Solving the Heating Problem of the Transparent Car

 One major consideration in having a transparent car in Malaysia is the amount of heat that can get in. People spend a lot of money to darken their glass windows to reduce the amount of heat going into the car.

Many are actually more interested in privacy than in cooling. For these people, they may just as well cover their windows completely with curtains or paint. It is not allowed in Malaysia even at the side windows.

Putting solar panels will reduce the amount of heat but solar panels only absorb 20% of the incoming energy, converting it into electricity. We still need a lot of energy to remove the rest of the heat. This is done by using the air-conditioning system after rolling down our windows.

For the Transparent Car, we just do the same thing. Cover our cars and turn on the air-conditioning. If we were to participate in the World Solar Challenge, we would be penalised heavily. That is why participants tolerate 50 degree heat inside their tiny cabins. For the transparent car, it will be even worse because more light gets in. Covering more places will defeat the safety aspect of transparency.

Because of the desire to improve the chances of winning, I have attempted to optimise the energy transfer in the car. We should be able to convert the extra heat energy into useful functions.

The principle of zero aerodynamic drag has always been applied in the design of the Transparent Car. The centrally frontal part of the car is open to allow high pressure air to enter and exit to the low pressure area at the back. Heat from the air-conditioning and cooling systems is used to heat the exiting air to provide extra propulsion energy while also filling the vacuum at the rear, thus reducing aerodynamic drag.

Initially there cannot be any air-conditioning because battery is supposed to be limited, but it turned out that in the cruiser class, battery is unlimited, so air-conditioning is possible. It will be installed for the comfort of the drivers or at least exhaust fans at the back to cool the inside of the car. But these methods just waste the solar energy by throwing them away.

One solution is to store the high air temperature inside a bottle at high pressure and thus increasing its temperature further. It is a method of storing energy which is allowed but need to inform the organiser. The extra energy can be released later when we need them for windscreen wiper and extra propulsion by mixing with the air from the cooling systems.

I did not realise that we can transfer energy by just mixing air. I though we need to provide a radiator, but when I was reading the notes in our Thermodynamics Textbook, I realised that we can mix air and it is a form of heat exchanger also. It should have been obvious but I just missed them completely.

While looking for fans to cool the cabin, I propose a ducted fan because it can produce propulsion by sending air at high speed. Ducted fans are used as jets for RC aeroplanes.

But the ducted fans can become generators also. For the World Solar Challenge, it is better to use them as generator rather than as a cooler because we want to gain the maximum energy efficiency.

The solution is therefore simple. When the car moves, the front port will be open to allow maximum air to mix with the hot cabin. The air thus gains energy that will pust out through the ducted fans and converted to electrical energy. There will be less propulsive force and less reduction of aerodynamic drag. The energy conversion should be better because propulsive force and aerodynamic drag reduction is difficult to achieve because of difficulty in isolating the air that exits to the rear.

Energy conversion should be more productive but it remains to be measured and compared. If the worst occur, we can just switch on the fan to gain cooling effect while getting more propulsive force and less aerodynamic drag. Both options need to be studied and used at the correct strategies during the World Solar Challenge.

To achieve this, the cabin needs to be air tight except at the front and rear parts of the car. No need to open windows, so maybe for this World Solar Challenge version, there is no need to provide sliding windows. Only doors which are tested for convenience points. Maybe we should just provide a small hinged window to pay tolls. For production cars, full sliding windows need to be provided. It will make the windows stronger as well.

Estimated conversion efficiency is given by the Carnot Heat Engine depends on the temperature difference:

Efficiency = 1 - TL/TH  = 1 - (273.15 + 30)/(273.15+50)  = 0.062.

We can only realistically get 5% of the heat energy collected by the body of the car to be converted to electrical energy. Better than nothing.

From Table A18, for Nitrogen:

At 320 K, internal energy, u is 6,645 kJ / kmol

At 300 K, internal energy, u is 6,229 kJ / kmol

With 20 degC temperature difference, the internal energy increased in 416 kJ / kmol.

From Table A2, for Nitrogen:

28.013 kg/kmol

I mole of ideal gas occupies 24 cu. dm, at room temperature, 20 degC.

https://www.chemguide.co.uk/physical/kt/idealgases.html

Instead of using kg. easier to use volume of the car to estimate the total amount of energy.

 1 kmole of air occupies 24000 cu. dm

1m = 10 dm

1 cu. m = 1000 cu. dm

1 kmole of air occupies 24 cu. m of air.

The dimenstion of the car is 1.7 m x 4 m x 1.3 m, but because of its shape it should be less than this rectangular shape. By usingg 1.5 x 4 x 1, we should get around 6 cu. m.

So the amount of internal energy is 416 kJ / 24 * 6 = 104 kJ

Using specific heat of air at constant pressure at 300K, of 1.005 kJ /kg. K,

and total mass of 6/24 kmol of air of 6/24* 28.013 kg=   7 kg,

the total energy gained with 20 degC temperature rise is 1.005 * 7 *20 kJ = 140 kJ.

At constant volue, the specific heat value is 0.718 kJ /kg. K, so internal energy is 100.5 kJ, closer to the value in the table.

Total solar energy per sq. m is estimated to be 1 kW per sq. m.

 With a surface area of 6 sq. m, the amount of energy entering the car is 6 kJ per second.

A solar panel that is 20% efficient only collects 1.5 kJ per second. We can still extract the other 3.5 kJ. The fact that the temperature inside a car is less than 50 degC is due to leaked air.

If we let the cabin to remain at 50 degC: 

The amount of energy collected should be 5% of 3.5 kJ per second. Or 0.17 kW.

If we let the cabin to remain at 70 degC: 

Carnot effficiency is   0.117, so practical efficiency can be 10%.

The amount of energy collected should be 10% of 3.5 kJ per second. Or 0.35 kW.

At least, instead of wasting energy switching on the fan, we can gain energy by using the cooling air which flows from the front of the car.

Energy from aerodynamic drag

This is to ignore the energy collected from the aerodynamic drag as a result of opening the front of the car and back of the car. The chance of collecting the energy is proportional the total frontal surface area that we open up. Too much opening will increase the total drag.

Realistically we can only hope for 10%. 90% loss is due to aerodyanamic drag. Of the 1 kW propulsion power required, 90% is spent to overcome aerodynamic drag. 90% of that should be 900 W, and if we can extract 10%, we can add 90 W to the battery.

Energy from Airconditioning

The principle is the same but instead of mixing the frontal air with the air inside the vehicle, the air is diverted to the rear of the vehicle and collects the heat from the radiator of the air-conditioner.

An air-conditioner is a refrigerator. It extracts heat from a low temperature to a high temperature.

Efficiency of Carnot Refrigerator = 1 / (TH/TL -1))

What happens if the inside temperature is 30 degC and the outside temperature is 30 degC also?

Efficiency is 1/(1-1)  = 1/0 = infinity

That is why it is not called efficiency, it is called COP, Coefficient of Performance

What it means is that there is no need for any energy input to get the heat to flow to the outside.

What if inside temperature, for a refrigerator it is called, TL = 50 degC.

Outside temperature, TH = 30 degC. What the value of COP?

COP.ref  = 1/[ (273 + 30) / (273 + 50) - 1 ] = 1/  [0.938  - 1] = - 16.15

What it means is the, there is no input energy required to send heat from inside the car to the outside. Remember that COP is just a ratio of output / input.

If we operate the aircon such that the cabin temperature is just 2 degree below the outside temperature, 

the COP.ref =  1/[ (273 + 30) / (273 + 28) - 1 ]= 150.5

If the refrigerator is able to handle the large energy input due to the solar energy entering the transparent body, the aircon compressor requires little energy to send that amount of energy to the rear part of the car to be mixed with the incoming air. 

The radiator temperature can be 70 degC because it is not in the cabin with the passengers, so the ducted fan generator can operate with an efficiency of 10 % in converting the solar hear to electrical power.

If the radiator is at 100 degC

The Carnot efficiency = 1 - TL/TH  = 1 - (273.15 + 30)/(273.15+100)  = 0.1875

Let us say the practical efficiency is 0.15.

The amount of charging energy 0.15 x 3.5 kJ / second = 525 W

You need to deduct the amount of energy used by the aircon. If the COP.ref is only 100,

the input power required by the aircon is only 3.5 kW /100 = 3.5 W

I do not thinnk that the aircon can operate with such low power. Even if it were 100 times more,

325 W, then there is still a net 525 W- 325 W= 200 W charging.

Even if there is zero gain, it is still worth it because the aircon does not drain the battery so much,, while allowing the driver and passenger comfortable ride.

It is certainly worthwhile to demonstrate this capability in the solar challenge if we have the time.