Please help me !!! for solution of this question
Water enters a horizontal pipe with a rectangular cross section at a speed of 1.00 m/s. The width of the pipe remains constant but the height decreases. 27.3 m from the entrance, the height is half of what it is at the entrance. If the water pressure at the entrance is 2921 Pa, what is the pressure 27.3 m downstream?
half the area is twice the speed
V1 = 1 m/s
V2 = 2 m/s
Bernoulli
(1/2) rho V1^2+P1 = (1/2) rho V2^2+P2
rho water =1000 kg/m^3
2921 Pa is very low by the way, like 1 atm = 100,000 Pa for example. Perhaps they mean gage pressure.
500 (1)^2 + 2921 = 500 (2)^2 + P2
2921 - P2 = 1500
P2 = 1421 Pa
You are welcome :)
very thanks
To solve this question, we can use Bernoulli's equation, which relates the pressure, velocity, and height of a fluid flowing in a pipe. Bernoulli's equation is given by:
P + 1/2ρv^2 + ρgh = constant
Where:
- P is the pressure of the fluid
- ρ is the density of the fluid
- v is the velocity of the fluid
- g is the acceleration due to gravity
- h is the height of the fluid above a reference point
1. Let's start by determining the velocity of the water at 27.3 m from the entrance. Since the width of the pipe remains constant, we can assume that the width does not affect the velocity. Therefore, the velocity of the water at 27.3 m is also 1.00 m/s.
2. We are told that the height at 27.3 m downstream is half of what it is at the entrance. Let's represent the height at the entrance as h1 and the height at 27.3 m downstream as h2. We can write the relationship between the two heights as h2 = 0.5*h1.
3. We are given the pressure at the entrance (P1 = 2921 Pa). We need to find the pressure at 27.3 m downstream (P2). Since the velocity and the width are constant, they do not affect the pressure.
4. Now, let's use Bernoulli's equation at both the entrance and 27.3 m downstream to solve for the pressure.
At the entrance:
P1 + 1/2ρv1^2 + ρgh1 = constant
At 27.3 m downstream:
P2 + 1/2ρv2^2 + ρgh2 = constant
Since the velocities and widths are constant, we can simplify the equations to:
P1 + ρgh1 = constant
P2 + ρgh2 = constant
5. Let's substitute the known values into the equations. We know that h2 = 0.5*h1, h1 = 0 m (since it is at the same height as the entrance), and P1 = 2921 Pa.
At the entrance:
2921 + ρg*0 = constant
2921 = constant
At 27.3 m downstream:
P2 + ρg*(0.5*h1) = constant
P2 + ρg*(0.5*0) = constant
P2 = constant
Since the constant value at both locations is the same, we can conclude that P1 = P2.
Therefore, the pressure 27.3 m downstream is 2921 Pa.