Question 4.25: The required hydraulic power to be extracted by an impulse t...

(a) In order to determine the design magnitude of the freely available head to be extracted by the turbine, h_{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:

\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}}
P_{tin}: = 700 kW                  Q: = 2 \frac{m^{3}}{sec}

Guess value:                    h_{turbine}: = 50 m

Given

P_{tin} = \gamma . Q . h_{turbine}
h_{turbine}: = Find (h_{turbine}) = 35.749 m

(b)–(e) In order to determine the design level of headwater in reservoir 1 to accommodate the design magnitude of the freely available head to be extracted by the turbine, h_{turbine} , the energy equation is applied between points 1 and a and between points b and 2. In order to determine the pressure at point a and the maximum allowable velocity of flow at the nozzle at point b without causing cavitation, the energy equation is applied between points a and b, assuming the pressure at point b is the vapor pressure of water at 20^{\circ} C . And, in order to determine the design pipe and nozzle size in order to accommodate the design magnitude of the required freely available discharge, the continuity equation is applied between points a and b. Thus, from Table A.2 in Appendix A, for water at 20^{\circ} C , the vapor pressure, p v is 2.34 × 10^{3} N/m^{2} abs. However, because the vapor pressure is given in absolute pressure, the corresponding gage pressure is computed by subtracting the atmospheric pressure as follows: p_{gage} = p_{abs} − p_{atm} , where the standard atmospheric pressure is 101.325 × 10^{3} N/m^{2}abs .

P_{1}: = 0 \frac{N}{m^{2}}                       V_{1: = 0 \frac{m}{sec} }                      Z_{a}: = 0 m                       Z_{b}: = 0 m

P_{2}: = 0 \frac{N}{m^{2}}                       V_{2: = 0 \frac{m}{sec} }                      Z_{2}: = 0 m

P_{V}: = 2.34 \times 10^{3} \frac{N}{m^{2}} – 101.325 \times 10^{3} \frac{N}{m^{2}} = -9.899 \times 10^{4} \frac{N}{m^{2}}
P_{b}: = P_{V} = – 9.899 \times 10^{4} \frac{N}{m^{2}}

k_{ent}: = 0.5                       k_{elbow}: = 1.5                       k_{nozzle}: = 0.02                  f: = 0.13                      L: = 200 m

Guess value:                        Z_{1}: = 100 m                      P_{a}: = 2 \times 10^{3} \frac{N}{m^{2}}                      V_{a}: = 1 \frac{m}{sec}                      V_{b}: = 2 \frac{m}{sec}
D_{a}: = 0.9 m                      D_{b}: = 0.4 m

h_{fmaj}: = 1 m                      h_{fent}: = 1 m                      h_{felbow}: = 1 m                      h_{fnozzle}: = 1 m

Given

\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}
\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_{fmaj} = f \frac{L}{D_{a}} \frac{V^{2}_{a}}{2.g}                       h_{fent}= k_{ent} \frac{V^{2}_{a}}{2.g}                       h_{felbow}= k_{elbow} \frac{V^{2}_{a}}{2.g}
h_{nozzle}= k_{fnozzle} \frac{V^{2}_{b}}{2.g}                       Q = V_{a} \frac{\pi .D^{2}_{a} }{4}                       Q = V_{b} \frac{\pi .D^{2}_{b} }{4}

\left ( \begin{matrix} Z_{1} \\ P_{a} \\ V_{a} \\ V_{b} \\ D_{a} \\ D_{b} \\ h_{fmaj} \\ h_{fent} \\ h_{felbow} \\ h_{fnozzle} \end{matrix} \right ) : = Find (Z_{1}, P_{a}, V_{a}, V_{b}, D_{a}, D_{b}, h_{fmaj}, h_{fent}, h_{felbow}, h_{fnozzle})
Z_{1} = 57.004 m                       P_{a}= 3.531 \times 10^{5} \frac{N}{m^{2}}                      V_{a} = 3.44 \frac{m}{sec}                      V_{b} = 29.996 \frac{m}{sec}
D_{a} = 0.85 m                      D_{b} = 0.291 m
h_{fmaj} = 18.226 m                       h_{fent}= 0.302 m                       2. h_{felbow}= 1.809 m                      h_{fnozzle}= 0.917 m

(f) The EGL and HGL are illustrated in Figure EP 4.23a.