heat load to warm up a truck

[JP123]

Hi,
I'm trying to find the heat load to warm up a 70 tons truck within 8 hours, from -40 deg C to 15 deg C. I found 246 kW*h. Is that ok?

How the heat should be distribute on the truck?

Any ideas?

Thanks!

 

[gepman]

First there is a difference between a rate and a quantity. kWh is a quantity and kW is a rate. If the total amount of heat that you require is 246 kWh over the 8 hour period then 246/8=rate in kW. This is about 31kW.

Second even if you know the rate at which you need to apply the energy it still does not mean it will heat up that fast since you have to use heat transfer equations to see if the heat will be absorbed by the truck at this rate.

The simplest way to calculate the total quantity of energy is to use the equation Q=mCp deltaT where Q is the total amount of energy, m is the mass, Cp is the heat capacity of the material (I would assume most of the mass is steel in your truck), and delta T for your case is 15 - -40 = 55 deg. C.

The heat transfer equation(s) will be in the form Q=UA deltaT where U is the overall heat transfer coefficient and A is the surface area. It will actually be a very complicated equation given the shape and thickness of the truck.

Distribute the heat over all the surfaces. I would assume this is some type of large truck in the artic. Probably a combination of radiation (infrared) and convection (forced hot air) would be the best.

 

[IRstuff]

70ton*[0.5kJ/(kg*K)]*55K = 485kWh

TTFN

FAQ731-376

 

[imok2]

If I have a truck of 70 tons X 2000 lbs = 140,000 lbs x 0.122(SP Heat)btu/lb/*F = 17,080 btu/*F x 131*F = 2,237,480 btu/8 hours = 279685 btu/hr/3412 btu/kw = 82kw/hr.
Please show me where I'm wrong in my caculations

 

[IRstuff]

The specific heat value you're using is about 2% higher than the one I'm using.

But your delta T of 131ºF is 72.8ºC, which is 30% higher than the stated delta T of 55ºC.

TTFN

FAQ731-376

 

[ChrisConley]

My guess is that 70 tons refered to capacity, not to truck weight.

 

[imok2]

So delta T should be 99*F or 40+59*F is that correct?
then 17080 btu x 99*F = 1690920 btu/8hrs = 211369btu/3412 = ~62kwh Thanks IRstuff

 

[imok2]

Well Chris he did say "to warm up a 70 ton truck" so your guess is as good as mine. The problem with many of the questions is that we have to make assumptions unless we are prepaired to ask a lot of questions. what?

 

[IRstuff]

Sure, lots of swags all around.

The calculated heat flow requires an infinitely conductive object. If there's a lot of thermal resistance, then you'd need a substantially higher heat flow to make up for the thermal resistance.

TTFN

FAQ731-376

 

[ChrisConley]

I ran into this situation with a structural engineer (fairly new to the industry) I thought he was going to have a heart attack when I told him I was putting a 350 ton chiller on the roof...

 

[imok2]

I ran into this situation with a structural engineer (fairly new to the industry) I thought he was going to have a heart attack when I told him I was putting a 350 ton chiller on the roof...
Now that's funny!!! Thanks for that Cris

 

[JP123]

the thing is that I don't want water to freeze on the truck when it's washing.

I think that some electric heaters blowing hot air on the truck will be appropriate. But what is the kw and cfm needed?

 

[IRstuff]

Didn't you pretty ask the same question here: thread391-197107 ?

TTFN

FAQ731-376