Areaction turbine extracts hydraulic power from water at 20ºC flowing from reservoir 1 though a 0.9-m galvanized iron pipe 150 m in length, as illustrated in Figure EP 4.24. The turbine is located 30 m below the elevation of the headwater at point 1 and 2 m above the elevation of the tailwater at point 2. The pipe entrance from reservoir 1 is square-edged, and the draft tube has a conical angle, θ= 9° . Assume a Darcy–Weisbach friction factor, f of 0.15; a minor head loss coefficient due to a squared-edged pipe entrance, k = 0.5; and a minor head loss coefficient due to the conical diffuser, k = 0.5. The pressure at point a in the pipe is measured using a piezometer. (a) Determine the pressure and velocity at points a and b. (b) Determine the magnitude of the freely available head tobeextracted by the turbine.(c) Determine themagnitudeof thefreely available discharge to be extracted by the turbine. (d) Determine the magnitude of the freely available hydraulic power to be extracted by the turbine. (e) Draw the energy grade line and the hydraulic grade line. (f) Assuming a typical turbine efficiency of 90%, determine the freely generated shaft power output by the turbine. (g) Assuming a typical generator efficiency of 85%, determine the freely generated electric power output by the generator. (h) Determine the resulting overall turbine–generator system efficiency.
Question 4.24: Areaction turbine extracts hydraulic power from water at 20^...
(a)–(b) In order to determine the pressure and velocity at points a and b and the magnitude of the freely available head to be extracted by the turbine, the energy equation is applied between points 1 and a, between points a and b, and between points b and 2, assuming z_{a} = z_{b} and the datum is at point 2. Furthermore, the velocities at points a and b are related by application of the continuity equation between points a and b as follows:
P_{1}: = 0 \frac{N}{m^{2}} V_{1}: = 0 \frac{m}{sec} Z_{1}: = 32 m Z_{a}: = 2 m Z_{b}: = 2 m
P_{2}: = 0 \frac{N}{m^{2}} V_{2}: = 0 \frac{m}{sec} Z_{2}: = 0 m
D_{a}: = 0.9 m A_{a}: = \frac{\pi . D^{2}_{a} }{4} = 0.636 m^{2}
D_{b}: = 0.9 m A_{b}: = \frac{\pi . D^{2}_{b} }{4} = 0.636 m^{2}
k_{ent}: = 0.5 k_{diffuser}: = 0.5 f: = 0.15 L: = 150 m
\rho : = 998 \frac{kg}{m^{3}} g: = 9.81 \frac{m}{sec^{2}} \gamma : = \rho .g = 9.79 \times 10^{3} \frac{kg}{m^{2}s^{2}}
Guess value: P_{a}: = 2 \times 10^{3} \frac{N}{m^{2}} P_{b}: = 1 \times 10^{3} \frac{N}{m^{2}} V_{a}: = 1 \frac{m}{sec}
V_{b}: = 2 \frac{m}{sec} h_{fmaj}: = 1 m h_{fent}: = 1 m h_{turbine}: = 20 m
h_{fdiffuser}: = 1 m
Given
\frac{P_{1}}{\gamma } + Z_{1} + \frac{V^{2}_{1} }{2.g} – h_{fmaj} – h_{fent} = \frac{P_{a}}{\gamma } + Z_{a} + \frac{V^{2}_{a} }{2.g}
\frac{P_{a}}{\gamma } + Z_{a} + \frac{V^{2}_{a} }{2.g} – h_{turbine} = \frac{P_{b}}{\gamma } + Z_{b} + \frac{V^{2}_{b} }{2.g}
\frac{P_{b}}{\gamma } + Z_{b} + \frac{V^{2}_{b} }{2.g} – h_{fdiffuser} = \frac{P_{2}}{\gamma } + Z_{2} + \frac{V^{2}_{2} }{2.g}
h_{fmaj}= f \frac{L}{D_{a}} \frac{V^{2}_{a}}{2.g} h_{fent} = k_{ent} \frac{V^{2}_{a}}{2.g} V_{a}.A_{a} = V_{b}.A_{b}
h_{fdiffuser} = k_{diffuser} \frac{V^{2}_{b}}{2.g}
\left ( \begin{matrix} P_{a} \\V_{a} \\h_{turbine} \\ P_{b} \\ V_{b} \\ h_{fmaj} \\ h_{fent} \\ h_{fdiffuser} \end{matrix} \right ): = Find (P_{a}, V_{a}, h_{turbine}, P_{b}, V_{b}, h_{fmaj}, h_{fent}, h_{fdiffuser})
P_{a}: = 1.639 \times 10^{5} \frac{N}{m^{2}} V_{a}: = 3.135 \frac{m}{sec} h_{turbine}: = 18.977 m
P_{b}: = -2.203 \times 10^{4} \frac{N}{m^{2}} V_{b}: = 3.135 \frac{m}{sec}
h_{fmaj}: = 12.522 m h_{fent}: = 0.25 m h_{fdiffuser}: = 0.25 m
Where
\frac{P_{a}}{\gamma } = 16.727 m \frac{P_{b}}{\gamma } = -2.25 m h_{fdiffuser}: = \frac{P_{a}}{\gamma } – \frac{P_{b}}{\gamma } = 18.977 m
(c)In order to determine the magnitude of the freely available discharge to be extracted by the turbine, the continuity equation is applied at point a as follows:
Q_{a}: = V_{a}.A_{a} = 1.997 \frac{m^{3}}{s}(d) In order to determine the magnitude of the freely available hydraulic power to be extracted by the turbine, Equation 4.187 h_{turbine} = \frac{(P_{t})_{in}}{\gamma Q} = \frac{(P_{t})_{out}/\eta _{turbine}}{\gamma Q} = \frac{wT_{shaft,out}/\eta _{turbine}}{\gamma Q} is applied as follows:
Q: = Q_{a} = 1.997 \frac{m^{3}}{s}
P_{tin}:= \gamma .Q. h_{turbine}= 370.529 kW
(e) The EGL and HGL are illustrated in Figure EP 4.24.
(f) Assuming a typical turbine efficiency of 90%, the freely generated shaft power output by the turbine is computed by applying Equation 4.167 \eta _{turbine}= \frac{shaft power}{hydraulic power} = \frac{\overset{\cdot }{W}_{turbine} }{(\overset{\cdot }{W}_{turbine} )_{e}} = \frac{\overset{\cdot }{W}_{shaft,out} }{(\overset{\cdot }{W}_{turbine} )_{e}} = \frac{wT_shaft,out}{(\overset{\cdot }{W}_{turbine} )_{e}} = \frac{P_{out of turbine}}{P_{in of turbine}} = \frac{(P_{t})_{out}}{(P_{t})_{in}} as follows:
Guess value: P_{tout}: = 350 kW
Given
\eta _{turbine} =\frac{ P_{tout}}{P_{tin}}
P_{tout}: = Find (P_{tout}) = 333.476 kW
(g) Assuming a typical generator efficiency of 85%, the freely generated electric power output by the generator is computed by applying Equation 4.174 \eta _{generator}= \frac{electricalpoweroutputfromgenerator}{shaftpowerinputfromturbine} = \frac{\overset{\cdot }{W}_{elect,out} }{(\overset{\cdot }{W}_{shaft,in} )_{e}} = \frac{P_{out of gen}}{P_{in to gen}} = \frac{(P_{g})_{out}}{(P_{g})_{in}} as follows:
P_{gin}: = P_{tout} = 333.476 kW \eta _{ generator} : = 0.85
Guess value: P_{gout}: = 350 kW
Given
\eta _{generator} = \frac{P_{gout}}{P_{gin}}
P_{gout}: = Find (P_{gout}) = 183.455 kW
(h) The resulting overall turbine–generator system efficiency is computed by applying Equation 4.176 \eta _{turbine – generator}= \frac{electricalpower}{hydraulic power} = \frac{\overset{\cdot }{W}_{elect,out} }{(\overset{\cdot }{W}_{turbine} )_{e}} = \frac{(P_{g})_{out}}{(P_{t})_{in}} = \eta _{turbine} \eta _{generator} as follows:
\eta _{ turbinegenerator}: = \frac{P_{gout}}{P_{tin}} = 0.765
\eta _{ turbinegenerator}: = \eta _{ turbine}.\eta _{generator}= 0.765