The figure shows a nozzle on a hose carrying water. If the pressure at Point 1 is 290 x 10^3 Pa, the velocity 0.64 m/s, and the diameter 20 cm.
Given that the diameter at point 2 is 1.4 cm, what is the velocity at point 2?
v_2 =
What is the pressure at point 2?
P_2 =
To find the velocity at point 2, we can use the principle of continuity. According to the principle of continuity, the flow rate of an incompressible fluid remains constant along a pipe or hose.
The equation for the principle of continuity is:
A1v1 = A2v2
Where A1 and A2 are the cross-sectional areas at point 1 and point 2 respectively, and v1 and v2 are the velocities at point 1 and point 2 respectively.
To find the velocity at point 2, we need to first calculate the cross-sectional area at point 1 and point 2.
The cross-sectional area at point 1 (A1) can be calculated using the formula for the area of a circle:
A1 = πr1^2
Given that the diameter at point 1 is 20 cm, we can calculate the radius at point 1:
r1 = d1/2 = 20 cm/2 = 10 cm = 0.1 m
Substituting the value of r1 into the formula, we get:
A1 = π * (0.1 m)^2 = 0.0314 m^2
The cross-sectional area at point 2 (A2) can be calculated in a similar way:
r2 = d2/2 = 1.4 cm/2 = 0.007 m
A2 = π * (0.007 m)^2 ≈ 0.000154 m^2
Now, we can substitute the values of A1 and A2 into the continuity equation and solve for v2:
A1v1 = A2v2
0.0314 m^2 * 0.64 m/s = 0.000154 m^2 * v2
Simplifying the equation, we find:
v2 = (0.0314 m^2 * 0.64 m/s) / 0.000154 m^2
v2 ≈ 130.129 m/s
Therefore, the velocity at point 2 is approximately 130.129 m/s.
Moving on to the pressure at point 2, we can use Bernoulli's equation. Bernoulli's equation states that the sum of the static pressure, dynamic pressure, and potential energy per unit volume is constant along a streamline of an incompressible, frictionless flow.
The equation for Bernoulli's equation is:
P1 + 1/2 * ρ * v1^2 + ρ * g * h1 = P2 + 1/2 * ρ * v2^2 + ρ * g * h2
Where P1 and P2 are the pressures at point 1 and point 2 respectively, ρ is the density of the fluid, v1 and v2 are the velocities at point 1 and point 2 respectively, g is the acceleration due to gravity, h1 and h2 are the heights at point 1 and point 2 respectively.
Since the problem does not provide information about the heights at point 1 and point 2, we can assume that they are at the same height, so h1 = h2.
The density of water (ρ) is approximately 1000 kg/m^3.
Now, we can substitute the given values into Bernoulli's equation:
P1 + 1/2 * 1000 kg/m^3 * (0.64 m/s)^2 = P2 + 1/2 * 1000 kg/m^3 * (130.129 m/s)^2
290,000 Pa + (1/2) * 1000 kg/m^3 * (0.64 m/s)^2 = P2 + (1/2) * 1000 kg/m^3 * (130.129 m/s)^2
Simplifying the equation, we find:
P2 = 290,000 Pa + (1/2) * 1000 kg/m^3 * (0.64 m/s)^2 - (1/2) * 1000 kg/m^3 * (130.129 m/s)^2
P2 ≈ 214,402.528 Pa
Therefore, the pressure at point 2 is approximately 214,402.528 Pa.