# A small crack occurs at the base of a 15.0 m high dam. The effective crack area through which water leaves is 1.30 *10-3 m2

## A small crack occurs at the base of a 15.0 m high dam. The effective crack area through which water leaves is 1.30 *10-3 m2

a) Ignoring viscous losses, what is the speed of water flowingthrough the crack?
b) How many cubic meters of water per second leave thedam?

## To calculate the volume flow rate of water leaving the dam through the crack, we can use the equation:

Q = A * v

Where:
Q is the volume flow rate (m³/s)
A is the cross-sectional area of the crack (m²)
v is the velocity of water flowing through the crack (m/s)

Given:
A = 1.30 × 10⁻³ m²
We need to determine the value of v.

To find the velocity of water, we can use Bernoulli's equation:

P₁ + ½ρv₁² + ρgh₁ = P₂ + ½ρv₂² + ρgh₂

Assumptions:
- Neglecting the air pressure (P₁ = P₂ = P)
- Neglecting the height difference (h₁ = h₂ = h)
- Neglecting the velocity and pressure at the surface of the reservoir, considering it to be zero (v₁ = 0 and P₁ = 0)

With these assumptions, the equation becomes:

½ρv² + ρgh = 0

1/2 * ρ * v² = -ρ * g * h

v² = -2gh

v = √(-2gh)

Where:
ρ is the density of water (1000 kg/m³)
h is the height of the water column above the crack (15.0 m)
g is the acceleration due to gravity (9.81 m/s²)

Now, we can calculate the velocity (v):

v = √(-2 × g × h)
v = √(-2 × 9.81 m/s² × 15.0 m)
v ≈ √(-294.3)
v ≈ 17.14 m/s

Now we have all the necessary information to calculate the volume flow rate (Q):

Q = A × v
Q = 1.30 × 10⁻³ m² × 17.14 m/s
Q ≈ 2.23 × 10⁻² m³/s

Therefore, the volume flow rate of water leaving the dam through the crack is approximately 2.23 × 10⁻² m³/s (or 0.0223 m³/s).

## To calculate the volume of water flowing through the crack, we need to multiply the area of the crack by the velocity of the water. The velocity of the water can be calculated using Bernoulli's equation.

Bernoulli's equation states that the sum of the pressure, kinetic energy per unit volume, and potential energy per unit volume of a fluid remains constant along a streamline.

For a small crack, we can assume that the pressure at the bottom of the dam is nearly equal to the atmospheric pressure, and the velocity of the water leaving the crack is small.

Therefore, Bernoulli's equation simplifies to:

P + ρgh + 1/2 * ρv^2 = constant

where P is the pressure, ρ is the density of water, g is the acceleration due to gravity, h is the height of the dam, and v is the velocity of the water leaving the crack.

Since the velocity of the water leaving the crack is small, we can ignore the kinetic energy term (1/2 * ρv^2). Also, the pressure at the bottom of the dam is equal to atmospheric pressure, so we can set P = 0.

Simplifying the equation further, we get:

ρgh = constant

Now we can plug in the values:

ρ = density of water = 1000 kg/m^3 (approximately)
g = acceleration due to gravity = 9.8 m/s^2 (approximately)
h = height of the dam = 15.0 m

Therefore, the constant value is (1000 kg/m^3) * (9.8 m/s^2) * (15.0 m) = 147,000 kg/(m^2s^2).

Since the crack area is given as 1.30 x 10^-3 m^2, the volume flow rate can be calculated by multiplying the constant value by the crack area:

Volume flow rate = constant * crack area
Volume flow rate = 147,000 kg/(m^2s^2) * 1.30 x 10^-3 m^2

Simplifying this expression will give you the volume flow rate of water through the crack.