Question

1. A surgeon inserted a needle in the patient’s vein. The needle was connected to a top-open plastic bag filled with therapeutic fluid and lifted to a height h. Find h , which is just sufficient for the fluid to enter the vein, if the pressure gauge inside the vein was 5980 Pa. The therapeutic fluid density was adjusted to be the same as that of blood, 1060kg/m3.

2. At the next stage of the treatment procedure, the surgeon used a syringe to administer a drug fluid into the same vein as in #1. The syringe was positioned at the same level as the vein. A force of 1.20 N was applied to the plunger of the cross-sectional area 0.81 cm² connected to the needle of the inner radius 0.6 mm. What was pressure difference between the proximal and distal parts of the needle, given that the proximal end opened into the vein? Use the same drug fluid density as in #1.

3. What volume of the drug fluid was injected into the vein in 2.5 s by the pressure difference calculated in #2? [Hint: neglect the fluid speed in the syringe].

Answer 1

(1) Pressure gauge inside the vain must be equal to the pressure applied by the fluid.

So, 5980 Pa = h * density of fluid * acceleration due to gravity

or, h * 1060 * 9.8 = 5980

or, h = 0.5757 m = 57.57 cm

Hence, the fluid is needed to be lifted at a height of 57.57 cm.

(2) Pressure difference between the proximal and distal parts of the needle

= Pressure inside the needle - Pressure applied by plunger

Now, cross - sectional area of inside part of needle = * ( 0.6 mm )² = 3.14 * ( 0.06 ) cm² = 0.0113 cm²

Hence, pressure difference between the proximal and distal parts of the needle

= ( 1.2 N / 0.0113 cm² ) - ( 1.2 N / 0.81 cm² )

= 104.7 N/ cm²

= 104.7 * 10⁴ N / m²

= 1.05 * 10⁶ Pa

(3)

Hence, total flow of fluid in 2.5 sec = ( 2.5 * 5 * 10^-5 ) m³ = 1.25 * 10^-4 m³