# How to calculate the horse power ?

1. May 22, 2010

### harvi101

how to calculate the horse power ? if the car weights 950 kg and its top speed is 144 km per hour. how can i size the engine from this data aerodynamic drag coffi = 0.65, tyre radius = 0.27m. please help.

2. May 22, 2010

### xxChrisxx

3. May 22, 2010

### harvi101

how can i size the engine for small hatch back car.

4. May 22, 2010

### xxChrisxx

What do you mean by size? If you mean have the power to do what you want then the aboev is your answer. In reality you'd need more to accelerate up to top speed, but 60ish hp seems sensible for a budget end hatchback.

5. May 22, 2010

### harvi101

engine size example
required power = 50 kw
A. choose 2 valve wedge/bathtub SI NA 4- stroke combustion chamber
B. bmep : 10 bar (torque peak and this value is assumed ) would be appropriate; also 20% back up
so bmep@power peak = 10/1.2 = 8.33 bar
C. allow Pmax (choose max port velocity) = 75 m/s
for 2 V wedge/bathtub, d/D = 0.45 ( d= valve dia, D= bore)
Mean piston speed, Ps = Pmax(d/D)(d/D)
so Ps = 15.19 m/s which is ok for life
D. Power = 1/4XPbxAxPs
50 = 1/4x8.33xAx15.19
A = 15.81/1000 meter square
E. Choose 4 cylinders each 15.81/(4x1000) = 3.952/1000 meter square =(pie/4)DxD
so D = 0.0709 m or 71 mm
F. choosing a stroke
(i) assume L/D = 1 , so L = 71 mm
Ps = 2 LN/60
N = 15.19x60/(2x0.071) = 6418 rev/min
(ii) assuming Nmax = 5500 rev/ min
therefore L = 15.19x60/(2x5500) = 82.9 mm ; L/D=1.167
compromise : make L = 77 mm and Nmax = 5918 rev/min
therefore Vs = 4x(pie/4)xDxDxL = 1.219 liters
(it might be worth setting to 1.2 or 1.3 liters)

now keeping the above example in mind how can i calculate the engine specification from the given data
Mass of vehicle (without passangers ) = 950 kg
Type of vehicle = small hatch back
drag coff = 0.65
tyre radius = 0.27 m
max velocity = 144 km/h
(i want for cylinder engine with four valves, is there a way to calculate he bmep and port velocity from the given data)

6. May 22, 2010

### xxChrisxx

No, that's not even remotely enough to attempt to specify an engine in that kind of detail. All you can specify is that you need 50hp.

To calcualte BMEP you'd need to know the displaced volume. Or you'd need to specify a BMEP to find a volume. There is a shed load of data you need to find out the flow at the port.

What is the point of this? Are you ment to be designing a new engine? Where did you get your example from?

Before you even start trying to specify something to that level you need to think about this engine. Is it going to be run on, carbs, multipoint injection, direct injection? What fuel? Is it going to be turbocharged? etc etc

Last edited: May 22, 2010
7. May 22, 2010

### harvi101

any good method to calculate the power required from engine to achieve the 144 kmph speed

from speed trap method
hp = weight x (velocity/234)x(velocity/234)x(velocity/234)
hp = 117.09844102277 horsepower ( without the passanger weight ) = 87.32 kw which is really high for a small car

8. May 22, 2010

### harvi101

its our assignment and this the example given by our teacher

9. May 22, 2010

### xxChrisxx

I said earlier.

You need to find the losses (aero drag, rolling resistance, and any internal resistance). I didn't think that you needed this much detail before so I just used the aero drag, which is by far the highest loss.

I also linked to wiki for the power required to travel a specific speed due to aero drag.

10. May 22, 2010

### harvi101

thanks xxChrisxx
i'll try to solve it my taking 40 kw as the required power

11. May 22, 2010

### xxChrisxx

Owing to the fact that you are using about 10bar as the BMEP, I assume this engine is a naturally aspirated petrol?

EDIT: Also note that that power required would be wheel power. You would have to make an estimate/guess of the losses due to the transmission. About 10-15% seems sensible.

So the aero drag indicates about 40 kW. So due to that alone you'd need approximately 47kW produced at the flywheel. Not only that you would need more to acutally accelerate up to that top speed.

At a guess, i'd estimate that that car would need about 55kW to acutally reach 144kph in a sensible time.

12. May 22, 2010

### harvi101

yes i have already mentioned it SI NA 4- stroke

13. May 22, 2010

### harvi101

but i don't know how he calculated d/D = 0.45 ( d= valve dia, D= bore) in step C.

14. May 22, 2010

### xxChrisxx

Ah so you did, just me skim reading again.

And he basically just made that d/D ratio up (it's not acutally calcualted), as it's looks a sensible size for a pent roof combustion chamber.

If you think about it, he specified it's 2 valves per cylinder. Obviously, you can't have 2 huge valves, as they would clash. Nor should you have two tiny valves. What he has said is make each lvavle slighlty smaller than half the dismeter of the bore so they are big, but still have clearence.

Your valves would have to be smaller as you want to use 4 of them. So the d/D ratio would be different.

15. May 22, 2010

### harvi101

thanks man 55kw seems ok . at least now i have starting point . i am thinking of taking bmep = 9 bar and port velocity = 70 m/s and three cylinder 2 valve engine with wedge shape.

16. May 22, 2010

### harvi101

losses
rolling resistance = 139.80
aero drag = 11403.05
gradient = 342
Total Losses = 11884.55 = 12 kw is it ok

17. May 22, 2010

### xxChrisxx

I've no idea how you got to those figures, but the aero drag is way too low. The others look pretty sensible.

18. May 28, 2010

### harvi101

this how i calculated the forces
Vehicle Data
Vehicle selected for engine sizing is Small Hatch back
Vehicle type Mass (kg) CdA (m2) rtypre(m) Vmax(Km/h)
Small hatch 950 (unladen) 0.61 0.27 144

Calculations for the payload
Assuming weight of one passenger = 70 kg
Then weight for the 5 passengers = 70 x 5 =350 kg
Assuming cargo volume = 200 liters = 200 kg
Then the total weight of the vehicle = 950+350+200 = 1500 kg

Calculations for the losses
1. Aerodynamic Drag Resistance

Given Cd A = 0.61 m2
ρ = 1.2 kg/m3
FD = ( Cd A ρ V2)/2 = (0.61x1.2x40x40)/2 = 586.6 = 586 N

2. Rolling Resistance

Given fr = 0.015

FR = m g fr = 1500x9.81x0.015 = 220.725 =221 N

3. Gradient Resistance

Given gradient = 36%

Tan θ = 0.36/1 => θ = Tanֿ (0.36/1) = 19.8º

FG = mg Sin θ = 1500x9.81x Sin 19.8o = 4984.528 N = 4985 N

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