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3 years ago

Reusable refrigerant containers under high-pressure must be hydrostatically tested how often?

Engineering

1 answer:

timurjin [86]3 years ago

4 0

Answer:

5 years

Explanation:

A hydrostatic testing machine is the equipment used to measure the strength or structural integrity of high pressure vessels that are designed for the transportation of liquids or gas. Among the containers that can be subjected to hydrostatic tests can be found boilers, gas cylinders (transport or storage) and pipes of water and gas systems.

In general, the use of this type of equipment is responsible for ensuring that the different types of containers do not leak in the entire container, pipes and connections, as well as that they are structurally safe to be operated by a person or Otherwise, the risk of leakage or explosion due to high pressure can be very high.

To perform hydrostatic tests on pipes or containers, it is necessary that the object to be evaluated be placed in a steel chamber that is filled with water at normal pressure; subsequently, the pressurized water is pumped into the item being tested, so that the container will expand, forcing the water out of the steel chamber and subsequently the pressure is released, which forces the water to return to The steel chamber.

From this test you can calculate the amount of water that goes out and the one that returns to the steel chamber, which is used to determine if the containers and pipes that are being tested pass or fail the hydrostatic test. From the tests it can be determined if the items are safe for use, if they require repairs or need replacing.

In the case of reusable high pressure refrigerant containers, the tests of said containers must be carried out at least every 5 years to avoid accidents and determine the conditions in which the container is located.

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A flywheel made of Grade 30 cast iron (UTS = 217 MPa, UCS = 763 MPa, E = 100 GPa, density = 7100 Kg/m, Poisson's ratio = 0.26) h

hram777 [196]

Answer:

N = 38546.82 rpm

Explanation:

$D_{1}$ = 150 mm

$A_{1}= \frac{\pi }{4}\times 150^{2}$

= 17671.45 $mm^{2}$

$D_{2}$ = 250 mm

$A_{2}= \frac{\pi }{4}\times 250^{2}$

= 49087.78 $mm^{2}$

The centrifugal force acting on the flywheel is fiven by

F = M ( $R_{2}$ - $R_{1}$ ) x $w^{2}$ ------------(1)

Here F = ( -UTS x $A_{1}$ + UCS x $A_{2}$ )

Since density, $\rho = \frac{M}{V}$

$\rho = \frac{M}{A\times t}$

$M = \rho \times A\times t$ $M = 7100 \times \frac{\pi }{4}\left ( D_{2}^{2}-D_{1}^{2} \right )\times t$

$M = 7100 \times \frac{\pi }{4}\left ( 250^{2}-150^{2} \right )\times 37$

$M = 8252963901$

∴ $R_{2}$ - $R_{1}$ = 50 mm

∴ F = $763\times \frac{\pi }{4}\times 250^{2}-217\times \frac{\pi }{4}\times 150^{2}$

F = 33618968.38 N --------(2)

Now comparing (1) and (2)

$33618968.38 = 8252963901\times 50\times \omega ^{2}$

∴ ω = 4036.61

We know

$\omega = \frac{2\pi N}{60}$

$4036.61 = \frac{2\pi N}{60}$

∴ N = 38546.82 rpm

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3 years ago

Why is concrete on its own not a good material to use

Inga [223]

Answer:

It has poor tensile strength despite having high compressive strength

Explanation:

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aleksandr82 [10.1K]

Answer:

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alekssr [168]

Answer:

d. All of the above would require an EIS.

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