Help with Mine Engineering Assignment Questions

[Poco Express]

Hi
I am doing a mine engineering degree by correspondence and I have an assignment to do and the lecture notes do not provide any assistance on some of the cals.
I am hoping some of you can help me
I will start with these two questions and see if you can help before posting more:
Q1)
Determine the reqired power to pump 11.36 l/s against a total dynamic head of 112.78m if the pump operates at 70% efficiency
Q2)
What is the required power to pump 4000 litres per minute against a total dynamic head of 120m if the pump operates at 75% efficiency

The formula the notes gives me is
P = Q.γ.H / 10 power 6
But I can't work out what this means

 

[mfb]

Power is distance*force/time, and force from gravity is mass*g with the gravitational acceleration $g=9.81\frac{m}{s^2}$. You know mass/time and distance, and g is given, so you can determine the true power required to lift that water. If that value is 70% of the input power of the pump, can you determine what 100% are?

 

[NascentOxygen]

http://imageshack.us/a/img811/5412/thgooglefriend1.gif Try http://www.engineeringtoolbox.com/pump-fan-power-d_632.html

 

[Poco Express]

Thanks mfb.
I will look at this in the morning. My quick look I am still confused but I am very tired too.
If you could give a working example I would greatly appreciate it.
Thanks again

 

[CWatters]

Raising 1L of water (weighing 1kg) by 2m in 3 seconds consumes energy...

Energy = mass * g * h
= 1 * 9.8 * 2
= 19.6 joules

Then Power = energy/time

= 19.6/3
= 6.5 Watts

If the pump is 70% eff then

(Power Out/Power In) * 100 = 70

so

Power In = Power Out * 100/70
= 6.5 * 100/70
= 9.3W

Pumping it against "a total dynamic head" of height h is the same as raising water height h. The expression "total dynamic head" includes any losses in the pipework. In the real world the "total dynamic head" might be made up of several things such as the actual height + pipe losses + losses in valves etc.

 

[Poco Express]

Thanks. I went back to my lecturer. Now I am struggling with the rest.

The answer was
P = Q.y.H
106
P = 11.36 x 9800 x 112.78
106
P = 12555572
1000000
P = 12.56 kW
Efficiency of the pump is 70%
Pump actual = P
E
Pump actual = 12.56
0.70
Pump actual = 17.93kW

 

[mfb]

That is a weird way to work with units (more precise, it is a weird way to work without units), but if you use gravitational acceleration in mm/s^2 and the result in kW (and ignore base SI units elsewhere), you get a factor of 10^6 (1000 both from mm<->m and W<->kW).
Apart from that, it is the same way as CWatters solved his example.

 

[CWatters]

I would do it like this..

Output Power required = 11.36 * 9.8 * 112.78 = 12.56 kW

Power in = 12.56 * 100/70 = 17.94 kW

 

[Poco Express]

Thanks again. I'm sure in reality there are calculators to do this stuff. And easier methods to cal it... But I think I will answer how the lecturer wants it... All about the marks at the moment. Lol.
Thanks heaps for your help though.