Question

How is the volume flow rate of water out of the tank, dVdt, related to the flow speed v ? Express your answer in terms of some, all, or none of the variables v , d, the acceleration due to gravity g, and the constant π.

Answer 1

Answer:

$\frac{dV}{dt}= \frac{\pi d^2}{4}v$

Explanation:

The rate of volume flow out of tank can be expressed as:

$\frac{dV}{dt} = A\frac{dL}{dt}$

where,

dV/dt = Volume flow rate

A = Cross-sectional area of outlet = πd²/4

d = diameter of circular outlet

dL = Displacement covered by water

dt = time taken

but we know that:

Velocity = υ = displacement/time = dL/dt

Substituting the values of "dL/dt" and "A" in the equation, we get:

$\frac{dV}{dt} = \frac{\pi d^2}{4}v$

This is the expression for volume flow rate dV/dt, on terms pf v, d.