Water is flowing down through the pipe shown in the drawing. Point A is 0.410 m higher than B. The speed of the water at A and B are vA = 5.00 m/s and vB = 3.80 m/s. Determine the difference PB - PA in pressures between B and A. The density of water is 1.00x10^3 kg/m3.
To determine the difference in pressures between points B and A in the pipe, we can use Bernoulli's equation, which states that the total energy of a fluid flowing through a system remains constant.
Bernoulli's equation is given by:
P + (1/2)ρv^2 + ρgh = constant
Where:
P = pressure
ρ = density of the fluid
v = velocity of the fluid
g = acceleration due to gravity
h = height of the fluid column
Let's calculate the pressures at points A and B using Bernoulli's equation:
At point A:
P + (1/2)ρvA^2 + ρg(hA) = constant
At point B:
P + (1/2)ρvB^2 + ρg(hA - hB) = constant
Since the two points are at the same horizontal level, the difference in height (hA - hB) is the same as the given height difference of 0.410 m.
Substituting the given values:
At point A:
P + (1/2)(1.00 x 10^3 kg/m^3)(5.00 m/s)^2 + (1.00 x 10^3 kg/m^3)(9.81 m/s^2)(0.410 m) = constant
At point B:
P + (1/2)(1.00 x 10^3 kg/m^3)(3.80 m/s)^2 + (1.00 x 10^3 kg/m^3)(9.81 m/s^2)(0.410 m) = constant
Since the constant term remains the same for both equations, we can subtract the equations to eliminate it:
[(1/2)(1.00 x 10^3 kg/m^3)(5.00 m/s)^2 + (1.00 x 10^3 kg/m^3)(9.81 m/s^2)(0.410 m)] - [(1/2)(1.00 x 10^3 kg/m^3)(3.80 m/s)^2 + (1.00 x 10^3 kg/m^3)(9.81 m/s^2)(0.410 m)]
Simplifying the expression, the difference in pressures PB - PA between points B and A is obtained.
To determine the pressure difference between points B and A, you can use Bernoulli's equation for fluid flow. The equation link the pressure, velocity, and elevation between two points in a fluid flow.
The equation is given as:
P1 + 0.5ρv1^2 + ρgh1 = P2 + 0.5ρv2^2 + ρgh2
Where:
P1 and P2 are the pressures at points A and B, respectively.
ρ is the density of the fluid, given as 1.00x10^3 kg/m^3 for water.
v1 and v2 are the velocities at points A and B, respectively.
g is the acceleration due to gravity, approximately 9.81 m/s^2.
h1 and h2 are the elevations at points A and B, respectively.
In this case, point A is 0.410 m higher than point B. Let's substitute the given values into the equation:
P1 + 0.5(1.00x10^3 kg/m^3)(5.00 m/s)^2 + (1.00x10^3 kg/m^3)(9.81 m/s^2)(0.410 m) = P2 + 0.5(1.00x10^3 kg/m^3)(3.80 m/s)^2 + (1.00x10^3 kg/m^3)(9.81 m/s^2)(0 m)
Simplifying the equation, we can cancel out the common terms:
P1 + 0.5(1.00x10^3 kg/m^3)(5.00 m/s)^2 + (1.00x10^3 kg/m^3)(9.81 m/s^2)(0.410 m) = P2 + 0.5(1.00x10^3 kg/m^3)(3.80 m/s)^2
P1 + 0.5(1.00x10^3 kg/m^3)(25.00 m^2/s^2) + (1.00x10^3 kg/m^3)(4.02 m^2/s^2) = P2 + 0.5(1.00x10^3 kg/m^3)(14.44 m^2/s^2)
Now, let's calculate the values:
P1 + 12,500 Pa + 4,020 Pa = P2 + 7,220 Pa
Rearranging the equation, we find:
P2 - P1 = 12,500 Pa + 4,020 Pa - 7,220 Pa
P2 - P1 = 9,300 Pa
Therefore, the pressure difference between points B and A is 9,300 Pa.
using the formula :P+rho*g*h + 1/2*rho*v^2=0
calculate: PA= -16518
PB = -7220
PB-PA=-7220-(-16518)=9298