find the velocity of escape of water through a small opening in a broiler where the pressure is 10^6 Pa above atmospheric pressure, assumed h1=h2 and v1=0
To find the velocity of escape of water through a small opening in a broiler, we can use Bernoulli's equation, which relates the pressure, velocity, and height of a fluid.
Bernoulli's equation is given by:
P1 + 1/2ρv1^2 + ρgh1 = P2 + 1/2ρv2^2 + ρgh2,
where P1 and P2 are the pressures at points 1 and 2, respectively,
ρ is the density of the fluid,
v1 and v2 are the velocities at points 1 and 2, respectively,
g is the acceleration due to gravity, and
h1 and h2 are the heights at points 1 and 2, respectively.
In this case, we have:
P1 = atmospheric pressure,
P2 = atmospheric pressure + 10^6 Pa (given),
h1 = h2 (given),
v1 = 0 (given).
Let's simplify the equation:
atmospheric pressure + 0 + 0 = atmospheric pressure + 10^6 Pa + 1/2ρv2^2 + 0.
Now, solve for v2, the velocity of escape through the small opening:
1/2ρv2^2 = 10^6 Pa.
Next, we need to determine the density of water (ρ). At standard conditions, the density of water is approximately 1000 kg/m^3.
Substituting the values, we have:
1/2 * 1000 kg/m^3 * v2^2 = 10^6 Pa.
To solve for v2, rearrange the equation:
v2^2 = (2 * 10^6 Pa) / (1000 kg/m^3).
Taking the square root of both sides, we get:
v2 = √(2 * 10^6 Pa / 1000 kg/m^3).
Simplifying the expression, we have:
v2 = √(2000 m^2/s^2).
Finally, the velocity of escape of water through the small opening in the broiler is approximately:
v2 ≈ 44.7 m/s.
So, the velocity of escape is approximately 44.7 m/s.