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A generator is to be driven by a small Pelton wheel with a head of 91.5m at inlet to the nozzle and discharge of 0.04m^3/s. The wheel rotates at 720rpm a n d the velocity coefficient of the nnozzle is 0.98. If the efficiency of the wheel (based on the energy available at entry to the nozzle) is 80 per cent and the ratio of bucket speed to jet speed is0.46, determine the wheel-to-.jet-di meter ratio at the centre-line of the buckets, and the speed of the wheel. What is the dimensionless power specific speed of the wheel?

Respuesta :

Answer:

ratio = 412

speed of wheel = 13.63

Explanation:

given data

head h = 91.5 m

discharge Q = 0.04 m³/s

wheel rotates N = 720 rpm

velocity coefficient Cv = 0.98

efficiency of the wheel η = 80 %

ratio of bucket speed to jet speed [tex]\frac{u}{v}[/tex] = 0.46

to find out

wheel to jet diameter ratio and power specific speed of the wheel

solution

we know that discharge = area × velocity .............1

here velocity = Cv√(2gh)

velocity = 0.98√(2×9.81×91.5) = 41.52 m/s

and area = [tex]\frac{\pi }{4} d^2[/tex]

so from equation 1

0.04 = [tex]\frac{\pi }{4} d^2[/tex] ×  41.52

d = 0.00123 m = 1.23 mm

we know bucket speed to jet speed = 0.46

so bucket speed u = 0.46 v

bucket speed u = 0.46 × 41.52 = 19.1 m/s

and bucket speed = [tex]\frac{\pi D N}{60}[/tex]      

so 19.1 = [tex]\frac{\pi D 720}{60}[/tex]  

so D = 0.506 m = 506.62 mm

so ratio is [tex]\frac{D}{d} = \frac{506.62}{1.23}[/tex]

ratio = 412

and

we know efficiency = [tex]\frac{outputpower}{inputpower}[/tex]

0.80 × ρQgh = output power

power output = 0.80 ×1000×0.04×9.81×91.5

power output P = 28.72 kW

so

speed of wheel is

speed of wheel = [tex]\frac{N\sqrt{P} }{h^{5/4}}[/tex]

speed of wheel = [tex]\frac{720\sqrt{28.72} }{91.5^{5/4}}[/tex]

speed of wheel = 13.63

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