Question

A Huge water tank is 2m above the ground if the water level on it is 4.9m high and a small opening is there at the bottom then the speed of efflux of non viscous water through the opening will be

Answer 1

Answer:

The speed of efflux of non-viscous water through the opening will be approximately 6.263 meters per second.

Explanation:

Let assume the existence of a line of current between the water tank and the ground and, hence, the absence of heat and work interactions throughout the system. If water is approximately at rest at water tank and at atmospheric pressure ($P_{atm}$), then speed of efflux of the non-viscous water is modelled after the Bernoulli's Principle:

$P_{1} + \rho\cdot \frac{v_{1}^{2}}{2} + \rho\cdot g \cdot z_{1} = P_{2} + \rho\cdot \frac{v_{2}^{2}}{2} + \rho\cdot g \cdot z_{2}$

Where:

$P_{1}$, $P_{2}$ - Water total pressures inside the tank and at ground level, measured in pascals.

$\rho$ - Water density, measured in kilograms per cubic meter.

$g$ - Gravitational acceleration, measured in meters per square second.

$v_{1}$, $v_{2}$ - Water speeds inside the tank and at the ground level, measured in meters per second.

$z_{1}$, $z_{2}$ - Heights of the tank and ground level, measured in meters.

Given that $P_{1} = P_{2} = P_{atm}$, $\rho = 1000\,\frac{kg}{m^{3}}$, $g = 9.807\,\frac{m}{s^{2}}$, $v_{1} = 0\,\frac{m}{s}$, $z_{1} = 6.9\,m$ and $z_{2} = 4.9\,m$, the expression is reduced to this:

$\left(9.807\,\frac{m}{s^{2}} \right)\cdot (6.9\,m) = \frac{v_{2}^{2}}{2} + \left(9.807\,\frac{m}{s^{2}} \right)\cdot (4.9\,m)$

And final speed is now calculated after clearing it:

$v_{2} = \sqrt{2\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (6.9\,m-4.9\,m)}$

$v_{2} \approx 6.263\,\frac{m}{s}$

The speed of efflux of non-viscous water through the opening will be approximately 6.263 meters per second.