A hypodermic needle is 1.5 cm long and has an inner radius of 0.016cm. What excess pressure is required along the needle so that the flow rate of water though it is 0.0013 kg/s? Use ç = 1.0 x 10-3 Ns/m2 for water.

the answet i have found is deltaP=758017 Pa

using the formula ...

delta P = R ( delta V/delta t)

This looks like a laminar pipe flow problem. I do not recognize your � symbol, nor the units Ns/m2. Is that the viscosity?

The friction factor is 64/(Reynolds number). You can either use pipe flow formulas with that friction factor, or special formulas for Poiseuille flow, which you can find at
http://galileo.phys.virginia.edu/classes/311/notes/fluids2/node6.html

75807pa

To solve this problem, we can use the formula for the excess pressure required to maintain a certain flow rate through a narrow pipe or needle. The formula is:

ΔP = R × (ΔV / Δt)

Where:
ΔP is the excess pressure required
R is the resistance (in this case, the resistance due to the needle)
ΔV is the change in volume (in this case, the volume of water passing through the needle)
Δt is the change in time (how long it takes for the water to pass through the needle)

Given information:
Needle length = 1.5 cm = 0.015 meters
Inner radius = 0.016 cm = 0.00016 meters
Flow rate = 0.0013 kg/s
Water viscosity (η) = 1.0 x 10^-3 Ns/m^2 (dynamic viscosity of water)

First, we need to find the resistance (R) of the needle. For a cylindrical pipe, the resistance is given by:

R = 8ηL / πr^4

Where:
η is the viscosity of the fluid (water in this case)
L is the length of the needle
r is the radius of the needle

Now let's substitute the given values into the equation:

R = (8 * (1.0 x 10^-3 Ns/m^2) * 0.015m) / (π * (0.00016m)^4)
≈ 1191960.6 Ns/m^3

Now, we need to find the change in volume (ΔV) per unit time (Δt). Given the flow rate (mass flow rate), we can use the equation:

ΔV / Δt = m / ρ

Where:
m is the mass flow rate
ρ is the density of water

The density of water is approximately 1000 kg/m^3. Substituting the given values:

ΔV / Δt = (0.0013kg/s) / (1000kg/m^3)
= 1.3 x 10^-6 m^3/s

Now, we can substitute the values of R and ΔV / Δt into the initial formula to find ΔP:

ΔP = (1191960.6 Ns/m^3) × (1.3 x 10^-6 m^3/s)
≈ 1549.54 Pa

Therefore, the excess pressure required to achieve a flow rate of 0.0013 kg/s through the needle is approximately 1549.54 Pa, or 1.55 kPa.

It seems that the answer you found (deltaP = 758017 Pa) is not correct. Double-check your calculations and make sure to use the correct values in the formula.