What is the exhaust pipe diameter required for the given exhaust speed?

[paulxu11]

Homework Statement

The problem:
Here is an approximate reaction for burning hydrocarbons in air:

(CH2)n + 1.5*n O2 + 6n N2 -> n CO2 + n H2O + 6n N2 (n is an integer)

The (CH2)n approximates the fuel, and everything on the right is the exhaust. The atomic masses of hydrogen, carbon, nitrogen and oxygen are in the ratio 1:12:14:16. (Hint: how many kg of exhaust for one kg of fuel?)

A car uses 4.0 kg of (CH2)n hydrocarbon fuel per 100 km while it travels at 100 k.p.h. in a straight line at constant speed. The gas comes out of the (horizontal) exhaust pipe at a speed of v = 50 m.s-1, measured with respect to the car.

The questions:
Q1: What is the magnitude of the force the exhaust gas exerts on the car? (Hint: the exhaust gasses are accelerated in the frame of the car, so ignore the momentum of the incoming gas.)

Q2: Because of the thrust, should you worry about the angle of your exhaust pipe?

Q3: The energy obtained from burning petrol is large: about 46 MJ/kg. At what rate is this car consuming chemical energy from its fuel?

Q4: Suppose the car has a frontal area of A = 2.1 m2 and a drag coefficient of CD = 0.36 and that the density of air is ρ = 1.2 kg/m3. Use without proof that the air resistance is ½CDρAv2. (Physclips has a section on car physics if you want to see why this is so.) What is the power used to overcome air resistance for this the car at this constant speed?

Q5: Make the rough approximation that rolling resistance is negligible in comparison with air resistance at this speed, so the latter is the only retarding force. What is the efficiency of the engine plus transmission? (i.e. what fraction of the available chemical energy is used to propel the car?)

Q6: The engine turns at 3000 rpm at this speed. The engine capacity has a displacement of 1.2 litres: this means that 1.2 litres of exhaust gas is expelled from the motor (into the exhaust manifold) every two* revolutions. Further assume that the gas leaves the exhaust pipe with the same volume** that it leaves the engine. What exhaust pipe diameter is required to obtain the given exhaust speed?

* Two revolutions because it is a four stroke engine: intake, compression, power and exhaust as the piston goes down, up, down and up.

** In practice, this would mean that the volumetric effect of the change in pressure approximately cancels that of the change in temperature.

Homework Equations

F=dI/dt
dI=dP=v*dm

The Attempt at a Solution

Q1: The force the exhaust gas exerts on the car:
F=dI/dt=v*(dm/dt)=1/18N=0.056N
where dt=100km/100kph=1h=3600s
dm/dt=4kg/3600s=1/900kg/s
v=50m/s

Q2: No

Q3: The rate of consuming energy;
P=dw/dt=184000KW/3600s=460/9KW=51KW
where dw=46MJ/kg * 4kg=184MJ=184000KJ
dt=3600s

Q4: The power to overcome;
P=(½CDρA)*v^3=1/2*0.36*1.2*(250/9)^3=9.7KW
where CD=0.36, ρ = 1.2 kg/m3, A = 2.1 m^2, v=100kph=250/9 m/s

Q5: The efficiency:
(51-9.7)/51 * % = 81%
where (total_energy - energy_to_overcome _air_resistance)/totaol_enetgy

Q6: Known:
rmp=3000;
v_exhaust=50 m/s;
engine_displacement=1.2 litres;
1.2 litres of exhaust gas is expelled from the motor (into the exhaust manifold) every two* revolutions.So I tried to solve all of the questions.
I don't really understand the last question.
The problem is quite lengthy.
Could someone be patient enough to suggest me if my answers are correct or not and help me with the Q6.
Thank you.
Paul

 

[BvU]

1.2 litres of exhaust gas is expelled from the motor (into the exhaust manifold) every two* revolutions

So what is the volume of gas coughed out every minute ? Say Y m³/min Convert to 'per second'. Say X m³/s.
Sppose the diameter of the single exhaust pipe is Z m, then what is the exhaust gas speed in the pipe ?

 

[paulxu11]

So what is the volume of gas coughed out every minute ? Say Y m³/min Convert to 'per second'. Say X m³/s.
Sppose the diameter of the single exhaust pipe is Z m, then what is the exhaust gas speed in the pipe ?

Thanks for replying
1.2 litres = 1.2 * 10^-3 m^3
rpm=3000 -> time_for_2revlotions=2/3000 mins=1/1500 *60 sec=0.04 sec
So volume_per_sec=dV=(1.2 * 10^-3 m^3) per (0.04 sec) =0.03 m^3/sec
speed_of_gas=50m/s from the question
If the diameter = Z m
Then Pi * (Z/2)^2 * 50 = 0.03 -> Z =0.02764 m =2.764 cm [Area * length(speed in one sec 50) = volume in one sec 0.03]
I followed your questions and got this.
Is it right?

 

[BvU]

If it's right I don't know But that's what I calculate (too).

I did not check your other answers -- they are yours and you should check them.
If you miss something, teacher will tell you. I should hope ... PF can't interfere too much in the intimate relationship between students and teachers

But when I do check anyway, some issues pop up:
For example: didn't you ask yourself what the hint hinted at ? What comes out of the exhaust ?

And when this efficiency is asked for, you divide the loss by the input ! Efficiency $\equiv$ useful / input !

 

[paulxu11]

If it's right I don't know But that's what I calculate (too).

I did not check your other answers -- they are yours and you should check them.
If you miss something, teacher will tell you. I should hope ... PF can't interfere too much in the intimate relationship between students and teachers

But when I do check anyway, some issues pop up:
For example: didn't you ask yourself what the hint hinted at ? What comes out of the exhaust ?

And when this efficiency is asked for, you divide the loss by the input ! Efficiency $\equiv$ useful / input !

That's a good new hh
Yes i will check them coz it is an online quiz, and once i submit it i get marked...
So I really want to make sure they are correct
For the efficiency , I think I did useful/input: ''[total_energy(input) - energy_air_resistance(loss) ]/total_energy(inpit) (51-7.9)/51 * 100%
Anyway this is my time to post a question. Thanks for replying

 

[BvU]

What you put in is not part of 'useful'. That way zero output would give you 100% efficiency !

If you pay 100 euro and you get back 20 euro worth of goods, your efficiency is not 80% !