Question

A pump with an 80% efficiency drives water up between two reservoirs through a piping system of total length L = 15 and circular cross -section diameter d = 7 cm. The reservoirs are open to the atmosphere at an ambient temperature of 20 C. The difference in elevation between the free surface can be accounted for by the local loss coefficients Kentrance  Kexit 1.0, Kelbow 0.4. If the volumetric flow rate in the system Q = 10 Liter/s, and the surface roughness of the pipe is  = 0.15 mm, calculate:
(a) The average water flow velocity in the pipe.
(b) The Reynolds number.
(c) The friction factor.
(d) The ratio between the friction head loss and other (local) head losses.
(e) The power required to drive the pump.

Answer 1

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

$\mathbf{K_{entrance}}$ ≈ $\mathbf{K_{exit}}$ ≈ 1.0, $\mathbf{K_{elbow}}$ ≈ 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;

$r = \mathbf{\dfrac{d}{2}}$

Which gives;

$r = \dfrac{0.07 \, cm}{2} = \mathbf{0.035 \, cm}$

$Average \ water \ velocity, \ v =\mathbf{ \dfrac{Q}{A}}$

Therefore;

$v = \dfrac{0.01}{ \pi \times 0.035^2} \approx 2.598$

• 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;

$Re = \mathbf{\dfrac{\rho \cdot V \cdot D}{\mu}}$

Which gives;

• $Re = \dfrac{998.58 \times 2.598 \times 0.07 }{0.001003} \approx \underline{181,058.58}$

(c) The friction factor is given by the following formula;

$\dfrac{1}{\sqrt{f} } = \mathbf{-2.0 \cdot log \left(\dfrac{\epsilon/D}{3.7} + \dfrac{5.74}{Re^{0.9}} } \right)}$

Which gives;

• f ≈ 0.025

(d) Friction head loss is given as follows;

$h_f = \mathbf{f \times \dfrac{L}{D} \times \dfrac{V^2}{2 \cdot g}}$

Which gives;

$h_f = 0.025 \times \dfrac{15}{0.07} \times \dfrac{2.598^2}{2 \times 9.81} \approx \mathbf{1.84}$

$Other \ head \ losses, \ h_l= \sum K \cdot \dfrac{V^2}{2}$

Which gives;

$h_l=(1 + 1+0.4) \times \dfrac{ 2.598^2}{2} \approx \mathbf{8.0995}$

Ratio between friction head loss and other head loss is therefore;

• $\dfrac{h_f}{h_l} \approx \dfrac{1.84}{8.0995} \approx \underline{0.227}$

• The ratio between friction head loss and other head loss is approximately 227:1000

(e) The power required P is found as follows;

$P= \mathbf{ \dfrac{\rho \cdot g \cdot Q \cdot H}{\eta}}$

Which gives;

$P= \dfrac{998.23 \times 9.81 \times 0.01 \times (1.84 + 8.0995)}{0.8} \approx \mathbf{ 1216.67}$

• The power required to drive the pump, P ≈ 1,216.67 W

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