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-Dominant- [34]
4 years ago
9

The high-pressure air system at OSU's Aerospace Research Center is fed by a set of two cylindrical tanks. Each tank has an outer

height of 50 ft and an outer diameter of 4.6 ft. The tanks are made of 0.1 ft thick steel (steel = 499 lbm/ft?) and store air at a maximum pressure of 2500 psi at -10 °F. How much load must the support structure at the base of the tanks carry?
Engineering
1 answer:
lana66690 [7]4 years ago
3 0

Answer:

179000 lb

Explanation:

The supports must be able to hold the weight of the tank and the contents. Since tanks are pressure tested with water, and the supports cannot fail during testing, we disregard the air and will consider the weight of water.

The specific weight of water is ρw = 62.4 lbf/ft^3

These tanks are thin walled because

D / t = 4.6 / 0.1 = 46 > 10

To calculate the volume of steel we can approximate it by multiplying the total surface area by the thickness:

A = 2 * π/4 * D^2 + π * D * h

The steel volume is:

V = A * t

The specific weight is

ρ = δ * g

ρs = 499 lbm/ft^3 * 1 lbf/lbm = 499 lbf/ft^3

The weight of the steel tank is:

Ws = ρs * V

Ws = ρs * A * t

Ws = ρs * (2 * π/4 * D^2 + π * D * h) * t

Ws = 499 * (π/2 * 4.6^2 + π * 4.6 * 50) * 0.1 = 37700 lb

The weight of water can be approximated with the volume of the tank:

Vw = π/4 * D^2 * h

Ww = ρw * π/4 * D^2 * h

Ww = 62.4 * π/4 * 4.6^2 * 50 = 51800 lb

Wt = Ws + Ww = 37700 + 51800 = 89500 lb

Assuming the support holds both tanks

2 * 89500 = 179000 lb

The support must be able to carry 179000 lb

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Answer:

Fluids

Explanation:

Fluids has special properties that allow forces and pressure to be distributed evenly within them.

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umka21 [38]

Technician A is correct. Technician B is wrong because a gear's transmission is used to increase or decrease torque.

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1 year ago
Koch traded Machine 1 for Machine 2 when the fair market value of both machines was $60,000. Koch originally purchased Machine 1
Mariana [72]

Answer:

Koch's adjusted basis in machine 2 after the exchange is $60,000

Explanation:

given data

fair market value = $60,000

originally purchased Machine 1 = $76,900

Machine 1 adjusted basis = $40,950

Machine 2 seller purchase = $64,050

Machine 2 adjusted basis = $55,950

solution

As he exchanged machine for another at $60,000

and this exchanged in fair market

so adjusted basis =  $50,000

Adjusted basis is the price of the item that affects the factors that are considered price. These factors usually include taxes, depreciation value, and other costs of acquiring and maintaining a given item. Adjusted basis is important so the right amount to sell

Adjusted basis increases when a person deducts expenses from factor taxes and operating statements

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3 0
3 years ago
There are two types of cellular phones, handheld phones (H) that you carry and mobile phones (M) that are mounted in vehicles. P
nexus9112 [7]

Answer:

A) P(W) = 0.5

B) P(MF) = 0.3

C) P(H) = 0.6

Explanation:

We are told that there are two types of cellular phones which are handheld phones (H) that you carry and mobile phones (M) that are mounted in vehicles.

Also, Phone calls can be classified by the traveling speed of the user as fast (F) or slow (W).

Thus, the sample space is combination of types and classification we are given and it is written as;

S = {HF, HW, MF, MW}

A) Now, phones can either be fast(F) or slow(W). Thus, we can write;

P(F) + P(W) = 1

We are given P(F) = 0.5

Thus;

0.5 + P(W) = 1

P(W) = 1 - 0.5

P(W) = 0.5

B) Now, from the problem statement, a phone call can either be made with a handheld(H) or mobile(M). Thus the sample space partition is {H, M} and we can express as;

P(H ∩ F) + P(M ∩ F) = P(F)

We are given P[F] = 0.5 and P[HF] = 0.2.

P(H ∩ F) is same as P[HF]

Also, P(M ∩ F) is same as P(MF)

Thus;

0.2 + P(MF) = 0.5

P(MF) = 0.5 - 0.2

P(MF) = 0.3

C) Similarly, mobile Phone calls can either be fast or slow. It means the sample space partition is {F, W}

Thus;

P(M) = P(MW) + P(MF)

P(M) = 0.1 + 0.3

P(M) = 0.4

Now, since cellular phones can either be handheld(H) or Mobile(M), then we can say;

P(H) + P(M) = 1

P(H) + 0.4 = 1

P(H) = 1 - 0.4

P(H) = 0.6

5 0
3 years ago
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