Water flows at a rate of 25L/min through a horizontal 8.0-cm-diameter pipe under a pressure of 5.7Pa . At one point, calcium deposits reduce the cross-sectional area of the pipe to 40cm2 .
Part A
What is the pressure at this point? (Consider the water to be an ideal fluid.)
To determine the pressure at the point where the cross-sectional area of the pipe is reduced, we can use the principle of continuity and Bernoulli's equation for ideal fluids.
The principle of continuity states that the mass flow rate of a fluid is constant in a pipe that is not changing in volume. This means that the product of the cross-sectional area and the velocity of the fluid is constant.
In this case, we are given the initial flow rate as 25 L/min, which we can convert to m^3/s by dividing by 60:
25 L/min = (25/1000) m^3/s = 0.4167 m^3/s
We can calculate the initial velocity, v1, by dividing the flow rate by the cross-sectional area of the initial pipe:
v1 = 0.4167 m^3/s / [(π/4) * (0.08 m)^2]
= 2.645 m/s
Now, using the principle of continuity, we can find the velocity, v2, at the point where the cross-sectional area is reduced to 40 cm^2:
v1 * A1 = v2 * A2
where A1 is the initial cross-sectional area and A2 is the reduced cross-sectional area. We can rearrange this equation to solve for v2:
v2 = (v1 * A1) / A2
Substituting the given values, we have:
v2 = (2.645 m/s * [(π/4) * (0.08 m)^2]) / [(π/4) * (0.04 m)^2]
= 10.58 m/s
Now, we can use Bernoulli's equation to find the pressure at the point of reduced cross-sectional area. Bernoulli's equation relates the pressure, velocity, and height of a fluid in a steady flow:
P1 + (1/2) * ρ * v1^2 + ρ * g * h1 = P2 + (1/2) * ρ * v2^2 + ρ * g * h2
In this case, we can assume the height and gravitational potential energy do not change, so we can omit the ρ * g * h terms:
P1 + (1/2) * ρ * v1^2 = P2 + (1/2) * ρ * v2^2
Since we are given that the fluid is ideal, we can assume the density, ρ, remains constant. Therefore, this equation simplifies to:
P1 + (1/2) * ρ * v1^2 = P2 + (1/2) * ρ * v2^2
Rearranging this equation to solve for P2, we have:
P2 = P1 + (1/2) * ρ * (v1^2 - v2^2)
Substituting the given values, we have:
P2 = 5.7 Pa + (1/2) * ρ * (2.645^2 - 10.58^2)
To obtain the value of P2, you would need to know the density of the water. Once you have that, substitute it into the equation along with the known values and calculate P2.
To determine the pressure at the point where the calcium deposits reduce the cross-sectional area of the pipe, we can make use of the equation for the flow rate of an ideal fluid through a pipe.
The equation to use is:
Q = A1v1 = A2v2
Where:
Q is the flow rate
A1 is the cross-sectional area of the pipe before the reduction
v1 is the velocity of the water before the reduction
A2 is the cross-sectional area of the pipe at the reduced point
v2 is the velocity of the water at the reduced point
Given information:
Q = 25 L/min = 25/60 m^3/s (since 1 L = 0.001 m^3 and 1 min = 60 s)
A1 = (πd1^2)/4 = (π(8cm)^2)/4 = (π(0.08m)^2)/4
A2 = 40 cm^2 = 0.004 m^2
First, we can calculate the velocity of the water before the reduction by rearranging the equation as follows:
v1 = Q/A1
Substituting the given values:
v1 = (25/60)/(π(0.08)^2)/4
Next, we can calculate the velocity of the water at the reduced point by rearranging the equation again:
v2 = Q/A2
Substituting the given values:
v2 = (25/60)/0.004
Finally, we can calculate the pressure at the reduced point by using Bernoulli's equation which states that the pressure plus the kinetic energy per unit volume plus the potential energy per unit volume is constant along a streamline.
P1 + 0.5ρv1^2 + ρgy1 = P2 + 0.5ρv2^2 + ρgy2
Since the point where the cross-sectional area is reduced is horizontal, the potential energy terms cancel each other out, and we're left with:
P1 + 0.5ρv1^2 = P2 + 0.5ρv2^2
Where:
P1 is the pressure before the reduction
P2 is the pressure at the reduced point
ρ is the density of the fluid
Rearranging the equation for P2:
P2 = P1 + 0.5ρv1^2 - 0.5ρv2^2
Substituting known values:
P2 = 5.7 Pa + 0.5ρv1^2 - 0.5ρv2^2
We are given that the water is an ideal fluid, which means it is incompressible and has a constant density. Therefore, ρ can be considered constant.
Substituting the previously calculated velocity values into the equation, we can obtain the pressure at the reduced point.
I get really irritated when a student types in bogus titles. this has nothing to do with buoyance, does it?
Specifically, it deals with Bernoulli's equation, density remains constant, and mass flow is conserved (mass flow in = mass flow out).
http://scienceworld.wolfram.com/physics/BernoullisLaw.html
P+1/2 rho*v^2+rho g h=constant
Not that is it always needed, but it is safe to change units to SI.
25liters=25l(1m^3/1000l)=.025m^2
1min= 60 sec
so 25l/min= .025/60 m^3/sec
Now, have two sides to work with:
Pi+1/2 rho*vi^2+rho g h=Pf+1/2 rho*vf^2+rho g h
the rho g h term can be lost frm both sides, you have Pi, vi, you know rho.
But vf: use the law of mass continuity
Vi*areainput=Vf*areaoutput solve for vf
Now, finally, solve for Pf