A viscometer of the Redwood type has an oil-containing cylinder 4.75cm in diam and an agate tube 0.17 cm in diameter and 1.2 cm long. The oil surface, when flow starts, is 9 cm above the outlet from the agate tube. To allow for the sudden contractions at entry to the tube, the effective length of the tube may be taken as the actual length plus the tube radius. Making allowance for the decreasing head of oil, the viscous resistance through the tube, and the KE (kinetic energy) of discharge, calculate using arithmetical integration, the time required for 50 cm3 of an oil of viscosity 0.5 poise and specific gravity 0.92 to flow through the viscometer.

How do you find the time?

From this question, I have tried to use quadratic equation to solve for $v$, but it is proving more difficult than I thought.

So far this is what I have: $$h=\frac{32\mu (l+d/2)v}{\rho g d^2}+\frac{v^2}{2g}$$

$$av^2 + bv + c = 0$$

where $a = 1/ 2g$, $b=\frac{32\mu (l+d/2)v}{\rho g d^2} and $c = $-h$

Using $v = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

I eventually got $v = - \frac{32\mu (l+d/2)v}{\rho d^2} \pm (\frac{32\mu (l+d/2)v}{\rho d^2} + \sqrt{2gh})$

In this expression $-b + b = 0$, so how do I deal with this?

Let the fall in level be $dh$ in a small interval of time $\text{d}t$, water area of cylinder be $A$ and area of tube be $a$. Therefore, continuity equation of tank and pipe yields quantity discharge in $\text{d}t$ seconds = $A \text{d}h$ = $a v \text{d}t$.

How do you integrate the expression for this time for the head level to fall from $H1$ to $H2$ for this question?

Does anyone have any ideas that can help me, so that I can solve this question?

  • $\begingroup$ I suggest you edit your equations to make it explicitly clear which values are in the denominator whenever you have inline divisions (either with parentheses or using \frac{}{}). $\endgroup$
    – Wasabi
    Oct 21 '16 at 17:49
  • $\begingroup$ may someone please reply to my question. $\endgroup$ Oct 28 '16 at 15:57

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