Heat transfer efficiency vs. coolant mass flow rate

[MRDAGERUS]

Hello Everybody,

Could you tell me if/where I'm wrong? Perhaps, you can point me toward other equations/formulae for heat transfer ratio applicable to air-to-water HE-s?
I'm trying to prove someone wrong.
This individual insists, that because of the higher coolant velocity (if I use a larger pump), the heat transfer in the HE/radiator is slower, because "the heat does not have enough time to convect out of the coolant, because it flows too fast".
IMO: a shear heresy.

The equations cited in this paper

http://jullio.pe.kr/fluent6.1/help/html/ug/node245.htm

show the only time dependent element, m, fluid mass flow rate (kg/s), is in the numerator, hence the heat flux q increases proportionally with fluid mass flow rate, (eq. 6.21-6). In another words, bigger the pump trough-put - the higher heat transfer and efficiency.

The same is true for the heat transfer coefficient, h, in eq. 6.21-7.

Do you agree with me?


Mater artium necessitas

 

[IRstuff]

Depends on what you mean by "heat transfer efficiency."

It could mean, "heat transferred per unit time," which is maximized by the flow rate
It could mean, "heat transferred per unit volume or mass," which would be maximized at a lower flow rate

heat transfer is maximized by the temperature difference, so the smallest amount of heat transferred per unit mass or volume maximizes the temperature difference, but that's expensive from the perspective of power consumed in circulating the fluid, and amount of coolant wasted.

TTFN (ta ta for now)
I can do absolutely anything. I'm an expert!

https://www.youtube.com/watch?v=BKorP55Aqvg

faq731-376 forum1529 Entire Forum list

http://www.eng-tips.com/forumlist.cfm

 

[CH5OH]

it is not that easy:
off coarse there is the basics:
gas outside of tube,liquid inside the tube (gas opposes the biggest heat transfer resistance,therefore on outside of tube, biggest contact area)
inside the tube:
boundry layer (google suggestion):a film of liquid in contact with the tube, comprising of liquid molecules with quite a different opinion with respect to flow direction (they may behave stationary or even counterflow)
sometimes this effect is counteracted by means of inserting a twisted tape inside the tube
google suggestions:laminair flow/turbulent flow,reynolds number,fluid dynamics
for a pure laminair flow, the bigger the flow, the better the cooling (lower delta t)

however, if this individual is your superior he is always right (the expression of tought:"you 1 braincell idiot", is not beneficial)

 

[EdStainless]

In many air applications the water flow rate is irrelevant because it is the air side coefficients that limit total heat transfer.
The point of a heat exchanger is not efficiency of heat transfer but the total amount of heat transfer.
Yes flowing slower allows the water gain the most heat, so if you had a very limited quantity this might be desirable, however it will greatly increase the temperature of the air being cooled.
The only efficiency that is usually an issue is how high can I push the flow velocity before the pumping costs become too high to justify.

= = = = = = = = = = = = = = = = = = = =
P.E. Metallurgy, Plymouth Tube

 

[LittleInch]

MRDAGERUS,

You need to read your attached paper more carefully. I believe the equation 6.21-3,4,5,6 etc are related to the mass of air flow through the radiator, not the water. Note it says "fluid" and not "air" or "water".

Which makes sense for a car radiator where mass flow of air will directly increase the heat transfer rate q (W/m2)

for something like a car radiator, the water flow rate will have an impact, but it's not a simple equation as it is also mixed up with the air supply. A radiator at low flow will transfer more heat per unit mass, because it has a longer residence time and ends up a the exit at a lower temperature than if the mass flow ( and nothing else) was doubled.

However in terms of actual heat energy transferred to the air per unit of time, the amount of heat would increase with higher flow (again with nothing else changing) because the temperature of the water would remain higher and hence heat flux, which is dependent on differential temperature, would overall be higher than at a lower flow.

SO I think you can be both correct, but you need to define what your argument is about. "Efficiency" as noted above, doesn't make much sense in this respect.

what I suspect your other person means is that all other things staying the same ( air velocity, area of radiator etc), if you increase mass flow of the coolant water, your outlet temperature will go up, but you're correct in saying the amount of heat transferred to the air will also go up.

Remember - More details = better answers
Also: If you get a response it's polite to respond to it.

 

[MortenA]

@MRDAGERUS,

Im with EdStainless - in most heat exchanges you will have a side that dominates the U value - and in an air cooler its usually the air side. Maybe you colleague is phrasing it in a wrong way?

By increasing the flow rate you may improve the heat transfer (move more joules/BTU's/calories or whatever)- but the temperature out of the cooler may still increase since the total flow through the HX increases more than your increase in heat transferred?

Best regards, Morten

 

[IRstuff]

As the other posters point out, and as described by your thermal network equations, heat flows through a series of thermal resistances, and making one resistance zero simply means that you are limited by the remaining resistances. For a radiator, there are essentially 3 thermal resistances, all in series:

> fluid to radiator convection
> radiator internal conduction
> radiator to air convection, in parallel with radiator to surrounds radiation

If you were to drive the fluid or the air to zero, you'd still have two other resistances to deal with, and the cost of driving one to zero is completely wasted if the others that you haven't reduced are large.

TTFN (ta ta for now)
I can do absolutely anything. I'm an expert!

https://www.youtube.com/watch?v=BKorP55Aqvg

faq731-376 forum1529 Entire Forum list

http://www.eng-tips.com/forumlist.cfm

 

[EdStainless]

There are actually five resistances since there are boundary layers on both sides also.
Raising flow velocities helps improve heat transfer mostly by lowering the boundary layer resistances.
It also helps by keeping the working delta T maximized.

= = = = = = = = = = = = = = = = = = = =
P.E. Metallurgy, Plymouth Tube

 

[davefitz]

If the radiator fluid is single phase ( ie not condensing steam) then the usual engineering treatment of the problem is using "compact heat exchanger" theory, sometimes called the e- NTU method. As the W*Cp of the 2 fluids vary, the effectiveness of the HX will vary. Likewise, the overall heat transfer coeficient U of each side of the HX will vary as the velocity varies, generally by the 0.8 power of velocity.

generally ,if W*Cp,a < W*Cp,w, then e= effectiveness= (Tao-Tai)/(Twi-Tai), q=W*Cp,a*(Tao-Tai) = W*Cp,w*( Twi-Two)
NTU= number of transfer units= U*A/(W*Cp, min), and we are assuming W*Cp, min = W*Cp, a.

The effectiveness of a counter flow HX can be calculated as a function of the value of NTU and the ratio R=W*Cp,a/ W*Cp,w. See texts by Kays and London titled "Compact heat exchangers"

"...when logic, and proportion, have fallen, sloppy dead..." Grace Slick

 

[Hurmanetar]

MRDAGERUS, you are correct. The coolant does not need "dwell time" inside the engine or inside the cooler/radiator for efficient operation of the cooling system. I'm guessing this myth has developed because a slower flow rate will likely result in a lower cooler outlet temp (but a higher average temp). I once heard the "dwell time" myth repeated in a Waukesha engine class. I had this very same argument with the instructor.

Q = m*Cp*dT
Heat load = mass flow rate * heat capacity of fluid * temperature differential of fluid across the cooler
If Q is fixed as in engine jacket water cooling, then if you double m, then dT is cut in half.

Q= U*A*EMTD
Heat load = overall heat transfer coefficient * surface area of cooler * effective mean temperature differential
In air cooled heat exchangers EMTD is the effective average temp above ambient air temp and is basically LMTD with a correction factor applied since geometry is somewhere between counterflow and cross-flow.

All other things being equal and taking temp regulator valves out of the equation, a higher flow rate should result in a more efficient heat transfer at a lower overall average temperature at the expense of pump power. However, there is a case where increasing the design coolant velocity results in a larger cooler design (explained below).

When considering how fluid flow rate affects cooling system efficiency, the main things to look at are turbulence, temperature differential, and pressure drop.

Increasing coolant velocity increases turbulence which increases tube-side heat transfer coefficient (which is one component of the overall heat transfer coefficient). Since coolant is not very viscous and is likely already in a turbulent flow regime this provides only a marginal increase in cooling efficiency. 50/50 EG/H2O coolant to air coolers typically have most of their thermal resistance on the air-side and so are more sensitive to airflow velocity but not very sensitive to coolant velocity changes.

On the other hand, oil is more viscous and so increasing the oil flow rate results in a less laminar and more turbulent flow regime, which can have an enormous impact on cooling efficiency. Oil to air coolers typically have more of their thermal resistance on the tube-side, so varying the airflow has less effect, but increasing oil flow rate can greatly increase thermal transfer efficiency thereby reducing the required size of the cooler. Since velocity and viscosity equate to pressure drop, a cooler design with more pressure drop will generally be more thermally efficient at the expense of pump power. 3-5 PSI differential across the cooler is normal for 50/50 EG/H2O and 10-50 PSI is normal for oil (specific applications may vary).

The other thing to consider is that changing the flow rate will change the temperature differential. For a fixed heat load, doubling the flow rate will cut the temperature differential across the cooler in half. All things being equal and disregarding temp control valves: a smaller temperature differential across the cooler means that the average coolant temp (EMTD) will drop closer to ambient; however, in designing the cooler, the heat exchanger application engineer must fix either the inlet or the outlet temp. If the cooler inlet temp is fixed, a higher flow rate will result in a higher EMTD and thus a smaller cooler design. If the cooler outlet temp is fixed, a higher flow rate will result in a lower EMTD and thus a larger cooler design. How big of an effect this has depends on how close the fluid temp is to ambient.

Take for example an Engine Jacket Water (EJW) section and a Turbo Aftercooler Water (TAW) section designed for 110 F ambient air temp:

The EJW is to be designed for 190 F at cooler inlet. At a given heatload and flow rate, let's say the cooler outlet temp is 170 F. If you double the design flow rate, the outlet temp is now 180 F. Since the average temp above ambient increased by about 8% and you get a little tubeside bonus from increased turbulence, the cooler can be designed maybe 10% smaller.

The TAW is to be designed for 130 F at cooler outlet. At a given heat load and flow rate, let's say the cooler inlet temp is 160 F. Doubling the design flow rate will reduce the design inlet temp to 145 F. The average temp above ambient decreased about 22%, but you got a tubeside bonus from the increased turbulence so your cooler design might be about 16% larger.

 

[MRDAGERUS]

Thank you all for your great responses.
No, I was not arguing w/my boss.
It was a Motorsport enthusiast, with whom I was discussing details of a water to air inter coolers.

Mater artium necessitas