Respiratory CalculationsRespiratory Calculations

• Gas Laws• Oxygen therapy• Humidity• Ventilator Management• Hemodynamics

Gas LawsGas 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)

Oxygen TherapyOxygen 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

Total FlowTotal Flow

FiO2

100

20 or21

If the FiO2Is .40 or >Use 20 (< .40 Use 21)

Subtract100 - FiO2

SubtractFiO2 - 20 (or 21)

These 2 valuesWillDetermine theAir: O2 ratio

Add the numbers of the ratio X flow rate = Total flow

Total Flow: ExampleTotal Flow: Example

A COPD patient is currently on a 40% aerosol face mask running at 10 LPM. Calculate the total flow.

40

20

100 20

60

1

3

(1 + 3) x 10 = 40 LPM

Tank DurationTank Duration

PSIG x Tank factor Flow rate

Pressure of the cylinder

E cylinder: .28H cylinder: 3.14

The flow the O2 Device is set at

Tank Duration: ExampleTank 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

(2200-200) x .28 4

152.6 minutes 60 min/hr

2.54 Hours

Arterial & Venous O2 ContentArterial & 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 tohemoglobin

O2 dissolved in bloodplasma

Comparison of CaO2 & CvO2Comparison of CaO2 & CvO2

CaO2 = (1.34 x Hgb x SaO2) + (PaO2 x .003)

CvO2 = (1.34 x Hgb x SvO2) + (PvO2 x .003)

Arterial O2 Content

Venous O2 Content

A constant Hemoglobin Arterial saturation A constant

PartialPressureOf arterialO2

Venoussaturation

PartialPressureOf venousO2

Arterial O2 Content: ExampleArterial O2 Content: Example

Given the following values, calculate the CaO2:

PaO2 = 93 mmHgPvO2 = 47 mmHg

SaO2 = 98%SvO2 = 77%

Hemoglobin = 16 g/dL

CaO2 = (1.34 x 16 x .98) + (93 x .003)

CaO2 = 21.01 + .279 = 21.29 vol %

Normal value for CaO2 is approximately 20 vol %

Venous O2 Content: ExampleVenous O2 Content: Example

Given the following values, calculate the CvO2:

PaO2 = 93 mmHgPvO2 = 47 mmHg

SaO2 = 98%SvO2 = 77%

Hemoglobin = 16 g/dL

CvO2 = (1.34 x 16 x .77) + (47 x .003)CvO2 = 16.51 + .141 = 16.65 vol %

Normal CvO2 is approximately 15 vol %

C(a-v) DifferenceC(a-v) Difference

The C(a-v) difference represents the difference between arterialAnd 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 = 21.29 vol %CvO2 = 16.65 vol %

To determine the C(a-v)O2, simply subtract:CaO2 - CvO2

21.29 - 16.65 = 4.64 vol %

Normal C(a-v)O2 = 5 vol %

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

C(a-v)O2 can be an important clinical indicator. Recall thatThe C(a-v)O2 reflects the amount of oxygen taken from arterialBlood to be used by body tissues. Refer to the diagram below:

Arterial: CaO2 = 20 vol% Venous: CvO2 = 15 vol%Tissues

O2

O2

O2

O2

5 vol% of O2Is extracted fromArterial blood

O2 that is NOT extractedFrom arterial blood entersVenous circulation

Arterial blood containsApprox 5 vol% of O2

C(a-v) Difference con’t…C(a-v) Difference con’t…

When blood flows through the body at a normal rate, approximately5 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 beginsTo pool and more O2 is taken from arterial blood. With the tissuesExtracting more O2, less O2 is present in the venous system. If youHave 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

Alveolar Air EquationAlveolar Air Equation

The Alveolar air equation represents the partialPressure of oxygen in the alveoli

PAO2

PaO2

Alveolus

Capillary

This is what weAre finding usingThe alveolarAir equation

A/C

MEMBRANE

O2O2

O2O2

O2O2

O2O2

diffusion

Alveolar Air Equation: Con’t…Alveolar Air Equation: Con’t…

PAO2 = [(PB-PH2O) FiO2] - PaCO2 / .8

Barometric pressureNormal is 760 mmHg

Water pressureConstant:47 mmHg

O2 concentration

ArterialCO2

Constant: RespiratoryQuotientCO2 removed/O2 consumed200 mL/ 250 mL= .8

Alveolar Air Equation: ExampleAlveolar Air Equation: Example

Given the following information, calculate the PAO2

PB = 760 mmHgFiO2 = .60PaCO2 = 40 mmHgPaO2 = 88 mmHgHgb = 14 g/dL

PAO2 = [(760 - 47).60] - 40 / .8

= 377.8 mmHg

P(A-a)O2 GradientP(A-a)O2 Gradient

P(A-a)O2 represents the difference between the partial pressureOf 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 diffusingInto the blood (refer to the diagram on the “alveolor airEquation slide for a better understanding)

If the P(A-a)O2 increases, it signals there is some problemwith the gas diffusion mechanism (shunting for example).

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

Using the PAO2 calculated earlier (377.8 mmHg), calculateThe P(A-a)O2 if the PaO2 is 80 mmHg

P(A-a)O2 = 377.8 - 80 297.8 mmHg

What does this number tell you?

This number indicates that a significant amount of the availableO2 is not diffusing into the blood, indicating a shunt is present

Heliox Flow RatesHeliox Flow Rates

Heliox is a mixture of helium and oxygen. Because helium is lessDense 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:

Helium : Oxygen 80 : 20 70 : 30

Multiply flowReading by A factor of 1.8To get actualflow

Multiply flowReading by A factor of 1.6 toGet actual flow

Heliox Flow Rates: ExampleHeliox 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 / 1.8Set flow rate = 18 LPM / 1.8Set flow rate = 10 LPM

In order to have an actual flow of 18 LPM, we need to set theFlow meter at 10 LPM (If this were a 70:30 mixture, replace1.8 with 1.6)

HumidityHumidity

• Body Humidity

Body HumidityBody Humidity

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

Relative Humidity: Humidity Deficit:

What is the relative humidityOf a gas saturated with 30 mg/LOf water at body temperature?

30 mg/L44 mg/L

What is the humidity deficitOf a gas saturated at 30 mg/LOf water at body temperature?

44 mg/L - 30 mg/L = 14 mg/L= 68%

This means that at 98.6 F (37 C) gas is saturated with44 mgHg or 44 mg/L of water vapor

Ventilator ManagementVentilator 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

Compliance Compliance

∆ Volume∆ Pressure

PIP(dynamic pressure)

Plateau pressure

Insp. Hold

PEEP

change-over frominsp to exp NEEP

I-Time E-Time

Pressure Orvolume

Graph of Mechanical Breath

Dynamic Static

Generic Equation

Dynamic ComplianceDynamic Compliance

Tidal Volume (mL) Peak Pressure - PEEP

Dynamic compliance measures the elasticity of the lungDuring air movement. It is a less reliable indicator of lungElasticity compared to static compliance

Note: Peak Pressure = PIP

Static ComplianceStatic Compliance

Tidal Volume (mL) Plateau Pressure - PEEP

Static compliance measures the elasticity of the lungWhen there is no air movement. It is the best indicatorOf the ability to ventilate the lungs.

Normal static compliance is: 60 - 70 mL/cmH2O

Note: Plateau pressure = PPL = Static Pressure

Understanding ComplianceUnderstanding Compliance

∆ Volume∆ Pressure

mL cmH2O

Compliance tells that for every1 cmH2O pressure the lungsCan hold X mL of air. The more mL of air that a lung can holdPer cmH2O, the more compliant the lung.

Example: Patient A:30 mL/cmH2O

Patient B:60 mL/cmH2O

Patient B has more compliant lungs. Patient A’s lungsCan only hold 30 mL of air for every cmH2O of pressure, Whereas patient B can hold 60 mL of air for every cmH2O.

Compliance Example 1Compliance Example 1

Calculate the static compliance given the followingInformation:

FiO2: .60 Rate: 12 bpmPeak Pressure: 38 cmH2O Plateau Pressure: 29 cmH2OVt: 600 mL PEEP: +5 cmH2O

Vt PPL - PEEP

600 29 - 5

25 mL/cmH2O

Compliance Example 2Compliance Example 2

Calculate the static compliance given the followingInformation:

FiO2: .60 Rate: 12 bpmPeak Pressure: 38 cmH2O Plateau Pressure: 29 cmH2OVt: 600 mL PEEP: +5 cmH2O

Vt PIP - PEEP

600 38 - 5

18.18 mL/cmH2O

Compliance Clinical ScenarioCompliance Clinical Scenario

Mr. J arrived to the ER in acute respiratory distress. He wasSubsequently intubated and placed on mechanical ventilationIn the ICU. Reviewing Mr. J’s ventilator sheet reveals theFollowing information:

8:00 a.m.Plateau Pressure:

12:00 p.m. 4:00 a.m.22 cmH2O 27 cmH2O 31 cmH2O

What does the information reveal about the compliance ofMr. J’s lungs?

PEEP: 5 cmH2O 5 cmH2O 5 cmH2O

Tidal Volume: 600 mL 600 mL 600 mL

Compliance Clinical ScenarioCompliance 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 resultsIn a decreased compliance

Airway Resistance (Raw)Airway Resistance (Raw)

Normal airflow Increased Raw

Airway resistance measures the force that opposes gas flowThrough the airway

Normal Raw is 0.6 - 2.4 cmH2O/L/Sec on a non-intubatedPatient, and 5 cmH2O/L/Sec on an intubated patient

Airway Resistance (Raw)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 placingIt in the equation

Example: Convert 60 L/min to L/sec

60 L/min60

1 L/sec

Airway Resistance ExampleAirway Resistance Example

Calculate the airway resistance, given the following

FiO2: .60 Rate: 12 bpmPeak Pressure: 38 cmH2O Plateau Pressure: 29 cmH2OVt: 600 mL PEEP: +5 cmH2OFlow: 40 LPM1st convert the flow 40 LPM

60.67 L/sec

PIP - PPL Flow

38 - 29 .67

13.43 cmH2O/L/Sec

I-time, Peak flow, VtI-time, Peak flow, Vt

The following generic equation can be used to findI-time, peak flow rate, and tidal volume

Tidal Volume (in L) I-time

Peak Flow(LPM) 60

=

Finding I-timeFinding I-time

I-time is the inspiratory portion of a breath. In other words,It is the amount of time spent on inspiration

I-time E-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

I-Time ExampleI-Time Example

Calculate the I-time given the following ventilator parameters

Vt: 600 cc Rate: 12 bmpPeak Flow: 60 LPM I:E = 1:2FiO2: .60

1st: determine the length of a single breathThere are 12 breaths in 1 minute and 60 seconds in 1 minute.Therefore 60 seconds / 12 breaths = 1 breath every 5 secondsTherefore, then legnth of 1 breath is 5 seconds

2nd: Use the I:E ratio to determine the length of the I-time1x + 2x = 53x = 5X = 5/3 or 1.67

1x equals the inspiratory portion of the Breath. 1 x 1.67 = 1.67 seconds

Finding Peak FlowFinding Peak Flow

Find the peak flow, given the following

VT = 750 cc RR = 15 I:E = 1:2.5

Tidal Volume (in L) I-time

Peak Flow(LPM) 60

=

First find the I-time (see the previous slide): 1.14 sec

.7501.14

= X 60

(.750)60 = 1.14X 45 1.14

39.47 LPM

Finding VtFinding 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

=

X 1.43

= 50 60

(1.43)50 = 60X 60X 71.5

X = 1.1917 L or 1191.7 mL

I:E RatioI:E Ratio

Determine the I:E ratio for a patient on a ventilator breathing20 bpm, Vt: 600 cc, Peak flow of 50 LPM.

1st, find the I-time:

2nd, Calculate the total breath time:

60 seconds 20

Tidal Volume (in L) I-time

Peak Flow(LPM) 60

=

. 6 X

= 50 60

.72 seconds

3 seconds

I:E RatioI:E Ratio

I-time: .72 secondsTotal breath time: 5 seconds

Remember that a total breath is composed of an inspiratoryTime and expiratory time, therefore:

Total time - I-time = E-time

3 - .72 = 4.28

I-time : E-time.72 : 2.28

Convert to a 1:X ratio.72 : 2.28 .72

1 : 3.2

Achieving correct CO2/Minute ventilation

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

8.57 L You would need to se the Ventilator with a rate and tidalVolume that equals 8.57 L.(e.g. rate of 10, Vt of 857 mL)

Achieving correct PaO2Achieving correct PaO2

Desired PaO2 x FiO2 Current PaO2

Example: A patient is currently hypoxic with a PaO2Of 60 on an FiO2 of .45. The physician orders to maintainA PaO2 of at least 80 mmHg and asks you to adjust theVentilator accordingly

80 mmHg x .45 60 mmHg

.60Increase the FiO2 to.60 to achieve a PaO2Of 80 mmHg

VD/VtVD/Vt

The VD/Vt equation illustrates the % of gas that does notParticipate in gas exchange. In other words, it reflectsThe % of gas that is deadspace.

PaCO2 -PeCO2 PaCO2

Deadspace refers to ventilation in the absence of perfusion

capillary

Alveoli

Blocked blood flow

O2

O2O2

O2

VD/Vt exampleVD/Vt example

Calculate the VD/Vt given the following:

PaO2: 88 mmHg Vt: 550 mLPaCO2: 40 mmHg PeCO2: 31 mmHg

PaCO2 -PeCO2 PaCO2

40 - 31 40

= 22.5%

To determine the actual volumeOf deadspace, just multiplyThe % deadspace by the givenTidal volume:

.225 x 550 = 123.75 mL

Normal deadspace: 20 - 40%, up to 60% on ventilator

Minute/Alveolar VentilationMinute/Alveolar Ventilation

Minute ventilation refers the volume of gas inhaled duringA 1 minute period.

Minute ventilation(VE) = Tidal volume x Respiratory rate

Normal Minute ventilation = 5 - 10 LPM

Alveolar ventilation refers the the volume of gas that actuallyParticipates in gas exchange.

Alveolar ventilation = (tidal volume - deadspace) x RR

1 mL/lb of body weightOr 1/3 of tidal volume

ExampleExample

Calculate the alveolar minute ventilation for a 150 lb maleWith a respiratory rate of 18 and tidal volume of 500 mL

Alveolar ventilation = (500 - 150) x 18= 6300 mL or 6.3 L

HemodynamicsHemodynamics

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