Water is flowing at 3 m/s in a horizontal pipe under a pressure of 200 kpa. The pipe, which narrows to half its original diameter, is laid out on a level ground.
a) What is the speed of flow in the narrow section?
b) What is the pressure in the narrow section?
c) How do the volume flow rates in the two sections?
To solve this problem, we can apply the principle of continuity, which states that the mass flow rate of an incompressible fluid remains constant along a pipe. This principle can be expressed as:
A1v1 = A2v2
Where A1 and A2 are the cross-sectional areas of the pipe before and after the narrowing, and v1 and v2 are the speeds of flow in those sections.
a) To find the speed of flow in the narrow section, we need to determine the relationship between the cross-sectional areas before and after the narrowing. Since the pipe narrows to half its original diameter, the cross-sectional area will be reduced by a factor of 4 (area is proportional to the square of the diameter). Therefore, we can calculate the speed in the narrow section using the equation:
(1/4)A1v1 = A2v2
Given that the speed of flow in the horizontal pipe is 3 m/s, we can substitute A1 = 1 and v1 = 3 into the equation to find A2v2.
(1/4) * 1 * 3 = A2v2
3/4 = A2v2
So, the speed of flow in the narrow section is 3/4 m/s.
b) To find the pressure in the narrow section, we can use Bernoulli's principle, which states that the total energy per unit volume of an incompressible fluid remains constant along a streamline. Bernoulli's equation can be written as:
P + (1/2)ρv^2 + ρgh = constant
Where P is the pressure, ρ is the density of the fluid, v is the speed of flow, g is the acceleration due to gravity, and h is the height above a reference level.
Since the pipe is laid out on level ground, the height above the reference level is the same in both sections. Therefore, we can neglect the ρgh term when comparing the pressures.
Using Bernoulli's equation for the horizontal pipe and the narrow section, we have:
P1 + (1/2)ρv1^2 = P2 + (1/2)ρv2^2
Given that the pressure in the horizontal pipe is 200 kPa, we can substitute P1 = 200 × 10^3 Pa and v1 = 3 m/s into the equation to find P2.
200 × 10^3 + (1/2)ρ(3^2) = P2 + (1/2)ρ(3/4)^2
200 × 10^3 + 4.5ρ = P2 + 0.5625ρ
Since the density, ρ, is the same for both sections, it cancels out. Solving for P2, we get:
199437.5 Pa = P2
So, the pressure in the narrow section is approximately 199.4 kPa.
c) The volume flow rate is given by the equation:
Q = Av
Where Q is the volume flow rate, A is the cross-sectional area, and v is the speed of flow.
Since the mass flow rate remains constant, the volume flow rate does as well. Therefore, the volume flow rates in the two sections will be equal.