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ExtremeBDS [4]
3 years ago
5

A rigid tank of 1 in3 contains nitrogen gas at 600 kPa, 400 K. By mistake someone lets 0.5 kg flow out. If the final temperature

is 375 K what is then the final pressure?

Engineering
1 answer:
tigry1 [53]3 years ago
6 0

Answer:

Final Pressure in the tank after letting 0.5 kg out of the tank is 504.6 kPa

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Liệt kê 10 quá trình sản xuất trong công nghiệp có sử dụng chất xúc tác
Ugo [173]

Answer: English please!

Explanation:

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Opposition to current flow, restricts or resists current flow
BlackZzzverrR [31]

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The answer is c-resistance

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Critical Thinking Scenario The Vice President for Compliance Management, Alexander Goodenuff, supports the position taken by Hen
Kisachek [45]

Answer:

True, True

Explanation:

1st Scenario: Purchase of new contamination equipment would indicate that the plant had been violating environmental regulations and reporting false data. So, Cathy and Henry must not purchase and install new contamination equipment unless the data indicates serious violatino of environmental regulations.

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8 0
3 years ago
A water jet that leaves a nozzle at 60 m/s at a certain flow rate generating power of 250 kW by striking the buckets located on
wlad13 [49]

Answer:

The mass flow rate of jet =69.44 kg/s

Explanation:

Given that

velocity of jet v= 60 m/s

Power  P=250 KW

As we know that force offered by water F

F=\rho\ A \ v^2

Power P= F.v

So now power given as

P=\rho\ A \ v^3

We know that mass flow rate = ρAv

P=mass\ flow\ rate\ \times v^2

250 x 1000 = mass flow rate x 3600

mass flow rate = 69.44 kg/s

So the mass flow rate of jet =69.44 kg/s

4 0
3 years ago
The compressed-air tank has an inner radius r and uniform wall thickness t. The gage pressure inside the tank is p and the centr
Sedaia [141]

Answer:

Explanation:

Given that:

The Inside pressure (p) = 1402 kPa

= 1.402 × 10³ Pa

Force (F) = 13 kN

= 13 × 10³ N

Thickness (t) = 18 mm

= 18 × 10⁻³ m

Radius (r) = 306 mm

= 306 × 10⁻³ m

Suppose we choose the tensile stress to be (+ve) and the compressive stress to be (-ve)

Then;

the state of the plane stress can be expressed as follows:

(\sigma_ x)  = \dfrac{Pd}{4t}+ \dfrac{F}{2 \pi rt}

Since d = 2r

Then:

(\sigma_ x)  = \dfrac{Pr}{2t}+ \dfrac{F}{2 \pi rt}

(\sigma_ x)  = \dfrac{1402 \times 306 \times 10^3}{2(18)}+ \dfrac{13 \times 10^3}{2 \pi \times 306\times 18 \times 10^{-3} \times 10^{-3}}

(\sigma_ x)  = \dfrac{429012000}{36}+ \dfrac{13000}{34607.78467}

(\sigma_ x)  = 11917000.38

(\sigma_ x)  = 11.917 \times 10^6 \ Pa

(\sigma_ x)  = 11.917 \ MPa

\sigma_y = \dfrac{pd}{2t} \\ \\ \sigma_y = \dfrac{pr}{t} \\ \\  \sigma _y = \dfrac{1402\times 10^3 \times 306}{18} \ N/m^2 \\ \\ \sigma _y = 23.834 \times 10^6 \ Pa \\ \\ \sigma_y = 23.834 \ MPa

When we take a look at the surface of the circular cylinder parabolic variation, the shear stress is zero.

Thus;

\tau _{xy} =0

3 0
3 years ago
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