# ODE, step change, Heat transfer

1. Apr 23, 2005

### Dynamo

Hi all.

I am doing some work with temperature equations. I have a book that gives an equation and then manipulates it. However, I can not follow what the author does, so can anyone help:

He starts with:

dT/dt = Q/MC - (T-O)/RMC

(in the following I use the text "delta" to represent lower case delta)

He then says he solves this for delta T (change in temperature), due to a step change in source temperature (delta O).

The equation then becomes:

delta T(t) = delta O (1- e^(-t/RMC))

I use this 2nd equation in my work, but just cant follow how the author jumps from the 1st eqn to this one.

I 'solve' the original ODE - but it comes no where near what he gets.

Any help would be great.

2. Apr 24, 2005

### saltydog

This is what I get:

For:

$$\frac{dT}{dt}=\frac{Q}{MC}-\frac{(T-O)}{RMC}$$

I get:

$$T(t)=RQ+O-Ke^{\frac{-t}{RMC}}$$

with K the integration factor.

Then the differential of T with respect to O is:

$$dT=\frac{dT}{dO}\Delta O$$

$$dT=\Delta O$$

See, not happening for me. Perhaps someone can help us.

3. Apr 26, 2005

### Dynamo

Thanks for having a look.

Thats pretty much what I get..

From: $$\frac{dT}{dt} = \frac{Q}{MC} - \frac{1}{RMC} (T- \theta)$$

I then diff with respect to theta, and get what you get,

I cant understand how he, and I quote...

"This equation can be solved for the change in temperature $$\delta T$$ due to a change in the temperature of the medium $$\delta\theta$$.

The result for a unit step change in $$\theta$$ is:

$$\delta T(t) = \delta \theta (1-e^{\frac{-t}{RMC}})$$"

I just can't get this (although I know that the first equation is correct, and so is the final equation - the 2nd equaion I have used a lot in my work, and it is correct).

Anyone else got any ideas?

(PS just realised the form can use latex).

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