Question

Calculate the pressure required to maintain the flow of water at the rate of 10 litres/s
through a horizontal tube 10 cm in diameter and 1 km in length. Viscosity of water=
0.001 Nsm-2​

Answer 1

Question:

Calculate the pressure required to maintain the flow of water at the rate of 10 litres/s through a horizontal tube 10 cm in diameter and 1 km in length. [Coefficient of Viscosity =0.001].

Given:

• Flow rate = 10 litres/s
• Diameter= 10cm => Radius=5cm
• Length= 1km
• Viscosity=0.001

Convert all the given quantities to its SI unit.

• Flow rate = 10 litres/s or $\sf 10^{-2} m^3/s$
• Radius = 5cm or $\sf 5 \times 10^{-2} m$
• Length = 1km or $\sf 10^3 m$
• Viscosity($\eta$) =0.001 or $\sf 10^{-3}$ [Constant]

To find:

Pressure(P) required to maintain the flow of water.

Solution:

Now, we know that:

$\sf Volume ~flow~rate= \frac{ \pi Pr^4}{8 \eta l} \\\\ \sf \implies 10^{-2}m^3/s = \frac{ 22 \times P \times (5 \times 10^{-2})^4m}{7 \times 8 \times 10^{-3} \times 10^{3}m} \\\\ \implies \sf P= \frac{10^{-2}m^3/s \times 7 \times 8 \times \cancel{10^{-3}} \times \cancel{10^{3}} m} {22 \times (5 \times 10^{-2})^4m} Pa\\\\ \sf \implies P= \frac{7 \times 8 \times 10^{-2}} {22 \times 5^4 \times 10^{-6}} Pa \\\\ \sf \implies P= \frac{56 \times 10^6}{22 \times 54} Pa \\\\ \sf \implies P= \frac{14 \times 10^6}{11 \times 27} Pa \\\\ \sf \implies P= \frac{1.27 \times 10^6}{27} Pa\\\\ \sf \implies P=4.70 \times 10^4 Pa$