Problem 32: Fluid Flow in Modified Pipe
1. Analogy to Electrical Resistance:
The problem states flow rate $q$ is proportional to pressure difference $\Delta P$. This allows us to use an analogy to Ohm’s Law where $\Delta P = q R$, where $R$ is fluid resistance. Resistance is proportional to length: $R \propto L$.
The problem states flow rate $q$ is proportional to pressure difference $\Delta P$. This allows us to use an analogy to Ohm’s Law where $\Delta P = q R$, where $R$ is fluid resistance. Resistance is proportional to length: $R \propto L$.
2. Original Setup:
A straight pipe of length $l$. $$ R_{original} = k l $$ $$ q_0 = \frac{\Delta P}{R_{original}} = \frac{\Delta P}{kl} $$
A straight pipe of length $l$. $$ R_{original} = k l $$ $$ q_0 = \frac{\Delta P}{R_{original}} = \frac{\Delta P}{kl} $$
3. Modified Setup:
A ring of radius $r$ is inserted in the middle. To insert the ring, a section of the straight pipe equal to the diameter of the ring ($2r$) is effectively removed and replaced by the ring structure.
Length of remaining straight pipe: $l – 2r$.
Resistance of Ring: The ring splits the flow into two semicircular paths. Length of one semicircle = $\pi r$. Resistance of one branch $R_{branch} = k (\pi r)$. Since the two branches are in parallel: $$ \frac{1}{R_{ring}} = \frac{1}{k\pi r} + \frac{1}{k\pi r} = \frac{2}{k\pi r} $$ $$ R_{ring} = \frac{k\pi r}{2} $$
A ring of radius $r$ is inserted in the middle. To insert the ring, a section of the straight pipe equal to the diameter of the ring ($2r$) is effectively removed and replaced by the ring structure.
Length of remaining straight pipe: $l – 2r$.
Resistance of Ring: The ring splits the flow into two semicircular paths. Length of one semicircle = $\pi r$. Resistance of one branch $R_{branch} = k (\pi r)$. Since the two branches are in parallel: $$ \frac{1}{R_{ring}} = \frac{1}{k\pi r} + \frac{1}{k\pi r} = \frac{2}{k\pi r} $$ $$ R_{ring} = \frac{k\pi r}{2} $$
4. Total New Resistance ($R_{new}$):
The straight sections and the ring are in series. $$ R_{new} = R_{straight} + R_{ring} $$ $$ R_{new} = k(l – 2r) + k\frac{\pi r}{2} $$ $$ R_{new} = k \left( l – 2r + \frac{\pi r}{2} \right) = \frac{k}{2} (2l – 4r + \pi r) $$
The straight sections and the ring are in series. $$ R_{new} = R_{straight} + R_{ring} $$ $$ R_{new} = k(l – 2r) + k\frac{\pi r}{2} $$ $$ R_{new} = k \left( l – 2r + \frac{\pi r}{2} \right) = \frac{k}{2} (2l – 4r + \pi r) $$
5. Calculate New Flow Rate ($q$):
$$ q = \frac{\Delta P}{R_{new}} $$ Taking the ratio $q / q_0$: $$ \frac{q}{q_0} = \frac{R_{original}}{R_{new}} = \frac{kl}{\frac{k}{2} (2l – 4r + \pi r)} $$ $$ q = q_0 \left( \frac{2l}{2l – 4r + \pi r} \right) $$
$$ q = \frac{\Delta P}{R_{new}} $$ Taking the ratio $q / q_0$: $$ \frac{q}{q_0} = \frac{R_{original}}{R_{new}} = \frac{kl}{\frac{k}{2} (2l – 4r + \pi r)} $$ $$ q = q_0 \left( \frac{2l}{2l – 4r + \pi r} \right) $$
Answer:
$$ q = q_0 \left( \frac{2l}{2l – 4r + \pi r} \right) $$