Recent content by campblor

Thanks,I'll go back and hit the books harder to get my basics up to scratch I'm studying Thermodynamics by correspondence - its a difficult subject as it is, but without the face to face interaction it makes it twice as hard. Thanks again for taking your time to help me out

Thanks mate, I really appreciate that!:smile: That definitely makes sense now. Its the little things that make a big difference!:cool:

or was i meant to have the ke2 & pe2 both be multiplied by (1kJ/kg / 1000m^2/s^2)

I'm getting the 42.022kJ/kg from the Sat. liquid hf@10C from table A-4 I'm completely lost with your second point sorry - I think I've been looking this for so long that my pea brain had turned to pea mush

Net heat transfer to the water is 0; delta Q = 0 -Wout = 0; Win = 2x(250MW x 0.98) = 490000kW m = 569444kg/s h1 = hf@10C = 42.022kJ/kg; h2=? ke2 = ½ v^2 = 1/2x(1.8m/s)^2 = 1.6m/s^2; ke1 = 0 pe2 = gz2 = 9.81m/s^2 x 70m = 686.7; pe1 = 0 Qin – Qout + Win – Wout = m(h2 – h1 + ke2 – ke1 + pe2 –...

I have this problem and have been pulling what's left of my hair out: The Wivenhoe hydroelectric power station requires water to be pumped from Wivenhoe Dam to Splityard Creek Dam, using two pumps (each consuming 250 MW of power at 98% efficiency) to elevate the water approximately 70 m. At...