Changing Units — Question 10

Question 10

Will the vessel reach its target before the deadline?

An initially empty cylindrical vessel has the inside dimensions shown. A pump delivers $0.750\ \mathrm{L/min}$ at a constant rate, but exactly $10.0\%$ of this volume is diverted before it enters the vessel. Ignore all other losses and any startup delay. The target must be reached within $75.0\ \mathrm{s}$, inclusive.

Use $V=\pi r^2 h$, exact $1\ \mathrm{cm}=10^{-2}\ \mathrm{m}$, $1\ \mathrm{L}=10^{-3}\ \mathrm{m^3}$, and $1\ \mathrm{min}=60\ \mathrm{s}$. Treat all given model data as exact; leave $\pi$ unrounded until the final numerical step.

1. Find the target liquid volume in $\mathrm{m^3}$ and litres. Show the distinction between diameter and radius.

2. Find the actual filling time in seconds and decide whether the deadline is met. Determine the minimum pump rate in $\mathrm{L/min}$ that would meet the deadline with the same diversion fraction.

3. A preliminary calculation converts all units correctly but ignores the diversion. What time and decision result? Would a dimensional check alone detect this error?

Question 10 - Solution

Will the vessel reach its target before the deadline?

(a) Convert lengths before forming the volume.

The radius is half the diameter: $r=6.0\ \mathrm{cm}=0.060\ \mathrm{m}$; $h=0.080\ \mathrm{m}$. Therefore $$V=\pi(0.060)^2(0.080)\ \mathrm{m^3} =0.000288\pi\ \mathrm{m^3}=0.288\pi\ \mathrm{L} \approx\boxed{0.904779\ \mathrm{L}}.$$ Using the diameter as radius would multiply the volume by four, a geometry error that unit cancellation would not reveal.

(b) Only the retained flow fills the vessel.

The entering rate is $(1-0.100)(0.750\ \mathrm{L/min})=0.675\ \mathrm{L/min}$. Thus $$t=\frac{0.288\pi}{0.675}\ \mathrm{min} =25.6\pi\ \mathrm{s}\approx\boxed{80.425\ \mathrm{s}}.$$ The deadline is not met: this exceeds $75.0\ \mathrm{s}$. That deadline is $1.25\ \mathrm{min}$, so the minimum pump rate is $$q_{\min}=\frac{0.288\pi\ \mathrm{L}}{(0.900)(1.25\ \mathrm{min})} =0.256\pi\ \mathrm{L/min}\approx\boxed{0.804248\ \mathrm{L/min}}.$$

(c) Correct units cannot replace a missing physical factor.

$$\frac{t_{\mathrm{actual}}}{t_{\mathrm{ignoring\ diversion}}}=\frac1{0.900}=\frac{10}{9},\qquad t_{\mathrm{actual}}-75\ \mathrm s\approx5.425\ \mathrm s.$$

Ignoring diversion gives $t=0.288\pi/0.750\ \mathrm{min} =23.04\pi\ \mathrm{s}\approx72.382\ \mathrm{s}$, incorrectly predicting success. Both computations have time units. The missing dimensionless retention factor is a physical-model error; a unit check alone cannot detect it.