A hypodermic syringe is attached to a needle that has an internal radius of 0.30

Question

A hypodermic syringe is attached to a needle that has an internal radius of 0.303 mm and a length of 3.02 cm. The needle is filled with a solution of viscosity 2.02 times 10-3 Pa · s; it is injected into a vein at a gauge pressure of 16.1 mm Hg. Ignore the extra pressure required to accelerate the fluid from the syringe into the entrance of the needle. What must the pressure of the fluid in the syringe be in order to inject the solution at a rate of 0.253 mL/s? Pa What force must be applied to the plunger, which has an area of 1.04 cm2? N

Explanation / Answer

the Poiseuille equation is :

dV / dt = pi/8*(R^4/n)*(p1 - p2 / L)

In this case, we have the velocity : 0.253 mL / s, the lenght, the internal radius, and the first pressure, so we can use :

viscosity = 2.53*10^-3 = n

L = 3.02*10^-2

R = 0.303*10^-3

v = 0.202*10^-6 m^3 / s

p1 = 16.1 mmHg = 16.1 / 760 atm = 16.1 / 760*10^5 Pa

0.202*10^-6 = pi/8*(0.303*10^-3)^4*(16.1 - p2) / 3.02*10^-2*2.53*10^-3

0.202*10^-6 * 7.4606*10^-5 * 8 = pi*(0.303*10^-3)^4*(16.1-p2)

12.056329 x10^-11=2.647936 x10^-14*(16.1-p2)

4553.1043 = p2 - 16.1/760*10^5

p2 = 6671.52 Pa

2)

1 Pa = 1 N/m2

Force = Pressure * Area

F= 6671.52 Pa *0.000104

F=0.6938 or 0.7 N

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