outlet temperatures on the hot side and

cold side, I am making a statement - for given values of Thi, The, Tci and Tce for the given

values, the delta Tm in counter flow is the highest. That is, for given values of this 4 quantities, if I calculate delta

Tm in parallel flow, if I calculate delta Tm in counter flow, in cross flow for any

of the 3 cases, what I will find is the delta Tm in counter flow is the highest of all flow

arrangements, amongst all flow arrangements; this is what we will find. Therefore, it follows

therefore since delta Tm in counter flow is always going to be the highest, it follows

that F - the correction factor F - which we have defined above here must be always in

the range 0 to 1.

It must be a fair number which will range between 0 and 1; 0 less than equal to F less

than or equal to 1 that is follows. And with this definition of F, if we are to calculate

the q for a cross flow arrangement

Therefore, if we are to calculate q for a cross flow arrangement, we will get q for

a cross flow arrangement is equal to U A delta Tm cross flow which is with the definition

of F equal to UA F delta Tm counter flow. So, when we have a cross flow problem, a situation

of cross flow, we will calculate the delta Tm in counter flow; get F which has been obtained

by numerical integration. Then which, I will tell you shortly how we get it from the various

charts or graphs available and substitute into this as our basic performance equation.

Now the quantity F - how do we get it? F based on the numerical integrations, F is plotted;

the results which have been obtained by numerical integration F is plotted as a function of

two parameters R and S, dimensionless parameters R and S. R is defined as, let me put down

the definitions R is equal to T1i minus T1e divided by T2e minus T2i and I will tell you

in a moment what is 1; what are the subscripts 1 and 2? And S is equal to T2e minus T2i divided

by T1i minus T2i. Now, the subscripts 1 and 2 correspond to the 2 fluids, the subscripts

1 and 2 correspond to the 2 fluids correspond to the 2 fluids.

Subscripts 1 and 2 correspond to the two fluids. For cases 1 and 3, what are cases 1 and 3?

Cases 1 and 3 is - case 1 is both fluids unmixed and case 3 is both fluids mixed. For cases

1 and 3 of cross flow, these 2 cases of cross flow it is immaterial which subscript corresponds to the hot side and which to the

cold side. It doesn't matter you can take, either way the results are valid. That means

for this situation, cases 1 and 3, these situation for cases 1 and 3 subscript 1 can be equal

to h and subscript 2 can be equal to c. Or we may have the reverse - subscript 1 can

be equal to c and subscript 2 can be equal to h. It doesn't matter which way you take

it; both are equally acceptable. The results - numerical results - come out right. It is

only for case 2 that you have to be a little careful; for case 2, let me put down what

you do.

For case 2 that is one fluid mixed and the other unmixed, care has to be taken; care

must be taken to see that the mixed fluid has subscript 1. The mixed fluid maybe the hot fluid or the cold fluid, it doesn't matter

which one but take care to see that the mixed fluid is given subscript 1 and the unmixed

fluid automatically becomes subscript 2. That is the only precaution we have to take. So,

these are 2 parameters in terms of which the values of F are found out. Now what are these

parameters? What is the dimensionless parameter R? If you look at it, it is nothing but the

ratio of the change of temperatures of the 2 fluids, that is what is R. And R by definition

therefore will be a positive quantity but unbounded greater than or equal to 0 and the

limit 0.

What is S? The parameter S the second dimensionless quantity - it is the ratio of change in temperature

of one of the fluids, ratio of change of temperature of one of the fluids to the difference of

inlet temperatures of the 2 fluids - Thi minus Tci, that is what it is. So, automatically

by definition you can see the parameter S is going to be a number which is going to

be between 0 and 1. The particular results that we get for R, the result that we get

for R - if you were to plot them. First just let me show the variation; it looks something

like this.

Suppose this is the results that are obtained based on the numerical integration for any

of the cases 1 to 3. If I plot F against S, S can range from 0 to 1, F can also range

from 0 to 1. Then, for a particular value of R a typical variation of S maybe something

like this. This is some particular value of R, some specified value of R. We will get,

this is the kind of variation which we get for F and these has been obtained based on

the numerical integration which I mentioned to you earlier. Now, let us look at actual

results which have been obtained. I am going to show you 3 charts for the 3 cases of cross

flow.

This is the chart which I am showing; you know the correction factor F for the mean

temperature difference in single pass cross flow both fluids unmixed - case 1, what we

have called. These are results based on the numerical integration to obtain the mean temperature

difference in cross and to get the correction factor F. So, on the y axis you have the correction

factor, on the x axis you have the parameter S and the different values for the different

graphs are the values of R ranging here from .2 to 4. The value of R equal to 0 will be

the horizontal line and the vertical line, this will correspond to R equal to 0. So,

this is these are the results for the correction factor F for the case of both fluids unmixed.

Now, let us look at the next results. I am just showing these results which are to be

used for actual calculations. The next figure shows the correct, let me read out, shows

the correction factor F for mean temperature difference in single pass cross flow fluid

with subscript 1 - mixed, fluid with subscript 2 - unmixed. The nature of the curves is of

course the same. But the values are very different for the different cases. Again it is F plotted

against S with R as the parameter varying from 0 or to 4. 0 is this case the vertical

line that is 0 then .2, .4, .6, .8 all the way up to 4. These are typically values encountered

in practice; R greater than 4 is probably not encountered very often so the values are

not plotted there at all.

And then finally, here is the correction factor F - results for correction factor F plotted

for mean temperature difference in single pass cross flow for the case of both fluids

mixed; similar results but of course numerical values are different. So, depending upon which

cross flow case you have, you will go to the appropriate figure and read off the values

of S that is the point I want to mean.

We will do a numerical example so that you will be to use, learn how to use these graphs.

Now, let us just talk briefly about some special cases before we do some numerical examples,

going to do a couple of numerical examples to illustrate all these ideas. First, let

us just look at two special cases; these two special cases are the following.

a m dot h Cph tending to infinity - this is one special case. The value of the product

of the flow rate and the specific heat on the hot side is very high tending to infinity

or automatically it follows that q is finite and m dot h Cph tends to infinity. Then, the

change of temperature on the hot side of the hot side fluid must be tending to 0. b is

the reverse case; the other case that is m dot c Cpc tends to infinity. The product of

the flow rate and the heat capacity on the cold side is tending to infinity and therefore

since it is tending to infinity Tce minus Tci is tending to 0. So, these are 2 special

cases.

Now, if you were to calculate the mean temperature difference for these 2 cases that is either

a or b, if you were to calculate the mean temperature difference for these 2 cases,

what you will find is the value of the mean temperature difference will be the same whether

the arrangement is parallel flow, counter flow or cross flow, any of the cases of cross

flow. You can do that yourself and convince yourself that that is what will happen. The

fluid temperature on one side doesn't change and as a result, the values of mean temperature

difference come out to be the same whatever be the flow arrangement. I would like you

to check it on your own and in your mind you should be able to explain why this is also

happening.

Cases a and b - incidentally one way of looking at them is to say that m dot h Cph is tending

to infinity. The other way of looking at case a is to say it is a case in which a saturated

vapor is condensing. Suppose I have a saturated vapor at some temperature Ts and it is condensing

at the same temperature. Obviously, since it is saturated vapor, the temperature will

not change as it changes from vapor to liquid so automatically the temperature Ts will not

change on the hot side. So, case a also corresponds to a saturated vapor condensing and vice versa.

Case b corresponds to a saturated liquid evaporating. So, remember case a and b can be interpreted

in two ways; one interpretation is the flow rates are very high on one or the other side,

the other interpretation is that case a corresponds to the case of saturated vapor condensing

and case b to a saturated liquid evaporating. So, these are 2 special cases and as I said

it is worth noting that the mean temperature difference is the same in these cases regardless

of the flow arrangement. Now, the same idea, again I have shown in a sketch here so I will

be sort of repeating things here.

This is the figure a corresponding to m dot h Cph tending to infinity; the temperature

on the hot side doesn't change or this could be a vapor condensing. The temperature on

the cold side is increasing. The b is the case of m dot c Cpc tending to infinity so

the temperature on the cold side doesn't change; this could be a saturated liquid evaporating

and the hot side fluid is giving up heat slowly like this. The way I have shown it, it is

parallel flow but remember - even if it is counter flow all that will happen is I will

get a mirror image kind of variation of Tc in the other direction. And therefore you

can see why the main temperature difference is the same whether it is counter flow, cross

flow or parallel flow. So, these are 2 special cases.

Now, let us do a couple of problems to illustrate all these ideas of mean temperature difference

for different flow arrangements. Let us take one example - a numerical example.

We have a parallel flow heat exchanger to be designed, a parallel flow heat exchanger

to be designed to cool 5 kilograms per second of air from 500 centigrade to 350 centigrade by an equal

flow rate of air entering at 90 degrees centigrade. 5 kg per second of air to be cooled from 500

to 350 on the cooling side, the same flow rate of air entering Tci being 90 degree centigrade.

We are given that the heat transfer coefficient on the hot side hh is equal to 60 and on the

cold side hc is equal to 30 - these are the values of h. Calculate the area A of the heat

exchanger, that is the example to be solved. Take Cp; we need the value of Cp; take Cp

of air on both sides to be 1020 joules per Kelvin per kilogram Kelvin, take this value of Cp. So, design

a parallel flow heat exchanger; that means find the area A of a parallel flow heat exchanger

for given conditions. Let us just draw a sketch of what we have, what we are asked to do.

This is, let us say - let me draw the sketch of the heat exchanger.

Let us say this is our heat exchanger, this is the heat transfer surface, the fluid on

the hot side, it is a parallel flow heat exchanger. m dot h equal to 5 kilograms per second, m

dot c is given to be the same - 5 kilograms per second. Thi, you are told the inlet temperature

on the hot side equal to 500 centigrade; Tci equal to 90 degrees centigrade, The given

to be 350 centigrade. The value of hh heat transfer coefficient on the hot side is equal

to 60 in the usual units, Watts per meter squared Kelvin, and the value of hc on the

cold side equal to 30 Watts per meter squared Kelvin. So, find the area, these are the exit

points; find the area A of the heat exchanger, the configuration is parallel flow.

We can sketch the temperature profiles - on the hot side going to be something like this,

on the cold side it is going to be something like this. This is 500, this is 90, this is

350. Now, the flow rates on the hot side and the cold side are equal and the Cps are given

to be equal. Therefore, it follows that the change of temperature on the hot side and

the change temperature on the cold side must be equal. So, the change of temperature on

the hot side is 150 so it follows that the change of temperature on the cold side must

be also 150. So, Tce must be equal to 240, that follows I don't have to even do any real

calculation for that. So, delta Ti is equal to 410 and delta Te is equal to 110 degree

centigrade. These are temperature profiles on the 2 sides; these are the values of delta

Ti and delta Te. It is relatively a very simple problem. We have to remember we have to substitute

into the expression q is equal to U A delta Tm. So, we need to get q, we need to get U,

we need to get delta Tm then we will get the value of A so let us get them one by one.

Let us get the value of, first, the value of q.

q is equal to 1020 that is the specific heat into the flow rate 5 into the change of temperature

on the hot side or cold side - it doesn't matter which side you take - 500 minus 350.

And that comes out to be 765 into 10 to the power of 3 Watts. This is the heat transfer

rate in the heat exchanger. The value of delta Tm in parallel flow is equal to delta Ti minus

delta Te divided by log to the base e delta Ti by delta Te. So, it is 410 minus 110 divided

by logarithm to the base e 410 divided by 110 and that comes out to be 2208.02 degrees

centigrade.

Now, we are given the values of hh and hc; we aren't told anything about fouling so we

will have to make an assumption. First of all about fouling and secondly about the thermal

resistance B by k - we are not told anything about that. But, notice the values of hh and

hc are low; they are low because we have got air flowing as the heat transfer medium. So,

values of hh and hc are bound to be of the order 50, 60, 100 something like that, not

going to be in thousands with air flows. Automatically, it follows that the thermal resistance of

any metal wall you can take, any typical wall, a millimeter thick, 2 millimeters thick, take

a conductivity of a metal of 10, 20. You will immediately see its thermal resistance will

be insignificant compared to 1 upon hh or 1 upon hc.

Similarly, if you look at any of the fouling factors that I have given you, you will remember

those fouling factors are for a situation involving liquids. With gases fouling factors

which are in the region of .30s, 130s, 230s, 4 in the usual units will be there, effect

will be negligible compared to the thermal resistance offered because of the heat transfer

coefficient. So, with gas flows the values of a neglecting fouling is a good assumption.

Neglecting the thermal resistance of the metal wall if that data is not available is also

a good assumption. And we can say in this situation 1 upon U is equal to 1 upon hh plus

1 upon hc. We just leave out B by k saying it is negligible and we will also not consider

fouling because it is bound to be negligible compared to the values of h that we are seeing

for gas flows. So, all that we get is this is equal to 1 upon 60 plus 1 upon 30 and therefore

the value of U comes out to be equal to 20 Watts per meter squared Kelvin. Now substitute

into our basic performance equation to get the value of A.

So, our basic performance equation is q is equal to U A delta Tm; in this case delta

Tm in parallel flow. So, we will get A is equal to q which is 10, q which is equal to

1020 into 5 into 150 divided by A UA. U is 20 sorry U delta Tm so U is 20 and delta Tm

is 2208.02. So, we get the area A of the heat exchanger to be 167.75 square meters, that

is the answer to the problem - 167.75 square meters for the given data that we have got.

Now, let us say instead of a parallel flow heat exchanger, the arrangement instead of

being parallel flow is counter flow. I ask you to calculate the area A of the heat exchanger

if the flow arrangement is counter flow; let us do a further calculation. Calculate the

area of the heat exchanger if the flow arrangement is counter flow. In this case, notice we again have to substitute into the same expression

q is equal to U A delta Tm. q is not going to change here is my basic expression, q is

equal to U A delta Tm. q is not going to change, U is not going to change, what is going to

change is the delta Tm. So, get a new value of delta Tm for counter flow situation and

again substitute into our basic equation. What is delta Tm in counter flow? Now let

us calculate that.

delta Tm in counter flow would be, notice this is a case of counter flow with m dot

h Cph equal to m dot c Cpc, notice that. Therefore, T in this case with m dot h Cph equal to m

dot c Cpc, the temperature profile on the 2 sides are going to be parallel lines like

this. So, we are going to have 500 here, hot side fluid entering temperature, 350 leaving.

Entering Tci 90, leaving Tce 240 and we will get the value of delta Ti and delta Te - both

delta Ti and delta Te. Both in this case will be 260 degree centigrade. So, delta Tm counter

flow will be equal to the same, delta Tm counter flow will also be equal to 260 degree centigrade.

This is the case of the 2 temperature profiles being parallel to each other; this is that

special case which we talked of earlier. As I said, q and U don't change therefore A will

be equal to the value of q which is 765000 Watts divided by the value of U and the value

of delta Tm - 260. So, we get 147.12 square meters; so this is the value for the area

A, the arrangement is counter flow.

I will make a statement here without proving it that if it is a cross flow arrangement,

any kind of cross flow arrangement, then you are going to get since this is the counter

flow arrangement is 147.1 2 and in parallel flow you have got 167.75. A cross flow arrangement,

any cross flow arrangement, would give an area between these 2 extremes; parallel flow

gives the highest, count flow gives the least. You will get a value for cross flow in between

these two; I am just making that as a statement. So, this is a good numerical example to illustrate

how we substitute into our basic performance equation - q is equal to U A delta Tm - and

calculate the area A of a heat exchanger for a given application. Now, let us do one more

problem, another numerical example.

The problem we are going to do now is the following - we say 1000 kilograms per hour

of water at 50 degrees centigrade enters a single pass cross flow heat exchanger and leaves at 40 degree centigrade. A hot

water stream, 1000 kilogram per hour of water, enters at 50, leaves at 40. The heat from

this hot water, the heat is transferred to cooling water entering at 35 centigrade and leaving at 40 degree centigrade. The cooling

water which is cooling this hot water enters at 35 and leaves at 40 degree centigrade.

Calculate the area of the heat exchanger if the fluids on both sides are unmixed.

Take U to be 1000 Watts per meter squared Kelvin with fouling included; that means the

effects of fouling are included in this value of U. Take the value of U to be 1000 Watts

per meter square Kelvin and the value of Cp on both sides, the Cp for water on both sides

to be 4174 Joules per kilogram Kelvin; take the value of Cp to be the following. It is

a straight forward example of calculating area, very similar to the earlier one excepting

that now the flow configuration is cross flow with both fluids unmixed. Now, let us just

draw a sketch.

Here is let us say a schematic diagram. This is the area, the cross flow situation; let

us say this is the heat transfer area A which we have to find out. This is the hot fluid

entering here, m dot h equal to 1000 kg per hour, Thi entering at a temperature of 50

degree centigrade, leaving at a temperature of - hot fluid leaving at a temperature of

40 degrees centigrade, cold fluid Tci entering with the temperature of 35 centigrade and

leaving with a temperature of 40 degrees centigrade. Calculate the area A.

Now again our basic performance equation is q is equal to U A delta Tm. In this case,

we are going to get to the correction factor F so we will write this as U A F which will

come from that graphs, multiplied by delta Tm if the arrangement had been counter flow.

So, if I want to get the value of A, I should get q, I should get U, I should get F; so

let us get each of the quantities. First of all - q; what is q? q is nothing but the flow

rate on the hot side 1000 divided by 3600 so many kilograms per second into 4174, is

the value of Cp, into the change of temperature on the hot side 50 minus 40 and that is equal

to 11594.4 so many Watts; so that is the value of q? What is delta Tm in counter flow? Let

us get the value of delta Tm in counter flow.

delta Tm will be equal to, this is the, let us draw a sketch for. If it is a counter flow

situation starting with 50 going out at 40 on the hot side, cold side entering at 35

and leaving at 40, so we have delta Ti is equal to 10 and delta Te equal to 5. So, we

have 10 minus 5 divided by logarithm 10 divided by 5 which comes out to be 7.21 degrees centigrade.

Now, let us go the charts since both fluids are unmixed; since both fluids are unmixed,

it is immaterial whether subscript 1 corresponds to the hot side

and subscript 2 to the cold side or vice versa. It doesn't matter which you take; so we say

take T1 equal to Th and T2 equal to Tc. Let us take that; it doesn't matter which you

take, you will get the same answers.

Therefore R is equal to Thi minus The divided by Tce minus Tci which is equal to 50 minus

40 upon 40 minus 35 which is equal to 2. And S is equal to Tce minus Tci upon Thi minus

Tci and that will be equal to for40 minus 35 divided by - Thi minus Tci - 50 minus 35

which is .333.

Now, we go to figure which I had shown you earlier for the case of the results obtained

for single pass cross flow; recall I had given you some situations there. And if you recall,

let us just show that figure again. Both fluids unmixed here; we have this is the situation

correction, factor F for both fluids unmixed. Now, in this case R is equal to 2 and S is

equal to .333 so this is the graph of R is equal to 2; here this is the graph of R is

equal to 2. Take S equal to .333, somewhere here, go up here to R is equal to 2 which

means somewhere out here you will go at this point, then go horizontally and read of the

value of F. So, here is the value of R equal to 2; get S equal to about .3, it will come

somewhere here. Go horizontally and you will get F is equal to .91 from the figure, F is

equal to .91. Therefore, now last step, area A is equal to q divided by U delta Tm counter

flow into F so it is equal to 11594.4 divided by the value of U is 1000. Value of delta

Tm in counter flow is 7.21 and the value of F is .91 so we get area A equal to 1.77 square

meters and that is the answer for this particular problem; that is the answer for this particular

problem.

So now, for different flow configurations through the help of these 2 examples, we have

illustrated how to obtain the area A for a given heat exchanger. Now this is one technique,

the q, using the equation q is equal to U A delta Tm - this is one technique for finding

out. The other technique which is also used which is the other method which is also used

for finding out the effect the area A of the heat exchanger or the performance of heat

exchanger is what is called as the effectiveness NTU method.

And the effectiveness NTU method is, there are certain reasons why this particular method

has been developed. So, next time we will look at the effectiveness NTU method; we will

discuss it, derive some relations for it and show how it is also applied for doing calculations

with heat exchanges.

The Description of Lecture - 27 Heat Exchangers - 3