How Do You Calculate Engine Power for Semi Trucks in Various Conditions?

[WumTrudo]

Summary: Calculating engine power required for semi trucks under different circumstances.

Hello everyone,

I would like to ask you for help if that is allowed.

For a current project I need to calculate engine power needed in watts under different circumstances.
This is something that I will have do more often, so I would like to make this into an excell sheet where I can simply input the prevailing conditions, and have it calculate the right value(s).

However, I underestimated the complexity of the formula due to all the different variables.

Can someone help me get to the right formula?

The value(s) are as follows:

Calculate the engine power needed for the following semi trucks in the described situations:

Semi Truck data:
Dimensions: L: 14m W: 2,55m H: 2,74
Front face of truck in m2: 7m2
Gross train weight: 40.000 kg
Overall drive train efficiency: 95%

CW value air resistance: 0,6

Situation 1:
Engine power (in kW), needed for the truck above to:
Drive up a slope of 3% with 80k/ph, while experiencing a headwind of 10kph.
Road Condition: Good asphalt with a rolling friction coefficient of 0.007.

Situation 2:
Engine power (in kW), needed for the truck above to:
Drive up a slope of 3% with 80k/ph, while experiencing a headwind of 15kph.
Road Condition: Sandy roads with a rolling friction coefficient of 0.2.

So far I have come up with for situation 1:

Formula for rolling resistance in Newtons:
9.81 * friction coefficient * vehicle weight in kilo’s * cos (angle of slope in degrees)
9.81 * 0.007 * 40000 * cos 1.72
= 2746 N

Formula for slope resistance in Newtons:
9.81 * vehicle weight in kilo’s * sin 1.72
9.81 * 40000 * sin 1.72
= 11778 N

Formula for air resistance in Newtons:
0.6 * 0.6 * 7 * 22,22^2
= 1244 N

Total resistance in Newton:
2746 N+ 11778 N + 1244 N = 15768 N

Formula for Required power without taking into account the driveline efficiency:
Total resistance * velocity in meters per second

15768 * 22,22 = 350364 watts (350,364kW)

Formula for Required power taking into account the driveline efficiency:
Power Required / driveline efficiency
350,364 / 0,95 = 368,804kW

This is how far I’ve come on my own, and I know I haven’t taken into account the headwind yet. I would like it if anyone could help me by looking over the calculations so far, and help me perfect them.

Kind Regards,

Wim

 

[jrmichler]

Is this homework? If so, we can move it to the homework forum.

Total resistance is the sum of rolling resistance on a level road plus slope resistance plus air drag. Those three are calculated separately, then added at the end. Also, be aware of significant digits. You do not get six digits of accuracy when starting with one digit inputs.

 

[WumTrudo]

I thought I posted it in homework. I changed the prefix to automotive, did I go wrong there?

 

[berkeman]

I thought I posted it in homework.

You posted it in the ME forum. I'll move it to the schoolwork forums for you now.

 

[WumTrudo]

You posted it in the ME forum. I'll move it to the schoolwork forums for you now.

Thanks!

 

[erobz]

Semi Truck data:
Dimensions: L: 14m W: 2,55m H: 2,74
Front face of truck in m2: 7m2
Gross train weight: 40.000 kg
Overall drive train efficiency: 95%

CW value air resistance: 0,6

Situation 1:
Engine power (in kW), needed for the truck above to:
Drive up a slope of 3% with 80k/ph, while experiencing a headwind of 10kph.
Road Condition: Good asphalt with a rolling friction coefficient of 0.007.Formula for air resistance in Newtons:
0.6 * 0.6 * 7 * 22,22^2
= 1244 N

Can you elaborate on the drag formula you used a bit more?

Typically, $F_d =C \rho A v^2$

$F_d =\frac{1}{2} C \rho A v^2$ (Forgot the factor of a half- now the calcs makes sense)

$C$ is the drag coefficient
$\rho$ is the density of the air
$A$ is the cross sectional area
$v$ is the velocity of the air as seen from the perspective of the truck.

 

[WumTrudo]

Can you elaborate on the drag formula you used a bit more?

Typically, $F_d =C \rho A v^2$

$C$ is the drag coefficient
$\rho$ is the density of the air
$A$ is the cross sectional area
$v$ is the velocity of the air as seen from the perspective of the truck.

In this case i’ve used:
drag coefficient (0,6) * Density of the air (0,6) * the cross section of the sectional area in m2 (7) * Speed of the truck in meters per second (22,22)^2

 

[jack action]

In this case i’ve used:
drag coefficient (0,6) * Density of the air (0,6) * the cross section of the sectional area in m2 (7) * Speed of the truck in meters per second (22,22)^2

Actually, you used $\frac{1}{2}\rho C_DAv^2$ where $\rho = 1.2\ kg/m^3$.

As for your calculations, everything looks OK to me.

What you may want to explore now is if your truck has enough traction to handle that much force. If the driven tires cannot produce enough friction force, the tires will just spin out and the truck won't move.

 

[erobz]

All you have to do to answer the initial question is account for the wind velocity $w$.