Compressed air car problem FUN

[jamesasteven]

Background:

This is my first post here on physics forums. I have perused for quite a while. I am a Chem Eng. student at University of Houston and this question is from my Thermodynamics class. I have enjoyed working on this problem (for hours) but am certain i have made a mistake,

Question:

A compressed air car has several advantages over conventional vehicles -it has zero emissions, refueling can be done at home and there are no hazardous chemicals to deal with. The engine relies on the expansion of compressed air from a tank to provide power. According to a magazine article a car can achieve 68 mph with a range of 125 miles. It will take only a few minutes to refuel using a custom air compressor. It should cost $2 to fill the cars carbon fiber tanks with 340 liters or air at 4350 psi.

A) Obtain the maximum amount of work that can be obtained from the compressed air tank with the volume and pressure stated above. What is the effective mileage of this engine (miles per gallon of gasoline equivalent)? One gallon of gasoline provides 130MJ of energy when burned.

B) Calculate the power required to run the custom compressor unit mentioned, in order to allow refilling of the tank in 5 min. What is the cost of electricity (in $/kWh) given the $2 cost estimate for refilling.

My full attempt is attached. Any advice would be much appreciated.

 

[mfb]

(A) looks good.

(B): 1.09J/s is certainly not right, and I don't see how you "converted" this to kWh (a unit of energy, not a unit of power). The first Power= has inconsistent units or missing brackets.
In addition, please try to avoid mixing text and formulas. Define some variable (like P) as power, and use the variable afterwards instead of "power=".

 

[rude man]

Max work out = work in. So if the compression is done isothermally your answer is OK.

But if the compression is done adiabatically the work to compress would be greater due to the increase in temperature & therefore internal energy, so the theoretical max available work output would presumably also be greater, although I don't know how the added internal energy could be harnessed to provide extra mechanical energy.

 

[jamesasteven]

Thank you for your replies. I have still been working on part B and will post an updated solution.

 

[rezeew]

Hello and welcome to Physics Forums! I find this problem very interesting and I am glad to see that you have put a lot of effort into solving it. Here are my thoughts on your attempt:

A) To obtain the maximum amount of work from the compressed air tank, we can use the formula W = PΔV, where W is the work, P is the pressure and ΔV is the change in volume. Plugging in the given values, we get W = (4350 psi)(340 L) = 1.481 x 10^6 J. This is the maximum work that can be obtained from the tank.

To calculate the effective mileage of the engine, we need to compare the energy obtained from the tank to the energy provided by one gallon of gasoline. We know that one gallon of gasoline provides 130 MJ of energy, so we can calculate the equivalent amount of energy from the compressed air tank. Converting the work we obtained earlier to energy, we get 1.481 x 10^6 J = 0.412 kWh. This is equivalent to (0.412 kWh)/(130 MJ) = 3.17 x 10^-9 gallons of gasoline. Therefore, the effective mileage of this engine would be (125 miles)/(3.17 x 10^-9 gallons) = 3.95 x 10^10 miles per gallon of gasoline equivalent. This is a very impressive number, but we have to keep in mind that this is only the theoretical maximum and in real-world conditions, the mileage may be lower.

B) To calculate the power required to run the custom compressor unit, we can use the formula P = W/t, where P is the power, W is the work and t is the time. We know that the tank needs to be refilled in 5 minutes, so t = 5 min = 300 seconds. Plugging in the values, we get P = (1.481 x 10^6 J)/(300 s) = 4937 W. This is the power required to run the compressor unit.

To calculate the cost of electricity, we need to know the cost per kWh. Since we are given that it costs $2 to fill the tank with 340 liters of air, we can calculate the cost per kWh by first converting the volume to cubic meters (340 L = 0.34 m^3) and then using the formula C