Question 16.SP.4: Refer to Fig. 16.1. Water is delivered from a reservoir thro......
Refer to Fig. 16.1. Water is delivered from a reservoir through a 4-ft-diameter pipe (f=0.025)10,000ft long to a nozzle that emits an 8-in-diameter jet that impinges on an impulse wheel. The surface of the reservoir is at an elevation 2420 ft higher than the nozzle. The impulse wheel is connected to a 20-pole generator in a 60-cycle system. The wheel has a diameter of 106 in and a bucket angle \beta_2=165°. Assuming the head loss through the nozzle can be expressed as 0.05Vj2/2g, where Vj is the jet velocity, find the following: velocity of flow in the pipe, jet velocity, flow rate, speed of the bucket value of u/V1, net head, force F on the bucket, torque on the wheel, shaft horsepower, horsepower at the base of the nozzle, and overall turbine efficiency Assume v2=v1, neglect minor losses, bearing friction, and windage.
Energy equation from surface of reservoir to jet neglecting minor loss at pipe entrance:
2420-0.025 \frac{10,000}{4} \frac{V^{2}}{2 g}-0.05 \frac{V_{j}^{2}}{2 g}=\frac{V_{j}^{2}}{2 g}
Continuity: V_{j}=(48 / 8)^{2} V, \quad V_{j}=36 \mathrm{~V}
\begin{aligned}& 2420-62.5 \frac{V^{2}}{2 g}-0.05 \frac{(36 V)^{2}}{2 g}=\frac{(36 V)^{2}}{2 g} \\\\& 2420=\frac{V^{2}}{2 g}(62.5+64.8+1296)=\frac{V^{2}}{2 g}(1423.3)\end{aligned}
V=\sqrt{2 g(2420 / 1423)}=10.47 \mathrm{ft} / \mathrm{sec}, velocity in the pipe.
V_{j}=36(10.47)=377 \mathrm{ft} / \mathrm{sec}
\begin{aligned}& Q=A V=\pi 2^{2}(10.47)=131.5 \mathrm{cfs} \end{aligned}
\begin{gathered}N=7200 / n \text { gives } n=7200 / N=7200 / 20=360 \mathrm{rpm} \\\\\omega=\frac{2 \pi n}{60}=\frac{6.28(360)}{60}=37.7 \mathrm{rad} / \mathrm{sec} \\\\u=\omega r=37.7\left(\frac{106 / 2}{12}\right)=166.5 \mathrm{ft} / \mathrm{sec} \end{gathered}
Hence: u / V_1=166.5 / 377=0.44
Eq. (16.5) for net head:
\begin{aligned}h & =\frac{p_{B}}{\gamma}+\frac{V_{B}^{2}}{2 g}=Y-h_{L}=2420-f \frac{L}{D} \frac{V^{2}}{2 g} \\\\& =2420-62.5 \frac{10.47^{2}}{2(32.2)}=2314 \mathrm{ft} \end{aligned}
Force on bucket
\begin{aligned}F & =p Q\left(V_{1}-u\right)(1-\cos \beta) ; \quad \beta=165^{\circ}, \quad \cos \beta=-0.966 \\\\& =1.94(131.5)(377-166.5)(1+0.966)=105,600 \mathrm{lb} \end{aligned}
Torque on wheel
T=F r=105,600(53 / 12)=466,300 \mathrm{ft} \cdot \mathrm{lb}
Shaft power \quad P=\frac{T \omega}{550}=\frac{466,300(37.7)}{550}=31,960 \mathrm{hp}
Power at base of nozzle (i.e., power input to entire turbine, including nozzle)
P=\frac{\gamma Q h}{550}=\frac{62.4(131.5) 2314}{550}=34,520 \mathrm{hp}
Overall efficiency
\eta=\frac{\text { Output (or shaft) power }}{\text { Input power }}=\frac{T \omega}{\gamma Q h}=\frac{31,960}{34,520}=0.926=92.6 \%
h=\frac{p_B}{\gamma}+\frac{V_B^2}{2 g}=Y-h_L (16.5)