Can a Simple Formula Estimate Train Energy Consumption?

[george95]

Homework Statement

Train consumption

Relevant Equations

v2 = a*s

Is there some rule of a thumb formula, which would be able to calculate the energy consumption of a train, based on the following parameters:

- mass of the train
- its speed
- inclination

It can be a simple approximation.

I would be grateful for any kind of reply.

 

[haruspex]

Relevant Equations:: v = a*s

Where v, a and s are what?

Per forum rules, you must show some attempt, or at the least your thoughts on the matter. What is the train doing that is likely to require energy?

 

[george95]

Hi haruspex,
I assumed (wrongly) that velocity would be equal:
v^2 = acceleration * distance
acceleration = gravitAccc * sin(inclinationAngle)
And hoped that by knowing it, the energy consumption would maybe be possible to be calculated?

 

[haruspex]

I assumed (wrongly) that velocity would be equal:

No, that was the right assumption to make. But isn't this stated in the problem? Please post the entire question as given to you.

v^2 = acceleration * distance

That isn't quite right, you left out a factor of 2, and you should specify that this assumes acceleration is constant and there is zero initial velocity. But it is irrelevant here since we are taking speed as constant.

acceleration = gravitAccc * sin(inclinationAngle)

That's closer to being relevant, but applies to an object moving up or down a ramp under the influence of gravity and nothing else - no friction, no engine.

What do you know that relates gravity and energy?

 

[george95]

Thank you for the reply Haruspex,
Sorry for the mistakes.
Is it possible to simplify the problem by not taking into account friction?
So only the energy consumption (not kinetic) while moving under the influence of gravity?

What do you know that relates gravity and energy?

Nothing to be honest.

 

[haruspex]

only the energy consumption (not kinetic) while moving under the influence of gravity?

Yes, based on the question as stated, that is all you need to consider.
If I lift a mass of 1kg through a height of 1m, how much work have I done?

 

[george95]

1 kg should be approx 10 Newtons (9.81)

So the work would be:
10N * 1m = 10 Nm (Jules)?

 

[jbriggs444]

1 kg should be approx 10 Newtons (9.81)

So the work would be:
10N * 1m = 10 Nm (Jules)?

Correct. (Spelled joules).

Can you calculate how many joules it would take to raise the moving train from its starting point to an ending point, one second later?

 

[george95]

I am not sure I know.
Are we now introducing the Power into equation?
Work = power * time

So if a train has 500 000 Watts (Joules/sec), the work in Joules would be:
Work = 500 000 Joules/sec * 1sec = 500 000 Joules
?

 

[jbriggs444]

So if a train has 500 000 Watts (Joules/sec), the work in Joules would be:
Work = 500 000 Joules/sec * 1sec = 500 000 Joules
?

I was not going for power just yet. Though that is where I was heading, yes.

We were given:

- mass of the train
- its speed
- inclination

Provide values for those parameters and calculate how many joules in one second. Then we can make the easy jump to how much power that is.

 

[george95]

Hi jbriggs444,

I can only assume:
mass of the train - 5000kg = 50000N
speed = 50 km/h -> 13.89 m/sec
inclination of the terrain: 10 degrees (train is climbing up the terrain)

But the previous example for work 10 Joules, was if the train was moving horizontally:
10N * 1m = 10 Nm (Joules)
Not under inclination? Or?

So if the length of the slope is 1meter and it is under 10 degrees of inclination, then the Work would be:

W = 50000N * (1m * cos(10) = 50000N * (1m * 0.9848) = 49 240 Joules

 

[jbriggs444]

I can only assume:
mass of the train - 5000kg = 50000N
speed = 50 km/h -> 13.89 m/sec
inclination of the terrain: 10 degrees (train is climbing up the terrain)View attachment 256718

Good. Nicely summarized.

But the previous example for work 10 Joules, was if the train was moving horizontally:
10N * 1m = 10 Nm (Joules)
Not under inclination? Or?

If the 1 kilogram mass were moving horizontally on a sheet of ice, the change in altitude is zero meters. The required force is zero Newtons. It takes no work at all to slide a mass across a sheet of ice.

So if the length of the slope is 1meter and it is under 10 degrees of inclination, then the Work would be:

W = 50000N * (1m * cos(10) = 50000N * (1m * 0.9848) = 49 240 Joules

Think about starting state and ending state. 50000 Newtons of weight is correct. But what is the change in altitude?

You could also worry about whether you are talking about the change in altitude over one meter of track or the change in altitude over one second of duration. You have not yet taken speed into account.

The cosine of 10 degrees is nearly one. Surely the slope is not steep like a vertical wall.

 

[jack action]

Homework Statement:: Train consumption
Relevant Equations:: v2 = a*s

Is there some rule of a thumb formula, which would be able to calculate the energy consumption of a train, based on the following parameters:

- mass of the train
- its speed
- inclination

It can be a simple approximation.

I would be grateful for any kind of reply.

Energy $E$ is equal to a force $F$ times a distance $d$, so the simple answer to your question is: No, you cannot calculate the energy consumption with the given parameters, because you are missing the distance traveled.

However, because power $P$ is the energy consumption per unit of of time $t$ and you know the speed $v$ of the train, you can find the energy consumption rate (i.e. power).

The relevant equations are:
$$E = Fd$$
$$P = \frac{E}{t}$$
$$v = \frac{d}{t}$$
Therefore, you can find that $P = \frac{Fd}{t} = Fv$.

Now, with the parameters given, you can identify only one force, the one due to the inclination of the track. To find it, you need to do a free body diagram. Once you found this force, you multiply it by the velocity and you will get the power required to maintain this force at this velocity.

Note that there will probably be other forces involve with a real train, such as: aerodynamic drag, rolling resistance or inertia under acceleration. The sum of all of these forces multiplied by the train velocity will give the total power needed to propel the train.

 

[george95]

If the 1 kilogram mass were moving horizontally on a sheet of ice, the change in altitude is zero meters. The required force is zero Newtons. It takes no work at all to slide a mass across a sheet of ice.

Why sheet of ice? Because we neglected the friction forces?

Think about starting state and ending state. 50000 Newtons of weight is correct. But what is the change in altitude?

If we use the length of the ramp to be 1m, then I guess the change should be:

sin(10) = altitude/1m
altitude = sin(10)*1m = 0.1736*1m = 0.18m

You could also worry about whether you are talking about the change in altitude over one meter of track or the change in altitude over one second of duration. You have not yet taken speed into account.

Maybe better the change of altitude over one meter of track.

The cosine of 10 degrees is nearly one. Surely the slope is not steep like a vertical wall.

Should we increase or decrease the slope?
=================

Thank you for the explanation Jack.
The "F" force from your final formula has two components: "Fn" and "Fs" from this screenshot:

Or am I mistaken?

 

[jack action]

Thank you for the explanation Jack.
The "F" force from your final formula has two components: "Fn" and "Fs" from this screenshot:

Or am I mistaken?

The one that counts is $F_S$, i.e the force in the direction of the displacement.

 

[jbriggs444]

Why sheet of ice? Because we neglected the friction forces?

Yes, just so.

If we use the length of the ramp to be 1m, then I guess the change should be:
View attachment 256727
sin(10) = altitude/1m
altitude = sin(10)*1m = 0.1736*1m = 0.18m

Should we increase or decrease the slope?

Back a couple of posts you'd calculated using the cosine of 10 degrees. That would correspond to an 80 degree slope -- near vertical. Here you have corrected yourself to the sine of 10 degrees.

You could take the next step and compute how much energy is required to lift a train weighing 50000 N a vertical distance of 0.18 m.

 

[george95]

Hi Jack action,

According to this video:
Fs = coefFriction * Fn

coefFriction = tan(10) = 0.176
Fn = mg*cos(10) = 50000kg * 9.81 m/s^2 * 0.9848 ~ 500 000N * 0.9848 ~ 492000N

Fs = 0.176 * 492000N ~ 87 000N

Is this wrong?

Back a couple of posts you'd calculated using the cosine of 10 degrees. That would correspond to an 80 degree slope -- near vertical. Here you have corrected yourself to the sine of 10 degrees.

Hi jbriggs444. Sorry for the mistake.
How do I continue from here?

 

[jbriggs444]

Hi jbriggs444. Sorry for the mistake.
How do I continue from here?

As I suggested (editted into previous post), figure how much energy it takes to lift a 50000 N train over 0.18 meters of vertical displacement.

Then figure how many of these 0.18 meter increments the train covers each second.

 

[jbriggs444]

The "F" force from your final formula has two components: "Fn" and "Fs" from this screenshot:
View attachment 256728
Or am I mistaken?

You show a net force from the tracks that is slanted forward. That will not turn out to be correct. You stipulated that the train's speed is constant. It is not accelerating along the tracks. It is most certainly not accelerating up away from or down into the tracks. Accordingly there can be no net acceleration in any direction.

The net force from the tracks must be directly upward and must be equal and opposite to gravity.

 

[george95]

As I suggested (editted into previous post), figure how much energy it takes to lift a 50000 N train over 0.18 meters of vertical displacement.

Then figure how many of these 0.18 meter increments the train covers each second.

I am not sure I know how to do that.

You show a net force from the tracks that is slanted forward. That will not turn out to be correct. You stipulated that the train's speed is constant. It is not accelerating. Accordingly there can be no net forward acceleration.

The "net force" is Fn?

 

[jbriggs444]

I am not sure I know how to do that.

You already did it for 1 kg moving vertically through 1 meter.

5000 kg moving vertically through 0.18 meters is the same thing.

The "net force" is Fn?

No. $F_n$ would typically denote the "normal" force.

[In physics speak, "normal" means "perpendicular to the surface". It has nothing to do with "typical", "average" or "expected"]

It is perhaps worth noting that @jack action and myself are working somewhat at cross purposes. He is trying to have you compute energy based on a small diagonal force pulling the train through a full diagonal distance. I am trying to have you compute energy based on a full vertical force lifting the train through a reduced vertical distance.

Both approaches are equivalent. Both involve a multiplier of sin(10).

 

[jack action]

Hi Jack action,

According to this video:
Fs = coefFriction * Fn

coefFriction = tan(10) = 0.176
Fn = mg*cos(10) = 50000kg * 9.81 m/s^2 * 0.9848 ~ 500 000N * 0.9848 ~ 492000N

Fs = 0.176 * 492000N ~ 87 000N

Is this wrong?

According to your video, $F_S = mg\sin \theta$ and that what is important to your train problem. It represents the portion of the gravity that is pulling your train downhill. Therefore, you need to produce an equivalent and opposite force ($F_S$) to counteract it. Multiplying that force with the train velocity, you get the power produced by the train.

 

[george95]

Thank you jack
Fs = mg*sin(10) = 50000kg * 9.81 m/s^2 * 0.173648 ~ 86000N
P = Fs*v = 86000N * 13.89 m/sec = 1194540 W = 1194 kW
So the energy consumption of a train per second when it travels uphill would be 1194kW?
Or is this per one meter length of travel, not per second?

What about if the train is going down the hill?

You already did it for 1 kg moving vertically through 1 meter.

5000 kg moving vertically through 0.18 meters is the same thing.

Thank you jbriggs.
W = 50000N * (1m * sin(10) = 50000N * (1m * 0.173648) = 8700 Joules
?

 

[jack action]

So the energy consumption of a train per second when it travels uphill would be 1194kW?
Or is this per one meter length of travel, not per second?

1194 kW equals 1194 kJ/s, so the energy consumption is 1194 kilojoules per second.

What about if the train is going down the hill?

Then the gravity is pushing the train downhill, so without any other energy input or braking, the train will accelerate ($a = \frac{F}{m}$).

 

[george95]

Thank you Jack,
a = F/m means that when train is going downhill, no energy is consumed?

 

[jack action]

Thank you Jack,
a = F/m means that when train is going downhill, no energy is consumed?

When you go downhill with a bicycle, do you need to use the pedals (i.e. consuming energy)?

But if you use the brakes with an equivalent force, you will not accelerate and an equivalent amount of energy will be transformed into heat by the brake system.

 

[george95]

When you go downhill with a bicycle, do you need to use the pedals (i.e. consuming energy)?

No.

But if you use the brakes with an equivalent force, you will not accelerate and an equivalent amount of energy will be transformed into heat by the brake system.

So if I assume that the train needs to keep a constant speed of 50 km/h -> 13.89 m/sec both uphill, and downhill - when going downhill an equal amount of energy needs to be consumed in order to prevent the train accelerating beyond 50km/h?
If this is so, then regardless of the slope sign (up hill or down hill), the energy consumption is always the same (the same for the same slope angle: +10 or - 10) - up hill to overcame the gravity, and downhill to prevent acceleration?

 

[jack action]

So if I assume that the train needs to keep a constant speed of 50 km/h -> 13.89 m/sec both uphill, and downhill - when going downhill an equal amount of energy needs to be consumed in order to prevent the train accelerating beyond 50km/h?

Yes.

If this is so, then regardless of the slope sign (up hill or down hill), the energy consumption is always the same (the same for the same slope angle: +10 or - 10) - up hill to overcame the gravity, and downhill to prevent acceleration?

When you go uphill, you have to produce energy - hence, consuming it - and when you go downhill you need to store or dissipate energy. A typical brake system convert the energy into heat and dissipates it, but you can also store it (in a battery for example) and reuse it at a later time.

 

[george95]

Thank you once again for the help and answers Jack!
This answers my initial post question.

 

[george95]

@jbriggs444 , thank you as well!