My Assignment

[satish]

A rear wheel drive car of mass 2000kg has a rolling resistance of 150 N, a frontal area of 2.5m2, a drag coefficient of 0.70, a transmission loss of 12% and a maximum engine output of 60kW. Assuming that the traction limit for each tyre is equal to the weight it carries and a 60/40 weight transfer to the back wheels under acceleration,

Take the density of air to be 1.2 Kg/m3 and g = 9.8 m/s2

a) What is the maximum power available at the back wheels?

b) What is its theoretical maximum speed in mph on a level road?

c) What is its theoretical maximum speed in mph up a 1 in 10 hill?

d) What is the maximum tractive force that each of the rear tyres could provide under acceleration before slipping occurs?

e) What would be the maximum, useable, tractive force at each of the rear tyres if full power was applied at 12.98 mph? (hint: is the useable tractive force limited by engine power or tyre grip?)

f) What would be the wind resistance at 12.98 mph?

g) What would be the acceleration (in mph/s) at 12.98 mph on the level?

h) What would be the acceleration (in mph/s) at 12.98 mph up a 1 in 10 hill?
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I was wondering if you guys could help me out please thank you.

[2turboimports]

on your first post? is this for some take home test or something? like everyone here has a degree in physics...

[2turboimports]

i did A - C but then i thought, you probably don't even know the answers.....so why should i waste my time? but it was a fun 15 minutes

[satish]

yes this is my first post i have been browsing this forum a long time now and i thought i would register...

2turboimports you did A-C? and if you think im just looking for answers NO i want to learn, so i want to know the FORMULAS on how i could work these questions out because i cant find any formulas on search engines.

Thank You

[ivymike1031]

Originally Posted by satish

A rear wheel drive car of mass 2000kg has a rolling resistance of 150 N, a frontal area of 2.5m2, a drag coefficient of 0.70, a transmission loss of 12% and a maximum engine output of 60kW. Assuming that the traction limit for each tyre is equal to the weight it carries and a 60/40 weight transfer to the back wheels under acceleration, Take the density of air to be 1.2 Kg/m3 and g = 9.8 m/s2

a) What is the maximum power available at the back wheels?
(engine power)*(drivetrain efficiency) = 60kW * (1-0.12) = 52.8kW

b) What is its theoretical maximum speed in mph on a level road?
You must balance (power @ rear wheels) = (losses)
52.8kW = (Aero drag) + (Rolling resistance) + (power into hills)

Aero drag * velocity = aero power loss = (from memory, forgive me if I'm wrong) = 0.5 * (density) * Cd * (Frontal Area) * Velocity^2 * Velocity
Rolling resistance power = (rolling resistance force) * (vehicle velocity)
Power into hills = 0 (level ground)
so maxP = (1/2 * dens * area * cd)*V^3 + (RR)*V + 0

To reduce the work I'd have to do, I'm going to solve by iteration:
Velocity (m/s) ... Power/Velocity ... Aero Force + RR + grav
10 m/s ... 5280 N ... 255 N
40 m/s ... 1320 N ... 1830 N
30 m/s ... 1760 N ... 1095 N
35.5 m/s ... 1487 N ... 1473 N
35.621 m/s ... 1482.3 N ... 1482.3 N

So 35.6 m/s is the approximate top speed on level ground.

c) What is its theoretical maximum speed in mph up a 1 in 10 hill?
Solve the above all over again, with non-zero power loss due to hill climbing.
Energy into the hill = mass*grav*height
Power into the hill = mass*grav*(change in height per unit time)
Better to look at it in force terms, as above. The forward force required to climb a 1 in 10 hill is the force due to gravity times sin(angle). Angle is atan(1/10), sin(atan(1/1)) = 1/10. Force due to gravity is mass*g*(1/10) = 1960N

Velocity (m/s) ... Power/Velocity (N) ... Aero Force + RR + grav (N)
20 ... 2640 ... 2530
20.6 ... 2563.1 ... 2555.6
20.645 ... 2557.5 ... 2557.5

So the max speed up the hill is 20.645 m/s

d) What is the maximum tractive force that each of the rear tyres could provide under acceleration before slipping occurs?

this is a "simple friction" calculation, so tractive force is assumed to be (friction coeff) * (downward force)
You didn't say whether 60/40 meant 60% Front, 40% Rear, or the other way around, so I'll assume 60% front.
(1) * (40% * 2000kg * g) = 7840N

e) What would be the maximum, useable, tractive force at each of the rear tyres if full power was applied at 12.98 mph? (hint: is the useable tractive force limited by engine power or tyre grip?)
Useable tractive force will be the minimum of (force due to engine power, tire traction force)
answer = min(52.8kw / 12.98mph , 7840N)
answer = min(9099.4 N, 7840N) = 7840N

f) What would be the wind resistance at 12.98 mph?
0.5 * (density) * Cd * (Frontal Area) * Velocity^2 = 35.353 N
wind power loss = 205.1 W

g) What would be the acceleration (in mph/s) at 12.98 mph on the level?
f = m*a
remaining propulsive force = tractive - aero - rolling
7840 N - 35.35 N - 150 N= 7655 N
acceleration = 3.827 m/s^2

h) What would be the acceleration (in mph/s) at 12.98 mph up a 1 in 10 hill?
remaining propulsive force = tractive - aero - rolling - grav
= 5695 N
acceleration = 2.847 m/s^2

[2turboimports]

Originally Posted by satish

yes this is my first post i have been browsing this forum a long time now and i thought i would register...

2turboimports you did A-C? and if you think im just looking for answers NO i want to learn, so i want to know the FORMULAS on how i could work these questions out because i cant find any formulas on search engines.

Thank You

cool stuff man, at least you want to learn. 'my bad'. they were just pretty darn specific questions..lol

thanks ivymike for the rest of those formulas, good stuff