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:

(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

(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.