Step 1: Understanding the Question:
The problem relates volume flow rate to power for a pump delivering fluid through a fixed pipe.
Twice the volume in the same time means the volume flow rate \(Q\) is doubled.
For a fixed pipe (fixed cross-sectional area \(A\)), the velocity \(v\) of the water must change to accommodate the increased flow.
Step 2: Key Formula or Approach:
Volume flow rate \(Q = Av \implies v \propto Q\).
Mass flow rate \(\frac{dm}{dt} = \rho Q = \rho Av \implies \frac{dm}{dt} \propto v\).
Power \(P = \frac{d}{dt} (\frac{1}{2}mv^2) = \frac{1}{2} \left(\frac{dm}{dt}\right) v^2\).
Step 3: Detailed Explanation:
Case 1: Flow rate is \(Q\). Velocity is \(v = Q/A\).
Power \(P_1 = \frac{1}{2} (\rho A v) v^2 = \frac{1}{2} \rho A v^3\).
Case 2: Volume is doubled in same time, so flow rate \(Q' = 2Q\).
Since area \(A\) is constant, new velocity \(v' = Q'/A = 2Q/A = 2v\).
Calculate new power \(P_2\):
\[ P_2 = \frac{1}{2} \rho A (v')^3 \]
\[ P_2 = \frac{1}{2} \rho A (2v)^3 \]
\[ P_2 = \frac{1}{2} \rho A (8v^3) \]
\[ P_2 = 8 \left( \frac{1}{2} \rho A v^3 \right) \]
\[ P_2 = 8 P_1 \]
Thus, power must be increased by a factor of 8.
Step 4: Final Answer:
Doubling the volume flow rate requires the velocity to double, which in turn causes the power requirement to increase by the cube of that factor (\(2^3 = 8\)).