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juin [17]
3 years ago
9

As shown, a load of mass 10 kg is situated on a piston of diameter D1 = 140 mm. The piston rides on a reservoir of oil of depth

h1 = 42 mm and specific gravity S = 0.8. The reservoir is connected to a round tube of diameter D2 = 5 mm and oil rises in the tube to height h2. Find h2. Assume the oil in the tube is open to atmosphere and neglect the mass of the piston.
Engineering
1 answer:
telo118 [61]3 years ago
7 0

Answer:

165 mm

Explanation:

The mass on the piston will apply a pressure on the oil. This is:

p = f / A

The force is the weight of the mass

f = m * a

Where a in the acceleration of gravity

A is the area of the piston

A = π/4 * D1^2

Then:

p = m * a / (π/4 * D1^2)

The height the oil will raise is the heignt of a colum that would create that same pressure at its base:

p = f / A

The weight of the column is:

f = m * a

The mass of the column is its volume multiplied by its specific gravity

m  = V * S

The volume is the base are by the height

V = A * h

Then:

p = A * h * S * a / A

We cancel the areas:

p = h * S * a

Now we equate the pressures form the piston and the pil column:

m * a / (π/4 * D1^2) = h * S * a

We simplify the acceleration of gravity

m / (π/4 * D1^2) = h * S

Rearranging:

h = m / (π/4 * D1^2 * S)

Now, h is the heigth above the interface between the piston and the oil, this is at h1 = 42 mm. The total height is

h2 = h + h1

h2 = h1 + m / (π/4 * D1^2 * S)

h2 = 0.042 + 10 / (π/4 * 0.14^2 * 0.8) = 0.165 m = 165 mm

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Answer:

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Explanation:

given data

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solution

we know that resistance range of the resistor is express as

V(t) = V_o \times e^{t\RC}    ...........1

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put here value

so for t min 11.5 μs

R = \frac{11.5}{0.227\times ln(\frac{5.03}{0.833})}

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A large tank is filled to capacity with 500 gallons of pure water. Brine containing 2 pounds of salt per gallon is pumped into t
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Answer:

A) A(t) = 10(100 - t) + c(100 - t)²

B) Tank will be empty after 100 minutes.

Explanation:

A) The differential equation of this problem is;

dA/dt = R_in - R_out

Where;

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We are given;

Concentration of salt in inflow = 2 lb/gal

Input rate of brine = 5 gal/min

Thus;

R_in = 2 × 5 = 10 lb/min

Due to the fact that the solution is pumped out at a faster rate, thus it is reducing at the rate of (5 - 10)gal/min = -5 gal/min

So, after t minutes, there will be (500 - 5t) gallons in the tank

Therefore;

R_out = (concentration of salt in outflow) × (output rate of brine)

R_out = [A(t)/(500 - 5t)]lb/gal × 10 gal/min

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So, we substitute the values of R_in and R_out into the Differential equation to get;

dA/dt = 10 - 10A(t)/(500 - 5t)

This simplifies to;

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Rearranging, we have;

dA/dt + 2A(t)/(100 - t) = 10

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Thus, the integrating factor is;

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Now, let's multiply the differential equation by the integrating factor 1/(100 - t)².

We have;

So, we ;

(1/(100 - t)²)(dA/dt) + 2A(t)/(100 - t)³ = 10/(100 - t)²

Integrating this, we now have;

A(t)/(100 - t)² = ∫10/(100 - t)²

This gives;

A(t)/(100 - t)² = (10/(100 - t)) + c

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A(t) = 1000 - 10t - 1000 + 20t - 0.1t²

A(t) = 10t - 0.1t²

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