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4 years ago

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Consider a 0.15-mm-diameter air bubble in a liquid. Determine the pressure difference between the inside and outside of the air

bubble if the surface tension at the air-liquid interface is a. 0.1 N/m. b. 0.12 N/m.

1 answer:

777dan777 [17]4 years ago

4 0

Answer:

a) $\Delta P= 2666.66 Pa$

b) $\Delta P= 3200 Pa$

Explanation:

Given that

Diameter ,d= 0.15 mm

We know that pressure difference is given as

$\Delta P=\dfrac{4\sigma }{d}$

Now by putting the values

When surface tension 0.1 N/m :

The surface tension , $\sigma=0.1\ N/m$

$\Delta P=\dfrac{4\times 0.1 }{0.15}\times 10^3\ Pa$

$\Delta P= 2666.66 Pa$

When surface tension 0.12 N/m :

The surface tension , $\sigma=0.12\ N/m$

$\Delta P=\dfrac{4\times 0.12 }{0.15}\times 10^3\ Pa$

$\Delta P= 3200 Pa$

a) $\Delta P= 2666.66 Pa$

b) $\Delta P= 3200 Pa$

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The denominator of a fraction is 4 more than the numenator. If 4 is added to the numenator and 7 is added to the denominator, th

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Answer:

$\frac{3}{7}$

Explanation:

Lets take the numerator of the fraction to be = x

So the denominator of the fraction is 4 more than the numerator = x+4

The fraction is ; $\frac{x}{4+x}$

Now add 4 to the numerator and add 7 to the denominator as;

$\frac{x+4}{4+x+7} =\frac{x+4}{x+11}$

This new fraction is equal to 1 half =1/2

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$\frac{x+4}{x+11} =\frac{1}{2}$

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3 years ago

A silicon carbide plate fractured in bending when a blunt load was applied to the plate center. The distance between the fractur

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Question in order:

A silicon carbide plate fractures in bending when a blunt load was applied to the plate center. The distance between the fracture origin and the mirror/mist boundary on the fracture surface was 0.796 mm. To determine the stress used to break the plate, three samples of the same material were tested and produced the following. What is the estimate of the stress present at the time of fracture for the original plate?

Mirror Radius (mm) Bending Failure Stress (MPa)

0.603 225

0.203 368

0.162 442

Answer:

191 MPa

Explanation:

Failure stress of bending is Inversely proportional to the mirror radius

$Bending Stress = \frac{1}{(Mirror Radius)^{n}}$

At mirror radius 1 = 0.603 mm Bending stress 1 = 225 Mpa..............(1)

At mirror radius 2 = 0.203 mm Bending stress 2 = 368 Mpa...............(2)

At mirror radius 3 = 0.162 mm Bending stress 3 = 442 Mpa...............(3)

comparing case 1 and 2 using the above equation

$\frac{Stress 1}{Stress 2} = ({\frac{Radius 2}{Radius 1}})^{n_1}$

$\frac{225}{368} = ({\frac{0.203}{0.603}})^{n_1}$

$0.6114 = (0.3366)^{n_1}$

Taking the natural logarithm of both side

ln(0.6114) = n ln(0.3366)

n₁ = ln(0.6114)/ln(0.3366)

n₁ = 0.452

comparing case 2 and 3 using the above equation

$\frac{Stress 2}{Stress 3} = ({\frac{Radius 3}{Radius 2}})^{n_2}$

$\frac{368}{442} = ({\frac{0.162}{0.203}})^{n_2}$

$0.8326 = (0.7980)^{n_2}$

Taking the natural logarithm of both side

ln(0.8326) = n ln(0.7980)

n₂ = ln(0.8326)/ln(0.7980)

n₂ = 0.821

comparing case 1 and 3 using the above equation

$\frac{Stress 1}{Stress 3} = ({\frac{Radius 3}{Radius 1}})^{n_3}$

$\frac{225}{442} = ({\frac{0.162}{0.603}})^{n_3}$

$0.5090 = (0.2687)^{n_3}$

Taking the natural logarithm of both side

ln(0.5090) = n ln(0.2687)

n₃ = ln(0.5090)/ln(0.2687)

n₃ = 0.514

average for n

$n = \frac{n_1 + n_2 + n_3}{3}$

$n = \frac{0.452 +0.821 + 0.514}{3}$

n = 0.596

Hence to get bending stress x at mirror radius 0.796

$\frac{Stress x}{Stress 3} = ({\frac{Radius 3}{Radius x}})^{0.596}$

$\frac{Stress x}{225} = ({\frac{0.603}{0.796}})^{0.596}$

$\frac{Stress x}{225} = 0.8475$

stress x = 191 MPa

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4 years ago

Oil with a density of 850 kg/m3 and kinematic viscosity of 0.00062 m2/s is being discharged by an 8-mm-diameter, 42-m-long horiz

sladkih [1.3K]

Answer:

The flow rate of oil through the pipe is 1.513E-7 m³/s.

Explanation:

Given

Density, ρ = 850 kg/m³

Kinematic viscosity, v = 0.00062 m²/s

Diameter, d = 8-mm = 0.008m

Length of horizontal pipe, L = 42-m

Height, h = 4-m.

We'll solve the flow rate of oil through the pipe by using Hagen-Poiseuille equation.

This is given as

∆P = (128μLQ)/πD⁴

Where ∆P = Rate of change of pressure

μ = Dynamic Viscosity

Q = Flow rate of oil through the pipe.

First, we need to determine the dynamic viscosity and the rate of change in pressure

Dynamic Viscosity, μ = Density (ρ) * Kinematic viscosity (v)

μ = 850 kg/m³ * 0.00062 m²/s

μ = 0.527kg/ms

Then, we calculate the rate of change of pressure.

Assuming that the velocity through the pipe is so small;

∆P = Pressure at the bottom of the tank

∆P = Density (ρ) * Acceleration of gravity (g) * Height (h)

Taking g = 9.8m/s²

∆P = 850kg/m³ x 9.8m/s² x 4m

∆P = 33320N/m²

Recall that Hagen-Poiseuille equation.

∆P = (128μLQ)/πD⁴ --- Make Q the subject of formula

Q = (πD⁴P)/(128μL)

By substituton;

Q = (π * 0.008⁴ * 33320)/(128 * 0.527 * 42)

Q = 0.00000015133693643099

Q = 1.513E-7 m³/s.

Hence, the flow rate of oil through the pipe is 1.513E-7 m³/s.

8 0

3 years ago