# Mass flow rate of water in a hydraulic turbine

#### oya

Hello again. I have another problem that I'm working on:

The intake to a hydraulic turbine installed in a dam is located at an elevation of 10 m above the turbine exit. Water enters at 20 C with negligible velocity and exits from the turbine at 10 m/s. The water passes through the turbine with no significant changes in temperature or pressure between the inlet and exit, and heat transfer is negligible. The acceleration of gravity is constant at g = 9.81 m/s2. If the power output at steady state is 500 kW, what is the mass flow rate of water in kg/s?

The mass flow rate is determined using the steady-state energy balance, dECV/dt=QCV−WCV+∑in[mi(hi+1/2Vi2+gzi)]−∑out[mO(hO+1/2VO2+gzO)]

I began with this:
0 = 0 - 500kW + ∑in [m (...

And this is where I stopped because I am not sure how to plug in the information that I do not have such as enthalpy, kinetic and potential energy, etc. Please help me figure this out!

#### studiot

Looks an awfully complicated equation your energy balance.

The gravitational potential energy is converted to two types of energy, if we neglect friction and second order effects like vena contracta.

1) The ??????? Energy of the moving water at the exit

2) The power takeoff of the turbine.

1) is determined by the water velocity v = √(2gH) by Torricelli's theorem. (This can also be derived from Bernoulli's theorem.)

2) The power takeoff is given.

Does this help?

#### oya

I think I got it... Can you please check to make sure it's correct?

Knowns
Height: Z1 - Z2 = 10 meters
Temperature: T1 = 20C = T2
Pressure: P1 = P2
Velocity: V1 = 0, V2 = 10 m/s
Heat transfer = 0
Gravity = 9.81 m/s2
Work flow rate = W.cv = 500kW

Assumption
This is a steady state, therefore m.i = m.o
Change in T = 0
Change in P = 0

Calculation
dE/dt = Q.CV−W.CV + m.[(h1 - h2)+(1/2V12 - 1/2V22)+g(z1 - z2)
Steady state so dE/dt = 0
Q.CV = 0 because no heat transfer
(h1 - h2) = 0 because no temperature or pressure changes
0 = 0 - 500kW + m.[(0 - (10 m/s)2 / 2) + (9.81 m/s2 x 10 m)]
500kW = m.(-50m2/s2 +98.1m2/s2)
500kW = m.(48.1m2/s2)
Conversion:
If 1000m2/s2 = 1kJ/kg, then 48.1m2/s2 = 0.0481kJ/kg
If 1 kW = 1 kJ/s, then 500kW = 500kJ/s
500kJ/s = m.(0.0481kJ/kg); therefore, m. = 500kJ/s / 0.0481kJ/kg ~ 10,395 kg/s