Looking for reliable answers? Westonci.ca is the ultimate Q&A platform where experts share their knowledge on various topics. Our Q&A platform offers a seamless experience for finding reliable answers from experts in various disciplines. Get immediate and reliable solutions to your questions from a community of experienced professionals on our platform.
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
We appreciate your visit. Hopefully, the answers you found were beneficial. Don't hesitate to come back for more information. Thanks for stopping by. We strive to provide the best answers for all your questions. See you again soon. Westonci.ca is committed to providing accurate answers. Come back soon for more trustworthy information.
