# Train-hill hp to pull, problem

1. Feb 9, 2009

### Ed Boon

1. The problem statement, all variables and given/known data
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.

2. Relevant equations

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

3. 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 cant relate them

any help is appriciated

Ed

2. Feb 9, 2009

### mgb_phys

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. Feb 9, 2009

### LowlyPion

Before you pick a speed, you might want to check the brake linings.

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

4. Feb 9, 2009

### Ed Boon

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. Feb 9, 2009

### Ed Boon

(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: Feb 9, 2009
6. Feb 9, 2009

### mgb_phys

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. Feb 10, 2009

### Ed Boon

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. Feb 10, 2009

### mgb_phys

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.