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3 years ago

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water at 20c discharges from a stroke tank through a 150m length of horizontal pipe. The pipe is smooth and has a diameter of 75

mm. The entrance of the pipe is square edged. What is depth of water needed to produce a volumetric flow rate of .1m3/s

1 answer:

inessss [21]3 years ago

7 0

Answer:

The dept of the water needed is 26.11m

Explanation:

Given the following parameters:

Length of the pipe = $150m$

Diameter of the pipe = $75mm = 0.75m$

Volumetric flow rate $1m^{3}/s$

Knowing that:

Volumetric flowrate $= area (A) * Velocity(V)$

==> $0.1 = A*V$ *Knowing that $V = \sqrt{2gh}$

==> $0.1 = \frac{\pi }{4}* (0.075)^{2} * \sqrt{2gh}$

==> $h = 26.11m$

Hence, the dept of water needed to produce a volumetric flow rate of $1m^{3}/s$ is 26.11m.

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2 years ago

Water is boiled in a pot covered with a loosely fitting lid at a location where the pressure is 85.4 kPa. A 2.61 kW resistance h

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Answer:

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Explanation:

The energy provided by the resistance heater must be equal to the energy required to boil the water:

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2 years ago

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marishachu [46]

Answer:

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Explanation:

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3 years ago

Read 2 more answers

A square-thread power screw is used to raise or lower the basketball board in a gym, the weight of which is W = 100kg. See the f

KIM [24]

Answer:

power = 49.95 W

and it is self locking screw

Explanation:

given data

weight W = 100 kg = 1000 N

diameter d = 20mm

pitch p = 2mm

friction coefficient of steel f = 0.1

Gravity constant is g = 10 N/kg

solution

we know T is

T = w tan(α + φ ) $\frac{dm}{2}$ ...................1

here dm is = do - 0.5 P

dm = 20 - 1

dm = 19 mm

and

tan(α) = $\frac{L}{\pi dm}$ ...............2

here lead L = n × p

so tan(α) = $\frac{2\times 2}{\pi 19}$

α = 3.83°

and

f = 0.1

so tanφ = 0.1

so that φ = 5.71°

and now we will put all value in equation 1 we get

T = 1000 × tan(3.83 + 5.71 ) $\frac{19\times 10^{-3}}{2}$

T = 1.59 Nm

so

power = $\frac{2\pi N \ T }{60}$ .................3

put here value

power = $\frac{2\pi \times 300\times 1.59}{60}$

power = 49.95 W

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as φ > α

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2 years ago

Water is pumped steadily through a 0.10-m diameter pipe from one closed pressurized tank to another tank. The pump adds 4.0 kW o

jekas [21]

Complete Question

Complete Question is attached below.

Answer:

$V'=5m/s$

Explanation:

From the question we are told that:

Diameter $d=0.10m$

Power $P=4.0kW$

Head loss $\mu=10m$

$\frac{P_1}{\rho g}+\frac{V_1^2}{2g}+Z_1+H_m=\frac{P_2}{\rho g}+\frac{V_2^2}{2g}+Z_2+\mu$

$\frac{300*10^3}{\rho g}+35+Hm=\frac{500*10^3}{\rho g}+15+10$

$H_m=(\frac{200*10^3}{1000*9.8}-10)$

$H_m=10.39m$

Generally the equation for Power is mathematically given by

$P=\rho gQH_m$

Therefore

$Q=\frac{P}{\rho g H_m}$

$Q=\frac{4*10^4}{1000*9.81*10.9}$

$Q=0.03935m^3/sec$

Since

$Q=AV'$

Where

$A=\pi r^2\\A=3.142 (0.05)^2$

$A=7.85*10^{-3}$

Therefore

$V'=\frac{0.03935m^3/sec}{7.85*10^{-3}}$

$V'=5m/s$

5 0

3 years ago