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3 years ago

What size discharge line ( pressure line ) would you select for a 20 gpm pump?

Engineering

1 answer:

Alenkasestr [34]3 years ago

3 0

Answer:

diameter= 1 in

Explanation:

Some authors have developed studies to find the recommended flow rate at the pump outlet, in order to avoid high pressure losses and cavitation.

Assuming that this pump carries water, the recommended speed is 2m / s, so knowing the flow rate = 20gpm we can find the diameter of the pipe.

Q=flow rate=20gpm=0.0012618m^3/s

V=speed=2m/s

A=Area

Q=VA

A=Q/V

$A=\frac{0.0012618m^3/s}{2m/s} =0.0006309m^2=630.2mm^2\\\\$

Taking into account that the flow is through a circular pipe we can use the equation to find the area of a circle, and then find the diameter

$A=\frac{\pi D^2 }{4} \\D=\sqrt[2]{\frac{4A}{\pi } } =\sqrt{\frac{4(630.8)}{\pi } } =23.34mm$

the result is 23.34mm, if we find the closest commercial diameter to this value, we find that it is 25.4mm = 1in

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A wing generates a lift L when moving through sea-level air with a velocity U. How fast must the wing move through the air at an

vredina [299]

Answer:

V1 = 1.721 * V2

Explanation:

To start with, we assume that both lift forces are equal, such that

L2 = L1

1 is that of the level at 10000 m, and 2 is that of the level at sea level.

Next, we try and substitute the general formula for both forces such that

C(l).ρ1/2.V1².A = C(l).ρ2/2.V2².A

On further simplification, we have

ρ1.V1² = ρ2.V2², making V1 subject of formula, we have

V1 = √(ρ2/ρ1). V2²

Using the values of density for air at 10000 m and at sea level(source is US standard atmosphere), we have

V1 = √(1.225/0.4135) * V2

V1 = √2.9625 * V2

V1 = 1.721 * V2

4 0

3 years ago

2. One of the many methods used for drying air is to cool the air below the dew point so that condensation or freezing of the mo

REY [17]

Answer:

0.5°c

Explanation:

Humidity ratio by mass can be expressed as

the ratio between the actual mass of water vapor present in moist air - to the mass of the dry air

Humidity ratio is normally expressed in kilograms (or pounds) of water vapor per kilogram (or pound) of dry air.

Humidity ratio expressed by mass:

x = mw / ma (1)

where

x = humidity ratio (kgwater/kgdry_air, lbwater/lbdry_air)

mw = mass of water vapor (kg, lb)

ma = mass of dry air (kg, lb)

It can be as:

x = 0.005 (100) / [(100 - 100)]

x = 0.005 x 100 / (100 - 100)

x = 0.005 x 100 / 0

x = 0.5°c

So the temperature to which atmospheric air must be cooled in order to have humidity ratio of 0.005 lb/lb is 0.5°c

6 0

3 years ago

Select the correct answer.

Lilit [14]

Answer:

Explanation:

Confidential data is not supposed to be shared amongst others.

8 0

3 years ago

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How large a force is required to accelerate a 1300 kg car from rest to a speed of 20 m/s in a distance of 80 m?

topjm [15]

F=m*a

F=80*20

F =1600 ans"

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3 years ago

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The inlet and exhaust flow processes are not included in the analysis of the Otto cycle. How do these processes affect the Otto

lara31 [8.8K]

Answer:

Suction and exhaust processes do not affect the performance of Otto cycle.

Explanation:

Step1

Inlet and exhaust flow processes are not including in the Otto cycle because the effect and nature of both the process are same in opposite direction.

Step2

Inlet process or the suction process is the process of suction of working fluid inside the cylinder. The suction process is the constant pressure process. The exhaust process is the process of exhaust out at constant pressure.

Step3

The suction and exhaust process have same work and heat in opposite direction. So, net effect of suction and exhaust processes cancels out. The suction and exhaust processes are shown below in P-V diagram of Otto cycle:

Process 0-1 is suction process and process 1-0 is exhaust process.

7 0

3 years ago