Train-hill hp to pull, problem

  • Thread starter Ed Boon
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In summary: This is the summary: In summary, to power a 15k tonne train up a 1 degree grade, ignoring rolling friction and wind resistance, we need to know a speed. Assuming a speed of 20km/h, the train will gain 349 meters vertically and require 5.1303E10 joules of potential energy. With one locomotive producing 1E10 joules per hour, it would require around 5 locomotives to power the train up the grade.
  • #1
Ed Boon
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Homework Statement


How many GE locomotives are needed to power a 15k tonne (note tonne not ton) train up a 1 degree grade. Ignore rolling friction and wind resistance. Locomotive traction power is 4400 HP/locomotive.


Homework Equations



1 hp = 745.7 J/s * 4400 hp = 3281080 J/s
and the grade force is
m*g*sin(1°)
so
15k (kg) *9.8*sin1=2565504 N
also




The Attempt at a Solution


I am assuming we need to know how fast they go so just say 15 mph.
(assuming 15 mph could you say at 4400 hp it would require 11816640000 n m/hr)
I am stuck between knowing each train has 3,236,195 W of power and knowing the train will take 2565504 N of force and somereason can't relate them

any help is appriciated

Ed
 
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  • #2
You need to know a speed - the power depends on how long you take to move a certain distance.

Just think in the upwards direction, how many m/s vertically is 15mph along a 1 deg slope?
Then how much potential energy (mgh) does the train gain per metre?
How many vertical m/s is it moving?
So how many joules/second = watts is this.

(ps pick a set of units to convert everything into)
 
  • #3
Before you pick a speed, you might want to check the brake linings.

15Mkg will be challenging to slow from 15m/h.
 
  • #4
LowlyPion said:
Before you pick a speed, you might want to check the brake linings.

15Mkg will be challenging to slow from 15m/h.



All the info in the question is typed above, which is why I'm kinda hesitant on some of "assumptions" because isn't it first required to know the speed and acceleration for these? I am aware that coal trains can reach up to 40-50 mph (if what I was reading was correct) so even if it were hard to stop should i still use 15 mph just for the problem's sake?
 
  • #5
mgb_phys said:
You need to know a speed - the power depends on how long you take to move a certain distance.

Just think in the upwards direction, how many m/s vertically is 15mph along a 1 deg slope?
Then how much potential energy (mgh) does the train gain per metre?
How many vertical m/s is it moving?
So how many joules/second = watts is this.

(ps pick a set of units to convert everything into)
(decided to use 20km/h for speed) so vertically it would be 20km*cos(1°) = 3.046km = .846 m (in an hour).
pot energy = 15Mkg*9.8*.846 = 124,362,000 kg m^2/s^2 =
(1 J = 1 kg m^2/s^2)
=1.24362 E8 J (required to move over one hour)

1 locomotive = 4400 hp = 3,281,080 J/s * 3600 s/h = 11811888000 = 1.1811888 E10 J/h

is this math right? the j/h of the locomotive seems to be higher so it would just take one?

EDIT*
for pot energy it would be 15Mkg*9.8*(.846m/h*1/3600h/s) = 34545 kg m^2/s^2 = as many joules (1 J = 1 kg m^2/s^2)
so it only needs 34545 J per hour to run? then it would be way under what the train produces, I must be missing something

Is that just the extra energy needed for the slope and I need to include flat land as well?
 
Last edited:
  • #6
20km*cos(1°) = 3.046km
Do you mean 20km sin(1°) = 349m vertically?
3.046km in 20km is 1:7, steep for a road never mind a train!

ps I would work in m/kg/s to avoid confusion
 
  • #7
mgb_phys said:
Do you mean 20km sin(1°) = 349m vertically?
3.046km in 20km is 1:7, steep for a road never mind a train!

ps I would work in m/kg/s to avoid confusion

yes, did it again and got 349m which would put the pot energy at 5.1303E10 J and with one locomotive giving about 1E10 J it would be around 5, which is what the prof said would be around 5-6
thanks
 
  • #8
pot energy at 5.1303E10 J and with one locomotive giving about 1E10 J
Be careful confusing energy and power. You can only talk about a potential energy change for a certain distance or time and you only have a POWER for the locomotive not an energy.
 

What is train-hill hp to pull?

Train-hill hp to pull refers to the amount of horsepower required for a train to pull a certain amount of weight up a hill. It is a measure of the train's pulling capacity.

Why is train-hill hp to pull important?

Knowing the train-hill hp to pull is important for determining the train's efficiency and performance on hilly terrain. It can also help in determining the appropriate size and number of locomotives needed for a particular route.

How is train-hill hp to pull calculated?

Train-hill hp to pull is calculated by taking into account the weight of the train, the grade of the hill, and the desired speed of the train. It is typically measured in horsepower, which is a unit of power equal to 550 foot-pounds per second.

What factors can affect train-hill hp to pull?

Several factors can affect train-hill hp to pull, including the condition of the track, the weather, and the type of locomotive being used. Inclines, curves, and altitude can also impact the train's pulling capacity.

How can train-hill hp to pull be optimized?

To optimize train-hill hp to pull, it is important to regularly maintain and upgrade the track, use efficient locomotives, and properly manage the weight and distribution of cargo on the train. Additionally, planning routes with less steep grades can also help optimize train-hill hp to pull.

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