Science question involving coolant through a motor vehicle

Lady Law

Hi all.

Need some help asap with an exam question in automotive science,

Coolant is passed through a motor vehicle at the rate of 28 litres per minute. The coolant enters the radiator at a temperature of 95 degrees celcius and exits at 75 degrees celcius. If the specific heat capacity of the coolantis 4.2 kJ/kgK, determine the quantity of heat given uo to the cooling air per minute ( take the value of the relative density of the coolant as 1)

Stallingrad

Does the water pump have a plastic or metal impeller?

Alanstrainor

What is this? The homework forum?!

I actually think I know the answer too:D

recyclebin

This reminds of first year of college and thermodynamics. I have forgotten most of it now. Try the engineering forum as you may get more help there

keithclancy

You better hope the guy correcting your paper doesn't google your answer ... boards.ie usually comes up first :pac::pac::pac:

recyclebin

I think the answer is 4.2*28*20 = 2352 kJ

langdang

recyclebin wrote: »

I think the answer is 4.2*28*20 = 2352 kJ

Ah now, you're assuming that solving the problem has nothing to do with cars or engines at all.;)

I say we can't answer this without the response to stalingrads impeller question!

Marcusm

If it enters at 95 and exits at 75 surely it is heating rather than cooling the air or have you changed the laws of thermodynamics?

WillyFXP

Marcusm wrote: »

If it enters at 95 and exits at 75 surely it is heating rather than cooling the air or have you changed the laws of thermodynamics?

The coolant is going from 95 to 75, the question relates to the amount of heat being transferred to the air cooling the radiator.

Lady Law

The question doesn't say anything about the type of pump...thanks for the comments so far. I've tried searching on the internet for anything that could help me with the calculations but all I got was how the system itself works. I understand it just can't apply it without any formula.

Joe 90

Lady Law wrote: »

The question doesn't say anything about the type of pump...thanks for the comments so far. I've tried searching on the internet for anything that could help me with the calculations but all I got was how the system itself works. I understand it just can't apply it without any formula.

I think the idea of the question is that YOU work out the formula. You got the volume, you got the SG or relative density, you got the specific heat, you got the change in temperature so all you have to do is tie it all together.
Your doing something that demands that you think rather than learn off by rote.

hi5

3000 joules?

Spook_ie

Mines an old air-cooled Volkswagen

kirving

Here you go!

If someone else could check the calculation and confirm it's correct.

Man of Aran

I think Kevin, the message to OP from other respondents was not to spoonfeed but get him/her to use own reasoning to solve for Q.
Guide them through it only, given that Specific Heat Capacity for water /kg was provided : the dT in Water in /out was provided as was Volume flow but so too was Density so volume flow could be converted to mass flow to plug into the SHC*Kg*dT formula.
Getting <units of measure> as in < minutes v seconds > or <litres to KG> or <1degree C= 1DegK> etc is all part of the reasoning too.

Mr Benevolent

39kw seems like a ridiculous amount of heat. Imagine standing next to a 39kw electric bar heater...

langdang

Yeah, the units practically spell it out for you.
If it was a race for the first correct answer recyclebin got there first through aptitude and common sense.

kirving

I know, but I already had the question done out as part of something for college, so it was just a case of changing the numbers.

I agree that students shouldn't be spoonfed but sometimes a more complete answer makes the whole thing click in your head.

Man of Aran

Ask that person OP in a week and I bet they may still not recall or understand how to arrive at the final value of Q.
Its the old saw "feed a man a fish versus teach a man to fish" applies IMO.

SilverBell

Confab wrote: »

39kw seems like a ridiculous amount of heat. Imagine standing next to a 39kw electric bar heater...

Its a lot of heat, but I think its pretty close. Please check if I have made a mistake in these rough and ready calc.

My car does 28mpg at around 60mph.
Thats 6.17 miles for 1 litre of fuel.
My speed is 1 minute per mile, so 6 minutes gets me 6 miles and uses 1 litre of fuel approx.

Say 40MJ in 1 litre of fuel. So I have burned 40 MJ of energy. Thats power of 40MJ/360s = 116.7kW But only a percentage of that goes to into tractive effort.
Depending on the efficiency of the engine (say 35%) that leaves 65% rejected to atmosphere as heat.
This is about 76kW for my rough calculation. Thats seriously high I thought, so I checked it using the drag equation, and the answer is probably somewhere in the middle.
Merc 300CE Cd is 0.36, and the frontal area is 2.369m^2.
using a rolling resistance of 1.1 I worked out it takes about 17kW for my car to do 60mph. This means around 34kW may be rejected to heat. So my fuel figures may be a bit off as I dont travel at 60mph constantly, or else my drag formulae is a bit low, but I think it gives a rough idea.

Joe 90

Confab wrote: »

39kw seems like a ridiculous amount of heat. Imagine standing next to a 39kw electric bar heater...

Surprisingly it seems pretty reasonable. 39Kw is about 52 bhp. If we are really lucky about a third of the energy in the fuel turns into mechanical energy and the rest turns into heat. Now some of the heat goes out by way of the exhaust and some by air flowing around the engine but most via the rad. So to keep it simple lets say it all goes through the rad and say 52 bhp as heap so 26 bhp at the flywheel. Allow a bit for transmission losses and it all looks very reasonable.

PS. Where I get really suspicious about claimed power losses is when people quote x bhp at the wheels on a rolling road is really x multiplied by some-silly-factor at the flywheel due to transmission losses. I ask just how many electric fires there are in the transmission.:)

Lady Law

Thanks for that. Can make sense of it now

hi5

Thats a lot of wasted energy, why not use the heat to drive a small steam turbine which could contribute to the motion of the car?

07Lapierre

hi5 wrote: »

Thats a lot of wasted energy, why not use the heat to drive a small steam turbine which could contribute to the motion of the car?

The Internal Combustion Engine has always been very inefficient. If memory* serves me correctly, a "good" engine transfers 3 - 4% of the heat energy created into motion. the other 95-96% is dissipated through the cooling system.

I'm sure a hybrid engine is a lot better as it used the braking forces to recharge the batteries, which would greatly improve things.

* note: I've a notoriously bad memory!

SilverBell

hi5 wrote: »

Thats a lot of wasted energy, why not use the heat to drive a small steam turbine which could contribute to the motion of the car?

Again a steam turbine is about 30-40% efficient, so 60 % of the waste heat would be lost again to atmosphere.

Add to that the bulk of the engine heat is around 80 or 90 degrees centigrade. So you couldnt use water ( practically cant make steam of it), so perhaps you'd have to use an alternative fluid, therefore more expense or complexity.

You'd have to find some way of rejecting all the heat from the turbine condenser. In a car, that would be air. Ambient temperatures are approx 20 degrees, so the max temperature difference in this situation is about 70 degrees. Carnot efficiency would be about 19%, (no matter what ideal fluid you could use)
The condenser would also need to be very large.

So for a lot of extra mechanical stuff under the bonnet, you'd only get about another 19% of the heat remaining in the fluid. In practice it would be a lot lower than 19%.

SilverBell

07Lapierre wrote: »

The Internal Combustion Engine has always been very inefficient. If memory* serves me correctly, a "good" engine transfers 3 - 4% of the heat energy created into motion. the other 95-96% is dissipated through the cooling system.

I'm sure a hybrid engine is a lot better as it used the braking forces to recharge the batteries, which would greatly improve things.

* note: I've a notoriously bad memory!

Its more between 30% and 40%....:)

07Lapierre

SilverBell wrote: »

Its more between 30% and 40%....:)

Wikipedia agrees with you:

Most steel engines have a thermodynamic limit of 37%. Even when aided with turbochargers and stock efficiency aids, most engines retain an average efficiency of about 18%-20%.[9] Rocket engine efficiencies are better still, up to 70%, because they operate at very high temperatures and pressures and can have very high expansion ratios.[10]

Popoutman

Joe 90 wrote: »

PS. Where I get really suspicious about claimed power losses is when people quote x bhp at the wheels on a rolling road is really x multiplied by some-silly-factor at the flywheel due to transmission losses. I ask just how many electric fires there are in the transmission.:)

Yep.
Though most cars I've driven do get hot gearboxes after a journey.

Done right, the rolling road determination of the drivetrain/etc losses can be done to quite a reasonable accuracy, if at the max rpm point reached the driver clutches and leaves the drivetrain slow down (as it would be driven by the freewheeling of the rolling road rollers) in an unpowered state. The rate that the rollers slow will be indicative of the friction in the car excluding the engine. This friction will be very very close to the same friction as under acceleration, the only real difference being different sides of the gears being in contact. The fudge factor should never be a simplistic multiplication factor - it should always be done on overrun.

The OP's exercise's numbers are pretty well thought out, and reasonable. You'd only need (ballpark) maybe 20bhp to run a car at 100kph, and if that were true you'd need slightly more than 240 bhp to run the same car at 200kph (aero drag increasing as the cube of speed increase) and at 100kph the car is likely to be dumping between maybe 45bhp (for a good diesel) and 80bhp (for a poorly tuned petrol) worth of heat to the surroundings.

I love these type of questions. Makes you think, or at least they should.

SilverBell

Popoutman wrote: »

Yep.
I love these type of questions. Makes you think, or at least they should.

Exactly! From school physics, engineering, and other interests its great to actually sit down and try to think things out in that way!

A friend of mine told me about his father wanting to mount a wind generator onto his mini which would then drive a motor to drive the car. The faster the mini would go, the more power would be generated and the faster the car would go again. My mate couldn't convince him it wouldn't work! :pac:

T-Maxx

42

Popoutman

SilverBell wrote: »

A friend of mine told me about his father wanting to mount a wind generator onto his mini which would then drive a motor to drive the car. The faster the mini would go, the more power would be generated and the faster the car would go again. My mate couldn't convince him it wouldn't work! :pac:

Though it is possible to design and run a rotor that would allow a wind-propelled car to move at a faster speed than the wind is blowing at. Slightly different, but certainly attainable none the less. See here: http://blueplanettimes.com/?p=3971

Makes sense in my head, justabout