5. In a hospital, a flexible \"IV\" (intravenous) solution bag contains a drug s

Question

5. In a hospital, a flexible "IV" (intravenous) solution bag contains a drug solution (1250. kg/m). If the upper part of the bag is flaccid, we can assume that it is in pressure equilibrium with the surrounding air, so the top surface of the solution inside the bag is at a pressure of 760.0 mm Hg (the hospital is located at sea level). The IV bag is connected by a flexible tube to a needle inserted into a patient's vein located at a height h below the top surface of the solution. a. (4 pts.) If the pressure inside the patient's vein is 790.0 mm Hg. find the minimum height h so that the fluid just barely flows into the patient (and so that the patient's blood does not flow back up the IV tube). (Assume that the top surface of the solution falls with negligible speed. Assune nearly zero flow speed of the blood in the vein.) Show your work. The patients just seem a lot calmer since we got these new V bags nstead just holds the IV ncedle in the air at the b. (2 pts) A nurse removes the needle from the patient's vein, and i same height h below the IV bag as calculated in panta. what is the speed of the solution as it flows out of the needle into the air? Show your work

Explanation / Answer

Hi,

In this case we have to apply Bernoulli's Equation in order to find the answer.

We will call A to a point on the surface of the solution inside the bag and B to the point where the needle connects with the vein of the patient.

The energy (in terms of pressure) should be equal in both points and they should have the following form:

E = P + pgh + (1/2)pv2 ; where P is the pressure, p is the density of the solution, h is the height above certain reference level and v is the speed of the solution.

a) At point A they tell us that the solution is at rest (v=0), and it is above point B by a certain distance (h), so:

EA = Po + pgh

As we can put the reference level for height at point B's height, the value of said parameter will be zero. Besides, as they are telling us that the speed of the blood is nerly zero (v=0), so:

EB = P

So the minimum value of height will be:

h = (P - Po)/pg = (790 mmHg - 760 mmHg)/ (9.8 m/s2 * 1250 kg/m3) = 30 mmHg / 12250 kg*m2/s2

h = 4000 Pa / 12250 kg*m2/s2 = 0.3265 m ::::::: h = 32.65 cm

Note: 1 atm = 760 mmHg = 101325 Pa ; 1 Pa = 1 N/m2

b) The difference in this case is that the point B has changed. Now, we have a pressure equal to the pressure at the point A (1 atm) but there is a certain speed:

EB = Po + (1/2)pv2

So, the speed of the solution will be:

v = (2gh)1/2 = (2*9.8 m/s2 * 0.3265 m) = 2.530 m/s

I hope it helps.

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