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**Respiratory Calculations**

Gas Laws Oxygen therapy Humidity Ventilator Management Hemodynamics

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**Gas Laws Dalton’s Law Fick’s Law of Diffusion**

Boyle’s Law, Charles Law, Gay-Lussac’s Combined Gas Law Graham’s Law Poiseuille’s Law Temperature Conversion (C to F and vice versa)

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**Oxygen Therapy Total Flow Tank Duration Arterial & Venous O2 Content**

[C(a-v)O2] difference Alveolar Air Equation P(A-a) O2 Gradient Heliox flow rates

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**Total Flow Subtract FiO2 - 20 (or 21) 100 These 2 values FiO2 Will**

Determine the Air: O2 ratio FiO2 20 or 21 Subtract 100 - FiO2 If the FiO2 Is .40 or > Use 20 (< .40 Use 21) Add the numbers of the ratio X flow rate = Total flow

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**Total Flow: Example (1 + 3) x 10 = 40 LPM**

A COPD patient is currently on a 40% aerosol face mask running at 10 LPM. Calculate the total flow. 20 1 100 40 (1 + 3) x 10 = 40 LPM 20 60 3

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**Tank Duration PSIG x Tank factor Flow rate E cylinder: .28**

H cylinder: 3.14 Pressure of the cylinder PSIG x Tank factor Flow rate The flow the O2 Device is set at

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**Tank Duration: Example**

A patient is currently on a 4 L nasal cannula. The patient needs to be transported using an E cylinder. The E cylinder reads 2200 psig on the Bourdon gauge. According to hospital policy, the tank should not be used once the pressure reading reaches 200 psig. Calculate how long the tank will last ( ) x .28 4 152.6 minutes 60 min/hr 2.54 Hours

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**Arterial & Venous O2 Content**

Arterial and venous. O2 content represents the amount of oxygen that is bound to hemoglobin and dissolved in the blood. The difference is that arterial O2 content represents the arterial system (high O2), and venous O2 content represents the venous system (low O2). CxO2 = (1.34 x Hgb x SxO2) + (PxO2 x .003) O2 carried/bound to hemoglobin O2 dissolved in blood plasma

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**Comparison of CaO2 & CvO2 CaO2 = (1.34 x Hgb x SaO2) + (PaO2 x .003)**

Arterial O2 Content CaO2 = (1.34 x Hgb x SaO2) + (PaO2 x .003) Partial Pressure Of arterial O2 A constant Hemoglobin Arterial saturation A constant Venous O2 Content CvO2 = (1.34 x Hgb x SvO2) + (PvO2 x .003) Partial Pressure Of venous O2 Venous saturation

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**Arterial O2 Content: Example**

Given the following values, calculate the CaO2: PaO2 = 93 mmHg PvO2 = 47 mmHg SaO2 = 98% SvO2 = 77% Hemoglobin = 16 g/dL CaO2 = (1.34 x 16 x .98) + (93 x .003) CaO2 = = vol % Normal value for CaO2 is approximately 20 vol %

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**Venous O2 Content: Example**

Given the following values, calculate the CvO2: PaO2 = 93 mmHg PvO2 = 47 mmHg SaO2 = 98% SvO2 = 77% Hemoglobin = 16 g/dL CvO2 = (1.34 x 16 x .77) + (47 x .003) CvO2 = = vol % Normal CvO2 is approximately 15 vol %

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C(a-v) Difference The C(a-v) difference represents the difference between arterial And venous oxygen content. It is a reflection of oxygen Consumption (oxygen used by tissues within the body) Recall the values from the 2 previous examples: CaO2 = vol % CvO2 = vol % To determine the C(a-v)O2, simply subtract: CaO2 - CvO2 = 4.64 vol % Normal C(a-v)O2 = 5 vol %

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**C(a-v) difference: Clinical Info**

C(a-v)O2 can be an important clinical indicator. Recall that The C(a-v)O2 reflects the amount of oxygen taken from arterial Blood to be used by body tissues. Refer to the diagram below: O2 O2 Arterial: CaO2 = 20 vol% Tissues Venous: CvO2 = 15 vol% O2 O2 Arterial blood contains Approx 5 vol% of O2 O2 that is NOT extracted From arterial blood enters Venous circulation 5 vol% of O2 Is extracted from Arterial blood

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**C(a-v) Difference con’t…**

When blood flows through the body at a normal rate, approximately 5 vol% of the O2 present in arterial blood is extracted by the tissues. The remaining O2 enters the venous system. When blood flows through the body slower than normal, blood begins To pool and more O2 is taken from arterial blood. With the tissues Extracting more O2, less O2 is present in the venous system. If you Have a lower venous O2 content, and subtract it from the CaO2, you Get a greater C(a-v)O2 difference An increase in the C(a-v)O2 difference = a decrease in cardiac output

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**Alveolar Air Equation The Alveolar air equation represents the partial**

Pressure of oxygen in the alveoli A/C M E B R A N This is what we Are finding using The alveolar Air equation Alveolus PAO2 O2 O2 O2 O2 diffusion PaO2 O2 O2 Capillary O2 O2

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**Alveolar Air Equation: Con’t…**

PAO2 = [(PB-PH2O) FiO2] - PaCO2 / .8 O2 concentration Barometric pressure Normal is 760 mmHg Water pressure Constant: 47 mmHg Arterial CO2 Constant: Respiratory Quotient CO2 removed/O2 consumed 200 mL/ 250 mL = .8

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**Alveolar Air Equation: Example**

Given the following information, calculate the PAO2 PB = 760 mmHg FiO2 = .60 PaCO2 = 40 mmHg PaO2 = 88 mmHg Hgb = 14 g/dL = mmHg PAO2 = [( ).60] - 40 / .8

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P(A-a)O2 Gradient P(A-a)O2 represents the difference between the partial pressure Of O2 in the alveoli and the partial pressure of O2 in the arteries. In other words, it reflects how much of the available O2 (PAO2) Is actually diffusing into the blood (PaO2). In a healthy individual, the P(A-a)O2 should be very small. In other words, the majority of the available O2 is diffusing Into the blood (refer to the diagram on the “alveolor air Equation slide for a better understanding) If the P(A-a)O2 increases, it signals there is some problem with the gas diffusion mechanism (shunting for example).

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**P(A-a)O2 Gradient: Example**

Using the PAO2 calculated earlier (377.8 mmHg), calculate The P(A-a)O2 if the PaO2 is 80 mmHg P(A-a)O2 = 297.8 mmHg What does this number tell you? This number indicates that a significant amount of the available O2 is not diffusing into the blood, indicating a shunt is present

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Heliox Flow Rates Heliox is a mixture of helium and oxygen. Because helium is less Dense than oxygen, it is used to carry oxygen past airway Obstructions. Because heliox is less dense than pure oxygen, It has a faster flow. 2 different heliox mixtures: Multiply flow Reading by A factor of 1.8 To get actual flow Helium : Oxygen : : Multiply flow Reading by A factor of 1.6 to Get actual flow

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**Heliox Flow Rates: Example**

A physician orders 80:20 heliox to be run at 18 LPM. At what flow rate should the flow meter be set? We know that Set Flow rate x 1.8 = actual flow of 80:20 heliox We can rearrange this equation to solve for the set flow rate: Set flow rate = Actual flow / 1.8 Set flow rate = 18 LPM / 1.8 Set flow rate = 10 LPM In order to have an actual flow of 18 LPM, we need to set the Flow meter at 10 LPM (If this were a 70:30 mixture, replace 1.8 with 1.6)

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Humidity Body Humidity

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**Body Humidity Normal body humidity is expressed as 44 mg/L or 47 mmHg**

This means that at 98.6 F (37 C) gas is saturated with 44 mgHg or 44 mg/L of water vapor Relative Humidity: Humidity Deficit: What is the relative humidity Of a gas saturated with 30 mg/L Of water at body temperature? What is the humidity deficit Of a gas saturated at 30 mg/L Of water at body temperature? 30 mg/L 44 mg/L 44 mg/L - 30 mg/L = 14 mg/L = 68%

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**Ventilator Management**

Compliance (dynamic vs. static) Resistance I-time, peak flow rate, vt I:E ratio Desired CO2 / VE Desired PaO2 VD/VT Minute Ventilation / Alveolar Ventilation

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**Compliance ∆ Volume ∆ Pressure Generic Equation**

Graph of Mechanical Breath ∆ Volume ∆ Pressure Dynamic Static

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**Dynamic Compliance Tidal Volume (mL) Peak Pressure - PEEP**

Dynamic compliance measures the elasticity of the lung During air movement. It is a less reliable indicator of lung Elasticity compared to static compliance Note: Peak Pressure = PIP

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**Static Compliance Tidal Volume (mL) Plateau Pressure - PEEP**

Static compliance measures the elasticity of the lung When there is no air movement. It is the best indicator Of the ability to ventilate the lungs. Normal static compliance is: mL/cmH2O Note: Plateau pressure = PPL = Static Pressure

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**Understanding Compliance**

mL cmH2O ∆ Volume ∆ Pressure Compliance tells that for every1 cmH2O pressure the lungs Can hold X mL of air. The more mL of air that a lung can hold Per cmH2O, the more compliant the lung. Patient B: 60 mL/cmH2O Example: Patient A: 30 mL/cmH2O Patient B has more compliant lungs. Patient A’s lungs Can only hold 30 mL of air for every cmH2O of pressure, Whereas patient B can hold 60 mL of air for every cmH2O.

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Compliance Example 1 Calculate the static compliance given the following Information: FiO2: Rate: 12 bpm Peak Pressure: 38 cmH2O Plateau Pressure: 29 cmH2O Vt: 600 mL PEEP: +5 cmH2O Vt PPL - PEEP 600 29 - 5 25 mL/cmH2O

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Compliance Example 2 Calculate the static compliance given the following Information: FiO2: Rate: 12 bpm Peak Pressure: 38 cmH2O Plateau Pressure: 29 cmH2O Vt: 600 mL PEEP: +5 cmH2O Vt PIP - PEEP 600 38 - 5 18.18 mL/cmH2O

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**Compliance Clinical Scenario**

Mr. J arrived to the ER in acute respiratory distress. He was Subsequently intubated and placed on mechanical ventilation In the ICU. Reviewing Mr. J’s ventilator sheet reveals the Following information: 8:00 a.m. 12:00 p.m. 4:00 a.m. Plateau Pressure: 31 cmH2O 22 cmH2O 27 cmH2O PEEP: 5 cmH2O 5 cmH2O 5 cmH2O Tidal Volume: 600 mL 600 mL 600 mL What does the information reveal about the compliance of Mr. J’s lungs?

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**Compliance Clinical Scenario**

600 mL 22 cmH2O - 5 cmH2O 600 mL 27 cmH2O - 5 cmH2O 600 mL 31 cmH2O - 5 cmH2O 35.29 mL/cmH2O 27.27 mL/cmH2O 23.08 mL/cmH2O Compliance is decreasing --> Increasing static pressure results In a decreased compliance

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**Airway Resistance (Raw)**

Airway resistance measures the force that opposes gas flow Through the airway Normal airflow Increased Raw Normal Raw is cmH2O/L/Sec on a non-intubated Patient, and 5 cmH2O/L/Sec on an intubated patient

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**Airway Resistance (Raw)**

Peak Pressure - Plateau Pressure Flow Flow must be in L/sec. If flow is given in L/min, Divide the flow by 60 seconds before placing It in the equation Example: Convert 60 L/min to L/sec 60 L/min 1 L/sec 60

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**Airway Resistance Example**

Calculate the airway resistance, given the following FiO2: Rate: 12 bpm Peak Pressure: 38 cmH2O Plateau Pressure: 29 cmH2O Vt: 600 mL PEEP: +5 cmH2O Flow: 40 LPM 40 LPM 60 1st convert the flow .67 L/sec PIP - PPL Flow .67 13.43 cmH2O/L/Sec

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I-time, Peak flow, Vt The following generic equation can be used to find I-time, peak flow rate, and tidal volume Tidal Volume (in L) I-time Peak Flow(LPM) 60 =

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Finding I-time I-time is the inspiratory portion of a breath. In other words, It is the amount of time spent on inspiration E-time I-time To find I-time 1st: determine the length of a single breath 2nd: Use the I:E ratio to determine the length of the I-time

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I-Time Example Calculate the I-time given the following ventilator parameters Vt: 600 cc Rate: 12 bmp Peak Flow: 60 LPM I:E = 1:2 FiO2: .60 1st: determine the length of a single breath There are 12 breaths in 1 minute and 60 seconds in 1 minute. Therefore 60 seconds / 12 breaths = 1 breath every 5 seconds Therefore, then legnth of 1 breath is 5 seconds 2nd: Use the I:E ratio to determine the length of the I-time 1x + 2x = 5 1x equals the inspiratory portion of the Breath. 1 x 1.67 = 1.67 seconds 3x = 5 X = 5/3 or 1.67

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**Find the peak flow, given the following**

Finding Peak Flow Find the peak flow, given the following VT = 750 cc RR = I:E = 1:2.5 First find the I-time (see the previous slide): 1.14 sec Tidal Volume (in L) I-time Peak Flow(LPM) 60 = X 60 .750 1.14 = 45 1.14 (.750)60 = 1.14X 39.47 LPM

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**Find the Vt given the following:**

Finding Vt Find the Vt given the following: PF = 50 LPM, RR = 14, I:E = 1:2 First, Find the I-time: 1.43 sec Tidal Volume (in L) I-time Peak Flow(LPM) 60 = 50 60 X 1.43 = 60X 71.5 X = L or mL (1.43)50 = 60X

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I:E Ratio Determine the I:E ratio for a patient on a ventilator breathing 20 bpm, Vt: 600 cc, Peak flow of 50 LPM. 1st, find the I-time: Tidal Volume (in L) I-time Peak Flow(LPM) 60 = . 6 X = 50 60 .72 seconds 2nd, Calculate the total breath time: 60 seconds 20 3 seconds

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**I:E Ratio I-time: .72 seconds Total breath time: 5 seconds**

Remember that a total breath is composed of an inspiratory Time and expiratory time, therefore: Total time - I-time = E-time = 4.28 I-time : E-time Convert to a 1:X ratio 1 : 3.2 : .72 : 2.28 .72

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**Achieving correct CO2/Minute ventilation**

Current VE x Current PaCO2 Desired PaCO2 Example: The doctor wants to decrease a patients PaCO2 from 50 mmHg to 35 mmHg. The doctor wants your advice on a proper minute ventilation. The current settings include a rate of 12 and a tidal volume of 500 mL. Current VE = 12 x 500 = 6000 mL or 6 L 6L x 50 35 You would need to se the Ventilator with a rate and tidal Volume that equals 8.57 L. (e.g. rate of 10, Vt of 857 mL) 8.57 L

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**Achieving correct PaO2 Desired PaO2 x FiO2 Current PaO2**

Example: A patient is currently hypoxic with a PaO2 Of 60 on an FiO2 of .45. The physician orders to maintain A PaO2 of at least 80 mmHg and asks you to adjust the Ventilator accordingly Increase the FiO2 to .60 to achieve a PaO2 Of 80 mmHg 80 mmHg x .45 60 mmHg .60

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**VD/Vt The VD/Vt equation illustrates the % of gas that does not**

Participate in gas exchange. In other words, it reflects The % of gas that is deadspace. PaCO2 -PeCO2 PaCO2 Deadspace refers to ventilation in the absence of perfusion O2 O2 Alveoli O2 O2 capillary Blocked blood flow

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**VD/Vt example Calculate the VD/Vt given the following:**

PaO2: 88 mmHg Vt: 550 mL PaCO2: 40 mmHg PeCO2: 31 mmHg PaCO2 -PeCO2 PaCO2 To determine the actual volume Of deadspace, just multiply The % deadspace by the given Tidal volume: 40 .225 x 550 = mL = 22.5% Normal deadspace: %, up to 60% on ventilator

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**Minute/Alveolar Ventilation**

Minute ventilation refers the volume of gas inhaled during A 1 minute period. Minute ventilation(VE) = Tidal volume x Respiratory rate Normal Minute ventilation = LPM Alveolar ventilation refers the the volume of gas that actually Participates in gas exchange. Alveolar ventilation = (tidal volume - deadspace) x RR 1 mL/lb of body weight Or 1/3 of tidal volume

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**Example Calculate the alveolar minute ventilation for a 150 lb male**

With a respiratory rate of 18 and tidal volume of 500 mL Alveolar ventilation = ( ) x 18 = 6300 mL or 6.3 L

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**Hemodynamics Shunt Pulmonary vascular resistance**

Systemic vascular resistance Mean pressure Pulse pressure Cardiac output (Fick's equation) Stroke Volume Cardiac Index

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