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Andrews [41]
3 years ago
6

A cylindrical tank is required to contain a gage pressure 520 kPa . The tank is to be made of A516 grade 60 steel with a maximum

allowable normal stress of 150 MPa . If the inner diameter of the tank is 2.6 m , what is the minimum thickness, t, of the wall? Express your answer with appropriate units to three significant figures.
Engineering
1 answer:
enot [183]3 years ago
4 0

Answer:

t= 4.5 mm

Explanation:

Given that

P = 520 KPa ( gauge)

Maximum allowable normal stress ,σ= 150

d= 2.6 m

Wall thickness = t

The normal stress for pressure vessel given as

\sigma=\dfrac{Pd}{2t}               ( hoop stress)

We always take maximum stress for safe design.

\sigma=\dfrac{Pd}{2t}

Now by putting the values

150\times 1000=\dfrac{520\times 2.6}{2t}

t= 4.5 mm

So the minimum thickness, t, of the wall is 4.5 mm

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Answer:

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lines need to send data for the fifth bit in an 8 bit system

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The time factor for a doubly drained clay layer
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Answer with Explanation:

Assuming that the degree of consolidation is less than 60% the relation between time factor and the degree of consolidation is

T_v=\frac{\pi }{4}(\frac{U}{100})^2

Solving for 'U' we get

\frac{\pi }{4}(\frac{U}{100})^2=0.2\\\\(\frac{U}{100})^2=\frac{4\times 0.2}{\pi }\\\\\therefore U=100\times \sqrt{\frac{4\times 0.2}{\pi }}=50.46%

Since our assumption is correct thus we conclude that degree of consolidation is 50.46%

The consolidation at different level's is obtained from the attached graph corresponding to Tv = 0.2

i)\frac{z}{H}=0.25=U=0.71 = 71% consolidation

ii)\frac{z}{H}=0.5=U=0.45 = 45% consolidation

iii)\frac{z}{H}=0.75U=0.3 = 30% consolidation

Part b)

The degree of consolidation is given by

\frac{\Delta H}{H_f}=U\\\\\frac{\Delta H}{1.0}=0.5046\\\\\therefore \Delta H=50.46cm

Thus a settlement of 50.46 centimeters has occurred

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3 years ago
Steep safety ramps are built beside mountain highways to enable vehicles with defective brakes to stop safely. A truck enters a
Virty [35]

Answer:

a) The additional time required for the truck to stop is <u>8.5 seconds</u>

b) The additional distance traveled by the truck is <u>230.05 ft</u>

Explanation:

Since the acceleration is constant, the average speed is:

(final speed - initial speed) / 2 = 0.75 v0

Since travelling at this speed for 8.5 seconds causes the vehicle to travel 690 ft, we can solve for v0:

0.75v0 * 8.5 = 690

v0 = 108.24 ft/s

The speed after 8.5 seconds is: 108.24 / 2 = 54.12 ft/s

We can now use the following equation to solve for acceleration:

v^2 - u^2 = 2*a*s

54.12^2 - 108.24^2 = 2*a*690

a = -6.367 m/s^2

Additional time taken to decelerate: 54.12/6.367 = 8.5 seconds

Total distance traveled:

v^2 - u^2 = 2*a*s

0 - 108.24^2 = 2 * (-6.367) * s

solving for s we get total distance traveled = 920.05 ft

Additional Distance Traveled: 920.05 - 690 = 230.05 ft

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3 years ago
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