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

ID: 1268910 • Letter: A

Question

A hypodermic syringe is attached to a needle that has an internal radius of 0.300 mm and a length of 3.02 cm. The needle is filled with a solution of viscosity 2.02 10-3 Pa · s; it is injected into a vein at a gauge pressure of 16.4 mm Hg. Ignore the extra pressure required to accelerate the fluid from the syringe into the entrance of the needle.

(a) 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? (Answer if Pa)

(b) What force must be applied to the plunger, which has an area of 1.02 cm2? (Answer in N)

Explanation / Answer

Ok, we know that the Poiseuille equation is :

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

In this case, we have the velocity : 0.252 mL / s, the lenght, the internal radius, and the first pressure, so we can use :

viscosity = 2.05*10^-3 = n

L = 3.01*10^-2

R = 0.3*10^-3

v = 0.252*10^-6 m^3 / s

p1 = 16.3 mmHg = 16.3 / 760 atm = 16.3 / 760*10^5 Pa

0.252*10^-6 = pi/8*(0.3*10^-3)^4*(16.3 - p2) / 3.01*10^-2*2.05*10^-3

0.252*10^-6*6.17*10^-5*8 = pi*(0.3*10^-3)^4*(16.3-p2)

4888.15 = p2- 16.3/760*10^5

p2 = 7032.88 Pa

p2 = 0.07032 atm

p2 = 53.44 mmHg
b)F=7032.88*1*10^-4=.7032N

Same question solved by me previously

Hope that helps

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