Question

Water at 25c is flowing through a 1.0 km long g.L. Pipe of 200 mm diameter at the rate of 0.07 m3/ s. If value of darcy friction factor for this pipe is 0.02 and density of water is 1000 kg/m3, the pumping power (in kw) required to maintain the flow is_____. Up to 1 decimal place

Answer 1

Answer:

sorry i d k

Explanation:

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Answer 2

The pumping power required to maintain flow is 17.2 kW.

Given:

Temperature of water = 25°C

Length of the galvanized pipe = L = 1.0 km = 1000 m

Diameter of the galvanized pipe = D = 200 mm = 0.2 m

Flow rate = Q = 0.07 $\bold{m^3/s}$

Friction factor = f = 0.02

Density of the water = ρ = 1000 $\bold{kg/m^3}$

To find:

Pumping power =P = ?

Formula used:

Power:

$\bold{P=\rho gQh_{f}}$

Pressure Drop:

$\bold{h_{f}=\frac{fLV^{2}}{2gD}}$

Where,

V is velocity of the flow

Solution:

Now, the pressure drop is:

$\bold{h_{f}=\frac{fLV^{2}}{2gD}}$

The velocity of the flow is given as:

$\bold{V=\frac{Q}{A}}$

Where,

Q is flow rate and A is cross sectional area of pipe

Cross section area of pipe is:

$\bold{A=\pi \frac{D^{2}}{4}}$

Now, the velocity of the flow is:

$\bold{V=\frac{Q}{\pi \frac{D^{2}}{4}}}$

$\bold{V=\frac{4Q}{\pi D^{2}}}$

Now, the formula of pressure drop becomes,

$\bold{h_{f}=\frac{fL\left ( \frac{4Q}{\pi D^{2}} \right )^{2}}{2gD}}$

$\bold{h_{f}=\frac{fL\left ( 16Q^{2} \right )}{2gD\left ( \pi^{2}D^{4} \right )}}$

$\bold{h_{f}=\frac{16Q^{2}fL}{2\pi^{2}gD^{5}}}$

$\bold{h_{f}=\frac{8Q^{2}fL}{\pi^{2}gD^{5}}}$

Now, on substituting the know values,

$\bold{h_{f}=\frac{8\times\left ( 0.07 \right )^{2}\times0.02\times1000}{\pi^{2}\times9.81\times\left ( 0.2 \right )^{5}}}$

$\bold{\therefore h_{f}=25.30 \ N/m^{2}}$

Now, the pumping power is:

$\bold{P=1000\times9.81\times0.07\times25.30}$

$\bold{P=17188.101 \ W}$

$\bold{\therefore P=17.2 \ kW}$