A pump is used to pump a liquid of density `rho` continuously through a pipe of cross, sectional area A. If liquid is flowing with speed v, then average power of pump is

Options

A. `(1)/(3)rho A V^(2)`
B. `(1)/(2)rhoAV^(2)`
C. `2rhoAV^(2)`
D. `(1)/(2)rhoAV^(3)`

Answer

To find the average power of the pump used to pump a liquid of density \( \rho \) through a pipe of cross-sectional area \( A \) at speed \( v \), we can follow these steps:

### Step 1: Calculate the mass flow rate
The mass flow rate \( \frac{dm}{dt} \) can be expressed in terms of the density \( \rho \), the cross-sectional area \( A \), and the speed \( v \) of the liquid.

Using the relationship:
\[ \frac{dm}{dt} = \rho \cdot \text{Volume flow rate} \]
The volume flow rate can be calculated as:
\[ \text{Volume flow rate} = A \cdot v \]
Thus, we have:
\[ \frac{dm}{dt} = \rho \cdot (A \cdot v) = \rho A v \]

### Step 2: Determine the kinetic energy of the liquid
The kinetic energy \( KE \) of a mass \( m \) moving with velocity \( v \) is given by:
\[ KE = \frac{1}{2} mv^2 \]
To find the kinetic energy per unit time (which relates to power), we need to express \( m \) in terms of the mass flow rate:
\[ KE = \frac{1}{2} \left(\frac{dm}{dt} \cdot t\right) v^2 \]
Substituting \( \frac{dm}{dt} = \rho A v \) into the equation gives:
\[ KE = \frac{1}{2} (\rho A v) v^2 \cdot t \]

### Step 3: Calculate average power
Average power \( P_{\text{average}} \) is defined as the energy per unit time. Therefore, we can express it as:
\[ P_{\text{average}} = \frac{KE}{t} = \frac{\frac{1}{2} (\rho A v) v^2 \cdot t}{t} \]
This simplifies to:
\[ P_{\text{average}} = \frac{1}{2} \rho A v^3 \]

### Final Result
Thus, the average power of the pump is:
\[ P_{\text{average}} = \frac{1}{2} \rho A v^3 \]

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