User Name Remember Me? Password

 Thermodynamics and Fluid Mechanics Thermodynamics and Fluid Mechanics Physics Help Forum Sep 6th 2017, 06:50 AM #1 Junior Member   Join Date: Sep 2017 Posts: 6 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!   Sep 6th 2017, 07:15 AM #2 Senior Member   Join Date: Apr 2015 Location: Somerset, England Posts: 1,035 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?   Sep 6th 2017, 08:34 AM #3 Senior Member   Join Date: Jun 2010 Location: NC Posts: 418 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   Sep 8th 2017, 10:06 AM #4 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  Tags conservation of energy, flow, hydraulic, hydraulic turbine, mass, mass flow rate, problem, rate, simple, thermodynamics, turbine, water Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Physics Forum Discussions Thread Thread Starter Forum Replies Last Post ling233 Advanced Mechanics 3 Jan 17th 2016 05:18 AM bquinn22 Thermodynamics and Fluid Mechanics 0 Jan 5th 2015 04:00 PM Padawan Thermodynamics and Fluid Mechanics 0 Oct 17th 2011 07:44 PM greg_900 Advanced Thermodynamics 1 Mar 17th 2009 09:13 AM Devlan Thermodynamics and Fluid Mechanics 0 Nov 10th 2008 08:48 AM