re: maths

Not without some material context. I'm guessing this is for something specific like, for example, a certain grade of coal? Tell us a little more please.

I do not understand your question: he asks ONLY for UNITS which totally independent of the kind of stuff materially speaking.

He makes an error since energy is in kWh and not in kW/h.

In fact he has only to look in units conversion tables via google where he finds the equivalence between BTU and J between lb and kg and so on.

For help MJ= 1000000J and 1kW=1000 w 1 w= 1 Nm/s = 1J/s

It's obvious he's looking for a conversion that requires knowledge of the combustion energy per mass of some specific substance.

It is obvious that he asked only to convert from US units to metric units and this conversion has noting to do with the process it can be burning, an other chemical reaction endo or exothermic.

sorry about the confusion, what I am trying eventually to do is see if it is worth while to power a steam turbine with the boiler fired by hydrogen for a boat, with the equivalent gasoline engine rated at 500hp, but as you can see I don't really have a great understanding of the maths or the route involve.

Regards

Terry

I would like to ask the group a similar question.

I have been trying to find a formula for how many watts of power or btus are needed to heat a specific weight of material to a specific temperature. I know that a watt is defined as a joule per second. Yet, I have not figured out how to convert power expended to a tempature. To put it simply, if a resistor is dissapating 5 watts, is there a way to calculate the temperature rise produced?

Tom Cornell

The answer is yes. You have only to make an energy balance:

the energy supplied by the resistor goes into the mass of material you heat which is able to accumulate it according to its physical properties. Since the piece you heat is not totally isolated from the environment when its temperature rises heat is also transmitted to the environment in general proportional to the contact area between the heated body and its surroundings.

It is a dynamic process and it is in the transient phase till a steady state is reached by a differential equation.

In the steady state the power coming in is equal to the power going out:

Pin (w)= S(m^2)x K(w/(°C x m^2) x (tbody - t environment)

you can thus compute the body temperatures as:

tbody = tenv + Pin/(S x K )

Thank You,

I have a tech degree so I understand the basics. But I am not clear on how your formula works. I am not sure what quantity is s and what is k. if you could explain this a little more I would appreciate it.

I realize that there are losses and finding an exact tempature is very difficult. This is not the point. If I have a heater or in this case a resisitor disapating a fixed amount of watts, we know the energy in joules. We as know the mass of the object and things like density and composition. It seems that if I am putting X amount of joules into a body, there must be a way to figure out the temperature rise. I seem to remember something from physics about the amount of energy needed to rase tbe tempature of water by one degree. There must be some analogy here.

Thank You.

Tom Cornell

Tom,

There are 2 parts to the answer.

The 1st part is "not exactly", i.e., you might be able to say the temperature is 150°F +/- 10°F, but you'll never be able to say that it is 151.2°F. How close you get depends on how carefully you set things up and control the experiment.

The 2nd part is figuring out how to do the calculation. There is a a fairly complicated, math-intensive way to do it, and then's there the quick and dirty way that most circuit board people use when they're doing small numbers of boards and can't afford a two-week FEA. Here, for example, is a link to one resistor manufacturer who gives a graph for the temperature rise.

http://www.ohmega.com/DesignResistor.html

Attention: the omega indications concern ONLY a flat resistor on a printed board. In such a case the dissipated heat goes per convection into the air and part of it goes by conduction in the board and is transfered to air by the other side . It is not the case you consider.

If you look at a round resistor it has a surface = ∏xdxL where d is the resistor diameter and L its length. Every type of resistor is rated for a power dissipation and dimensions are determined so that the resistor value will not change too much when used at full power. Consider the basics i gave to you in the other message.

Nick Name,

Thanks for the reply. Generally when I think of resistors, I think of circuit boards and I answered accordingly.

Most resistors on pcbs are flat these days but, even if I use a relatively small round resistor (so that it lies very nearly against the board - the usual way), the effective radiating area (which is the area used in thermal calculations) is about the same for flat as for round since you basically use the x-y projection of the round ones.

But still, when all is said and done, thermal calculations remain only an approximation. You still have to put the finished board in a long term test fixture and actually measure temperatures on all the components (although most people do an IR picture and then pick which ones to measure).

I'm not disagreeing with you. What you said was fine, but that's not the way many people do resistor temperature rise in a design environment.

http://www.simetric.co.uk/