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2 years ago

18. Bela is doing a continuity test. What's he checking?

Engineering

1 answer:

murzikaleks [220]2 years ago

8 0

Answer:

the bela is doing a continuity test he checking current

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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

maria [59]

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 $\frac{u}{v}$ = 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 = $\frac{\pi }{4} d^2$

so from equation 1

0.04 = $\frac{\pi }{4} d^2$ × 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 = $\frac{\pi D N}{60}$

so 19.1 = $\frac{\pi D 720}{60}$

so D = 0.506 m = 506.62 mm

so ratio is $\frac{D}{d} = \frac{506.62}{1.23}$

ratio = 412

and

we know efficiency = $\frac{outputpower}{inputpower}$

0.80 × ρQgh = output power

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

power output P = 28.72 kW

speed of wheel is

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

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

speed of wheel = 13.63

7 0

4 years ago

Gawain 2 talasalitaan

sukhopar [10]

N dnkdicwnjxiwkxdndjdjdj

7 0

3 years ago

Water at 20 °C is flowing with velocity of 0.5 m/s between two parallel flat plates placed 1 cm apart. Determine the distances f

Basile [38]

Answer:

The distance from the entrance at which the boundary layers meet is 0.516m

The distance from the entrance at which the thermal boundary layers meet is 1.89m

Explanation:

For explanation, look at the attached file

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3 years ago

What is the work required to deflect a linear spring (k=32 kN/m) by 120 cm?

Anit [1.1K]

Answer:

the work done by the linear spring will be equal to 23.04 k J

Explanation:

given,

k = 32 k N /m

deflected spring = 120 cm

= 1.2 m

work done by spring can be calculate as

= $\dfrac{1}{2}kx^2$

= $\dfrac{1}{2}\times 32 \times 1.2^2$

= 23.04 k J

hence, the work done by the linear spring will be equal to 23.04 k J

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3 years ago

Which term refers to the impurities found during the welding process ?

Aleks04 [339]

Answer:

idk

Explanation:

idk

6 0

3 years ago