Go Back   Physics Help Forum > High School and Pre-University Physics Help > Thermodynamics and Fluid Mechanics

Thermodynamics and Fluid Mechanics Thermodynamics and Fluid Mechanics Physics Help Forum

Reply
 
LinkBack Thread Tools Display Modes
Old Sep 6th 2017, 07:50 AM   #1
oya
Junior Member
 
Join Date: Sep 2017
Posts: 6
Post Mass flow rate of water in a hydraulic turbine

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!
oya is offline   Reply With Quote
Old Sep 6th 2017, 08:15 AM   #2
Senior Member
 
Join Date: Apr 2015
Location: Somerset, England
Posts: 685
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?
studiot is online now   Reply With Quote
Old Sep 6th 2017, 09:34 AM   #3
Senior Member
 
Join Date: Jun 2010
Location: NC
Posts: 360
oya, Hi...

Hi,

Yours is a basic problem. My example: Niagara Power is very similar to what you have. Take a look!
Niagara Power, Inc | THERMO Spoken Here!

Also here is a proof of Torricelli's Theorem:
Torricelli's Theorem | THERMO Spoken Here!

Good Luck with your studies...
TSH
THERMO Spoken Here is offline   Reply With Quote
Old Sep 8th 2017, 11:06 AM   #4
oya
Junior Member
 
Join Date: Sep 2017
Posts: 6
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
oya is offline   Reply With Quote
Reply

  Physics Help Forum > High School and Pre-University Physics Help > Thermodynamics and Fluid Mechanics

Tags
conservation of energy, flow, hydraulic, hydraulic turbine, mass, mass flow rate, problem, rate, simple, thermodynamics, turbine, water



Thread Tools
Display Modes


Similar Physics Forum Discussions
Thread Thread Starter Forum Replies Last Post
volume flow rate ling233 Advanced Mechanics 3 Jan 17th 2016 06:18 AM
Water Flow Rate Through 1/2 Cu Tube bquinn22 Thermodynamics and Fluid Mechanics 0 Jan 5th 2015 05:00 PM
Mass flow rate question Padawan Thermodynamics and Fluid Mechanics 0 Oct 17th 2011 08:44 PM
mass flow air rate of a compressor greg_900 Advanced Thermodynamics 1 Mar 17th 2009 10:13 AM
Flow rate Devlan Thermodynamics and Fluid Mechanics 0 Nov 10th 2008 09:48 AM


Facebook Twitter Google+ RSS Feed