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A tube of radius 5 cm is connected to tube of radius 1 cm as shown above. Water is forced through the tube at a rate of 10 liters/min. The pressure in the 5 cm tube is 1×105 Pa. The density of water is 1000 kg/m3. Assume that the water is nonviscous and uncompressible.
(a) What is the velocity of the water in the 5 cm radius tube?
v1 = m/s
(b) What is the velocity of the water in the 1 cm radius tube?
v2 = m/s
(c) What is the water pressure in the 1 cm radius tube?
P2 = N/m2
A) V1= 10L/((pie)(.05^2)(1000)(60))=.0212m/s
B) V2= 10L/((pie)(.01^2)(1000)(60)=.5307m/s
C)P=P1 - P2 = 1000*[(.0212)²/2 - (.5307)²/2]
P1 - P2 = 140.5 Pa
P2 = 10^5 - 140.5 = 99859.5 Pa.
given;
r2 = 0.05 m
r1 = 0.01 m
Q = 10 liters/min = 10 E-3/60 = 1.67 E-4 m³/s
P2 = 10^5 Pa
ρ = 1000 kg/m³
no. # (a)
Q = (A2)(v2)
Q = π(r2)²)(v2)
1.67 E-4 = π(0.05)²)(v2)
V2 = 0.0212 m/s
no. # (b)
V1 = (r2/r1)² (V2) = (5/1)² (0.0212)
V1 = 0.5316 m/s
no. # (c)
using Bernoulli's equation in dynamics fluid
P1 + ½ ρ V1² + ρ g h = P2 + ½ ρ V2² + ρ g h
I assume there's no level differences and no losses of energy.
P1 + ½ ρ V1² = P2 + ½ ρ V2²
P1 + ½ (1000) (0.5316)² = 10^5 + ½ (1000) (0.0212)²
P1 = 99859 Pa
To answer these questions, we can use the principle of conservation of mass and Bernoulli's equation.
(a) To find the velocity of the water in the 5 cm radius tube (v1), we can use the equation of continuity, which states that the mass flow rate is constant in a pipe. The equation is given by:
A1 * v1 = A2 * v2
where A1 and A2 are the cross-sectional areas of the 5 cm radius tube and the 1 cm radius tube respectively, and v2 is the velocity of water in the 1 cm radius tube.
Since the area of a circle is given by A = π * r^2, we can calculate the ratios:
A1 = π * (5 cm)^2 = 25π cm^2
A2 = π * (1 cm)^2 = π cm^2
Substituting the values into the equation of continuity:
25π cm^2 * v1 = π cm^2 * v2
Simplifying the equation:
v1 = v2 * (1/25)
(b) To find the velocity of the water in the 1 cm radius tube (v2), we can use the principle of conservation of mass, as mentioned above.
Since we're given that the water is forced through the tube at a rate of 10 liters/min, we can convert this to a flow rate in m^3/s:
10 liters/min = 10/1000 m^3/s = 0.01 m^3/s
We know that the flow rate (Q) is given by:
Q = A * v
where Q is the flow rate, A is the cross-sectional area of the pipe, and v is the velocity of water.
Substituting the values into the equation:
0.01 m^3/s = π cm^2 * v2
Converting the units to m^2:
0.01 m^3/s = π * (0.01 m)^2 * v2
Simplifying the equation:
v2 = 1/π m/s
(c) To find the water pressure in the 1 cm radius tube (P2), we can use Bernoulli's equation, which states that the total energy of fluid particles remains constant along a streamline.
Bernoulli's equation is given by:
P1 + (1/2) * ρ * v1^2 = P2 + (1/2) * ρ * v2^2
where P1 is the pressure in the 5 cm radius tube, ρ is the density of water, and v1 and v2 are the velocities of water in the 5 cm and 1 cm radius tubes respectively.
Substituting the given values:
1×10^5 Pa + (1/2) * 1000 kg/m^3 * v1^2 = P2 + (1/2) * 1000 kg/m^3 * v2^2
Since the pressure in the 5 cm radius tube (P1) is given as 1×10^5 Pa, we can simplify the equation:
(1/2) * 1000 kg/m^3 * v1^2 = P2 + (1/2) * 1000 kg/m^3 * v2^2
Simplifying further:
v1^2 = 2 * (P2 - 1×10^5 Pa) / (1000 kg/m^3)
Hence, the water pressure in the 1 cm radius tube (P2) can be found by rearranging the equation and plugging in the value of v1 obtained in part (a).
To find the velocity of the water in the 5 cm radius tube, you can use the principle of continuity, which states that the volume flow rate of an incompressible fluid remains constant along a tube.
The volume flow rate can be calculated using the formula: flow rate = A1 * v1 = A2 * v2, where A1 and A2 are the cross-sectional areas of the tubes, and v1 and v2 are the velocities of the water in the tubes.
Given that the radius of the 5 cm tube is 5 cm and the radius of the 1 cm tube is 1 cm, we can calculate the cross-sectional areas:
A1 = π * (5 cm)^2
A2 = π * (1 cm)^2
Now, since flow rate = 10 liters/min and 1 liter = 1000 cm^3, we can convert the flow rate to cubic meters per second (m^3/s):
flow rate = 10 liters/min * (1000 cm^3 / 1 liter) * (1 m^3 / 10^6 cm^3) * (1 min / 60 s)
Now we can substitute the known values into the continuity equation and solve for v1:
A1 * v1 = A2 * v2
(π * (5 cm)^2) * v1 = (π * (1 cm)^2) * v2
(25π cm^2) * v1 = π cm^2 * v2
v1 = (π cm^2 * v2) / (25π cm^2)
v1 = v2 / 25
Since we know the pressure in the 5 cm tube is 1×10^5 Pa, we can use the Bernoulli's equation to relate the velocity and pressure:
P1 + (1/2) * ρ * v1^2 = P2 + (1/2) * ρ * v2^2
Given that the density of water is 1000 kg/m^3, we can calculate P2:
P1 = 1×10^5 Pa
P2 = P1 + (1/2) * ρ * v1^2 - (1/2) * ρ * v2^2
Now you can substitute the known values and solve for v2:
P2 = (1×10^5 Pa) + (1/2) * (1000 kg/m^3) * (v1^2 - v2^2)
P2 = (1×10^5 Pa) + (1/2) * (1000 kg/m^3) * ((v2 / 25)^2 - v2^2)
(v2^2 / 625) - v2^2 = ((2 * P2 - 2 * 1×10^5 Pa) / (1000 kg/m^3))
Solving this equation will give you the value of v2, the velocity of the water in the 1 cm radius tube.
Keep in mind that these calculations are based on the assumptions of non-viscous and incompressible fluid, so the results may not be exact in real-world scenarios.