# Flow rate and pressure of air in a constant diameter hose vs. flow rate and pressure of air in a hose of varying diameter

Say we have Hose A, a straight hose with constant diameter D. Also we have Hose B, a hose with a varying diameter, but whose average diameter throughout the length of the hose is D. (diameter of hose A = average diameter through length of hose B) 1. Consider if pressurized air is input at X1 through each hose at a given flow rate and pressure. Will the pressure and flow rate at X2 in each hose be identical given that the input conditions are identical?

2. Essentially, will the two hoses function identically? Why or why not?

Will the pressure and flow rate at X2 in each hose be identical given that the input conditions are identical?

If the input flow rate, at $$X_1$$, is the same for both hoses, the output flow rate, at $$X_2$$, in the steady state, has to be the same as well.

The relationship between flow rate and pressure is described by Poiseuille's law:

$$F=\frac {P_{X_1}-P_{X_2}} R$$

where $$F$$ is flow rate and $$R$$ is resistance to flow. From here:

$$P_{X_2}=P_{X_1}-FR$$.

Since $$P_{X_1}$$ and $$F$$ are the same for both hoses, $$P_{X_2}$$ will be determined by the difference in the resistance to the flow, $$R$$.

The resistance to the flow could be estimated as:

$$R=\frac {8\eta L} {\pi r^4}$$

where $$\eta$$ is viscosity, $$L$$ is the length of a hose and $$r$$ is its radius.

Since the hoses have the same length, the difference in their resistance could be caused by the difference in their radii.

The average radius of the hoses is the same, so, if the radius of hose A is $$r=\frac D 2$$, for every section $$\Delta L$$ of hose B with radius $$r+\Delta r$$, there will be the same length section with radius $$r-\Delta r$$. Therefore, we need to compare the combined series resistance of these two sections $$\big (\frac {8\eta \Delta L} {\pi (r+\Delta r)^4}+ \frac {8\eta \Delta L} {\pi (r-\Delta r)^4}\big)$$ with the resistance of the same length section of hose A $$\big (\frac {2 \times 8\eta \Delta L} {\pi r^4}\big )$$.

It could be shown (for instance, by induction) that, for any $$m>1$$, $$\frac 1 {(r+\Delta r)^m}+ \frac 1 { (r-\Delta r)^m} > \frac 2 {r^m}$$, which means that the resistance to the flow for hose B is greater than the resistance to the flow for hose A.

Therefore, the pressure at $$X_2$$ will be greater for hose A than for hose B.