Undergrad Car, gas and total energy calculation

Click For Summary
The discussion centers on the principle of conservation of energy in relation to a car's motion and fuel consumption. Participants explore the total energy equation, including kinetic energy and the energy from gasoline and exhaust. Questions arise about how to algebraically represent the energy from gas and exhaust in the total energy calculation, with some suggesting that the chemical potential energy of exhaust is negligible. The conversation also touches on the energy required to accelerate air and the role of exhaust gases in energy calculations. Ultimately, it is noted that a car moving at constant speed on flat ground does not conserve energy, as all energy is lost to various forms of drag and inefficiencies.
jeanmarc69
Messages
4
Reaction score
0
Hi,

What I understood about the principle of conservation of energy: Et = Ep + Ec = constant.
For example: Et = 1/2mv^2 + mgh (h = height).

Consider a car moving at speed v.

For example: Et = 1/2mv^2 + E(gas + exhausts). Indeed, I include the exhausts, otherwise with the drop in the quantity of gas necessary to drive the car, there is an imbalance, therefore no conservation.

My question: What would be the algebra for E(gas + exhaust) in the context of Et?

Thank you for your answers.
 
Physics news on Phys.org
Wow, it's hard for me to decode your post. What are ##E_p## and ##E_c##? What are you trying to calculate? Are you trying to add the chemical potential energy of gasoline into a total energy calculation for a car? If so, the chemical potential energy of the exhaust can be taken as zero. If you have a turbofan driving the car, then I suppose you could include the exit momentum of the exhaust in the energy equations, but otherwise it makes little sense to worry about the exhaust gasses from a car, IMO.
 
jeanmarc69 said:
My question: What would be the algebra for E(gas + exhaust) in the context of Et?
The air, taken in through the air filter of a car, must be first accelerated to the speed of the car. The energy required would normally appear as part of the aerodynamic drag of the vehicle. The fuel in the tank of the car would appear to be part of the mass of the vehicle.

The nitrogen in the air would be heated before being released as exhaust. The oxygen would be combusted with fuel, to be exhausted as hot H2O and CO2.

The exhaust gasses of piston-engined aircraft were directed backwards to gain a few HP of forward thrust. That 'jet' effect would also be true for cars, but is not as important as it once was in military aircraft.
 
Your math makes little sense to me too, but please note a car moving at constant speed on flat ground is not a situation where energy is conserved: 100% of it is lost.
 
I built a device designed to brake angular velocity which seems to work based on below, i used a flexible shaft that could bow up and down so i could visually see what was happening for the prototypes. If you spin two wheels in opposite directions each with a magnitude of angular momentum L on a rigid shaft (equal magnitude opposite directions), then rotate the shaft at 90 degrees to the momentum vectors at constant angular velocity omega, then the resulting torques oppose each other...

Similar threads

  • · Replies 30 ·
2
Replies
30
Views
3K
  • · Replies 10 ·
Replies
10
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
Replies
65
Views
5K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 3 ·
Replies
3
Views
8K
  • · Replies 5 ·
Replies
5
Views
2K
Replies
4
Views
1K
  • · Replies 5 ·
Replies
5
Views
2K
Replies
2
Views
2K