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Thepotemich [5.8K]
4 years ago
10

Which of the following is NOT true about hydraulic valves? A. Directional control valves determine the path of a fluid in a give

n circuit B. Pressure reducing valve is another name for pressure relief valve C. A pressure relief valve is used to limits the maximum pressure in a hydraulic system D. Pilot check valves can permit flow in the normally blocked opposite direction E. All of the above
Engineering
1 answer:
Lelechka [254]4 years ago
5 0

Answer: Option D is not true of hydraulic valves. A hydraulic valve is a device that can change the opening degree of liquid flow path

Explanation:

The pilot check valve allows flow of liquid in one direction and blocks flow in the opposite direction

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Who developed the process of blueprinting?
VikaD [51]
Answer: C.) John Herschel
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3 years ago
Which tool ensures that a fastener has the proper amount of tightness
Sidana [21]

A torque wrench tool is a tool that ensures that a fastener has the proper amount of tightness.

<h3>What is the torque wrench used for?</h3>

The torque wrench tool is used to ensure screws and bolts are properly tightened. When performing home repairs and maintenance of equipment it is quite important that a torque wrench is used in other to prevent a scenario where a fastener (screws and bolts) does not become loose leading to equipment failure or damage. Because of its many advantages, this tool is often found in the possession of construction workers.

You can learn more about the benefits of a torque wrench tool here

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7 0
2 years ago
What was the first prototype of the artificial tree like?
olga_2 [115]

Answer:

An artificial Christmas tree is an artificial pine or fir tree manufactured for the specific purpose of use as a Christmas tree. The earliest artificial Christmas trees were wooden, tree-shaped pyramids or feather trees, both developed by Germans. Most modern trees are made of polyvinyl chloride (PVC) but many other types of trees have been and are available, including aluminum Christmas trees and fiber-optic illuminated Christmas trees.

Explanation:

7 0
4 years ago
Given: A graphite-moderated nuclear reactor. Heat is generated uniformly in uranium rods of 0.05m diameter at the rate of 7.5 x
sineoko [7]

Answer:

The maximum temperature at the center of the rod is found to be 517.24 °C

Explanation:

Assumptions:

1- Heat transfer is steady.

2- Heat transfer is in one dimension, due to axial symmetry.

3- Heat generation is uniform.

Now, we consider an inner imaginary cylinder of radius R inside the actual uranium rod of radius Ro. So, from steady state conditions, we know that, heat generated within the rod will be equal to the heat conducted at any point of the rod. So, from Fourier's Law, we write:

Heat Conduction Through Rod = Heat Generation

-kAdT/dr = qV

where,

k = thermal conductivity = 29.5 W/m.K

q = heat generation per unit volume = 7.5 x 10^7 W/m³

V = volume of rod = π r² l

A = area of rod = 2π r l

using these values, we get:

dT = - (q/2k)(r dr)

integrating from r = 0, where T(0) = To = Maximum center temperature, to r = Ro, where, T(Ro) = Ts = surface temperature = 120°C.

To -Ts = qr²/4k

To = Ts + qr²/4k

To = 120°C + (7.5 x 10^7 W/m³)(0.025 m)²/(4)(29.5 W/m.°C)

To = 120° C + 397.24° C

<u>To = 517.24° C</u>

5 0
4 years ago
An 80-percent-efficient pump with a power input of 20 hp is pumping water from a lake to a nearby pool at a speed of 10 ft/s thr
Sladkaya [172]

Answer:

Part A

The mass plow rate, is approximately 97.0 lbm/s

Part B

The power used to overcome friction, is approximately 1.9 hp

Explanation:

The efficiency of the pump, η = 80%

The power input to the pump, P = 20 hp

The speed of the water through the pipe, v = 10 ft./s

The diameter of the pipe, d = 5.2 inches = 13/30 ft.

The free surface of the pool above the lake, h = 80 ft.

The density of the water, ρ = 62.4 lbm/ft.³

Part A

The mass plow rate, \dot m = Q × ρ

Where;

ρ = 62.4 lbm/ft³

Q = A × v

A = The cross-sectional area of the pipe

∴ Q = π·d²/4 × v = π × ((13/30 ft.)²)/4 × 10 ft.s ≈ 1.4748 ft.³/s

∴ The mass plow rate, \dot m ≈ 1.4748 ft.³/s × 62.4 lbm/ft.³ = 97.02752 lbm/s

The mass plow rate, \dot m ≈ 97.0 lbm/s

Part B

The power to pump the water at the given rate, P_w =  \dot m·g·h

∴ P_w =  97.02752 lbm/s × 32.1740 ft./s² × 80 ft. ≈ 14.1130725 Hp

P_w ≈ 14.1130725 Hp

The power output of the pump, P_{out} = 0.8 × 20 hp = 16 hp

Therefore, the power used to overcome friction, P_f = P_{out} - P_w

∴ P_f ≈ 16 hp - 14.1130725 Hp ≈ 1.8869275 hp

The power used to overcome friction,  P_f ≈ 1.9 hp

7 0
3 years ago
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