11 replies to this topic

[satendra.deshmukh]

Dear all,

I have basic qurerry regarding how heat exchanger works...

Consider heat exchanger with cold fluid on one side ang hot fluid on other side.. heat transfer willl be from hot side to cold side due to their enthalpy difference...

My question is by keeping one side flow constant and varying other side what will be its effect on outlet temperature..

what i understood is if we change mass flow rate heat exchanger duty will change ( Q= M*Cp*dT)

Suppose i decrease hot side flow by keeping cold side flow constant .. it will result in decrease in heat duty of heat exchanger.. but i am sure that hot side outlet temperature will be less than previous one.. and also if i decrease cold side flow by keeping hot side flow constant cold side outlet temperature will be greater than previous one as there will more heat gain per kg ( though its heat duty will decrease)..

Now my question is how i can i prove this phenonmeno therotically... or please correct me if i have some misunderstanding reagarding this....

 

My collegues has different understanding over this they say that if i will decrease cold side flow by keeping hot side flow constant.. cold side outlet temperature can increase or decrease.... but according to me cold side temperature will increase...

 

 

Please help...

 

Regards,
Satendra Deshmukh

Edited by satendra, 04 May 2013 - 06:52 AM.

[alcir grohmann]

Dear Satendra Deshmukh

 

What you have observed is correct. In one case we have : 1) keeping cold side with constant flow while decreasing the hot side flow.

So, what will happen ?

 

Let's suppose WATER as cold side changing its temperatura from 20 oC to 30 oC, with a flow of 50 m3/h.

The heat received in 1h is :   Q = m*c*detal(t) =  50.000 kg *  1 kcal/kg oC * (30-20) oC = 500.000 kcal.

 

Let's suppose WATER again as hot side, changing ist temperature from 60 oC to 40 oC with a massa flow of 25 m3/h.

The heat delivered in 1h is :  Q = m*c*delta(t) =  25.000 kg *  1 kcal/kg oC * (60-40) = 500.000 kcal 

 

Surely both heat received and delivered are equals.

 

Now, as mentioned in your question, let's suppose the cold side flow is kept constant, while the hot side flow is decreased.

If its flow is decreased, then greater amount of heat will be delivered because the time to loose heat will be bigger.

 Let's suppose its temperature in this case changes from  60 oC to 35 oC...

 

Here, in this point, you made a comment : (sic) "...I am sure that the hot side outlet temperature will be less than the previous one !! "

 

I agree with your comment...BUT I don't agree with your other comment : (sic)"...it will result in decrease of heat duty of heat exchanger !! " 

 

This is not true...because with MORE TIME to loose its temperature, the amount of energy lost will be greater...and the heat lost will be bigger...and the temperature will be lesser than the previous...That's why I cannot agree with your sentence  !!

 

Now...you ask to PROVE this with an equation. How could I prove this...?? All I could say for now is that : WITH DECREASED FLOW, MORE TIME WILL BE LEFT TO THE HOT SIDE TO LOOSE ITS ENERGY AND - SURELY - TO GET LESSER OUTLET TEMPERATURE. The hot side will tend to have the same temperature of the cold side...

 

Are you happy with my trial to explain this phenomenon ??

 

Best Regards from

 

Alcir Grohmann

[alcir grohmann]

Dear Satendra Deshmukh

 

My e-mail :   alcir.grohmann@yahoo.com.br

 

Feel free to reply !!

 

Alcir Grohmann

[Dacs]

Do we expect an increase or decrease in the cold fluid temperature if we reduce the flow of the hot fluid? The answer is it's not that simple to deduce because while the decrease in hot fluid flow will bring an increase in residence time (since you have a fixed heat exchanger), it contains a lower amount of heat to expend to the cold fluid in the first place. And that will result in a lower delT between the hot and cold fluid and thereby will result in a lower heat flux.

 

Going the other way around, if you increase the hot fluid flow, you're more likely to maintain a larger delT between the hot and cold fluid inside the exchanger and this will result in a higher heat flux but at the expense of a shorter residence time.

 

So to qualitatively answer the question, we have to consider the (1) effective residence time in contact between the fuids and (2) the resulting heat flux that is governed by the temperature delta (delT) and the resistance to heat flow (U) in your exchanger. Basically (1) dictates how much time you can have a heat exchange between fluids and (2) dictates how much heat you can transfer from the hotter fluid to the colder one.

 

For instance, if you have a double pipe HEx with fluids flowing co-currently. If we increase the hot fluid flow, the rate of temperature drop of the hot fluid is lower (thereby having a higher delT and a higher heat flux) but with the increase in flow, the time of heat exchange is decreased (because higher flow = lower residence time). Same goes for the other case, lowering the hot fluid flow will increase the residence time, but as you go father along the double pipe HEx, you'd have lower delT (and lower heat flux).

 

I have not included the effect of velocity to the heat transfer coefficient (U) into the mix and this needs consideration as well.

 

So which is which then? IMHO you have to do basically a "rating" of the heat exchanger in question to really know the answer. Knowing the flows won't be enough. You have to consider the factors mentioned above to really ascertain what will happen.

 

My 2 cents

[alcir grohmann]

I would like to make a comment about the answer sent by our colleague where he says : (sic) "..., it contains a lower amount of heat to expend to the cold fluid in the first place. And that will result in a lower delT between the hot and cold fluid and thereby will result in a lower heat flux."

I agree with the fact that the heat flux will be lower, and I agree also with the fact that the delta(T) will be lower...but this is what will result in a lower outlet temperature of the hot fluid. A comparison between the initial situation and the proposed situation will show that the relationship  between both masses  (= mass of hol fluid / mass of cold fluid ) will be bigger after the hot fluid flow decreases. Or - in other words - a greater mass of cold fluid was used, leading to a lesser hot fluid outlet temperature. 

Surely other factors should be considered as paralell, counter-current or mixed flow, as well as effect of fluid velocity and global heat transfer coefficient. But these factors should be of little importance in comparison with the hot fluid flow reduction.

 

Best regards from

 

Alcir Grohmann

[Dacs]

By the way, I'm talking about heat duty (heat transfer between fluids), not necessarily the outlet temperature of both fluids. It's better this way since this is what heat exchanger is built for.

 

A comparison between the initial situation and the proposed situation will show that the relationship  between both masses  (= mass of hol fluid / mass of cold fluid ) will be bigger after the hot fluid flow decreases. Or - in other words - a greater mass of cold fluid was used, leading to a lesser hot fluid outlet temperature.

I think that while you'd have a correspondingly higher amount of cold fluid per a certain mass of hot fluid (when you reduce the hot fluid flow), it doesn't necessarily mean that you'd achieve the same amount of total heat transfer, in which your sentence implies (lesser hot fluid outlet temperature if you keep the cold fluid outlet temperature the same as if you have 100% flow of hot fluid temperature)

 

Yeah, it's possible that your hot fluid outlet temperature is lower at lower flowrate (and I think this is most likely to happen), but it doesn't necessarily mean that you achieved the same amount of heat duty. The fact that you have a smaller amount of heat being carried by the lesser hot fluid flow doesn't help.

 

So just to reflect my previous post, it's possible that (1) you'd have better heat transfer because of a higher hot fluid residence time or (2) you'd have worse heat transfer since the rate of cooling of hot fluid will be faster because of a smaller heat content (which lowers the heat flux). And I don't think we can answer it at face value. A rating of HEx is warranted.

 

All my 2 cents. Feel free to correct me if I got something wrong.

 

Edited by Dacs, 06 May 2013 - 10:22 PM.

[satendra.deshmukh]

Dear dacs and alcir,

Thanks for answering my query,

 Dear dacs I am somewhat agree with u that "outlet temperature will depend on rating of heat exchanger"

 Take example...

Hot fluid at 500 kg/hr getting cooled from 40 C to 32 C, and 500 kg/hr of cold fluid is getting heated from 8 C to 16 C.

Heat lost by hot fluid = 500 * (40-32) *1 = 4,000 kcal/hr

Heat Gain by cold fluid = 500 * (16-8) * 1 = 4,000 kcal/hr

Enthalpy at inlet of hot fluid= 500*1*(40-0) = 20,000 kcal and Enthalpy of hot side fluid at outlet = 500*1*(32-0) = 16,000 kcal/hr

Enthalpy of cold side fluid at inlet = 500*1*(8-0) = 4,000 kcal, at outlet = 500*16*1 = 8,000 kcal

Enthalpy difference at inlet conditions = 20,000 - 4,000 = 16,000 kcal

Ideally Maximum heat transfer should be equal to 8,000 kcal to reach equilibrium

20,000 – 8,000= 12,000 kcal and 4,000 + 8,000 = 12,000 kcal

(But practically it not possible, it will be heat transfer area, residence time and other things...

In our case for given set of conditions heat transfer is 4,000 kcal

  Now consider we have decreased hot side fluid flow to 200 kg/hr, now heat content of hot side fluid is lesser than previous case...

Heat content at inlet 200*1*(40-0) = 8,000 kcal

Will keep cold side conditions unchanged, heat content at cold side inlet = 500*8*1 = 4,000 kcal

Enthalpy difference between hot and cold side = 8000-4000 = 4000 kcal

Max possible heat transfer = 2,000 kcal

Once hot fluid will lose 2,000 kcal of heat to cold side both will be in equilibrium and there will be no heat transfer

4000+2000 = 6000 kcal and 8000-2000 = 6000 kcal

 

But due to heat losses and heat exchanger limitation heat transfer will be less than 2,000 kcal

 Let’s take any value less= 1,950 kcal

So hot side outlet temperature will be= x

 1,950 kcal = 200 kg/hr * (40-X) *1

X =30.25 C

Hot side outlet temperature 30.25 C is now less than hot side outlet temperature of first case (32 C).....

Now you can see.

Maximum possible Heat exchanger duty is 2,000 kcal. Which means by decreasing hot side inlet flow heat exchanger duty is decreased.

Total heat gain by cold side fluid in first case

=4000/500

Heat gain per Kg =8 kcal

 

Total heat gain by cold side fluid in second case

= 1,950 kcal

Heat gain per kg = 1950/200 = 9.75 kcal

 

 

So in second case though heat exchanger duty is decreasing, heat lost per kg to cold fluid is increased..

 

Now I am still confused? As Dacs is saying it is depend on rating of heat exchanger …

Can I consider above example applicable for any case?

Still I am not getting perfect and satisfactory answer for my query...

Edited by satendra, 18 May 2013 - 08:21 AM.

[Dacs]

I have to admit I can't follow your method (I'll try to digest it when I have more time), but just to reiterate what I said before, it's not as straightforward as it sound when dealing with this.

[satendra.deshmukh]

Okay.. i will find perfect answer....Thanks dacs..

[narendrasony]

Dear Satendra,

Potential for heat transfer is not the difference in the Enthalpy values but the temperature difference. Enthalpies are calculated at a reference temperature of generally 0⁰C but it can be any temperature as well. 

Extending your example, suppose hot water (Inlet T = 40⁰C) flow is 75 Kg/Hr. Cold water (Inlet T = 80⁰C) flow is unchanged to 500 Kg/Hr.

So heat content (based on 0⁰C refernce) is 3,000 Kcal/Hr and 4,000 Kcal/Hr for hot and cold streams respectively. So, according to your proposition heat transfer can not take place now from hot to cold stream since  heat content for cold water is now more than that for hot water. But you know that heat will transfer from hot to cold fluid - in this case also. You can re-write heat balance equation bases on enthalpy balance for an arbitrary reference temperature T_ref as follows:

 

75 * 1 * (40 -T_ref )  +  500 * 1 * (8 - T_ref )  =  75 *1* (T_ho - T_ref )   +  500* 1* (T_co - T_ref )          

where T_ho  and T_co   are outlet temperaures for hot and cold streams.

 

You can notice here that T_ref  terms cancel out individually for hot and cold streams and has no effect on overall heat balance (assuming constant heat capacities from inlet to outlet).

Based on T_ref = 0⁰C, sum of inlet enthalpy = 7,000 = sum of outlet enthalpy. So enthalpy for hot stream may reduce from 3,000 to 2000 and same for cold stream may increase from 4,000 to 5,000. Enthalpy balance doesn't mean that heat should flow from higher heat content to lower heat content fluid. Heat flows from higher temperature to lower temperature - always.

 

Infact if hot stream flow is held constant,  increasing cold stream flow will increase overall heat transfer coefficient U and heat transfer (Q = U.A.dTm where dTm = LMTD correction factor Ft * LMTD). But the outlet temperature will always reduce. Similarly reducing the cold side flow  will increase the temperature but it  will become assymptotic. How much - depends on exchanger geometry including no of passes , physical properties , inlet flow and temperatures.

 

Regards

Narendra

[alcir grohmann]

Dear Narendra

You gave good tips, congratulations. Just to help, you mentioned 80 oC at the beginning, but used the correct meant value of  8 oC as cold water temperature.

Regards

Alcir Grohmann

[satendra.deshmukh]

Dear narendra,

Thanks for correcting me...

 

 

Satendra