Oxygen therapy uses various devices to provide oxygen to patients having difficu
ID: 492667 • Letter: O
Question
Oxygen therapy uses various devices to provide oxygen to patients having difficulty getting sufficient amounts from air through normal breathing. Among the devices are nasal cannula, which transport oxygen through small plastic tubes from a supply tank to prongs placed in the nostrils. Consider a specific configuration in which the supply tank (volume = 7.00 ft3) was filled to a pressure of 2100.0 psig at a temperature of -30.0°F. The patient is in an environment where the ambient temperature is 40.0°F. When the tank is used, the pressure begins to decrease as oxygen flows at 5.00 L/min through a tube to the cannula where it is discharged into the nostrils.
1) Estimate the original mass of oxygen in the tank using the compressibility-factor equation of state.
2) What is the initial pressure when the temperature is 40.0°F?
3) How long will it take for the gauge on the tank to read 50.0 psig at an atmospheric pressure of 1 atm and the stated discharge?
Explanation / Answer
a)
Accentric factor of oxygen, = 0.022
Critical temperature, Tc = 154.6 K
T = -30 °C = 243.15 K
Tr = T / Tc = 243.15/154.6 = 1.57
Critical pressure, Pc = 5050 kPa = 5050000 Pa
P = 2100 psig = 14478.99 kPa-g
= 101.325 + 14478.99 = 14580.315 kPa
= 14580315 Pa
R = 8.314 J/mol-K
a = 0.42747 R2 Tc2 / Pc = 0.14
b = 0.08664 R Tc/ Pc = 2.2 x 10-5
= (1 + (0.48508 + 1.55171 – 0.15613 2) (1 – Tr0.5))2
= 0.99
A = a P / (R2 T2) = 0.495
B = b P / (R T) = 0.159
Equation of state:
Z3 – Z2 + Z *(A – B – B2) – A*B = 0
Z3 – Z2 + 0.31 Z – 0.079 = 0
Solving we get, Z = 0.72
Volume of tank, V = 7 ft3
= 0.198 m3
P V = Z n RT
n = P V / (Z R T)
= 1983.4 moles
Molecular mass of O2 = 32 g/mol
Mass of oxygen = 1983.4 * 32 = 63467.4 g
= 63.47 kg
b)
The patient is at ambient conditions at 40 °F
Initial pressure = ambient pressure = 1 atm
c)
Final pressure of oxygen, P = 50 psig = 446.063 kPa = 446063 Pa
Solving for Z by repeating part a) with P = 446063 Pa we get,
Z = 0.99
P V = Z n RT
n = P V / (Z R T)
= 44.2 moles
Molecular mass of O2 = 32 g/mol
Final mass of oxygen = 44.2 * 32 = 1414 g
= 1.414 kg
Rate of oxygen flow rate at discharge, Q = 5 L/min
P = 101325 Pa = 1 atm
T = 40 °F = 277.6 K
Mole flow rate = P Q / (R T) = 1 * 5 / (0.0821 * 277.6)
= 0.22
Mass flow rate = 0.22 * 32 / 1000 kg/min
= 0.007 kg/min
Time taken = (Initial mass – Final mass)/Mass flow rate
= (63.47 – 1.414)/0.007 min
= 8840min = 147.3 hours
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.