Pressure Question: Intravenous infusions are often made under gravity. Assuming the fluid has a density of 1.00g/cm^3, at what height h should the bottle be placed so the liquid pressure is 55mm-Hg?
Atmospheric pressure, Po, = 101kPa = 101000Pa density of fluid, ρ, = 1g/cm^3 = 1000kg/m^3 density of Hg, ρ, = 13.6x10^3 P=ρgh P=Po + pgh
Pressure of Hg, P= ρgh = (13.6x10^3) x 9.8 x (5.5x10^-3) = 733.04 Pa \[P=Po + pgh\] \[h=\frac{ P - Po }{ pg }\] \[h= \frac{ 733.04 - 101000 }{ 1000 x 9.8 }\] \[h= -10.23m\] Therefore, the bag must be 10.23m below whatever the reference point is. Which is obviously wrong! The answer gives nearly what I've got but I don't understand one aspect of it. It says... \[h= \frac{ DeltaP }{ gh }\] Where they have the calculation for delta P as being... \[(55mm-Hg)\left(\begin{matrix}133Pa \\ 1mm-Hg\end{matrix}\right)\] I don't understand where the 133Pa came from and why they are doing this calculation. Can someone please explain?
They are doing a conversion between 55mm Hg and Pa. Bc 1mm Hg is about 133Pa
Otherwise your answer looks good to me. Just make sure you always use the same pressures (gage or abs)
But I thought I did the conversion? Pressure of Hg, P= ρgh = (13.6x10^3) x 9.8 x (5.5x10^-3) = 733.04 Pa Then to get deltaP you minus this from atmospheric pressure?
Sorry, minus atmospheric pressure from the pressure of Hg. Not the other way round.
So I could use my original calculation is correct I just converted to metres wrong?
Uh... no. Using 10^-2 instead of 10^-3 in my calculation gives the final answer as -9.56m. Which is still wrong. Help??
i do not think that atmospheric pressure plays any role here. In order to infuse some liquid in to the human body , it should be at sufficient pressure to enter the blood in order to counter the force applied by blood pressure. From what we are given ,it seems that blood pressure is 55mmHg. Now the liquid is in some sort of bottle and the atmospheric pressure at the top is zero and the other end may be a needle inside the body and the atmospheric pressure is again zero there. So the total pressure at the end will just be because of the weight of the liquid which is hdg Equate hdg = 55mmHg We must convert 55 mmHg into N/m2 or pascals The conversion factor is 760 mmHg = 10^5 Pascals
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