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6. There is a gas cylinder filled with oxygen in front of you. It is 1.5 meters tall, and 30 centimeters in diameter. The oxygen within the cylinder has dropped to a gauge pressure of 2.52 x 106Pa at a temperature of 20°C. This pressure is insufficient to ventilate a patient in respiratory distress. You do not have another oxygen cylinder present; and, without oxygen the patient will die.
a) To maintain minimal ventilation, the pressure must be raised to a gauge pressure of 4.0 x 106 Pa. The only parameter you can vary is the temperature (by one of several means). What must the new temperature in the tank be to attain this minimal pressure?
b) What is the new absolute pressure in the tank?
c) How much gas is left in the tank? Give this value in two ways: the number of moles, and the more useful quantity of the number of Liters.
d) How long do you have before the oxygen in the tank is exhausted?
e) If the tank nozzle/outlet is a "Christmas Tree" fitting with an inside diameter of 0.4 cm, and the tube to the mask has an inside diameter of 0.5 cm, what is the flow rate inside the tubing?
f) Characterize the flow in the tube by calculating its Reynold's Number.
g) If the top of the tank, where the flow meter says 15L/min is 0.25 m above the patient's bed, what is the actual flow rate the patient is experiencing? You may assume the pressure in the supply line to the mask is the same as the tank; and, the pressure in the patient's mask to be atmospheric pressure.
h) As the gas exits the tube into the nonrebreather mask, the Venturi effect becomes important. The pressure in the mask is approximately atmospheric pressure (you can't get don't want, an air tight seal). What is the flow rate of oxygen into the mask? The inside diameter of the supply tube is 0.5 cm, and the effective cross-sectional area of the mask/mouth is 15 cm2.
Patm=1.013 x 105 Pa 6.895 x 10+ 3 Pa/psi Density of O2(approximate)=1.3 kg/m3
Oxygen=O2=32g/mole 1cm3 = 1ml Viscosity of O2 (approximate)= 1.0 x 10-3
Length of the Gas cylinder = l = 1.5 m
radius of the cylinder = r = 30/2 cm = 15 cm = 0.15 m
Initial gauge pressure Pgi = 2.52*10^6 Pa
Atmospheric pressure Patm = 1.013 *10^5 Pa
hence, initial absolute pressure = Pi = Pgi+Patm
=> Pi = 2.52*10^6 + 1.013*10^5 = 26.213 *10^5 Pa
Final gauge pressure = Pgf = 4.0*10^6 Pa
Final absolute pressure = Pf = 41.013*10^5 Pa
Initial temperature = Ti = 20+273 = 293 K
Final temperature = Tf = ?
Pi/Ti = Pf/Tf
=> Tf = Pf*Ti/Pi = ...
With full formulas, all parts of the problem are solved.