PROBLEM 30: An old pipe is to be replaced by a new one with the same diameter and length. If the friction factor of the new is $$ 2 / 3 $$ of the old, compute the ratio of flow of the new to the old pipe. $$ \begin{array}{ll} ext { a. } 0.85 & ext { b. } 1.41 \ ext { c. } 1.22 & ext { d. } 0.81\end{array} $$

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PROBLEM 30: An old pipe is to be replaced by a new one with the same diameter and length. If the friction factor of the new is $$ 2 / 3 $$ of the old, compute the ratio of flow of the new to the old pipe. $$ \begin{array}{ll}	ext { a. } 0.85 & 	ext { b. } 1.41 \ 	ext { c. } 1.22 & 	ext { d. } 0.81\end{array} $$

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To find the ratio of flow rates between the new and old pipes, we can use the Darcy-Weisbach equation, which relates the pressure drop due to friction in a pipe to the flow rate. The flow rate $$ Q $$ can be expressed in terms of the friction factor $$ f $$, pipe diameter $$ D $$, length $$ L $$, and the pressure drop $$ \Delta P $$.

The Darcy-Weisbach equation for head loss $$ h_f $$ due to friction is given by:

$$
h_f = \frac{f L v^2}{2 g D}
$$

where:
- $$ h_f $$ is the head loss,
- $$ f $$ is the friction factor,
- $$ L $$ is the length of the pipe,
- $$ v $$ is the flow velocity,
- $$ g $$ is the acceleration due to gravity,
- $$ D $$ is the diameter of the pipe.

The flow rate $$ Q $$ can be expressed as:

$$
Q = A v = \frac{\pi D^2}{4} v
$$

where $$ A $$ is the cross-sectional area of the pipe.

Now, let's denote:
- $$ f_{old} $$ as the friction factor of the old pipe,
- $$ f_{new} = \frac{2}{3} f_{old} $$ as the friction factor of the new pipe.

The head loss for the old pipe is:

$$
h_{f, old} = \frac{f_{old} L v_{old}^2}{2 g D}
$$

And for the new pipe:

$$
h_{f, new} = \frac{f_{new} L v_{new}^2}{2 g D} = \frac{\left(\frac{2}{3} f_{old}\right) L v_{new}^2}{2 g D}
$$

Assuming the same pressure drop $$ \Delta P $$ for both pipes, we can set the head losses equal to each other:

$$
\frac{f_{old} L v_{old}^2}{2 g D} = \frac{\left(\frac{2}{3} f_{old}\right) L v_{new}^2}{2 g D}
$$

We can cancel out common terms:

$$
f_{old} v_{old}^2 = \frac{2}{3} f_{old} v_{new}^2
$$

Dividing both sides by $$ f_{old} $$ (assuming $$ f_{old} \neq 0 $$):

$$
v_{old}^2 = \frac{2}{3} v_{new}^2
$$

Taking the square root of both sides:

$$
v_{old} = \sqrt{\frac{2}{3}} v_{new}
$$

Now, we can express the flow rates in terms of the velocities:

$$
Q_{old} = \frac{\pi D^2}{4} v_{old} = \frac{\pi D^2}{4} \sqrt{\frac{2}{3}} v_{new}
$$

$$
Q_{new} = \frac{\pi D^2}{4} v_{new}
$$

Now, we can find the ratio of the flow rates:

$$
\frac{Q_{new}}{Q_{old}} = \frac{\frac{\pi D^2}{4} v_{new}}{\frac{\pi D^2}{4} \sqrt{\frac{2}{3}} v_{new}} = \frac{1}{\sqrt{\frac{2}{3}}} = \sqrt{\frac{3}{2}} = \frac{\sqrt{3}}{\sqrt{2}} \approx 1.2247
$$

Thus, the ratio of the flow of the new pipe to the old pipe is approximately $$ 1.22 $$.

The correct answer is:

**c. 1.22**
The ratio of the flow of the new pipe to the old pipe is approximately 1.22.

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