In an underground coal mine, the air quantities \( Q_A \) and \( Q_B \) pass through ventilation districts A and B, respectively, as shown, with a pressure drop of \( \Delta P \) across them. The air quantity in district B is to be increased to \( Q_B' \) by installing a booster fan, as shown, without affecting the air quantity in ventilation district A. The capacity of the booster fan \( (P_B) \) can be expressed as
Show Hint
Write the pressure balance for the two parallel airways and see what extra head the fan must supply once B's flow goes up.
\( P_B = \Delta P \left( \dfrac{Q_B}{Q_B'} \right)^2 - \Delta P \)
\( P_B = \Delta P \left( \dfrac{Q_B'}{Q_B} \right)^2 - \Delta P \)
\( P_B = \Delta P - \Delta P \left( \dfrac{Q_B'}{Q_B} \right)^2 \)
\( P_B = \Delta P + \Delta P \left( \dfrac{Q_B'}{Q_B} \right)^2 \)
Show Solution
The Correct Option isB
Solution and Explanation
$\textbf{Work with a ratio instead of the raw quantities.}$
Let $k = Q_B' / Q_B$ be how many times bigger the new airflow in district B is compared to the old one. The job is to express the booster fan pressure $P_B$ in terms of $\Delta P$ and $k$, then swap $k$ back for $Q_B'/Q_B$ at the end.
$\textbf{Recall why A and B share the same pressure drop.}$
Districts A and B connect the same two junctions, so whatever the total pressure difference is between those junctions, both airways feel that same value. Since $Q_A$ is not allowed to change, that shared pressure difference has to stay fixed at $\Delta P$, exactly what it was before the fan was added.
$\textbf{Express B's own resistance to airflow.}$
The square law for airway B originally reads $\Delta P = R_B Q_B^2$, so $R_B = \Delta P / Q_B^2$. This number depends only on the airway's shape and surface, not on how much air moves through it, so it stays the same when the fan is added.
$\textbf{Find what pressure the new flow needs.}$
At the new flow $Q_B' = kQ_B$, the airway resists with:
\[ R_B (Q_B')^2 = \frac{\Delta P}{Q_B^2}(kQ_B)^2 = \Delta P \, k^2 \]
So pushing $k$ times as much air through the same airway needs $k^2$ times the pressure, that is $\Delta P k^2$ in total.
$\textbf{Subtract what nature already gives you.}$
The junctions still only supply $\Delta P$ on their own, fixed by district A. The fan has to make up the shortfall:
\[ P_B = \Delta P k^2 - \Delta P \]
Substituting $k = Q_B'/Q_B$ back in:
\[ P_B = \Delta P \left( \frac{Q_B'}{Q_B} \right)^2 - \Delta P \]
$\textbf{Final answer.}$
This matches option (B). Since $Q_B' > Q_B$, the ratio $k > 1$, so $P_B$ comes out positive, which makes physical sense: a booster fan adds pressure, it does not remove it.
\[ \boxed{P_B = \Delta P \left( \dfrac{Q_B'}{Q_B} \right)^2 - \Delta P} \]
