# 14.7 Viscosity and turbulence  (Page 4/14)

 Page 4 / 14

## Using flow rate: air conditioning systems

An air conditioning system is being designed to supply air at a gauge pressure of 0.054 Pa at a temperature of $20\phantom{\rule{0.2em}{0ex}}\text{°}\text{C}.$ The air is sent through an insulated, round conduit with a diameter of 18.00 cm. The conduit is 20-meters long and is open to a room at atmospheric pressure 101.30 kPa. The room has a length of 12 meters, a width of 6 meters, and a height of 3 meters. (a) What is the volume flow rate through the pipe, assuming laminar flow? (b) Estimate the length of time to completely replace the air in the room. (c) The builders decide to save money by using a conduit with a diameter of 9.00 cm. What is the new flow rate?

## Strategy

Assuming laminar flow, Poiseuille’s law states that

$Q=\frac{\left({p}_{2}-{p}_{1}\right)\pi {r}^{4}}{8\eta l}=\frac{dV}{dt}.$

We need to compare the artery radius before and after the flow rate reduction. Note that we are given the diameter of the conduit, so we must divide by two to get the radius.

## Solution

1. Assuming a constant pressure difference and using the viscosity $\eta =0.0181\phantom{\rule{0.2em}{0ex}}\text{mPa}\cdot \text{s}$ ,
$Q=\frac{\left(0.054\phantom{\rule{0.2em}{0ex}}\text{Pa}\right)\left(3.14\right){\left(0.09\phantom{\rule{0.2em}{0ex}}\text{m}\right)}^{4}}{8\left(0.0181\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-3}\phantom{\rule{0.2em}{0ex}}\text{Pa}\cdot \text{s}\right)\left(20\phantom{\rule{0.2em}{0ex}}\text{m}\right)}=3.84\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-3}\phantom{\rule{0.2em}{0ex}}\frac{{\text{m}}^{3}}{\text{s}}.$
2. Assuming constant flow $Q=\frac{dV}{dt}\approx \frac{\text{Δ}V}{\text{Δ}t}$
$\text{Δ}t=\frac{\text{Δ}V}{Q}=\frac{\left(12\phantom{\rule{0.2em}{0ex}}\text{m}\right)\left(6\phantom{\rule{0.2em}{0ex}}\text{m}\right)\left(3\phantom{\rule{0.2em}{0ex}}\text{m}\right)}{3.84\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-3}\phantom{\rule{0.2em}{0ex}}\frac{{\text{m}}^{3}}{\text{s}}}=5.63\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{4}\phantom{\rule{0.2em}{0ex}}\text{s}=15.63\phantom{\rule{0.2em}{0ex}}\text{hr}.$
3. Using laminar flow, Poiseuille’s law yields
$Q=\frac{\left(0.054\phantom{\rule{0.2em}{0ex}}\text{Pa}\right)\left(3.14\right){\left(0.045\phantom{\rule{0.2em}{0ex}}\text{m}\right)}^{4}}{8\left(0.0181\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-3}\phantom{\rule{0.2em}{0ex}}\text{Pa}\cdot \text{s}\right)\left(20\phantom{\rule{0.2em}{0ex}}\text{m}\right)}=2.40\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-4}\phantom{\rule{0.2em}{0ex}}\frac{{\text{m}}^{3}}{\text{s}}.$

Thus, the radius of the conduit decreases by half reduces the flow rate to 6.25% of the original value.

## Significance

In general, assuming laminar flow, decreasing the radius has a more dramatic effect than changing the length. If the length is increased and all other variables remain constant, the flow rate is decreased:

$\begin{array}{ccc}\hfill \frac{{Q}_{A}}{{Q}_{B}}& =\hfill & \frac{\frac{\left({p}_{2}-{p}_{1}\right)\pi {r}_{A}^{4}}{8\eta {l}_{A}}}{\frac{\left({p}_{2}-{p}_{1}\right)\pi {r}_{B}^{4}}{8\eta {l}_{B}}}=\frac{{l}_{B}}{{l}_{A}}\hfill \\ \hfill {Q}_{B}& =\hfill & \frac{{l}_{A}}{{l}_{B}}{Q}_{A}.\hfill \end{array}$

Doubling the length cuts the flow rate to one-half the original flow rate.

If the radius is decreased and all other variables remain constant, the volume flow rate decreases by a much larger factor.

$\begin{array}{ccc}\hfill \frac{{Q}_{A}}{{Q}_{B}}& =\hfill & \frac{\frac{\left({p}_{2}-{p}_{1}\right)\pi {r}_{A}^{4}}{8\eta {l}_{A}}}{\frac{\left({p}_{2}-{p}_{1}\right)\pi {r}_{B}^{4}}{8\eta {l}_{B}}}={\left(\frac{{r}_{A}}{{r}_{B}}\right)}^{4}\hfill \\ \hfill {Q}_{B}& =\hfill & {\left(\frac{{r}_{B}}{{r}_{A}}\right)}^{4}{Q}_{A}\hfill \end{array}$

Cutting the radius in half decreases the flow rate to one-sixteenth the original flow rate.

## Flow and resistance as causes of pressure drops

Water pressure in homes is sometimes lower than normal during times of heavy use, such as hot summer days. The drop in pressure occurs in the water main before it reaches individual homes. Let us consider flow through the water main as illustrated in [link] . We can understand why the pressure ${p}_{1}$ to the home drops during times of heavy use by rearranging the equation for flow rate:

$\begin{array}{ccc}\hfill Q& =\hfill & \frac{{p}_{2}-{p}_{1}}{R}\hfill \\ \hfill {p}_{2}–{p}_{1}& =\hfill & RQ.\hfill \end{array}$

In this case, ${p}_{2}$ is the pressure at the water works and R is the resistance of the water main. During times of heavy use, the flow rate Q is large. This means that ${p}_{2}-{p}_{1}$ must also be large. Thus ${p}_{1}$ must decrease. It is correct to think of flow and resistance as causing the pressure to drop from ${p}_{2}$ to ${p}_{1}$ . The equation ${p}_{2}-{p}_{1}=RQ$ is valid for both laminar and turbulent flows. During times of heavy use, there is a significant pressure drop in a water main, and p 1 supplied to users is significantly less than p 2 created at the water works. If the flow is very small, then the pressure drop is negligible, and p 2 ≈ p 1 .

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