Suppose we have a tube whose cross-sectional area narrows from [itex]A_1[/itex] to [itex]A_2[/itex]. We attach this to a tank which supplies fluid a constant flow rate [itex]Q[/itex]. Assuming the fluid is nonviscous, we can apply Bernoulli's equation to the narrow and wide ends to calculate the pressure drop [itex]\Delta p = p_1 - p_2[/itex].(adsbygoogle = window.adsbygoogle || []).push({});

[itex]p_1 + \frac{1}{2}\rho v_1^2 = p_2 + \frac{1}{2}\rho v_2^2[/itex]

[itex]\Delta p = \frac{1}{2}\rho(v_2^2 - v_1^2)[/itex].

where v_1 and v_2 can be determined using the continuity equation.

Now assume that the fluid is viscous. Could you please point out any holes in the following reasoning?

1. By the continuity equation, the fluid speeds in the wide and narrow sections of the tube are the same as in the nonviscous case: ie [itex]v_1[/itex] and [itex]v_2[/itex] respectively.

2. Since the fluid is viscous, energy is lost from the fluid as it moves. To keep track of the lost energy, I will write

[itex]p_1 + \frac{1}{2}\rho v_1^2 = p_2 + \frac{1}{2}\rho v_2^2 + [\textrm{viscous head}][/itex]

[itex]\Delta p = \frac{1}{2}\rho(v_2^2 - v_1^2) + [\textrm{viscous head}][/itex]

which means that the pressure drop is even larger when viscosity is included. This contradicts the answer to an old exam question I attempted recently. See

https://www.physicsforums.com/showthread.php?t=99620

Thanks in advance.

James

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# Conceptual fluid mechanics question

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