Question

An incompressible viscous fluid flows through a pipe with a flow rate of 1 mL/s. The pipe has a uniform diameter D0 and a length L0. A pressure difference of P0 between the ends of the pipe is required to maintain the flow rate. What would be the flow rate if the pressure difference was increased to 2P0 and the diameter was increased to 2D0

Answer 1

Answer:

$Q_2 = 32$ mL/s

Explanation:

Given :

The flow is incompressible viscous flow.

The initial flow rate, $Q_1$ = 1 mL/s

Initial diameter, $D_1= D_0$

Initial length, $L_1=L_0$

The initial pressure difference to maintain the flow, $P_1=P_0$

We know for a viscous flow,

$\Delta P = \frac{32 \mu V L}{D^2}$

$\Delta P = \frac{32 \mu Q L}{\frac{\pi}{4}D^4}$

$Q \propto \Delta P \times D^4$

$\frac{Q_1}{Q_2}= \frac{P_1}{P_2} \times \left( \frac{D_1}{D_2} \right)^4$

$\frac{1}{Q_2}= \frac{P_0}{2P_0} \times \left( \frac{D_0}{2D_0} \right)^4$

$\frac{1}{Q_2}= \frac{1}{2} \times \left( \frac{1}{2} \right)^4$

$\frac{1}{Q_2}= \frac{1}{32}$

∴ $Q_2 = 32$ mL/s

Answer 2

The flow rate if the pressure difference was increased to 2P0 and the diameter was increased to 2D0 is; Q2 = 32 mL/s

We are given;

Initial flow rate; Q1 = 1 mL/s

Initial uniform diameter; D0

Initial Length; L0

Initial Pressure difference; P0

Relationship between pressure, flow rate and diameter for vicious flow is given by;

Q1/Q2 = (P1/P2) × (D1/D2)⁴

Where;

Q1 is initial flow rate

Q2 is final flow rate

P1 is initial pressure difference

P2 is final pressure difference

D1 is initial diameter

D2 is final diameter

We are told that the pressure difference was increased to 2P0 and the diameter was increased to 2D0. Thus;

P2 = 2P0

D2 = 2D0

Thus;

1/Q2 = (P0/2P0) × (D0/2D0)⁴

>> 1/Q2 = ½ × (½)⁴

1/Q2 = 1/32

Q2 = 32 mL/s

Read more about vicious flow at; https://brainly.com/question/2684299