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3 years ago

Write down the types and tasks of the pressure control valves ?

Engineering

1 answer:

Yuki888 [10]3 years ago

8 0

Answer:

There are 6 types of pressure control valves and their function is to regulate the pressure below a threshold level within safe limits and to maintain and control pressure of a particular circuit.

Explanation:

The six type of Pressure valve with their functions are given below:

a. Unloading Valve:

These type of pressure valve are used to pour fluid into the container at very low or no pressure.

b. Safety valve:

These are used when the pressure within the vessel is in excess as inside temperature is near about preset [point point then these valves are open to release the extra pressure and are closed once normal conditions are regained.

c. Pressure Reducing Valve:

These are basically used for the control of the pressure in downstream not exceeding the design limits.

d. Pressure Relief Valves:

These are basically used to limit and regulate the pressure of any system.

e. Counter Balance Valve:

These are used to develop pressure in the reverse direction at the actuator's return line in order to keep the load under control.

f. Sequence Valve:

These are used to maintain sequence or order in the operations of two parts or branches.

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g A smooth pipeline with a diameter of 5 cm carries glycerin at 20 degrees Celsius. The flow rate in the pipe is 0.01 m3/s. What

earnstyle [38]

Answer:

The friction factor is 0.303.

Explanation:

The flow velocity ( $v$ ), measured in meters per second, is determined by the following expression:

$v = \frac{4\cdot \dot V}{\pi \cdot D^{2}}$ (1)

Where:

$\dot V$ - Flow rate, measured in cubic meters per second.

$D$ - Diameter, measured in meters.

If we know that $\dot V = 0.01\,\frac{m^{3}}{s}$ and $D = 0.05\,m$ , then the flow velocity is:

$v = \frac{4\cdot \left(0.01\,\frac{m^{3}}{s} \right)}{\pi\cdot (0.05\,m)^{2}}$

$v \approx 5.093\,\frac{m}{s}$

The density and dinamic viscosity of the glycerin at 20 ºC are $\rho = 1260\,\frac{kg}{m^{3}}$ and $\mu = 1.5\,\frac{kg}{m\cdot s}$ , then the Reynolds number ( $Re$ ), dimensionless, which is used to define the flow regime of the fluid, is used:

$Re = \frac{\rho\cdot v \cdot D}{\mu}$ (2)

If we know that $\rho = 1260\,\frac{kg}{m^{3}}$ , $\mu = 1.519\,\frac{kg}{m\cdot s}$ , $v \approx 5.093\,\frac{m}{s}$ and $D = 0.05\,m$ , then the Reynolds number is:

$Re = \frac{\left(1260\,\frac{kg}{m^{3}} \right)\cdot \left(5.093\,\frac{m}{s} \right)\cdot (0.05\,m)}{1.519 \frac{kg}{m\cdot s} }$

$Re = 211.230$

A pipeline is in turbulent flow when $Re > 4000$ , otherwise it is in laminar flow. Given that flow has a laminar regime, the friction factor ( $f$ ), dimensionless, is determined by the following expression:

$f = \frac{64}{Re}$

If we get that $Re = 211.230$ , then the friction factor is:

$f = \frac{64}{211.230}$

$f = 0.303$

The friction factor is 0.303.

4 0

2 years ago

Acoke can with inner diameter(di) of 75 mm, and wall thickness (t) of 0.1 mm, has internal pressure (pi) of 150 KPa and is suffe

NemiM [27]

Answer:

All 3 principal stress

1. 56.301mpa

2. 28.07mpa

3. 0mpa

Maximum shear stress = 14.116mpa

Explanation:

di = 75 = 0.075

wall thickness = 0.1 = 0.0001

internal pressure pi = 150 kpa = 150 x 10³

torque t = 100 Nm

finding all values

∂1 = 150x10³x0.075/2x0,0001

= 0.5625 = 56.25mpa

∂2 = 150x10³x75/4x0.1

= 28.12mpa

T = 16x100/(πx75x10³)²

∂1,2 = 1/2[(56.25+28.12) ± √(56.25-28.12)² + 4(1.207)²]

= 1/2[84.37±√791.2969+5.827396]

= 1/2[84.37±28.33]

∂1 = 1/2[84.37+28.33]

= 56.301mpa

∂2 = 1/2[84.37-28.33]

= 28.07mpa

This is a 2 d diagram donut is analyzed in 2 direction.

So ∂3 = 0mpa

∂max = 56.301-28.07/2

= 14.116mpa

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