This means that once the patient has started using the ventilator, 5 mL of the set tidal volume will be lost in the tubing for every 1 cm H₂O registering on the manometer.

EXAMPLE:

Tubing compliance, 5 mL/cm H₂O

VT, 800 mL

Peak inspiratory pressure, 20 cm H₂O

Lost volume = tubing compliance × peak inspiratory pressure

Set VT is 800 mL

e. Lost volume is affected by the water level in the humidifier. Lower water levels allow more compressed volume into the walls of the humidifier; hence, more volume is lost or less volume is delivered to the patient.

f. Volume is also lost in the patient’s conducting airways (e.g., trachea and bronchi) as a result of resistance to gas flow. This part of the patient’s airway is called the anatomic dead space. It is the part of the airway where no gas exchange occurs, and it is often called “wasted air.”

Exam Note

Anatomic dead space is equal to 1 mL/lb of the patient’s IBW, but it is reduced by 50% in the intubated patient. Thus, anatomic dead space is equal to about 1 mL/kg.

EXAMPLE:

An intubated 75-kg (165-lb) patient has an anatomic dead space of approximately 75 mL. This means that 75 mL of the patient’s VT does not reach the alveoli to take part in gas exchange. This may also be subtracted from the ventilator VT setting to obtain a corrected VT.

When the formula for selecting ventilator VT as 10 to 12 mL/kg of body weight is used, this lost volume is taken into account, and this is usually an adequate VT. Some ventilators compensate for volume lost in the tubing by delivering a higher VT automatically.

8. Exhaled VT is measured by an exhaled volume digital display. Exhaled VT is most accurately measured with a respirometer placed between the ET tube and ventilator Y adapter or at the exhalation valve. To most accurately measure the VT delivered by the ventilator, place a respirometer directly on the ventilator outlet.

9. Volume may be lost from other causes.

a. Loose humidifier jar or tubing connections cause low exhaled VT readings.

b. Leakage around the ET tube cuff results in low exhaled VT readings.

C. Respiratory Rate Control

1. Normal initial setup is 8 to 12 breaths/min.

2. Adjusting the rate control alters the expiratory time, therefore altering the I:E ratio.

a. Increasing the rate decreases expiratory time.

b. Decreasing the rate increases expiratory time.

3. Adjusting the rate alters the minute volume.

a. Increasing the rate increases minute volume.

b. Decreasing the rate decreases minute volume.

4. Adjusting the rate affects the PaCO₂ level.

a. Increasing the rate decreases PaCO₂.

b. Decreasing the rate increases PaCO₂.

Exam Note

Adjusting the rate to alter the patient’s PaCO₂ level is more beneficial for patients receiving controlled ventilation or SIMV. When the ventilator is on assist/control, the patient may obtain as many machine breaths as needed, no matter what rate is set, so the PaCO₂ is most effectively altered by adjustment of the VT control.

D. Inspiratory Flow Control

1. Normal setting: 40 to 60 L/min.

2. Adjusting the flow rate alters the inspiratory time, therefore altering the I:E ratio.

a. Increasing the flow rate decreases inspiratory time.

b. Decreasing the flow rate increases inspiratory time.

E. I:E ratio

1. A comparison of the inspiratory time with the expiratory time

2. Normal I:E ratio for the adult is 1 : 2. This means that expiration is twice as long as inspiration.

3. Normal I:E ratio for the infant is 1 : 1.

4. The I:E ratio is established by the use of three ventilator controls on the volume ventilator.

a. VT control

(1) Increasing the VT increases inspiratory time (makes inspiratory time longer).

(2) Decreasing the VT decreases inspiratory time (makes inspiratory time shorter).

b. Flow rate control

(1) Increasing the flow rate decreases inspiratory time.

(2) Decreasing the flow rate increases inspiratory time.

c. Respiratory rate control

(1) Increasing the respiratory rate decreases expiratory time.

(2) Decreasing the respiratory rate increases expiratory time.

5. Calculate the I:E ratio with the following formula

EXAMPLE:

VT	800 mL (0.8 L)
Rate	12/min
Flow rate	40 L/min

6. Calculation of inspiratory time

EXAMPLE:

Calculate the inspiratory time if the I:E ratio is 1 : 2 and the ventilator rate is 10/min.

7. Inspiratory time should not exceed expiratory time, except in specific situations. This is referred to as an inverse I:E ratio. Inverse ratio ventilation is sometimes utilized to increase the PaO₂ when FiO₂ and PEEP levels are already high. It may greatly compromise venous blood return to the heart and increase intrathoracic pressure.

Exam Note

When given a scenario on the exam where a ventilator patient has a normal PaCO₂ with hypoxemia and the PEEP and FiO₂ levels are high, increase the inspiratory time to increase the PaO₂. This allows for a longer time for oxygen to diffuse across the alveolar capillary membrane, thereby increasing the PaO₂.

a. If the I:E ratio alarm is sounding on the ventilator indicating an inverse I : E ratio, three controls may be altered to correct it.

(1) Rate: decrease to lengthen expiratory time.

(2) Volume: decrease to shorten inspiratory time.

(3) Flow: increase to shorten inspiratory time.

Exam Note

Increasing flow is the most common adjustment to correct for an inverse I:E ratio.

F. O₂ Percentage Control

1. Adjustable from 21% to 100% to maintain normal PaO₂ levels

2. O₂ percentage should be increased to a maximum of 60% to maintain normal PaO₂ levels. Once 60% is reached, PEEP should be added or increased.

3. O₂ percentage should be reduced first to a level of 60% before decreasing PEEP levels in hyperoxygenated patients.

4. The initial setting should be what FiO₂ level the patient was receiving before the initiation of mechanical ventilation.

Exam Note

The major ventilator controls have been discussed thus far in the chapter, so let’s summarize how to set the variables to initiate mechanical ventilation.

Mode: A/C or SIMV

VT: 8 to 12 mL/kg IBW

Ventilator rate: 8 to 12/min

FiO₂: whatever the FiO₂ the patient was receiving before the initiation of ventilation

G. Sensitivity Control

1. This determines the amount of patient effort required to cycle the ventilator into inspiration.

2. Should be set so that the patient generates −0.5 to −2.0 cm H₂O pressure.

3. If the ventilator self-cycles, the sensitivity is too high. Decrease the sensitivity.

4. If it takes more than −2.0 cm H₂O pressure to cycle the ventilator into inspiration, increase the sensitivity.

5. In the control mode of ventilation, the sensitivity is turned off and does not allow the patient to trigger a machine breath.

H. Sigh controls

1. The sigh rate should be set at 6 to 12 sighs/h.

2. The sigh volume should be set 1.5 to 2.0 times the VT.

3. Sighs aid in preventing atelectasis.

4. Usually not functional in the SIMV mode

Exam Note

Although sighs are not commonly used in practice in many areas, the exam may still offer questions on the subject.

I. Inflation Hold Control

1. Adjustable from 0 to 2 s

2. The mechanism keeps the exhalation valve closed, causing the ventilator VT to be held in the lungs for a preset time.

3. Used to improve oxygenation by reducing atelectasis and shunting and increasing the diffusion of gases.

4. Using an inspiratory hold causes an increase in intrathoracic and mean airway pressure, which may result in reduced venous return and cardiac output.

5. Used to obtain a plateau pressure to calculate static lung compliance.

J. Expiratory Retard (Expiratory Resistance)

1. Used to prevent premature airway collapse during expiration.

2. Increases the expiratory time, therefore altering the I:E ratio and increasing intrathoracic pressure.

3. Expiratory retard causes an increase in intrathoracic and mean airway pressure, which may result in reduced venous return and cardiac output.

Exam Note

Although expiratory resistance is rarely an option on ventilators, exam questions may appear on the subject.

K. Positive End-Expiratory Pressure

1. Used to maintain positive pressure in the airway after a ventilator breath.

FIGURE 11-1 Positive end-expiratory pressure (PEEP) curves compared with other curves.

2. Indications for PEEP

a. Atelectasis

b. Hypoxemia with the use of 60% or more O₂

c. Decreased functional residual capacity (FRC)

d. Lowering O₂ percentage to safe levels (less than 60%)

e. Decreased lung compliance

f. Pulmonary edema

3. Hazards of PEEP

a. Barotrauma

b. Decreased venous return

c. Decreased cardiac output

d. Decreased urinary output

4. Excessive PEEP levels may lead to decreases in PaO₂ and lung compliance by overdistending already open alveoli and shunting blood to collapsed alveoli.

5. A decreased cardiac output (or cardiac index) caused by PEEP is indicated by a decrease in blood pressure and PvO₂ values.

6. Optimal PEEP: the level of PEEP that improves lung compliance without decreasing the cardiac output.

7. A mixed venous PO₂ (PvO₂) level may be obtained from the pulmonary artery by means of a pulmonary artery catheter and reflects changes in the cardiac output.

a. Normal PvO₂ is 35 to 45 mm Hg.

b. A PvO₂ of less than 35 mm Hg indicates a possible decrease in cardiac output. If the PvO₂ decreases after initiation of PEEP, it is an indicator of reduced venous return and cardiac output caused by PEEP.

c. The PvO₂ value indicates the adequacy of tissue oxygenation.

d. Use the PEEP level that provides the best lung compliance and PvO₂ value.

EXAMPLE:

Determining optimal PEEP

Which of the following represents optimal PEEP?

PEEP (cm H₂O)	PaO₂ (mm Hg)	PvO₂ (mm Hg)
4	68	34
6	74	37
8	78	33
10	82	32

Notice how the PvO₂ decreased after the increase in PEEP from 6 cm H₂O to 8 cm H₂O. This indicates a drop in cardiac output with this PEEP change. Therefore, return to the PEEP level that maintains the highest PvO₂. In this example, the optimal PEEP level is 6 cm H₂O.

Remember that the PaO₂ does not determine optimal PEEP level. Even though the PaO₂ in this example continued to increase at increasing PEEP levels, it did so at the expense of a decreasing cardiac output. This indicates a worsening oxygenation status.

Another method of determining optimal PEEP levels is the determination of upper and lower inflection points on a pressure/volume curve. The lower inflection point indicates the alveolar critical opening pressure. The upper inflection point indicates the point of alveolar overdistention.

FIGURE 11-2 Best PEEP and PIP levels.

Setting the PEEP level at the lower inflection point ensures that the alveoli remain open at end exhalation. This reduces damage to the alveoli by preventing the opening and closing of the alveoli with each breath, which results in shear force damage. Not allowing the peak inspiratory pressure to exceed the upper inflection point ensures that the lowest airway pressure is being used to ventilate the lungs adequately without overdistending the alveoli and causing alveolar damage and a possible pneumothorax. Lower airway pressure also reduces the potential for cardiac side effects.

Allowing PIP above the upper inflection point results in higher PIP levels with little or no increase in delivered tidal volume.

Figure 11-2 shows the lower inflection point (best PEEP level) to be about 10 cm H₂O and the upper inflection point (best PIP level) to be about 28 to 30 cm H₂O. Using a PEEP of 10 cm H₂O keeps the alveoli from collapsing at end exhalation. Using a PIP of 28 to 30 cm H₂O ventilates the lungs effectively while preventing overdistention of the alveoli.

III. VENTILATOR ALARMS AND MONITORING

CRT Exam Content Matrix: IA7d-e, IB9c,m, IB10c, IC6, IC9, IIIE4e, IIIE7b, IIIE9, IIIF2i7, IIIG3g

RRT Exam Content Matrix: IA7d-e, IB9c,m, IB10c, IIIE4b, IIIE5, IIIE7b

A. Low-Pressure Alarm

1. Should be set 5 to 10 cm H₂O below peak inspiratory pressure.

2. This alarm is activated by leaks in the ventilator circuit, leaks around the chest tube, a ruptured ET tube cuff, inadequate cuff pressure, or patient disconnection.

B. High-Pressure Alarm

1. Should be set 5 to 10 cm H₂O above peak inspiratory pressure.

2. When the high-pressure limit is reached on a volume ventilator, inspiration ends prematurely, decreasing delivered VT.

3. This alarm may be activated by

a. Decreasing lung compliance

b. Increasing airway resistance caused by

(1) Airway secretions

(2) Bronchospasm

(3) Water in the ventilator tubing

(4) Kink in the ventilator tubing

(5) Patient coughing

Exam Note

The PIP level should be maintained at less than 35 to 40 cm H₂O to prevent lung tissue damage.

C. Low PEEP/CPAP Alarm

1. Should be set 2 to 4 cm H₂O below the baseline level.

2. This alarm is activated for the same reasons as stated previously for the low-pressure alarm.

D. Apnea Alarm

1. Set according to the patient’s respiratory rate.

2. This alarm is activated after a preset time passes with no inspiratory flow through the tubing.

3. Important alarm for patients receiving ventilation in the CPAP mode or with low SIMV rates

E. Low Tidal Volume Alarm

1. Should be set approximately 10% below the set tidal volume.

2. This alarm is activated for the same reasons as stated previously for the low-pressure alarm.

F. Mean Airway Pressure (

) Monitoring

is the average pressure applied to the airway over a specific time.

is directly affected by

a. Ventilator rate

b. Peak inspiratory pressure

is commonly measured by a digital reading on a

monitor, or it may be calculated with the use of the following equation

4. Optimal

is the level that improves oxygenation and ventilation without resulting in cardiovascular side effects and barotrauma.

5. Studies have shown that

levels above 12 cm H₂O result in an increased risk of barotrauma.

6. If there has been no change in dynamic compliance or ventilator variables and if

decreases, it indicates less pressure required to ventilate the patient’s lungs or an increased static lung compliance. An increased

indicates higher pressure required to ventilate or a decreased static lung compliance.

G. End-Tidal CO₂ Monitoring (Capnography)

1. Capnography is a technique by which exhaled CO₂ is measured. This measurement is obtained by the use of a mass spectrometer. (End-tidal CO₂ is abbreviated PETCO₂.)

2. Normal PETCO₂ is approximately the same as alveolar CO₂, which is equal to arterial PCO₂. PETCO₂ therefore is a noninvasive technique to obtain the patient’s PaCO₂ level.

3. PETCO₂ may be expressed as partial pressure or a percentage. Normal PETCO₂ is 35 to 45 mm Hg or 4.5% to 5.5%.

4. There is a difference of approximately 2 to 5 mm Hg between normal PaCO₂ and PETCO₂.

5. PETCO₂ readings may decrease as a result of any of the following. (A low reading results from decreased perfusion to the pulmonary capillaries, rendering an inaccurate PETCO₂ reading. PaCO₂ may actually be increasing.)

a. Hyperventilation

b. Apnea (reading falls to zero)

c. Total airway obstruction (reading falls to zero)

d. Conditions in which perfusion is decreased (e.g., hypotension, pulmonary embolism, decreased cardiac output)

6. PETCO₂ readings may increase as a result of the following

a. Hypoventilation

b. Hyperthermia (increased CO₂ production)

7. Continuous PETCO₂ monitoring with tracings is becoming an important technique in monitoring critically ill patients. The tracing records CO₂ readings during inspiration (which should be zero because little CO₂ is in inspired air) and during expiration, when CO₂ begins to increase.

8. A PETCO₂ measurement by itself should not be used to predict PaCO₂ in patients with left ventricular failure (decreased cardiac output), pulmonary embolism, or COPD because of inaccurate readings under these conditions.

FIGURE 11-3 Capnography.

From Wilkins RL, Stoller JK, Kacmarek R: Egan’s fundamentals of respiratory care, ed 9, St Louis, 2009, Mosby.

IV. INDICATIONS FOR MECHANICAL VENTILATION

CRT Exam Content Matrix: IA7c,e, IB9d-e,j, IB10d,f

RRT Exam Content Matrix: IA7c,e, IB9d-e,j, IB10d,f

A. Apnea

B. Acute Ventilatory Failure

1. PaCO₂ of greater than 50 mm Hg indicates ventilatory failure and a need for mechanical assistance.

2. To determine ventilatory failure in a patient with COPD who chronically retains CO₂, note whether the pH is below 7.30. If so, there may be a need for ventilator assistance.

C. Impending Acute Ventilatory Failure

1. Sometimes normal ABG levels can be deceiving. A patient may have normal ABG levels, but the respiratory rate may be 30/min to 40/min to achieve this normal PCO₂ level. This indicates pending ventilatory failure. The patient is likely to tire soon, which would result in an increasing PaCO₂ level and ventilatory failure.

2. A patient with a neuromuscular disease, such as Guillain-Barré syndrome, must be monitored closely for lung muscle involvement. Measurement of the patient’s maximal inspiratory pressure (MIP) and vital capacity will help determine the lung status.

D. Oxygenation

1. The patient’s lungs may be ventilating adequately but oxygenating poorly.

2. Mechanical ventilation is indicated if O₂ deficiency is directly related to an abnormal ventilatory pattern or an increased work of breathing.

3. A patient using an O₂ mask at 60% or more whose lungs are being ventilated well (normal or low PaCO₂) but are not being oxygenated adequately (low PaO₂) is probably exhibiting a large intrapulmonary shunt. This may be corrected with CPAP. Mechanical ventilation may not be necessary initially.

V. COMMON CRITERIA FOR INITIATION OF MECHANICAL VENTILATION

CRT Exam Content Matrix: IA7b,c,e, IB9d-e,j,m, IB10d-f

RRT Exam Content Matrix: IA7b,c,e, IB9d-e,j,m, IB10d-f

A. Vital capacity (VC) of less than 10 to 15 mL/kg; normal is 65 to 75 mL/kg.

B. P(A−a)O₂ of greater than 450 mm Hg with the use of 100% O₂; normal is 25 to 65 mm Hg.

1. When the PaO₂ is low and the P(A−a)O₂ is normal for ambient conditions and the patient’s age, hypoxemia is most likely the result of hypoventilation.

2. When the PaO₂ is low and the P(A−a)O₂ is high, hypoxemia is most likely the result of either a V/Q mismatch, diffusion defect, or shunting. In this situation, the patient may be hyperventilating to compensate for the hypoxemia.

C. VD/VT ratio of greater than 60%; normal is 25% to 35%.

D. Unable to obtain a MIP of at least −20 cm H₂O; normal is −50 to −100 cm H₂O.

E. PEP of less than 40 cm H₂O; normal is 100 cm H₂O.

Exam Note

An MIP of greater than −20 cm H₂O or a PEP of less than 40 cm H₂O indicates that the patient cannot generate an adequate cough to maintain secretion clearance.

F. Respiratory rate of greater than 35/min; normal is 10/min to 20/min.

VI. COMPLICATIONS OF MECHANICAL VENTILATION

CRT Exam Content Matrix: IA7a, IB2-4, IB7a-b, IIIE6

RRT Exam Content Matrix: IA7a, IB2-4, IIIE6

A. Barotrauma (resulting from excessive airway pressure)

1. Pneumothorax: may be characterized by subcutaneous emphysema (air in the subcutaneous tissues)

2. Pneumomediastinum

3. Pneumopericardium

B. Pulmonary Infection

1. Debilitated patients have lower resistance.

2. Contaminated equipment

3. Improper airway care (e.g., tracheostomy care, suctioning)

4. Retained secretions as a result of ET tube and poor ability to cough

5. Ciliary dysfunction caused by ET tube

6. Ventilator-associated pneumonia (VAP). (See section XVII of this chapter.)

Exam Note

The most cost-effective method of preventing cross-contamination of patients and equipment is proper hand washing techniques.

C. Atelectasis

1. Using a minimum VT of 8 to 10 mL/kg of body weight prevents this (5 to 6 mL/kg IBW should be used for patients with ARDS).

2. Use of PEEP to prevent alveolar collapse

D. Pulmonary O₂ Toxicity

1. May result in ARDS.

2. Results from the use of high O₂ concentrations for prolonged periods.

3. Characterized by

a. Impaired surfactant production

b. Capillary congestion

c. Edema

d. Fibrosis

e. Thickening of alveolar membranes

f. Decreased lung compliance, resulting in high peak pressures being required to ventilate the patient’s lungs. (See Chapter 12 on disorders of the respiratory system.)

E. Tracheal Damage

1. Usually at the cuff site

F. Decreased Venous Blood Return to the Heart

1. Results from the positive airway pressures being transferred onto the large veins returning blood to the heart, which causes decreased pulmonary blood flow, decreased cardiac output, and decreased blood pressure.

G. Decreased Urinary Output

1. Results from decreased renal blood flow (caused by decreased cardiac output).

2. Also results from an increased production of antidiuretic hormone (ADH).

a. ADH production increases because of baroreceptors in the atria of the heart, which sense the decreased venous return.

b. These receptors send a message to the hypothalamus, which stimulates the pituitary gland to secrete more ADH, thereby reducing urinary output.

H. Lack of Nutrition

1. Malnutrition may lead to

a. Difficulty weaning from the ventilator as a result of weakened respiratory muscles.

b. Reduced response to hypoxia and hypercarbia.

c. Impaired wound healing.

d. Decreased surfactant production.

e. Infection.

f. Pulmonary edema from decreased serum albumin levels.

2. Because oral feeding is not possible, nasogastric feedings should be implemented. Other feeding routes include an IV line or enteral feedings through a catheter in the stomach.

3. High-protein, high-carbohydrate diets are recommended.

VII. DEAD SPACE VOLUME (VD)

CRT Exam Content Matrix: IB9k, IB10i, IB10k

RRT Exam Content Matrix: IB10i

A. VD is that portion of the VT that does not take part in gas exchange.

B. Types of VD

1. Anatomic VD (discussed earlier) consists of the conducting airways from the nose and mouth to the terminal bronchioles (i.e., air that does not reach the alveolar epithelium where gas exchange occurs).

a. Anatomic VD = 1 mL/lb body weight

b. A tracheostomy decreases the anatomic VD by bypassing the upper airway.

2. Alveolar VD

a. Air reaches the alveoli but does not take part in gas exchange.

b. Results from lack of perfusion to air-filled alveoli.

c. May result from hyperinflated alveoli where blood is not able to use all the air.

3. Physiologic VD

a. The sum of anatomic and alveolar VD.

b. The most accurate measurement of VD.

4. Mechanical VD

a. Ventilator circuits have a certain amount of VD, ranging from 75 to 150 mL.

b. Because anatomic VD decreases when a patient has an ET tube or tracheostomy tube, the VD created by the circuit is balanced out.

c. Additional mechanical dead space (VD) may be added to the ventilator circuit between the ventilator Y adapter and the ET tube adapter to increase PaCO₂ levels. This results from some of the patient’s exhaled air (which contains CO₂) getting trapped in the excess tubing and being rebreathed with each ventilator breath.

(1) For every 100 mL of dead space added, the PaCO₂ increases approximately 5 mm Hg.

(2) Mechanical dead space (VD) may be added to the circuits of patients using control or assist/control modes only. Never add dead space if the patient is receiving SIMV, PSV, or CPAP, which are modes the patient breathes spontaneously. Dead space would increase airway resistance and work of breathing in these modes.

Exam Note

Although adding dead space to the ventilator circuit is rarely used clinically, questions regarding dead space are occasionally asked on the exams.

Exam Note

If a ventilator patient has an elevated PaCO₂ and the question indicates the patient has dead space added to the ventilator circuit, always remove the dead space first. Do not increase the VT or rate.

VIII. LUNG COMPLIANCE (CL)

CRT Exam Content Matrix: IA7d, IB9m, IB10m, IC6

RRT Exam Content Matrix: IA7d, IB9m, IB10n, IC7

A. Lung compliance is defined as the ease with which the lung expands. It varies inversely with the pressure required to move a specific volume of air.

1. The higher the compliance, the easier it is to ventilate the lung. (The lung requires less pressure to ventilate.)

2. The lower the compliance, the stiffer the lung is and the harder it is to ventilate. (The lung requires more pressure to ventilate.)

3. Normal total lung compliance (sum of the compliance of lung tissue and thoracic cage) is 0.1 L/cm H₂O (100 mL/cm H₂O).

B. Calculation of Lung Compliance

1. Dynamic compliance

a. Formula

EXAMPLE:

Given the following data, calculate the patient’s dynamic lung compliance.

VT	600 mL
PIP	35 cm H₂O
PEEP	5 cm H₂O

b. Dynamic compliance is measured as air is flowing through the circuit and airways; therefore, it is actually a measurement of airway resistance, or R_AW and lung compliance

c. Dynamic compliance changes with changes in R_AW caused by

(1) Water in the ventilator tubing

(2) Bronchospasm

(3) Airway secretions

(4) Mucosal edema

d. Dynamic compliance is not an accurate measurement of lung compliance.

2. Static compliance

a. Is a more accurate measurement of lung compliance because it is measured with no air flowing through the circuit and airways (i.e., under static conditions).

b. Airflow may be stopped with the volume remaining in the lungs by adjustment to a 1- to 2-s inspiratory hold.

c. Once the flow has stopped, a plateau pressure or static pressure occurs after peak pressure has been reached. The drop from peak to plateau pressure represents the pressure that is being generated as gas is flowing through the tubing, rubbing against the sides of the tubing, ET tube, and airways.

d. Static compliance is calculated as follows

EXAMPLE:

VT	800 mL
Plateau pressure	25 cm H₂O
PEEP	5 cm H₂O
Peak pressure	45 cm H₂O

Calculate the static lung compliance.

Exam Note

Sometimes on the exams, the data provided to calculate static lung compliance will include not only the ventilator tidal volume but also the actual or exhaled volume. Always use the actual volume in the equation if it is provided in the question.

C. Important Points Concerning Lung Compliance

1. Increasing plateau pressures indicate that the lung compliance is decreasing, or the lungs are harder to ventilate.

2. If the peak pressures are increasing but the plateau pressure remains the same, then lung compliance is not decreasing. An increased R_AW is occurring from bronchospasm, secretions, coughing, tubing obstructions, and so on.

EXAMPLE:

Time	Peak Pressure	Plateau Pressure
6:00 AM	28 cm H₂O	10 cm H₂O
7:00 AM	34 cm H₂O	10 cm H₂O
8:00 AM	42 cm H₂O	10 cm H₂O

In this example, the peak pressures are increasing while the plateau pressures remain stable. This indicates an increase in R_AW and not a decreasing lung compliance.

EXAMPLE:

Time	Peak Pressure	Plateau Pressure
1:00 PM	34 cm H₂O	16 cm H₂O
2:00 PM	40 cm H₂O	22 cm H₂O
3:00 PM	44 cm H₂O	26 cm H₂O

In this example, the plateau pressures are increasing along with the peak pressures. This indicates a decreasing lung compliance.

3. Decreasing static lung compliance results from

4. Normal static lung compliance in the patient receiving ventilation is 60 to 70 mL/cm H₂O.

Exam Note

Calculation of lung compliance is also important in the determination of optimal PEEP level.

EXAMPLE:

Optimal PEEP is represented by which of the following?

Remember that optimal PEEP is the level of PEEP that produces the highest static lung compliance. There is no need to calculate the compliance for all four PEEP levels in this problem. Because the VT is the same at all PEEP levels, simply subtract the PEEP level from the plateau pressure. The lowest number you get after doing this is the optimal PEEP level because the lower the number divided into the VT, the higher the lung compliance. In other words, a PEEP of 4 resulted in a plateau pressure of 20, which is a difference of 16. A PEEP of 6 also resulted in a difference of 16; a PEEP of 8, a difference of 15; a PEEP of 10, a difference of 17. The lower the number divided into the VT, the higher the compliance result; therefore, the optimal PEEP level in this example is 8 cm H₂O.

Exam Note

If compliance data and cardiac output are both given in a question regarding optimal PEEP, select the PEEP level before the drop in cardiac output even if a higher PEEP level resulted in the best lung compliance. For example, the static lung compliance increases when the PEEP is increased from 10 cm H₂O to 12 cm H₂O, but the cardiac out drops. Choose 10 cm H₂O as the optimal PEEP. Remember, optimal PEEP is the level of PEEP that results in the best static lung compliance without decreasing the cardiac output.

D. Calculation of Airway Resistance (R_AW). For an intubated ventilator patient, the amount of pressure being delivered to the airways and how much is going to the alveoli must be known. The difference between these two is the amount of pressure lost as a result of R_AW. Determine this loss by subtracting the plateau pressure from the PIP. The closer the plateau pressure is to the peak pressure, the lower the pressure loss caused by R_AW, and vice versa. When this pressure difference is divided by the inspiratory flow, R_AW can be measured

Normal R_AW in nonintubated individuals is 0.6 to 2.4 cm H₂O/L/s, based on a flow rate of 30 L/min or 0.5 L/s. Normal R_AW in a ventilator patient is approximately 5 cm H₂O/L/s.

Because R_AW is measured in cm H₂O/L/s, flow rate must be converted from L/min to L/s by dividing flow by 60.

EXAMPLE:

The following data have been collected from a patient using a volume ventilator.

Peak inspiratory pressure	35 cm H₂O
Plateau pressure	20 cm H₂O
Flow rate	60 L/min = 1 L/s

Because normal R_AW in an intubated patient is around 5 cm H₂O/L/s, this example reflects a high R_AW.

IX. VENTILATION IN THE PATIENT WITH HEAD TRAUMA

Note: Although there is nothing in either the CRT or RRT Exam Content Matrix specific to this topic, the exams have questions regarding it.

A. Higher than normal flow rates should be used to make inspiratory time shorter, which would lessen the time of positive pressure in the airways. The longer the time of positive pressure in the airways, the more impedance of blood flow from the head, which increases ICP.

B. Maintain the PaCO₂ between 25 and 30 mm Hg to reduce ICP by vasoconstriction of cerebral vessels. ICP should be maintained below 15 mm Hg; normal ICP is <10 mm Hg.

Exam Note

Studies indicate that hyperventilation to reduce cerebral blood flow may be detrimental because of inadequate perfusion to healthy brain tissue, but the exams consider it acceptable provided the question indicates the patient has an elevated ICP.

Exam Note

To decrease the PaCO₂ level, always increase the ventilator rate. Do not increase the tidal volume or inspiratory pressure because this may cause an increased intrathoracic pressure obstructing venous blood flow from the upper body and head back to the heart. Therefore, ICP may increase as the venous blood returning to the heart decreases.

C. If the ICP begins to increase, the patient should be hyperventilated (preferably with a resuscitation bag) to reduce the cerebral blood flow, thus lowering ICP.

D. Caution must be exercised when suctioning because this tends to increase ICP as a result of hypoxemia.

X. WEANING FROM MECHANICAL VENTILATION

CRT Exam Content Matrix: IIID7, IIIF2i12, IIIG1g

RRT Exam Content Matrix: IIID6, IIIF2e10, IIIG1g

A. Criteria for Weaning

1. VT equal to three times the body weight in kilograms

2. VC >10 to15 mL/kg, or twice the VT

3. Ability to achieve an MIP of at least −20 cm H₂O

4. VD/VT <0.60

5. P(A−a)O₂ <350 mm Hg with the use of 100% O₂

6. Rapid shallow breathing index (RR/VT) <105

7. Respiratory rate <25/min

8. PEEP of 10 cm H₂O or less

9. PaO₂/FiO₂ >200

10. Underlying disease or condition stable or improving

11. Alert patient who is able to follow commands

12. Patient not taking any medications that may hinder spontaneous ventilation

13. No life-threatening situations, such as shock or hypotension

14. No anemia, fever, or electrolyte imbalances

B. Weaning Techniques

1. SIMV to decrease the number of mechanical ventilator breaths while allowing for more spontaneous breathing

a. Patient should be using 40% O₂ or less before extubation.

b. Patient should be receiving an SIMV rate of 4/min or less before removal from the ventilator.

c. Postoperative patients may be weaned as they begin waking up. When patients have normal blood gas levels and are beginning to awaken, the SIMV rate can begin to be decreased.

2. Time On–Time Off Method

a. The patient stops using the ventilator periodically, is given T-tube flow-by for a specific length of time, then begins using the ventilator again.

b. The time without the ventilator is gradually increased until the patient is spending more time without the ventilator than he or she is using it.

c. The patient is often returned to the ventilator while asleep.

d. The patient must be monitored closely (e.g., blood pressure, VT, heart rate, respiratory rate, ABG levels, SaO₂) while receiving flow-by.

e. Often, the O₂ percentage is increased 10% while receiving flow-by.

XI. HIGH-FREQUENCY VENTILATION

CRT Exam Content Matrix: IIID4

RRT Exam Content Matrix: IIA2c, IIID4

A. High-frequency ventilation (HFV) refers to breathing rates that are four times the normal rate (60/min in the adult). A smaller than normal VT is used, sometimes less than the anatomic VD. HFV has been shown to improve gas exchange without the barotrauma and cardiovascular problems associated with conventional volume-limited and pressure-limited ventilation in both infants and adults.

The ventilator rate or frequency is measured in Hertz (Hz) which is equal to one breath per second. For example, a rate of 60/min is equal to 1 Hz. A rate of 300/min is equal to 5 Hz, or 5 breaths/s.

B. Three Classifications of HFV

1. High-frequency positive pressure ventilation

a. Gas delivery to the patient occurs with the use of a time-cycled or pressure- or volume-limited device through ventilator tubing that has a low compressible volume. This ensures that little volume is lost in the tubing.

b. Gas is directed through an insufflation catheter that is placed through the ET tube, and the rapid opening and closing of the exhalation valve determines gas flow into and out of the lungs.

c. The exhalation valve opens and closes in response to a pneumatic or electric source. Some pneumatic units incorporate fluidic gates to accomplish the opening and closing of the exhalation valve.

d. Ventilatory rates with high-frequency positive pressure ventilation (HFPPV) are between 60/min and 100/min (11.67 Hz), and the VT is small (between 3 and 5 mL/kg of body weight). I:E ratio of 1 : 3 or less are normally used.

e. Delivery of a small VT results in lower peak inspiratory pressure and lower mean airway pressure at a much higher rate than conventional ventilation and an improved distribution of gas.

f. PEEP may also be used with HFPPV to improve oxygenation and cause fewer cardiovascular side effects than conventional positive pressure ventilation and PEEP.

g. HFPPV has also been used during thoracic surgery, such as lobectomy or pneumonectomy, in which conventional ventilation may not be effective.

2. High-frequency jet ventilation (HFJV)

a. A high-pressure gas source injects short, rapid bursts of gas through a jet catheter, usually incorporated into a special ET tube. Air is entrained through a separate channel during inspiration, which increases the flow to the patient.

b. Frequency rates vary from about 100 to 600 cycles per minute (1.6710 Hz) at I:E ratio of 1 : 1 to 1 : 4. Peak inspiratory pressures are usually about 8 to 10 cm H₂O above baseline. The VT is usually a little larger than the anatomic VD.

c. A low-pressure alarm set 2 to 3 cm H₂O below peak inspiratory pressure should be incorporated to indicate power loss or system leaks.

d. A high-pressure alarm should be set 5 to 10 cm H₂O above peak inspiratory pressure. This alarm may be activated by a plugged ET tube, air-trapping (resulting from short exhalation time at high rates), pneumothorax, or airway secretions requiring suctioning.

e. The gas is most effectively humidified with the use of a heat exchanger that is designed to withstand high pressure. Water and the jet gas mix then pass through the exchanger. This warms the gas and also humidifies it.

f. An infusion pump may also be used to humidify the inspired gas. The pump places water in front of the jet nozzle, and as the gas passes through the jet, it combines with the water just outside the jet, humidifying the gas.

g. Because a low VT is delivered with HFV, resulting in an increased potential for atelectasis, PEEP should be employed so that the incidence of atelectasis is reduced.

3. High-frequency oscillation

a. High-frequency oscillation (HFO) requires an oscillating device that forces small impulses of gas into and out of the patient’s airway.

b. Three types of oscillating devices are used.

(1) Piston: As the piston moves inward, a small volume of gas is delivered to the patient. As the piston withdraws, the same amount of gas is drawn away from the patient (exhalation). A sine-type flow wave is usually produced.

(2) Diaphragm: Audio loudspeakers have been used to accomplish HFO. As the diaphragm vibrates, the rapid movement, both forward and backward, moves a volume of gas into and out of the patient’s lungs.

(3) Flow interrupter: As flow is being delivered to the patient’s airways, it passes through a rotating bar with a hole in it. Gas flow periodically passes through the hole and is delivered in small bursts to the patient. Exhalation occurs by normal passive recoil of the lung.

c. Oscillations occur at a rate of 60 to 3600/min (160 Hz) at a VT of less than anatomic VD.

d. HFO may be beneficial in treating patients with large degrees of intrapulmonary shunting. It is being used currently to ventilate the lungs of infants with respiratory distress syndrome.

Exam Note

To lower the PaCO₂ using high frequency ventilation/oscillation, increase the frequency or the oscillatory amplitude.

C. Potential Advantages of HFV over Conventional Ventilation

1. Reduced risk of barotrauma

2. Reduced risk of cardiac side effects

3. Less fluctuation in ICP

4. Improvement of mucociliary clearance

XII. ESTIMATING DESIRED VENTILATOR VARIABLE CHANGES

CRT Exam Content Matrix: IIID2b, IIIF2i2-3, IIIG3b-c

RRT Exam Content Matrix: IIID2b, IIIF2e2-3, IIIG3b-c

A. Making Changes in FiO₂

EXAMPLE:

The data below are from a patient using a volume ventilator.

To increase this patient’s PaO₂ to 80 mm Hg, to what level must the FiO₂ be changed?

B. Making Changes in the Ventilator Rate

EXAMPLE:

Data collected from a patient using a volume ventilator in the control mode.

To raise the patient’s PaCO₂ to 35 mm Hg, the ventilator rate should be adjusted to what level?

C. Making Changes in Minute Volume (

)

EXAMPLE:

Below are data collected from a patient using a volume ventilator in the control mode.

Which of the following ventilator settings would decrease the patient’s PaCO₂ to 45 mm Hg?

In this question, we use the equation above to derive the necessary to decrease the PaCO₂ to the desired level. Then choose the answer with the appropriate that has been calculated.

The required to decrease the PaCO₂ to 45 mm Hg is 8.4 L. In the example, the second choice gives a minute volume of 8.4 L (700 mL × 12/min).

D. Making Changes in Alveolar Ventilation

EXAMPLE:

Below are data from a patient who is using a volume ventilator in the control mode.

Which of the following ventilator settings would decrease the patient’s PaCO₂ to 40 mm Hg?

In this question, use the equation above to derive the alveolar ventilation necessary to bring the PaCO₂ down to 40 mm Hg. It is the same equation used in the previous problem, except that we correct for anatomic VD (in this case, 150 mL), which is subtracted from the VT. This value is then multiplied by the respiratory rate to determine the .

Therefore, an alveolar volume of 9.75 L is required to decrease the PaCO₂ to 40 mm Hg. Now choose the appropriate choice from the table above; in this case, it is the third one.

XIII. PRACTICE VENTILATOR PROBLEMS

CRT Exam Content Matrix: IIID2b, IIIF2i2-3, IIIG3b-c

RRT Exam Content Matrix: IIID2b, IIIF2e2-3, IIIG3b-c

A. A 75-kg male patient is using a volume ventilator in the control mode. Appropriate data from his chart are as follows

Based on this information, the respiratory therapist should recommend which of the following ventilator changes?

1. Increase PEEP to 10 cm H₂O.

2. Add 100 mL of mechanical VD.

3. Increase the FiO₂ to 0.60.

4. Increase the VT to 800 mL.

Answer: The fourth choice. The patient is hypoventilating as a result of an inadequate VT. The patient weighs 75 kg; therefore, use 10 to 12 mL/kg IBW to determine an adequate VT. The ventilator should be set on a VT between 750 and 900 mL.

B. A patient is using a volume ventilator in the control mode of ventilation. Pertinent data follow.

What is the most appropriate ventilator change to recommend at this time?

1. Decrease the FiO₂ to 0.50.

2. Increase the VT to 900 mL.

3. Decrease the rate to 10/min.

4. Decrease the PEEP to 6 cm H₂O.

Answer: The fourth choice. The patient’s lungs are well ventilated but are hyperoxygenated with the use of 60% O₂. On the board exams, begin weaning the patient from PEEP if the O₂% is 60% or less.

C. A 46-year-old, 80-kg (176-lb) man’s lungs are being mechanically ventilated with a volume ventilator in the assist/control mode. Data follow.

What is the most appropriate recommendation at this time?

1. Decrease the rate to 8/min.

2. Decrease VT to 750 mL.

3. Increase the FiO₂ to 0.50.

4. Begin PEEP at 8 cm H₂O for the patient.

Answer: The third choice. This patient is slightly hyperventilating as a result of hypoxemia. As the PaO₂ increases, hyperventilation should subside. Increasing the FiO₂ or adding PEEP both will elevate the PaO₂, but because the FiO₂ is 0.30, we can safely increase it to as high as 0.60 before adding PEEP.

D. A patient is using the control mode of ventilation with the following settings

VT	800mL
Rate	10/min
FiO₂	0.35

ABG values

pH	7.50
PaCO₂	29 mmHg
PaO₂	97 mmHg
HCO₃⁻	25 mEq/L

What would be the most appropriate ventilator change to make at this time?

1. Increase the inspiratory flow.

2. Add 5 cm H₂O PEEP.

3. Increase the FiO₂ to 0.50.

4. Decrease VT to 700 mL.

Answer: The fourth choice. This patient is hyperventilating, which is resulting in a low PaCO₂. Because the PaO₂ is normal, hypoxemia is not the cause of the hyperventilation. Minute ventilation is too high and can be reduced by decreasing VT, thereby increasing PaCO₂.

E. A 60-kg (132-lb) female patient is using a volume ventilator in the control mode with the following settings:

What is the appropriate ventilator change at this time?

1. Increase the FiO₂ to 0.50.

2. Add PEEP of 5 cm H₂O.

3. Decrease rate to 6/min.

4. Decrease VT to 600 mL.

Answer: The fourth choice. The VT setting is more than 12 mL/kg, which is causing the patient to hyperventilate. Decreasing the VT will increase the PaCO₂. Decreasing the rate will also increase the PaCO₂, but a rate of 6/min (third choice) in the control mode is too low.

XIV. VENTILATOR FLOW, VOLUME, AND PRESSURE WAVEFORMS

CRT Exam Content Matrix: IIID3, IIIF2i8, IIIG3h,k

RRT Exam Content Matrix: IIID3, IIIF2e7, IIIG3g,j

A. Flow Waveforms

1. Square wave (constant flow)

FIGURE 11-4

a. With this flow pattern, the flow remains constant throughout inspiration.

b. Changes in R_AW and compliance do not change the flow pattern under normal circumstances.

c. This type of flow pattern is beneficial to patients with increased respiratory rates.

2. Sine (sinusoidal) wave

FIGURE 11-5

a. The flow gradually accelerates from the beginning of inspiration then decelerates toward the end of inspiration.

b. This flow pattern benefits patients with increased R_AW because airway turbulence produced by this flow is decreased.

3. Decelerating ramp wave

FIGURE 11-6

a. The initial flow is high and begins decelerating as inspiration continues.

b. The flow never decreases more than 50% to 55% of initial flow.

c. The pattern benefits patients with low compliance by allowing ventilation to occur at a decreased pressure.

d. This flow usually results in optimal distribution of gas, lower PIP, reduced work of breathing, and improved patient comfort.

4. Accelerating ramp wave

FIGURE 11-7

a. The flow is initially slow and accelerates to a peak flow by the end of inspiration.

b. This pattern creates less turbulence of flow in the beginning of inspiration; therefore, more volume may be delivered through narrowed or obstructed airways.

B. Volume Waveform

1. This is a typical volume waveform showing volume in milliliters on the vertical axis and time in seconds on the horizontal axis. Notice how the waveform returns to zero (baseline) during exhalation.

FIGURE 11-8 Typical volume waveform.

Modified from Branson R, Hess D, Chatburn R: Respiratory care equipment, Philadelphia, 1999, Lippincott, Williams, and Wilkins.

2. If the volume tracing does not return to baseline at the end of exhalation, it is an indication of leaks in the ventilator circuit or around the ET tube cuff or chest tube, or it may result from air-trapping.

FIGURE 11-9 Atypical volume waveform.

3. If the volume tracing goes below the baseline, this is an indication of auto-PEEP or the patient may be coughing or agitated.

FIGURE 11-10 Typical pressure waveform.

4. Obstructive lung disease causes the tracing to be flat during exhalation as a result of decreased expiratory flows.

FIGURE 11-11

C. Pressure Waveform

1. This is a typical pressure waveform that is shown during volume ventilation; pressure is shown on the vertical axis, and time is on the horizontal axis. (When pressure ventilation is used, the pressure waveform is almost square.)

FIGURE 11-12

Lung compliance and auto-PEEP. Modified from Branson R, Hess D, Chatburn R: Respiratory care equipment, Philadelphia, 1999, Lippincott, Williams, and Wilkins.

2. Peak pressure rises during inspiration and is determined by VT, R_AW, lung compliance, and inspiratory flow.

3. The difference between peak pressure and plateau pressure is R_AW.

4. If the waveform does not return to baseline during exhalation, the patient is receiving PEEP, or auto-PEEP is occurring.

XV. SUMMARY OF VENTILATOR ADJUSTMENTS ACCORDING TO ABG RESULTS

CRT Exam Content Matrix: IIID2b, IIIF2i2-3, IIIG3b-c

RRT Exam Content Matrix: IIID2b, IIIF2e2-3, IIIG3b-c

ABG Abnormality	Ventilator Adjustment
Decreased pH, increased PaCO₂, decreased or normal PaO₂	Increase VT (Maintain 8 to 12 mL/kg) or increase rate
Increased pH, decreased PaCO₂, normal or increased PaO₂	Decrease VT, decrease rate or add mechanical VD
Normal pH, increased PaCO₂, PaO₂ of 50 to 65 mm Hg, increased HCO₃⁻	No adjustment needed; this is a patient with COPD. May begin weaning if other weaning criteria are met.
Normal or increased pH, normal or decreased PaCO₂, decreased PaO₂	Increase FiO₂ if <0.60. Add or increase PEEP if FiO₂ is 0.60 or more.
Normal pH, normal PaCO₂, increased PaO₂	Decrease FiO₂ if it is higher than 0.60; decrease PEEP if FiO₂ is 0.60 or lower.
Decreased pH, normal PaCO₂, decreased HCO₃⁻	No ventilator changes are necessary; administer HCO₃⁻.