Question

A bag of saline for intravenous infusion is held up at 80 cm above the level of patient's vein. The bag is connected through a 4 mm diameter tube to a catheter with 0.45 mm diameter. Unfortunately, the end of the catheter gets What is the downward force of the saline solution on the clot at the tip of the catheter? How much does that force increases if the bag is lifted to 2m?

Answer 1

Part A)

the force is F = P*A

where

P = p*g*h,

p = rho = density = 1000 kg/m^3,

g = 9.81 m/s^2,

h = 80 cm = 80*10^-2 m

the area is A = pi*(d/2)^2

d = 0.45 mm = 0.45*10^-3 m

F = P*A = p*g*h*pi*(d/2)^2

F = 1000*9.81*(80*10^-2)*3.14*[(0.45*10^-3)/2]^2

F = 1.25*10^-3 N

Part B)

the new pressure is

P1 = p*g*h1

where

h1 = 2 m

P1 = 1000*9.81*2 = 19620 Pa

F' = P1*A = 19620*3.14*[(0.45*10^-3)/2]^2

F' = 3.12*10^-3 N

Delta F = F' - F = 1.87*10^-3 N

The Increase in force if the bag is lifted to 2m, is 1.87*10^-3 N