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Sagot :
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 8° 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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