An impulse turbine extracts hydraulic power from water at 20^{\circ} C flowing from reservoir 1 though a 0.75-m riveted steel pipe, 200 m in length, fitted with two regular 90^{\circ} threaded elbows in order to change the direction of the flow, and finally with a 0.3-m nozzle in order to create a forced jet, as illustrated in Figure EP 4.23a. The turbine is located 50 m below the elevation of the headwater at point 1 and just slightly above the elevation of the tailwater at point 2. The pipe entrance from reservoir 1 is square-edged, and the nozzle has a conical angle, θ = 30^{\circ} . Assume a Darcy–Weisbach friction factor, f of 0.13; a minor head loss coefficient due to a squared-edged pipe entrance, k = 0.5; a minor head loss coefficient due to an elbow, k= 1.5; and a minor head loss coefficient due to the nozzle, k = 0.02. The pressure at point a in the pipe is measured using a piezometer. (a) Determine the pressure and velocity at point a in the pipe and the velocity at point b at the nozzle. (b) Determine the magnitude of the freely available discharge to be extracted by the turbine. (c) Determine the magnitude of the freely available head 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 4.23: An impulse turbine extracts hydraulic power from water at 20...
(a) In order to determine the pressure and velocity at point a in the pipe and the velocity at point b at the nozzle, the energy equation is applied between points 1 and a and between points a and b, assuming the datum is through points a and b. 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}: = 50 m Z_{a}: = 0 m Z_{b}: = 0 m
P_{b}: = 0 \frac{N}{m^{2}} D_{a}: = 0.75 m A_{a}: = \frac{\pi . D^{2}_{a} }{4} = 0.442 m^{2}
D_{b}: = 0.3 m A_{b}: = \frac{\pi . D^{2}_{b} }{4} = 0.071 m^{2}
k_{ent}: = 0.5 k_{elbow}: = 1.5 k_{nozzle}: =0.02 f: = 0.13 L: = 200 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}} V_{a}: = 1 \frac{m}{sec} V_{b}: = 2 \frac{m}{sec}
h_{fmaj}: = 1 m h_{fent}: = 1 m h_{felbow}: = 1 m h_{fnozzle}: = 1 m
\frac{P_{1}}{\gamma } + Z_{1} + \frac{V^{2}_{1} }{2.g} -h_{fmaj} – h_{fent} – 2. h_{felbow} = \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_{fnozzle} = \frac{P_{b}}{\gamma } + Z_{b} + \frac{V^{2}_{b} }{2.g}
h_{fmaj} = f \frac{L}{D_{a}} \frac{V^{2}_{a}}{2.g} h_{fent} = k_{fent} \frac{V^{2}_{a}}{2.g} h_{felbow} = k_{felbow} \frac{V^{2}_{a}}{2.g}
h_{fnozzle} = k_{fnozzle} \frac{V^{2}_{b}}{2.g} V_{a}. A_{a} = V_{b}. A_{b}
P_{a}: = 2.437 \times 10^{5} \frac{N}{m^{2}} V_{a}: = 3.546 \frac{m}{sec} V_{b}: = 22.163 \frac{m}{sec}
h_{fmaj}: = 22.219 m h_{fent}: = 0.32 m 2.h_{felbow}: = 1.923 m h_{fnozzle}: = 0.501 m
(b) In order to determine the magnitude of the freely available discharge to be extracted by the turbine, the continuity equation is applied at point b at the nozzle as follows:
Q_{b}: = V_{b}. A_{b}= 1.567 \frac{m^{3}}{s}(c) In order to determine the magnitude of the freely available head to be extracted by the turbine, the energy equation is applied between points b and 2 as follows:
P_{2}: = 0 \frac{N}{m^{2}} V_{2}: = \frac{m}{sec} Z_{2}: = 0 m
Guess value: h_{turbine}: = 20 cm
Given
\frac{P_{b}}{\gamma } + Z_{b} + \frac{V^{2}_{b} }{2.g} – h_{turbine} = \frac{P_{2}}{\gamma } + Z_{2} + \frac{V^{2}_{2} }{2.g}
h_{turbine}: = Find (h_{turbine}) = 25.037 m
(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_{b} = 1.567 \frac{m^{3}}{s}
P_{tin}: = \gamma .Q. h_{turbine} = 384.014 kW
(e) The EGL and HGL are illustrated in Figure EP 4.23a.
(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}) = 345.613 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} = 345.613 kW \eta _{generator}: = 0.85
Guess value: P_{gout}: = 350 kW
Given
\eta _{generator}= \frac{P_{gout}}{P_{gin}} P_{gout}: = Find (P_{gout}) = 293.771 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
The turbine–generator system is illustrated in Figure EP 4.23b.