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3 years ago

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Water is pumped from one large reservoir to another at a higher elevation. If the flow rate is 2.5 ft3 /s and the pump delivers

14 hp, find the energy loss associated with pumping from section 1 to section 2

1 answer:

4vir4ik [10]3 years ago

6 0

Answer:

2132hp ed enregia

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Explanation:

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Prelest! Introduction to Engineering and Technology 1 Select the correct answer. What technological invention allowed for the pr

satela [25.4K]

Answer:

I would think OD. Computers

Explanation:

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3 years ago

(20 points) A 1 mm diameter tube is connected to the bottom of a container filled with water to a height of 2 cm from the bottom

zzz [600]

Solution :

Given :

h = 2 cm

Diameter of the tube , d = 1 mm

Diameter of the hose, D = 6 mm

Between 1 and 2, by applying Bernoulli's principle, we get

As point 1 is just below the free surface of liquid, so

$P_1=P_{atm} \text{ and} \ V_1=0$

$\frac{P_{atm}}{\rho g}+\frac{v_1^2}{2g} +h = \frac{P_2}{\rho g}$

$\frac{101.325}{1000 \times 9.81}+0.02 =\frac{P_2}{\rho g}$

$P_2 = 111.35 \ kPa$

Therefore, 111.325 kPa is the gas supply pressure required to keep the water from leaking back into the tube.

Velocity at point 2,

$V_2=\sqrt{\left(\frac{111.135}{\rho g}+0.02}\right)\times 2g$

= 1.617 m/s

Flow of water, $Q_2 = A_{tube} \times V_2$

$=\frac{\pi}{4} \times (10^{-3})^2 \times 1.617$

$1.2695 \times 10^{-6} \ m^3/s$

Minimum air flow rate,

$Q_2 = Q_3 = A_{hose} \times V_3$

$V_3 = \frac{Q_2}{\frac{\pi}{4}D^2}$

$V_3 = \frac{1.2695 \times10^{-6}}{\pi\times 0.25 \times 36 \times 10^{-6}}$

= 0.0449 m/s

b). Reynolds number in hose,

$Re = \frac{\rho V_3 D}{\mu} = \frac{V_3 D}{\nu}$

υ for water at 25 degree Celsius is $8.9 \times 10^{-7} \ m^2/s$

υ for air at 25 degree Celsius is $1.562 \times 10^{-5} \ m^2/s$

$Re_{hose}=\frac{0.0449 \times 6 \times 10^{-3}}{1.562 \times 10^{-5}}$

= 17.25

Therefore the flow is laminar.

Reynolds number in the pipe

$Re = \frac{V_2 d}{\nu} = \frac{1.617 \times 10^{-3}}{8.9 \times 10^{-7}}$

= 1816.85, which is less than 2000.

So the flow is laminar inside the tube.

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2 years ago

Find the dryness fraction, specific volume and internal energy of steam at 7bar nd enthalpy 2600kj/kh. (0.921,0.2515m³/kg , 2420

Blizzard [7]

Answer:

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Explanation:

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2 years ago

List the parts of a manual transmission <br><br> List the parts of a typical clutch assembly?

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Answer:

Explanation: Clutch Plate.

Clutch Cover.

Clutch Bearing (Release bearing)

Release Fork (clutch fork)

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2 years ago

Gas is kept in a 0.1 m diameter cylinder under the weight of a 100 kg piston that is held down by a spring with a stiffness k =

Artyom0805 [142]

Answer:

The spring is compressed by 0.275 meters.

Explanation:

For equilibrium of the gas and the piston the pressure exerted by the gas on the piston should be equal to the sum of weight of the piston and the force the spring exerts on the piston

Mathematically we can write

$Force_{pressure}=Force_{spring}+Weight_{piston}$

we know that

$Force_{pressure}=Pressure\times Area=300\times 10^{3}\times \frac{\pi \times 0.1^2}{4}=750\pi Newtons$

$Weight_{piston}=mass\times g=100\times 9.81=981Newtons$

Now the force exerted by an spring compressed by a distance 'x' is given by $Force_{spring}=k\cdot x=5\times 10^{3}\times x$

Using the above quatities in the above relation we get

$5\times 10^{3}\times x+981=750\pi \\\\\therefore x=\frac{750\pi -981}{5\times 10^{3}}=0.275meters$

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3 years ago