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Sagot :
The losses in the pipe increases the power requirement of the pipe to
maintain a given flowrate.
Responses (approximate value);
(a) 2.598 m/s
(b) 181,058.58
(c) 0.025
(d) 227:1000
(e) 1,216.67 W
Which methods can be used to calculate the pressure head in the pipe?
The given parameters are;
Pump efficiency, η = 80%
Length of the pipe, L = 15 m
Cross-sectional diameter, d = 7 cm
Reservoir temperature, T = 20°C = 293.15 K
[tex]\mathbf{K_{entrance}}[/tex] ≈ [tex]\mathbf{K_{exit}}[/tex] ≈ 1.0, [tex]\mathbf{K_{elbow}}[/tex] ≈ 0.4
Volumetric flow rate, Q = 10 Liters/s = 0.01 m³/s
Surface roughness, ∈ = 0.15 mm
(a) The cross sectional area of the pipe, A = π·r²
Where;
[tex]r = \mathbf{\dfrac{d}{2}}[/tex]
Which gives;
[tex]r = \dfrac{0.07 \, cm}{2} = \mathbf{0.035 \, cm}[/tex]
[tex]Average \ water \ velocity, \ v =\mathbf{ \dfrac{Q}{A}}[/tex]
Therefore;
[tex]v = \dfrac{0.01}{ \pi \times 0.035^2} \approx 2.598[/tex]
- The average velocity of the water, v ≈ 2.598 m/s
(b) The viscosity of water at 20°C is 0.001003 kg/(m·s) given as follows;
Density of water at 20°C, ρ = 998.23 kg/m³
Reynolds' number, Re, is found as follows;
[tex]Re = \mathbf{\dfrac{\rho \cdot V \cdot D}{\mu}}[/tex]
Which gives;
- [tex]Re = \dfrac{998.58 \times 2.598 \times 0.07 }{0.001003} \approx \underline{181,058.58}[/tex]
(c) The friction factor is given by the following formula;
[tex]\dfrac{1}{\sqrt{f} } = \mathbf{-2.0 \cdot log \left(\dfrac{\epsilon/D}{3.7} + \dfrac{5.74}{Re^{0.9}} } \right)}[/tex]
Which gives;
- f ≈ 0.025
(d) Friction head loss is given as follows;
[tex]h_f = \mathbf{f \times \dfrac{L}{D} \times \dfrac{V^2}{2 \cdot g}}[/tex]
Which gives;
[tex]h_f = 0.025 \times \dfrac{15}{0.07} \times \dfrac{2.598^2}{2 \times 9.81} \approx \mathbf{1.84}[/tex]
[tex]Other \ head \ losses, \ h_l= \sum K \cdot \dfrac{V^2}{2}[/tex]
Which gives;
[tex]h_l=(1 + 1+0.4) \times \dfrac{ 2.598^2}{2} \approx \mathbf{8.0995}[/tex]
Ratio between friction head loss and other head loss is therefore;
- [tex]\dfrac{h_f}{h_l} \approx \dfrac{1.84}{8.0995} \approx \underline{0.227}[/tex]
- The ratio between friction head loss and other head loss is approximately 227:1000
(e) The power required P is found as follows;
[tex]P= \mathbf{ \dfrac{\rho \cdot g \cdot Q \cdot H}{\eta}}[/tex]
Which gives;
[tex]P= \dfrac{998.23 \times 9.81 \times 0.01 \times (1.84 + 8.0995)}{0.8} \approx \mathbf{ 1216.67}[/tex]
- The power required to drive the pump, P ≈ 1,216.67 W
Learn more about flow in pipes here:
https://brainly.com/question/7246532
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