[Dolphinsiu]

I meet one question that is very difficult although the answer is given!

Q. Water is stored in an elevated reservoir. To generate power, water flows from this reservoir down through a large conduit to a turbine and then through a similar-sized conduit. At a point in the conduit 89.5 m above the turbine, the pressure is 172.4 kPa, ans at a level 5 m below the turbine, the pressure is 89.6 kPa. The water flow rate is 0.800 m3/s. The output of the shaft of the turbine is 658 kW.The water density is 1000 kg/m3. If the efficiency of the turbine in converting the mechanical energy given up by the fluid to the turbine shaft is 89% (n_t = 0.89), calculate the friction loss in the turbine in J/kg. Note that in the mechanical-energy balance equation, is equal to the output of the shaft of the turbine over n_t.                         Ans. F = 85.3 J/kg

By Mechanical-energy balance equation,

(p2 - p1)/ρ + 1/2 (u2^2 - u1^2) + g(z2 - z1) - F = n_tWp
-82.8 + 0 + 9.8 (5 - 89.5) - F = - 0.89 (658 x 10^3) / (1000 x 0.8 )
                                      F = - 178.9 J/kg

What's wrong with my calculation?? Why it is equal to 85.3 J/kg in text book answer?

[Donaldson Tan]

Your application of the equation is wrong.

Energy Transferred to Turbine = Difference in Mechanical Energy of Water.

Since the turbine is 89% efficient, then 11% of the energy transferred to the turbine must be consumed by friction, thus:

F = (1 - n_t)(p₂ - p₁ + 0.5ρ(u₂² - u₁²) + g(z₂- z₁))

[Dolphinsiu]

Sorry! I haven't typed the bold word in last time. Will I still apply it correctly?

[BenH]

I know this is an old post, but I just encountered myself with this problem also. I think the balance you are considering must include the n (efficency) dividing the total work.

On the other hand, you should consider the distance 1 = -89.5 m, since it is above the turbine.

Does this makes sense to anyone?