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4 years ago

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

Engineering

1 answer:

timurjin [86]4 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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Ammonia gas is diffusing at a constant rate through a layer of stagnant air 1 mm thick. Conditions are such that the gas contain

fiasKO [112]

Answer:

The solution to this question is 5.153×10⁻⁴(kmol)/(m²·s)

That is the rate of diffusion of ammonia through the layer is

5.153×10⁻⁴(kmol)/(m²·s)

Explanation:

The diffusion through a stagnant layer is given by

$N_{A} = \frac{D_{AB} }{RT} \frac{P_{T} }{z_{2} - z_{1} } ln(\frac{P_{T} -P_{A2} }{P_{T} -P_{A1} })$

Where

$D_{AB}$ = Diffusion coefficient or diffusivity

z = Thickness in layer of transfer

R = universal gas constant

$P_{A1}$ = Pressure at first boundary

$P_{A2}$ = Pressure at the destination boundary

T = System temperature

$P_{T}$ = System pressure

Where $P_{T}$ = 101.3 kPa $P_{A2} =0$ , $P_{A1} =y_{A}$ , $P_{T} =$ 0.5×101.3 = 50.65 kPa

Δz = z₂ - z₁ = 1 mm = 1 × 10⁻³ m

R = $\frac{kJ}{(kmol)(K)} ,$ T = 298 K and $D_{AB}$ = 1.18 $\frac{cm^{2} }{s}$ = 1.8×10⁻⁵ $\frac{m^{2} }{s}$

$N_{A} = \frac{1.8*10^{-5} }{8.314*295} *\frac{101.3}{1*10^{-3} }* ln(\frac{101.3-0}{101.3-50.65})$ = 5.153×10⁻⁴ $\frac{kmol}{m^{2}s }$

Hence the rate of diffusion of ammonia through the layer is

5.153×10⁻⁴(kmol)/(m²·s)

5 0

3 years ago

The 5-kg collar has a velocity of 5 m>s to the right when it is at A. It then travels along the smooth guide. Determine its s

Gnoma [55]

Answer:

The speed at point B is 5.33 m/s

The normal force at point B is 694 N

Explanation:

The length of the spring when the collar is in point A is equal to:

$lA=\sqrt{0.2^{2}+0.2^{2} }=0.2\sqrt{2}m$

The length in point B is:

lB=0.2+0.2=0.4 m

The equation of conservation of energy is:

$(Tc+Ts+Vc+Vs)_{A}=(Tc+Ts+Vc+Vs)_{B}$ (eq. 1)

Where in point A: Tc = 1/2 mcVA^2, Ts=0, Vc=mcghA, Vs=1/2k(lA-lul)^2

in point B: Ts=0, Vc=0, Tc = 1/2 mcVB^2, Vs=1/2k(lB-lul)^2

Replacing in eq. 1:

$\frac{1}{2}m_{c}v_{A}^{2}+0+m_{c}gh_{A}+ \frac{1}{2}k(l_{A}-l_{ul}) ^{2}=\frac{1}{2}m_{c}v_{B}^{2}+0+0+\frac{1}{2}k(l_{B}-l_{ul}) ^{2}$

Replacing values and clearing vB:

vB = 5.33 m/s

The balance forces acting in point B is:

Fc-NB-Fs=0

$\frac{m_{C}v_{B}^{2} }{R}-N_{B}-k(l_{B}-l_{ul})=0$

Replacing values and clearing NB:

NB = 694 N

6 0

4 years ago

Read 2 more answers

9.For a single-frequency sine wave modulating signal of 3 kHz with a carrier frequency of 36 MHz, what is the spacing between si

nikdorinn [45]

The spacing between sidebands is equal to 6 kHz.

<u>Given the following data:</u>

Modulating signal = 3 kHz.

Carrier frequency = 36 MHz.

<h3>What is a sideband?</h3>

A sideband can be defined as a band of frequencies that are lower or higher than the carrier frequency due to the modulation process. Thus, it will either be lower than or higher than the carrier frequency.

Generally, the frequency of the modulating signal is equal to the spacing between the sidebands. Therefore, a modulating signal of 3 kHz simply means that the lower sideband is <u>3 kHz</u> higher while the upper sideband is <u>3 kHz</u> lower.

Spacing = 3 kHz + 3 kHz = 6 kHz.

Read more on frequency here: brainly.com/question/3841958

8 0

2 years ago

Please help i give brainliest

Mazyrski [523]

Answer:

A mock-up

Explanation:

It is made of cheap and easy to access parts.

5 0

3 years ago

Tech A says never use a water hose to clean up dust after a repair. Tech B says never use a floor scrubber to clean up dust afte

DENIUS [597]

Answer:

Tech B

Explanation:

You ruin your mop that way. it will clump up and than you'll have a mud mess.

5 0

4 years ago