Poiseuille’s equation relates the volumetric flow rate Q (volume per time unit) to the
pressure differential ΔP that drives the flow:
Q = (πr^4)/(8η) x (ΔP/L)
Here η is the viscosity of the fluid and the tube has length L and radius r. A fire brigade
needs to triple its hose length to be able to serve a new neighborhood where the fire
hydrants will be further from each other. The old hoses had a radius of r=3 cm.
(a) Determine the new hose radius required to maintain the same volume flow rate as
the original hoses.
(b) What must the minimum water pressure at the fire hydrant be to ensure a flow
rate of 300 gallons per minute through a 300 foot long new hose? The viscosity of
water at room temperature is approximately 1.
10-3 Pascal second.
To determine the new hose radius required to maintain the same volume flow rate as the original hoses, we can use the Poiseuille's equation.
Let's denote the old hose radius as r1 and the new hose radius as r2.
According to the equation, the volumetric flow rate Q is directly proportional to the fourth power of the radius, Q ∝ r^4.
Therefore, to triple the hose length while maintaining the same volume flow rate, we need to find the new radius (r2) such that (r2^4)/(r1^4) = 3.
(a) Solving for the new radius (r2):
(r2^4)/(r1^4) = 3
r2^4 = 3(r1^4)
r2 = (3^(1/4)) * r1
Given that the old hose radius r1 = 3 cm, we can substitute this value and solve for r2:
r2 = (3^(1/4)) * 3
r2 ≈ 3.860 cm
Therefore, the new hose radius required to maintain the same volume flow rate as the original hoses is approximately 3.860 cm.
(b) To determine the minimum water pressure required to achieve a flow rate of 300 gallons per minute through a 300-foot long new hose, we can rearrange Poiseuille's equation to solve for the pressure differential ΔP:
Q = (πr^4)/(8η) * (ΔP/L)
Given:
Q = 300 gallons per minute ≈ 1135.62 liters per minute ≈ 0.0189 cubic meters per second
r = radius of the new hose (in meters) ≈ 3.860 cm or 0.0386 meters
η = viscosity of water ≈ 1.10^-3 Pascal second
L = length of the new hose ≈ 300 feet ≈ 91.44 meters
Substituting these values into the equation, we can solve for ΔP:
0.0189 = (π(0.0386^4))/(8 * 1.10^-3) * (ΔP/91.44)
Solving for ΔP:
ΔP = (0.0189 * 8 * 1.10^-3 * 91.44) / (π(0.0386^4))
ΔP ≈ 263.44 Pascal
Therefore, the minimum water pressure at the fire hydrant should be approximately 263.44 Pascal to ensure a flow rate of 300 gallons per minute through a 300-foot long new hose.
To find the new hose radius required to maintain the same volume flow rate as the original hoses, we can use Poiseuille's equation:
Q = (πr^4)/(8η) x (ΔP/L)
Let's assume that the original hose length is L1, the original hose radius is r1, and the new hose length is L2. Since the fire brigade needs to triple the hose length, we have L2 = 3L1.
(a) To find the new hose radius (r2), we need to compare the volume flow rates of the original and new hoses:
Q1 = (πr1^4)/(8η) x (ΔP/L1) ---> Equation 1
Q2 = (πr2^4)/(8η) x (ΔP/L2) ---> Equation 2
We want to find r2 such that Q2 = Q1. Since ΔP and η are constant, we can set Equation 1 equal to Equation 2 and solve for r2:
(πr1^4)/(8η) x (ΔP/L1) = (πr2^4)/(8η) x (ΔP/L2)
Canceling out ΔP and η from both sides, and substituting L2 = 3L1, we get:
r2^4 / r1^4 = L2 / L1 = 3
Taking the fourth root of both sides:
r2 / r1 = ∛3
Finally, we can solve for r2:
r2 = ∛3 * r1
Substituting r1 = 3 cm, we can calculate r2:
r2 = ∛3 * 3 cm = 3^(1/3) * 3 cm ≈ 3.3 cm.
Therefore, the new hose radius required to maintain the same volume flow rate as the original hoses is approximately 3.3 cm.
(b) To find the minimum water pressure at the fire hydrant to ensure a flow rate of 300 gallons per minute through a 300-foot long new hose, we need to rearrange Poiseuille's equation to solve for ΔP:
Q = (πr^4)/(8η) x (ΔP/L)
First, let's convert the flow rate from gallons per minute to cubic meters per second:
1 gallon ≈ 0.00378541 cubic meters
1 minute ≈ 60 seconds
Therefore, 300 gallons per minute ≈ 300 * 0.00378541 cubic meters ≈ 1.13562 cubic meters per minute ≈ 1.13562 / 60 ≈ 0.01893 cubic meters per second.
Now we can plug in the values:
0.01893 m^3/s = (πr^4)/(8 * 1.10^-3) x (ΔP/300 ft)
Canceling out the units and solving for ΔP:
ΔP = (0.01893 m^3/s) x (8 * 1.10^-3) x (300 ft) / (πr^4)
Converting the ft term to meters:
1 ft ≈ 0.3048 meters
ΔP = (0.01893 m^3/s) x (8 * 1.10^-3) x (300 ft * 0.3048 meters/ft) / (πr^4)
Simplifying further:
ΔP = (0.01893 x 8 x 1.10^-3 x 300 x 0.3048) / (πr^4) m^-3
Substituting η = 1.10^-3 Pascal second and r = 3.3 cm = 0.033 meters:
ΔP = (0.01893 x 8 x 1.10^-3 x 300 x 0.3048) / (π(0.033)^4) m^-3
Evaluating the expression:
ΔP ≈ 61.203 Pascal
Therefore, the minimum water pressure at the fire hydrant should be approximately 61.203 Pascal to ensure a flow rate of 300 gallons per minute through a 300-foot long new hose.