Question 4.27: The required hydraulic power to be extracted by a reaction t...

(a) In order to determine the design magnitude of the freely available discharge, 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}: = 800 kW                  h_{turbine}: = 25 m

Guess value:                           Q: = 2 \frac{m^{3}}{sec}

Given

P_{tin} = \gamma .Q. h_{turbine}
Q: = Find (Q)= 3.269 \frac{m^{3}}{s}

(b)–(e) In order to determine the design level of headwater in reservoir 1 in order to
accommodate the design magnitude of the freely available head to be extracted
by the turbine, h{_turbine} and the pressure and velocity at point a, the energy equation is applied between points 1 and a and between points a and b, assuming the pressure at point b is the vapor pressure of water at 20^{\circ} C . In order to determine the maximum allowable height of the draft tube, Z_{b} without causing cavitation, the energy equation is applied between points b and 2, assuming the pressure at point b is the vapor pressure of water at 20^{\circ} C . And, in order to determine the design pipe 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}                   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_{diffuser}: = 0.5                  f: = 0.15                  L: = 150 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}: = 0.5 m                  D_{b}: = 0.5 m                   Z_{a}: = 2 m                  Z_{b}: = 2 m
h_{fmaj}: = 1 m                  h_{fent}: = 1 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}                   h_{ fdiffuser} = k_{diffuser}\frac{V^{2}_{b}}{2.g}

Q = V_{a} \frac{\pi . D^{2}_{a} }{4}                   Q = V_{b} \frac{\pi . D^{2}_{b} }{4}                   Z_{a} = Z_{b}                       D_{a} = D_{b}

Z_{1}: = 57.405 m                   P_{a}: = 1.458 \times 10^{5} \frac{N}{m^{2}}                  V_{a}: = 4.982 \frac{m}{sec}
V_{b}: = 4.982 \frac{m}{sec}
D_{a}: = 0.914 m                  D_{b}: = 0.914 m                   Z_{a}: = 9.478 m                  Z_{b}: = 9.478 m
h_{fmaj}: = 31.14 m                  h_{fent}: = 0.632 m                  h_{ fdiffuser}: = 0.632 m

(f) The EGL and HGL are illustrated in Figure EP 4.24.