2) Oil with ρ= 876 kg/m3 and μ= 0.24 kg/m · s is flowing through a 1.5 cm diameter pipe that discharges into the atmosphere at 88 kPa. The absolute pressure 15 m before the exit is measured to be 135 kPa. Determine the flow rate of oil through the pipe if the pipe is (a) horizontal, (b) inclined 8° upward from the horizontal, and (c) inclined 8° downward from the horizontal.

Answers

Answer 1

The pressure in a fluid flowing with laminar flow through a pipe is given by

Hagen-Poiseuille equation.

The correct responses are;

(a) If the pipe is horizontal, the flow rate is approximately 1.622 × 10⁻⁵ m³/s(b) If the pipe is inclined upwards, the flow rate is approximately 1.003 × 10⁻³ m³/s(c) If the pipe is inclined 8° downwards, the flow rate is approximately 2.24 × 10⁻⁵ m³/s

Reasons:

When the flow is a steady incompressible flow through pipe, the flow rate

can be derived from the Hagen-Poiseuille equation as follows;

[tex]\displaystyle \dot V = \mathbf{\frac{\left[\Delta P - \rho \cdot g \cdot L \cdot sin\left(\theta \right) \right] \cdot \pi \cdot D^4 }{128 \cdot \mu \cdot L}}[/tex]

ΔP = 135 kPa - 88 kPa = 47 kPa

The density of the oil, ρ = 876 kg/m³

μ = 0.24 kg/(m·s)

L = 15 m

The diameter of the pipe, D = 1.5 cm = 0.015 m

(a) When the pipe is horizontal, we have;

θ = 0°

Which gives;

[tex]\displaystyle \dot V = \mathbf{\frac{\left[47 \times 10^3 - 876 \, kg/m^3 \times 9.81 \, m/s^2 \times 15 \, m \cdot sin\left(0^{\circ} \right) \right] \times \pi \times (0.015 \, m)^4 }{128 \times 0.24 \, kg/(m \cdot s) \times 15 \, m}}[/tex]

[tex]\displaystyle \dot V = \frac{\left[47 \times 10^3\, Pa - 128903.4\, Pa \cdot sin\left(0^{\circ} \right) \right] \times \pi \times (0.015 \, m)^4 }{128 \times 0.24 \, kg/(m \cdot s) \times 15 \, m}[/tex]

[tex]\displaystyle \dot V = \frac{\left[47 \times 10^3\, Pa - 128903.4 \, Pa \cdot sin\left(0^{\circ} \right) \right] \times \pi \times (0.015 \, m)^4 }{128 \times 0.24 \, kg/(m \cdot s) \times 15 \, m} =\frac{0.002379357\cdot \pi}{460.8}[/tex]

[tex]\displaystyle \dot V=\frac{0.002379357\cdot \pi}{460.8} = \mathbf{1.622 \times 10^{-5}}[/tex]

The flow rate when the pipe is horizontal, [tex]\displaystyle \dot V[/tex] = 1.622 × 10⁻⁵ m³/s

(b) When the pipe is inclined 8°, we have;

[tex]\displaystyle \dot V = \mathbf{\frac{\left[47 \times 10^3\, Pa - 128903.4\, Pa \cdot sin\left(8^{\circ} \right) \right] \times \pi \times (0.015 \, m)^4 }{128 \times 0.24 \, kg/(m \cdot s) \times 15 \, m}} = 1.003 \times 10^{-5} \, m^3/s[/tex]

The flow rate of oil through the pipe if the pipe is inclined 8° upwards from the horizontal, [tex]\displaystyle \dot V[/tex] = 1.003 × 10⁻⁵ m³/s

(c) If the pipe is inclined 8° downward from the horizontal, we have;

[tex]\displaystyle \dot V = \frac{\left[47 \times 10^3\, Pa - 128903.4\, Pa \cdot sin\left(-8^{\circ} \right) \right] \times \pi \times (0.015 \, m)^4 }{128 \times 0.24 \, kg/(m \cdot s) \times 15 \, m} = 2.24\times 10^{-5} \, m^3/s[/tex]

If the pipe is inclined 8° upwards from the horizontal, the flow rate of oil through the pipe is, [tex]\displaystyle \dot V[/tex] = 2.24 × 10⁻⁵ m³/s

Learn more about flow through pipes here:

https://brainly.com/question/6858718


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Answer:

f(x) = 8.917x³ + 2x² + 0.083x - 7

Step-by-step explanation:

f(x)= y

Let the coefficients of x³, x², x be a, b, c and the constant be d

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When y = -14, x = -1

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When y = -7, x = 0

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When y = 4, x = 1

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when y = 78, x = 2

78 = a(2)³ + b(2)² + c(2) + d

78= 8a + 4b + 2c + d ...eq. 5

From eq. 3, d is - 7

putting d as - 7 in equations 1, 2, 4 and 5

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-65= -8a + 4b - 2c - 7

-58= -8a + 4b - 2c ...eq. 6

In eq. 2

-14= -a + b - c - 7

-7= -a + b - c ...eq. 7

In eq. 4,

4= a + b + c - 7

11= a + b + c ...eq. 8

In eq. 5

78= 8a + 4b + 2c - 7

85= 8a + 4b + 2c ...eq.9

Subtracting eq. 9 from eq. 6

-143 = -16a - 4c... eq 10

subtracting eq. 8 from eq 7

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putting a as 9 - c in eq. 10

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132 = 107 + 12b + 1

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= 2

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