Respiratory Calculations

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

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)

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

Total Flow

100 Subtract FiO2 - 20 (or 21) If the FiO2 Is .40 or > Use 20 (< .40 Use 21) 20 or 21 FiO2 Subtract 100 - FiO2 These 2 values Will Determine the Air: O2 ratio

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

Total Flow: Example

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

20 1 40 (1 + 3) x 10 = 40 LPM 20 60 3

Tank Duration

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

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 (2200-200) x .28

4 152.6 minutes 60 min/hr 2.54 Hours

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).

C x O2 = (1.34 x Hgb x S x O2) + (P x O2 x .003)

O2 carried/bound to hemoglobin O2 dissolved in blood plasma

Comparison of CaO2 & CvO2

Arterial O2 Content C a O2 = (1.34 x Hgb x S a O2) + (P a O2 x .003)

A constant Hemoglobin Arterial saturation Partial Pressure Of arterial O2 A constant

Venous O2 Content C v O2 = (1.34 x Hgb x S v O2) + (P v O2 x .003)

Venous saturation Partial Pressure Of venous O2

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 = 21.01 + .279 =

21.29 vol %

Normal value for CaO2 is approximately 20 vol %

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 = 16.51 + .141 =

16.65 vol %

Normal CvO2 is approximately 15 vol %

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 = 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 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: Arterial: CaO2 = 20 vol% Arterial blood contains Approx 5 vol% of O2 O 2 O 2 O 2 Tissues O 2 5 vol% of O2 Is extracted from Arterial blood Venous: CvO2 = 15 vol% O2 that is NOT extracted From arterial blood enters Venous circulation

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

Alveolar Air Equation

The Alveolar air equation represents the partial Pressure of oxygen in the alveoli A/C This is what we Are finding using The alveolar Air equation P

a

O2 P

A

O2 O2 O2 O2 O2

diffusion

O2 O2 O2 O2 Alveolus Capillary M E M B R A N E

Alveolar Air Equation: Con’t…

P A O2 = [(P B -P H2O ) FiO2] - P a CO 2 / .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

Alveolar Air Equation: Example

Given the following information, calculate the P A O2 PB = 760 mmHg FiO2 = .60

PaCO2 = 40 mmHg PaO2 = 88 mmHg Hgb = 14 g/dL PAO2 = [(760 - 47).60] - 40 / .8 =

377.8 mmHg

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).

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 = 377.8 - 80

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

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 80 : 20 70 : 30 Multiply flow Reading by A factor of 1.6 to Get actual flow

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)

Humidity

• Body Humidity

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:

What is the relative humidity Of a gas saturated with 30 mg/L Of water at body temperature?

30 mg/L 44 mg/L = 68%

Humidity Deficit:

What is the humidity deficit Of a gas saturated at 30 mg/L Of water at body temperature?

44 mg/L - 30 mg/L =

14 mg/L

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

Compliance

Generic Equation

∆ Volume ∆ Pressure Pressure Or volume Dynamic Static Graph of Mechanical Breath PIP(dyna mic pressure) Plateau pressure Insp . Hold PEEP chang e-over from insp to exp NEEP I-Tim e E-Time

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

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: 60 - 70 mL/cmH2O

Note: Plateau pressure = PPL = Static Pressure

Understanding Compliance

∆ Volume ∆ Pressure mL cmH2O 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.

Example:

Patient A: 30 mL/cmH2O Patient B: 60 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.

Compliance Example 1

Calculate the static compliance given the following Information: FiO2: .60 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

Compliance Example 2

Calculate the static compliance given the following Information: FiO2: .60 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

Compliance Clinical Scenario

Plateau Pressure: PEEP:

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:

Tidal Volume: 8:00 a.m.

22 cmH2O 5 cmH2O 600 mL

12:00 p.m.

27 cmH2O 5 cmH2O 600 mL

4:00 a.m.

31 cmH2O 5 cmH2O 600 mL What does the information reveal about the compliance of Mr. J’s lungs?

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

Airway Resistance (Raw)

Airway resistance measures the force that

opposes

Through the airway gas flow

Normal airflow Increased Raw

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

Airway Resistance (Raw)

Peak Pressure - Plateau Pressure Flow

Flow must be in Divide the flow by 60 seconds before placing It in the equation

L/sec.

If flow is given in L/min, Example: Convert 60 L/min to L/sec 60 L/min 60

1 L/sec

Airway Resistance Example

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

13.43 cmH2O/L/Sec

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

Finding 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 Example

Calculate the I-time given the following ventilator parameters Vt: 600 cc Peak Flow: 60 LPM I:E = 1:2 FiO2: .60

Rate: 12 bmp

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 3x = 5 X = 5/3 or 1.67

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

1.67 seconds

Finding Peak Flow

Find the peak flow, given the following V T = 750 cc RR = 15 I:E = 1:2.5

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

1.14 sec

Tidal Volume (in L) I-time .750

1.14

= = Peak Flow(LPM) 60 X 60 (.750)60 = 1.14X

45 1.14

39.47 LPM

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 X 1.43

= = Peak Flow(LPM) 60 50 60 (1.43)50 = 60X 60X 71.5

X = 1.1917 L or 1191.7 mL

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 . 6 X = Peak Flow(LPM) 60 = 50 60

.72 seconds 2nd, Calculate the total breath time:

60 seconds 20

3 seconds

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

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

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 tidal Volume that equals 8.57 L.

(e.g. rate of 10, Vt of 857 mL)

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 80 mmHg x .45

60 mmHg

.60

Increase the FiO2 to .60 to achieve a PaO2 Of 80 mmHg

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 O2 O2 Alveoli capillary Blocked blood flow

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 40 - 31 40

= 22.5%

To determine the actual volume Of deadspace, just multiply The % deadspace by the given Tidal volume: .225 x 550 =

123.75 mL

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

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 = 5 - 10 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

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 = (500 - 150) x 18

= 6300 mL or 6.3 L

Hemodynamics

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