Question 9.4.1: A Pelton wheel nozzle, for which velocity coefficient Cν = 0......
A Pelton wheel nozzle, for which velocity coefficient C_ν = 0.97, is 400 m below water level of a lake. The jet diameter is 80 mm, the pipe diameter D is 60 cm, its length is 4 km and f = 0.0032. The bucket deflects the jet trough 165° and they run at 0.48 times the jet speed, bucket friction reducing the relative velocity at outlet by 15% of the relative velocity at inlet. Mechanical efficiency = 90%. Determine (a) flow rate and (b) shaft power developed by the turbine.
Given:
\begin{matrix}d=80 mm,& \mathrm{Pipe dia}, D=60 cm &\mathrm{Pipe length}, L=4 km & f=0.0032\\K_f=0.97 &K_u=0.48 &\eta _{\mathrm{mech}}=0.9 & K=1-0.15 =0.85 \end{matrix}
z= height of water surface in lake above nozzle axiz =400 m
Steps: This problem explains how to take into account the head loss in pipe, nozzle and to calculate shaft horse power. Following steps will do the needful.
1. We have to first determine jet velocity so that discharge through nozzle may be determined while taking into account the friction loss in pipe. It can be obtained by using (i) mass conservation between two sections one in pipe and another at nozzle outlet (ii) modified Bernoulli’s equation between points on free surface in lake and exit of nozzle. This gives, if A and C are pipe area and velocity of flow in pipe respectively,
AC=a C_1 \mathrm{(Mass conservation)} \Rightarrow C = (d/D)^2 C_1=0.0177 C_1 (A)
p_a/w+0+z=p_1/w + {C_1}^2/2g+0+h_L+h_n, (B)
Where h_L \text{and} h_n are head loss in pipe and nozzle, pa is atmospheric pressure at free surface in lake, velocity head at free surface is neglected, being small. In Pelton turbine
p_1 = p_a, since jet is issued in atmosphere. Let C_1^{\prime} is theoretical velocity without any loss in nozzle then C_ν = C_1/C_1^{\prime}. Therefore,
h_n=(C_1^{\prime2}-{C_1}^2)/2g=(1/C_\upsilon ^2-1) C_1^2/2g (C)
h_c is obtained from Darey equation h_L=\frac{4fL}{D}\times \frac{C^2}{2g}, which gives h_L terms of C and in turn
in C_1, using Eq. A. Substituting the values in Eq. B, we get after simplification C_1 = 83.48 m/s and Q = discharge = π/4 × d^2 × C_1 = 0.419 m^3/s
2. To find out power developed by turbine, first hydraulic power will be determined from the velocity diagram, which is shown in Fig. 9.4.3. Let us first find out all that is required to draw it. C_1 has been obtained and therefore u = K_u × C_1 = 40.07 m/s, since K_u is given. Inlet triangle gives Cr_1 = C_1 – u = 43.4 m/s \text{and then} Cr_2 = 0.85 × Cr_1 = 310.89 m/s. It may be noted that bucket velocity coefficient is not given rather loss in relative velocity is given. From this bucket velocity coefficient comes out to 0.85. Also angle of deflection of jet is 165°, which means β_2 = 180 – 165 = 15°. With this data, outlet velocity triangle can be completed. This gives, assuming runner is fast (see Fig.9.4.3):
Cw_2 = u – Cr_2 \cos β_2 = 4.44 m/s,
Hydraulic power = (wQ/g) (Cw_1 – Cw_2) u = 1327 kW
Shaft power = 0.9 × 1327 = 1194.3 kW
Note a positive value of Cw_2 means that our assumption is correct. It would not affect our calculations had it been zero or negative as the value was to be used in turbine equation taking the sign as well.
