Consider fully developed laminar flow in a circular pipe. If the viscosity of the fluid is reduced by half by heating while the flow rate is held constant, how will the pressure drop change?

Short Answer

Expert verified
Answer: The pressure drop will decrease by half.

Step by step solution

01

Understand the Hagen-Poiseuille equation

The Hagen-Poiseuille equation describes the pressure drop in laminar flow through a circular pipe. The equation is given by: \[\Delta P = \frac{8 \mu LQ}{\pi R^4}\] where: - ΔP: Pressure drop - µ: Fluid viscosity - L: Length of the pipe - Q: Flow rate - R: Radius of the pipe
02

Determine the initial pressure drop

Using the Hagen-Poiseuille equation, let's find the initial pressure drop with the given fluid viscosity. Keep the values as variables since the question asked is about how the pressure drop would change and not specific values. The initial pressure drop is: \[\Delta P_1 = \frac{8 \mu_1 LQ}{\pi R^4}\]
03

Determine the final pressure drop

Now, let's find the final pressure drop after the fluid viscosity is reduced by half while keeping the flow rate constant. The new viscosity is: \[\mu_2 = \frac{\mu_1}{2}\] Using the Hagen-Poiseuille equation with the reduced viscosity, the final pressure drop is: \[\Delta P_2 = \frac{8 \mu_2 LQ}{\pi R^4} = \frac{8 (\frac{\mu_1}{2}) LQ}{\pi R^4} = \frac{4 \mu_1 LQ}{\pi R^4}\]
04

Compare the initial and final pressure drops

To find the change in pressure drop, we can compare the initial and final pressure drops. Dividing the final pressure drop by the initial pressure drop gives: \[\frac{\Delta P_2}{\Delta P_1} = \frac{\frac{4 \mu_1 LQ}{\pi R^4}}{\frac{8 \mu_1 LQ}{\pi R^4}} = \frac{1}{2}\]
05

Conclusion

The pressure drop in the laminar flow through the circular pipe will decrease by half after reducing the fluid viscosity by half while keeping the flow rate constant.

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