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abruzzese [7]
2 years ago
13

There are two identical oil tanks. The level of oil in Tank A is 12 ft and is drained at the rate of 0.5 ft/min. Tank B contains

8 ft of oil and is drained at the rate of 0.25 ft/min. After how many minutes will the level of oil in the two tanks be the same?
Engineering
1 answer:
Luba_88 [7]2 years ago
4 0

Answer:

  16 minutes

Explanation:

This is an example of a class of problems in which two quantities start with different initial values and change at different rates. In such problems, the rates of change are generally ones that cause the values to converge.

The question usually asks when the values will be the same. The generic answer is, "when the difference in rates makes up the difference in initial values."

Here the tanks differ in initial fill height by 12 -8 = 4 ft. The rates of change differ by 0.5 -0.25 = 0.25 ft/min. The more filled tank is draining faster (important), so the fill heights will converge after ...

  (4 ft)/(0.25 ft/min) = 16 min

The level in the two tanks will be the same after 16 minutes.

__

<em>Additional comment</em>

The oil levels at that time will be 4 ft.

You can write two equations for height:

  y = 12 -0.5x . . . . . . . height in feet after x minutes (tank A)

  y = 8 -0.25x . . . . . .  height in feet after x minutes (tank B)

These will be equal when ...

  y = y

  12 -0.5x = 8 -0.25x

  4 = 0.25x . . . . . . . . . . add 0.5x -8

  16 = x . . . . . . . . . . . . multiply by 4 . . . . time to equal height

The graph shows when the tanks will have equal heights and when they will be drained.

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Elza [17]

Assumptions:

  • Steady state.
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  • Ideal gas.
  • Reversible process.
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Explanation:

Otto cycle is a thermodynamic cycle widely used in automobile engines, in which an amount of gas (air) experiences changes of pressure, temperature, volume, addition of heat, and removal of heat. The cycle is composed by (following the P-V diagram):

  • Intake <em>0-1</em>: the mass of working fluid is drawn into the piston at a constant pressure.
  • Adiabatic compression <em>1-2</em>: the mass of working fluid is compressed isentropically from State 1 to State 2 through compression ratio (r).

        r =\frac{V_1}{V_2}

  • Ignition 2-3: the volume remains constant while heat is added to the mass of gas.
  • Expansion 3-4: the working fluid does work on the piston due to the high pressure within it, thus the working fluid reaches the maximum volume through the compression ratio.

         r = \frac{V_4}{V_3} = \frac{V_1}{V_2}

  • Heat Rejection 4-1: heat is removed from the working fluid as the pressure drops instantaneously.
  • Exhaust 1-0: the working fluid is vented to the atmosphere.

If the system produces enough work, the automobile and its occupants will propel. On the other hand, the efficiency of the Otto Cycle is defined as follows:

           \eta = 1-(\frac{1}{r^{\gamma - 1} } )

where:

           \gamma = \frac{C_{p} }{C_{v}} : specific heat ratio

Ideal air is the working fluid, as stated before, for which its specific heat ratio can be considered constant.

           \gamma = 1.4

Answer:

See image attached.

5 0
3 years ago
Convert 250 lb·ft to N.m. Express your answer using three significant figures.
vfiekz [6]

Answer:

It will be equivalent to 338.95 N-m

Explanation:

We have to convert 250 lb-ft to N-m

We know that 1 lb = 4.45 N

So foe converting from lb to N we have to multiply with 4.45

So 250 lb = 250×4.45 =125 N

And we know that 1 feet = 0.3048 meter

Now we have to convert 250 lb-ft to N-m

So 250lb-ft=250\times 4.45N\times 0.348M=338.95N-m

So 250 lb-ft = 338.95 N-m

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3 years ago
The dam cross section is an equilateral triangle, with a side length, L, of 50 m. Its width into the paper, b, is 100 m. The dam
lisabon 2012 [21]

Answer:

Explanation:

In an equilateral trinagle the center of mass is at 1/3 of the height and horizontally centered.

We can consider that the weigth applies a torque of T = W*b/2 on the right corner, being W the weight and b the base of the triangle.

The weigth depends on the size and specific gravity.

W = 1/2 * b * h * L * SG

Then

Teq = 1/2 * b * h * L * SG * b / 2

Teq = 1/4 * b^2 * h * L * SG

The water would apply a torque of elements of pressure integrated over the area and multiplied by the height at which they are apllied:

T1 = \int\limits^h_0 {p(y) * sin(30) * L * (h-y)} \, dy

The term sin(30) is because of the slope of the wall

The pressure of water is:

p(y) = SGw * (h - y)

Then:

T1 = \int\limits^h_0 {SGw * (h-y) * sin(30) * L * (h-y)} \, dy

T1 = \int\limits^h_0 {SGw * sin(30) * L * (h-y)^2} \, dy

T1 = SGw * sin(30) * L * \int\limits^h_0 {(h-y)^2} \, dy

T1 = SGw * sin(30) * L * \int\limits^h_0 {(h-y)^2} \, dy

T1 = SGw * sin(30) * L * \int\limits^h_0 {h^2 - 2*h*y + y^2} \, dy

T1 = SGw * sin(30) * L * (h^2*y - h*y^2 + 1/3*y^3)(evaluated between 0 and h)

T1 = SGw * sin(30) * L * (h^2*h - h*h^2 + 1/3*h^3)

T1 = SGw * sin(30) * L * (h^3 - h^3 + 1/3*h^3)

T1 = 1/3 * SGw * sin(30) * L * h^3

To remain stable the equilibrant torque (Teq) must be of larger magnitude than the water pressure torque (T1)

1/4 * b^2 * h * L * SG > 1/3 * SGw * sin(30) * L * h^3

In an equilateral triangle h = b * cos(30)

1/4 * b^3 * cos(30) * L * SG  > 1/3 * SGw * sin(30) * L * b^3 * (cos(30))^3

SG > SGw * 4/3* sin(30) * (cos(30))^2

SG > 1/2 * SGw

For the dam to hold, it should have a specific gravity of at leas half the specific gravity of water.

This is avergae specific gravity, including holes.

6 0
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gregori [183]

Answer:

A: Agricultural Engineer

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