# LMTD is lower and heat transfer is more

1. May 3, 2014

### knight92

I have tested two radiators with 90 degrees celcius of hot water going through it. The radiators were placed in a wind tunnel with the ability to control air speed.

The bigger radiator cools the water more than the smaller radiator and the air temperature rise after the air has gone through the bigger radiator is also more than that of the smaller radiator.

That means I have a bigger heat transfer rate for the bigger radiator at say 10 ms-1 of air speed than the smaller radiator however calculating the LMTD and corrected LMTD shows that the bigger radiator has almost half the LMTD values for the smaller radiator. The smaller radiator however has less cooling effect on the hot water going in.

I thought the larger the LMTD value the more heat is transferred but my results are contradicting themselves. Its true the bigger radiator is dissipating more heat but LMTD values oppose the reality ?

I am confused about what is going on. Can there be any reason why the LMTD values might be lower for the bigger radiator ?

Equation for LMTD I used:

T(LMTD) = ( (Air In temp - Water outlet temp) - (Air out temp - water inlet temp) ) / LN ( (Air In temp - Water outlet temp) / (Air out temp - water inlet temp) )

One of the results for bigger radiator:

Coolant/Water side:
Water inlet temp = 363.15 K
Water outlet temp = 325.15 K

Air side:
Air in temp = 295.15 K
Air out temp = 335.15 K

Thank you.

2. May 3, 2014

### Staff: Mentor

The LMTD was derived based on counter current flow in a heat exchanger. In your situation, you have air cross flow perpendicular to the radiator tubes. So the LMTD is not the appropriate temperature difference to use. You need to rederive the differential heat balance, and then integrate it analytically from one end of the radiator to the other to figure out the correct mathematical form of the temperature difference to use.

Chet

3. May 3, 2014

### knight92

can you please tell me how do I rederive the differential heat balance ?

4. May 3, 2014

### knight92

I have not come across the differential heat balance equation. Searching on google gives many equations I am not sure which one you are referring to ?

5. May 3, 2014

### Staff: Mentor

Before I do that, please try something else first. In your equation for the LMTD, try using the inlet air temperature in both terms. I'd like to see if this makes a difference in your interpretation.

Chet

6. May 3, 2014

### knight92

by both terms do you mean using inlet temperature of air for water aswell ? another thing I am confused about is my LMTD values are negative. so for the temperature values I wrote above the LMTD value for the bigger radiator is -28.99 and for the smaller radiator is -39.83 for temperatures of:
Water inlet temp = 363.15 K
Water outlet temp = 334.48 K

Air side:
Air in temp = 295.15 K
Air out temp = 323.15 K

Anyway inputting inlet air temperature for both inlet temperature terms I got a value of +38.99 for LMTD of bigger radiator as opposed -28.99. Also the Corrected LMTD becomes infinite if I do this as it is divided by zero because of the term called 'S' for calculating the correction factor becomes 0.

where S = (Water outlet temperature - water inlet temperature)/(Air inlet temperature - Water inlet temperature) and 'S' is a measure of the temperature efficiency of the heat exchanger.

Last edited: May 3, 2014
7. May 3, 2014

### Staff: Mentor

The differential heat balance goes is done on the section of radiator pipe between location x and location x + dx. It yields:

$$WC_p\frac{dT}{dx}=-πDU(T-T_a)$$

where W is the mass flow rate of water, Cp is the heat capacity of the water, T is the temperature of the water at location x, D is the diameter of the pipe, Ta is the inlet temperature of the air being blown over the tubes, and U is the overall heat transfer coefficient. The initial condition on this equation is T = Twi at x = 0. The solution to the equation is:

$$\frac{T(x)-T_{air}}{T_{win}-T_{air}}=exp\left(-\frac{πDxU}{WC_p}\right)$$

$$Q=WC_p(T_{win}-T(L))$$

From the previous equation,
$$WC_p=-\frac{πDLU}{\ln\left(\frac{T(L)-T_{air}}{T_{win}-T_{air}}\right)}$$

Combining the previous two equations gives;

$$Q=\frac{πDLU(T_{win}-T(L))}{\ln\left(\frac{T_{win}-T_{air}}{T(L)-T_{air}}\right)}=πDLUΔT_{LM}$$
where
$$ΔT_{LM}=\frac{((T_{win}-T{air})-(T(L)-T{air}))}{\ln\left(\frac{T_{win}-T_{air}}{T(L)-T_{air}}\right)}$$

Chet

8. May 3, 2014

### knight92

by radiator pipe do you mean one of the tubes in the radiator through which water passes within the radiator ?

9. May 3, 2014

### Staff: Mentor

Yes.

10. May 3, 2014

### Staff: Mentor

For this data:
Water inlet temp = 363.15 K
Water outlet temp = 334.48 K

Air side:
Air in temp = 295.15 K
Air out temp = 323.15 K

I would use: LMTD = (363-334)/ln((363-295)/(334-295))

But, note that the heat transfer coefficient is going to be different for the two radiators, as are the tube diameters, the tube lengths, and the water flow rates. So you have to expect that the LMTD is not going to tell the whole story, as far as the heat load is concerned.

chet

11. May 3, 2014

### knight92

one of the reasons why I wanted to find LMTD was to use the equation Q = U×A×LMTD to find the overall heat transfer coefficient, U. This is because I have all the data I need to calculate heat transfer, Q = Volumetric flow rate x Density x Specific heat capacity x ΔT. Since you say LMTD wont tell me the whole story, would using the LMTD in the heat transfer equation be wrong ? Thanks

12. May 3, 2014

### knight92

Also calculating the overall heat transfer coefficient, U for the radiators I get a larger value for U for the bigger radiator as compared to the smaller radiator which seems to be right since U depends on the geometry of the radiators too.

13. May 3, 2014

### Staff: Mentor

No. What you are trying to do is right on target. Just use the LMTD equation I gave, rather than the one you were using. There are good strategies for experimentally determining the effect of water flow rate and air flow rate on the overall heat transfer coefficient by carrying out experiments over a range of these parameters and expressing the overall heat transfer coefficient in terms of separate contributions from the air side, the water side, and the tube wall. Probably, the main resistance to heat transfer is going to be on the air side.

Chet

14. May 3, 2014

### knight92

Thank you for your answer. I just have one last question you said I would use: LMTD = (363-334)/ln((363-295)/(334-295))

Which tells me that Twin = Temp at water inlet
T(L) = temp at water outlet
Tair = Temperature of air at inlet

Am I right ?

15. May 3, 2014

### Staff: Mentor

Yes.

16. May 3, 2014

### knight92

Thank you for your help. Can you just tell me where you know the first two equations from as I need to reference those ?

17. May 3, 2014

### Staff: Mentor

A good reference on heat transfer is Transport Phenomena by Bird, Stewart, and Lightfoot. You can find the derivation of these equations there. Another good reference is Heat Transmission by McAdams.

Chet

18. May 3, 2014

### knight92

Great, Cheers.