Pressure with fluid problem

[preluda97]

1.) A hypodermic needle is 3.2 cm in length and 0.32 mm in diameter. What excess pressure is required along the needle so that the flow rate of water through it will be 1 g/s?

answer should be in units Pa

2.) When a person inhales, air moves down the bronchus (windpipe) at 14 cm/s. The average flow speed of the air doubles through a constriction in the bronchus. Assuming incompressible flow, determine the pressure drop in the constriction.

answer should be in units Pa

3.) A pipe carrying water from a tank 25.0 m tall must cross 3.15 102 km of wilderness to reach a remote town. Find the radius of the pipe so that the volume flow rate is at least 0.0500 m3/s. (Use the viscosity of water at 20°C.)

answer should be in units of Meters

 

[berkeman]

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[kyle8921]

1.) To calculate the excess pressure required for a flow rate of 1 g/s through a hypodermic needle, we can use the Hagen-Poiseuille equation:

Q = (πΔPr^4)/(8ηL)

Where Q is the flow rate, ΔP is the pressure difference, r is the radius of the needle, η is the viscosity of water, and L is the length of the needle.

Rearranging the equation, we get:

ΔP = (8ηQL)/(πr^4)

Substituting the given values, we get:

ΔP = (8 * 0.001 * 0.032)/(π * 0.00016^4) = 318,309.89 Pa

Therefore, an excess pressure of approximately 318,309.89 Pa is required along the needle for a flow rate of 1 g/s.

2.) According to the Bernoulli's equation, the pressure drop in a constriction can be calculated as:

ΔP = (ρV^2)/2

Where ρ is the density of air and V is the velocity of air.

Since the average flow speed of air doubles through the constriction, we can write:

V = 2 * 14 cm/s = 28 cm/s = 0.28 m/s

Substituting the values, we get:

ΔP = (1.225 kg/m^3 * 0.28^2)/2 = 0.048 Pa

Therefore, the pressure drop in the constriction is approximately 0.048 Pa.

3.) The volume flow rate through a pipe can be calculated as:

Q = (πr^4ΔP)/(8ηL)

Where Q is the volume flow rate, r is the radius of the pipe, ΔP is the pressure difference, η is the viscosity of water, and L is the length of the pipe.

Rearranging the equation, we get:

r = ∛((8ηLQ)/(πΔP))

Substituting the given values, we get:

r = ∛((8 * 0.00089 * 0.050)/(π * 3.15 * 10^5 * 25)) = 0.00113 m

Therefore, the radius of the pipe should be approximately 0.00113 m to achieve a volume