Blood Transfusion: Calculating Height of Bag

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The discussion focuses on calculating the height at which a flexible blood transfusion bag must be placed above a needle to achieve a required flow rate of 4.0 x 10^-6 m^-3 per minute. Key parameters include the needle's diameter of 0.40 mm, blood density of 1.06 x 10^3 kg/m^3, and viscosity of 2.084 x 10^-3 Pa.s. The calculation involves determining the pressure exerted by the blood column and equating it to the pressure in the vein, which is 2.4 kPa above atmospheric pressure. Additionally, the implications of using a rigid glass bottle instead of a flexible bag are discussed, highlighting the need to account for atmospheric pressure in that scenario.

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Hi, i don't know how to do this question - some help would be appreciated.

Q: A patient is to be given a blood transfusion. The blood is to flow through a tube from a raised flexible bag to a needle inserted in the vein. The inside diameter of the 40mm-long needle is 0.40mm and the required flow rate is 4.0*10^-6 m^-3 of blood per minute. [density of blood = 1.06*10^3 kg.m^-3], [Viscosity (eta) of blood = 2.084*10^-3 Pa.s]

a) How high should the bag be placed above the needle? Assume the blood pressure in the vein is 2.4 kPa above atmospheric pressure. :confused:

b)What would happen if the blood were to be transfused from an inverted glass bottle rather than a flexible bag?

Could some please show me how to do part a)? :rolleyes:

Thanks!
 
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ussrasu said:
Hi, i don't know how to do this question - some help would be appreciated.

Q: A patient is to be given a blood transfusion. The blood is to flow through a tube from a raised flexible bag to a needle inserted in the vein. The inside diameter of the 40mm-long needle is 0.40mm and the required flow rate is 4.0*10^-6 m^-3 of blood per minute. [density of blood = 1.06*10^3 kg.m^-3], [Viscosity (eta) of blood = 2.084*10^-3 Pa.s]

a) How high should the bag be placed above the needle? Assume the blood pressure in the vein is 2.4 kPa above atmospheric pressure. :confused:
Think of the blood as a column above the needle into the vein. The diameter of the needle is 0.40 mm= 0.00040 m so the area is \pi (0.00040)^2square m. If the bag is x m above the needle then that column has volume \pi (0.00040)^2 x m^3. You are told the density of blood so you can calculate the weight of that volume of blood.
Divide by the area calculated above to find the pressure at the needle.
(I don't see why "flow rate" or "viscosity" are needed since you are told to "assume blood pressure in the vein is 2.4 kPa above atmospheric pressure.)

b)What would happen if the blood were to be transfused from an inverted glass bottle rather than a flexible bag?

With a flexible bag, the atmosphere can press on the blood itself. That's why, in (a), you only had to allow for the pressure "above atmospheric pressure". If you use a glass bottle, the atmosphere presses against the rigid bottle, not the blood. Now, you would have to make the height enough to equal the entire pressure: atmospheric pressure and the additional 2.4 kPa.
 
how does this find the height? I thought this involved Poisieulles Equation to find L?
 
I don't think you need Poisieulles' equation for this- that applies to flow through a vessel and here you are only asked what the height must be to get the given pressure.
I told you how to find the height before. Think of the tube as a cylinder of blood above the needle as a cylinder having base area the same as the needle opening (you could calculate it from the data given but let's just call it "A") and height h (meters). The volume is Ah and so the weight of the blood is "mass times acc due to gravity times volume"= (1.06*10^3 kg/m3)(9.81 m/s2)(Ah). The pressure is that downward force divided by the the area, A: (1.06*10^3)(9.81)h Newtons/m2. Convert that to "kiloPascals", set it equal to 2.4 and solve for h.

(Oh, I just checked: 1 Pascal is 1 N/m2 so to convert to kPa just divide by 1000: in other words, just drop that "10^3".)
 
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I know it will be nearly negligible, but we have ignored the effects of friction here. I'm not sure if you are in introductory fluid flow right now, or if you have covered head losses and the like.

To calculate the additional height you will need to raise the bag, find the Reynolds number. I will assume that the flow is laminar since flow is so small and diameter is small as well. Now, if I remember correctly, friction factor for laminar flow through circular pipes is f=64/Re. Now, apply your head loss equation to obtain an additional height required. Add that on to the height found earlier.

Also, the question does ask for a height given a required flow, so you will need to take into consideration frictional losses.
 
Hi;
Just a quick addendum to Halls' response to b): You would have to vent the bottle to prevent it from vacuum-locking (I know that's not the right term, but I don't know what else to call it).
 

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