Water flows through a 4.0-cm-diameter horizontal pipe at a speed of 1.3m/s . The pipe then narrows down to a diameter of 2.0cm .
Ignoring viscosity, what is the pressure difference between the wide and narrow sections of the pipe?
To find the pressure difference between the wide and narrow sections of the pipe, we can use Bernoulli's equation. This equation relates the pressure, velocity, and height of a fluid at two different points in a flow.
Bernoulli's equation is given by:
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 points 1 and 2,
v1 and v2 are the velocities at points 1 and 2,
ρ is the density of the fluid (water in this case),
g is the acceleration due to gravity,
h1 and h2 are the heights at points 1 and 2 (which we can assume to be the same).
In this case, the fluid is flowing horizontally, so there is no change in height (h1 = h2). We can also ignore the effect of gravity since it is not mentioned in the problem statement. So the equation simplifies to:
P1 + 1/2 * ρ * v1^2 = P2 + 1/2 * ρ * v2^2
Now, let's substitute the given values into the equation:
P1 + 1/2 * ρ * (1.3 m/s)^2 = P2 + 1/2 * ρ * (v2)^2
The density of water (ρ) is approximately 1000 kg/m^3.
Now, we can focus on the change in the velocity of the fluid. Since the pipe narrows down, the velocity will increase at the narrow section (v2 > v1). We can relate the velocities at the two sections using the equation of continuity:
A1 * v1 = A2 * v2
Where A1 and A2 are the cross-sectional areas of the wide and narrow sections of the pipe, respectively.
The cross-sectional area of a pipe is given by:
A = π * r^2
So, we can rewrite the equation of continuity as:
π * (4.0 cm/2)^2 * (1.3 m/s) = π * (2.0 cm/2)^2 * v2
Simplifying the equation and converting the units to consistent ones:
(0.02 m)^2 * 1.3 m/s = (0.01 m)^2 * v2
0.52 * m^2/s = 0.01 * m^2 * v2
Now we can solve for v2:
v2 = (0.52 * m^2/s) / (0.01 * m^2)
v2 = 52 m/s
Now, substituting this value back into the Bernoulli's equation:
P1 + 1/2 * ρ * (1.3 m/s)^2 = P2 + 1/2 * ρ * (52 m/s)^2
P1 + 0.5 * (1000 kg/m^3) * (1.69 m^2/s^2) = P2 + 0.5 * (1000 kg/m^3) * (2704 m^2/s^2)
P1 + 845 N/m^2 = P2 + 1352 N/m^2
Now, we can subtract P1 from both sides of the equation to isolate the pressure difference (P2 - P1):
P2 - P1 = 1352 N/m^2 - 845 N/m^2
P2 - P1 = 507 N/m^2
Therefore, the pressure difference between the wide and narrow sections of the pipe is 507 N/m^2.