# Heat transfer: lumped system

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1. Jul 28, 2016

### JohnJohn8

Hello all,

I have a question regarding heat transfer. When the Biot number is small (<0.1) a mass can be regarded as lumped. Which means that the temperature is the same everywhere in the mass. Now if the Biot number is larger than 0.1 (say around 0.5-0.8) and I still do a lumped system analysis. Is there a relation which gives the error? And is there like a factor which compensates for this error?

2. Jul 28, 2016

### Mech_Engineer

The Biot number is used to determine if the lumped capacitance method is a valid approximation of a transient problem that involves convection about a solid. When the Biot number is much less than 1, the resistance to conductive heat transfer within the solid is much less than the resistance to convective heat transfer across the fluid boundary layer.

Put another way, if the Biot number is much less than 1 then the temperature gradient across the solid is much less than the temperature gradient across the fluid boundary layer. Given two temperature gradients, if one is much larger than the other then the smaller one can be assumed to have a negligible effect on the system, and so it may be disregarded.

To answer your question, I'm not aware of any "correction factor" for using the lumped capacitance method with larger Biot numbers. In the case that you have a larger Biot number and accuracy is critical, you may be better off using the conduction equation to calculate the temperature gradient in the solid...

3. Jul 28, 2016

### Staff: Mentor

There is a trick way to do what you want to do. If the Biot number is on the low side (as in the cases you have identified), then the surface temperature is changing very slowly, and thus the heat flux to the surface is changing very slowly. So the object is receiving nearly a constant heat flux over long intervals of the process time. You can look up the asymptotic internal Nussult number for your object under constant heat flux conditions. This is tabulated in many books. You then use the corresponding internal heat transfer coefficient for the object in conjunction with the external heat transfer coefficient from the Biot number to obtain an overall heat transfer coefficient. You then use this overall heat transfer coefficient with your lumped parameter model. Try this out and see how the results compare with the results in books. I think that you will be very pleased that it gives pretty accurate results over a much larger range of Biot numbers than obtained by just including the external heat transfer coefficient, and with very little additional effort.

4. Jul 29, 2016

### JohnJohn8

Thank you very much for your answer, but what do you mean by the internal asymptotic Nusselt number?

Last edited: Jul 29, 2016
5. Jul 29, 2016

### Staff: Mentor

In the case of conductive heat transfer to a cylinder (for example) under a constant wall heat flux, the temperature gradient at the wall at any instant of time is related to the heat flux at the wall by:$$q=-k\left(\frac{\partial T}{\partial r}\right)_{r=R}$$It is also possible to define an instantaneous internal heat transfer coefficient h for this system by:$$q=h(T_w-\bar{T})$$ where $T_w$ is the instantaneous temperature at the wall at time t, and $\bar{T}$ is the instantaneous average temperature of the cylinder. The instantaneous internal Nussult number at any time t can be defined as:$$Nu=\frac{hD}{k}$$ In this system, after a relatively short time, the difference between the wall temperature and the average bar temperature $(T_w-\bar{T})$ is found to a approach a constant (asymptotic) value. This then implies that h and Nu also approach constant values. If I remember correctly, for conductive heat transfer to a cylinder under a constant wall heat flux, the asymptotic internal Nussult number is equal to 48/11.