# Thread: Differential Equation for an engineering situation

1. ## Differential Equation for an engineering situation

During the manufacture of a steel component it is often necessary to quench them in a large bath of liquid in order to cool them down. This reduces the temperature of the component to the temperature of the liquid. If T is the temperature of the component in excess of the liquid temperature, the rate of change of the component proportion to the temperature of the component. Take the proportional constant as K. K depends upon the volume and surface area of the component, its specific heat capacity, and the heat transfer coefficient between the component and the liquid.

a) Formulate a differential equation for the above engineering solution

b) Determine a solution for the formulated formula using numerical methods and power series, the initial condition that t+0 the temperature excess is 200.

2. ## Re: Differential Equation for an engineering situation

What ideas have you had so far?

3. ## Re: Differential Equation for an engineering situation

Well; Newton's law of cooling states that the rate at which a hot body loses heat to its surroundings is proportional to the temperature difference between the body and its surroundings. So, if the instantaneous temperature of the body is θ(t) and the temperature of the surroundings is θs (assumed constant)... we end up with something like:

dθ/dt = -k(θ-θs)

Where k is a positive constant.

If that is correct, im not sure what to do from there...??

4. ## Re: Differential Equation for an engineering situation

You have the right idea, but you should formulate your differential equation in the variables specified by the original problem statement. Once you've done that, you'll have finished part a, I think. What about part b?

5. ## Re: Differential Equation for an engineering situation

I guess in part b) i am required to test my formulated formula with the methods stated, starting at a temperature of 200?

so would part a be sometimg like:

dT/dt = -K(T-Ts)

6. ## Re: Differential Equation for an engineering situation

As for your part a, I think the problem is just a hair trickier. T already is the difference between the steel component and the liquid bath. How does that change your formulation?