Changing Units — Question 8

Question 8

Changing graph axes changes the numbers, not the delivery

A pump’s volume flow rate $q$ increases linearly as shown. For a linear ramp, the delivered volume equals the average of the endpoint flow rates multiplied by duration. The ramp’s slope is $\Delta q/\Delta t$. No calculus is needed.

Use exact $1\ \mathrm{L}=10^{-3}\ \mathrm{m^3}$ and $1\ \mathrm{min}=60\ \mathrm{s}$. Treat all graph values as exact.

1. Find the delivered volume in litres. Convert to $\mathrm{m^3}$.

2. Express both endpoint flow rates in $\mathrm{m^3/s}$ and the duration in seconds. Recalculate the delivered volume in these units.

3. Find the ramp slope first in $\mathrm{L/min^2}$, then in $\mathrm{m^3/s^2}$.

4. On converting the horizontal axis to seconds and the vertical axis to $\mathrm{m^3/s}$, the time numbers are multiplied by 60 and the flow numbers by $1/60\,000$. Explain why the numerical area changes by their product, but the numerical slope by their quotient.

Question 8 - Solution

Changing graph axes changes the numbers, not the delivery

(a) Average the endpoints over the duration.

$$V=\frac{2.0+8.0}{2}\ \mathrm{L/min}\,(3.0\ \mathrm{min}) =\boxed{15.0\ \mathrm{L}=0.0150\ \mathrm{m^3}}.$$

(b) Convert the complete flow-rate unit.

$1\ \mathrm{L/min}=(10^{-3}/60)\ \mathrm{m^3/s}$. Thus $q_0=1/30\,000\ \mathrm{m^3/s}$, $q_1=1/7500\ \mathrm{m^3/s}$, and $t=180\ \mathrm{s}$. Their average is $1/12\,000\ \mathrm{m^3/s}$; multiplying by $180\ \mathrm{s}$ again gives $0.0150\ \mathrm{m^3}$.

(c) A time factor occurs twice in the slope unit.

$$\frac{\Delta q}{\Delta t}=\frac{8-2}{3}\ \mathrm{L/min^2} =2\ \mathrm{L/min^2} =2\frac{10^{-3}}{60^2}\ \mathrm{m^3/s^2} =\boxed{\frac{1}{1\,800\,000}\ \mathrm{m^3/s^2}}.$$

(d) Area multiplies axes; slope divides them.

$$\frac{t_s}{t_{\min}}=60,\quad\frac{q_{\mathrm{m^3/s}}}{q_{\mathrm{L/min}}}=\frac1{60000};\qquad A\mapsto\frac A{1000},\quad m_{\mathrm{slope}}\mapsto\frac{m_{\mathrm{slope}}}{3600000}.$$

The numerical trapezoid area is multiplied by $(1/60\,000)(60)=1/1000$: 15 becomes 0.015. The slope is multiplied by $(1/60\,000)/60=1/3\,600\,000$: 2 becomes $1/1\,800\,000$. These different numerical changes restore the same physical volume and the same physical rate of change when their respective new units are attached.