Cairo: Pilbeam's Mechanical Ventilation, 6th Edition

Chapter 01: Basic Terms and Concepts of Mechanical Ventilation

WorkBook Answer Key

Key Terms Crossword Puzzle

Chapter Review Questions
1.	Ventilation refers to the movement of gas into and out of the lungs. Respiration is the movement of gas molecules across a cell membrane.
2.	External respiration, internal respiration.
3.	The pressure at point A needs to be higher than the pressure at point B to create a pressure gradient.
4.	Intrapleural pressure, 5 cm H2O, 10 cm H2O.
5.	a. A = Pressure at the airway opening (Pawo).
B = Airway pressure (Paw).
C = Pressure at the body’s surface (Pbs).
D = Alveolar pressure (PA).
E = Intrapleural pressure (Ppl).
b. Mouth pressure (Pm), airway opening pressure (Paw), proximal airway pressure, and mask pressure.
6.	Intrapleural pressure (Ppl) is estimated using the esophageal pressure (PES). PES is obtained by placing a balloon in the esophagus and monitoring pressure changes in the balloon.
7.	Transairway pressure (PTA), conductive airways, PTA = Paw  PA.
8.	Transthoracic pressure (PW), PW = PA  Pbs.
9.	Transpulmonary pressure (PL), PL = PA  Ppl.
10.	Transrespiratory pressure (PTR), PTR = Paw  Pbs.
11.	a. Paw  Pbs = Transrespiratory pressure (PTR), which is the pressure required to inflate the lungs and airways during positive pressure ventilation.
b. PA  Pbs = Transthoracic pressure (PW), which is the pressure needed to expand or contract the lungs and chest wall at the same time.
c. PA  Ppl = Transpulmonary pressure (PL), which is the pressure required to maintain alveolar inflation.
d. Paw  PA = Transairway pressure (PTA), which is the pressure needed to overcome airway resistance or to cause movement in the conducting airways.
12.	

13.	Negative-pressure ventilation (NPV) attempts to mimic the function of the respiratory muscles to allow breathing through normal physiological mechanisms. With this device, the patient’s head and neck are exposed to ambient pressure while the thorax and the rest of the body are enclosed in an airtight container that is subjected to negative pressure (i.e., pressure less than atmospheric pressure). Negative pressure generated around the thoracic area is transmitted across the chest wall, into the intrapleural space, and finally into the intraalveolar space.
With negative-pressure ventilators, as the intrapleural space becomes negative, the space inside the alveoli becomes increasingly negative in relation to the pressure at the airway opening (atmospheric pressure). This pressure gradient results in the movement of air into the lungs. In this way, negative-pressure ventilators resemble normal lung mechanics. Expiration occurs when the negative pressure around the chest wall is removed. The normal elastic recoil of the lungs and chest wall causes air to flow out of the lungs passively.
14.	Advantages of using negative pressure ventilators include (a) not requiring artificial airways (e.g., an endotracheal tube); (b) patients are able to eat and speak; and (c) this mode has fewer physiologic disadvantages in patients with normal cardiovascular function than does positive pressure ventilation.
15.	Transairway pressure (PTA) = Mouth pressure (Pm)  Intraalveolar pressure (PA): 25 cm H2O  5 cm H2O = 20 cm H2O.
16.	 

17.	

18.	Peak inspiratory pressure, peak airway pressure.
19.	Baseline pressure.
20.	Positive end expiratory pressure.
21.	Plateau pressure, selecting inspiratory pause or inflation hold maneuver.
22.	Elastic, frictional forces.
23.	Compliance, elastance.
24.	Change in volume that corresponds to the change in pressure, C = ∆V ÷ ∆P.
25.	0.1 L/cm H2O or 100 mL/cm H2O, 0.05-0.17 L/cm H2O.
26.	40-50 mL/cm H2O, 35  45 mL/cm H2O.
27.	CSTAT = ∆V ÷ (Pplateau  EEP).
28.	When more pressure is required to deliver a specific volume, the lung compliance is decreasing.
29.	CSTAT = 750 mL ÷ (27  10 cm H2O) = 44 mL/cm H2O.
30.	CSTAT = 575 mL ÷ (35  5 cm H2O) = 19 mL/cm H2O.
31.	CSTAT = 650 mL ÷ (18  0 cm H2O) = 36 mL/cm H2O.
32.	As the lungs become harder to ventilate, PIP will increase. As the lungs become harder to ventilate, lung compliance will decrease.
33.	Resistance is defined as frictional forces associated with ventilation.
34.	Raw = (PIP  Pplateau) ÷ Flow (L/sec).
35.	At flow rates of 0.5 L/sec, Raw is 0.6-2.4 cm H2O/L/sec.
36.	Emphysema can cause increased compliance due to tissue destruction and loss of elastic recoil. It also can cause airway resistance due to airway inflammation, small airway obstruction, and bronchospasm.
37.	PTA = PIP − Pplateau 27 cm H2O  20 cm H2O = 7 cm H2O.
38.	The amount of pressure needed to overcome airway resistance or the amount of pressure lost to airway resistance is equal to the transairway pressure. PTA = 30 cm H2O  20 cm H2O = 10 cm H2O.
39.	5 cm H2O.
40.	Raw = (PIP  Pplateau) ÷ Flow (L/sec)
= (48  30 cm H2O) ÷ (40 L ÷ 60 sec)
= 18 cm H2O ÷ 0.67 L/sec
Raw = 27 cm H2O/L/sec, this is an extremely high Raw.
41.	Flow rate of 60 L/min = 1 L/sec; Raw = (25  15 cm H2O) ÷ 1 L/sec
Raw = 10 cm H2O/L/sec.
42.	The characteristics of the lung are not homogeneous, because each acinus may have entirely different values for compliance and resistance.
43.	A low compliance unit, because it is stiff, receives less volume than a normal unit; however, it fills and empties more rapidly than the normal unit. A unit with high airway resistance fills slowly and takes longer to empty than a normal lung unit. The faster the respiratory rate, the less volume enters a unit with high airway resistance.
44.	The movement of structures, including the lungs, abdominal organs, rib cage, and diaphragm; gas viscosity; gas density; the length and diameter of the airways; and the flow rate of the gas are all factors that contribute to airway resistance.
45.	Clinical factors that contribute to airway resistance include bronchospasm, airway secretions, mucosal edema, small endotracheal tubes, and airway inflammation.
46.	CSTAT × Raw = Time constant; 25 mL/cm H2O = 0.025 L/cm H2O; 0.025 L/cm H2O × 30 cm H2O/L/sec = 0.75 sec.
47.	Step 1: Calculate CSTAT. CSTAT = 0.6 L ÷ 24 cm H2O = 0.025 L/cm H2O.
Step 2: Calculate Raw. First convert L/min to L/sec. 60 L/min = 1 L/sec.
Then calculate the resistance, Raw = (30  24 cm H2O) ÷ 1 L/sec = 6 cm H2O/L/sec.
Step 3: The product of Raw and CSTAT is the time constant. 6 cm H2O/L/sec × 0.025 L/cm H2O = 0.15 seconds. The time constant is 0.15 seconds.
48.	1—63%, 2—86%, 3—95%, 4—98%, 5—100%.
49.	The time constant for patient 1 is very short, meaning that this patient’s lungs are noncompliant, or stiff. Patient 1 would receive less volume than normal lungs. The time constant for patient 2 is very long, meaning that there is increased airway resistance. This patient’s lungs will fill very slowly and receive less volume than normal lungs unless slower rates are used. Patient 3 has a normal time constant; this patient’s lungs will receive the most volume of the three patients.
50.	0.055 L/cm H2O × 6 cm H2O/L/sec = 0.33 second.
51.	The third time constant will have 95% of the volume emptied. Therefore, 0.33 × 3 = 0.99 second.
52.	Patients with increased airway resistance have longer time constants. This means that their lungs take longer to fill and empty. Rapid respiratory rates reduce the expiratory time, causing gas to remain in the lungs. The patient is unable to exhale fully. This incomplete emptying of the lungs increases the FRC (air-trapping) and causes auto-PEEP.
53.	CSTAT = 0.7 L ÷ (18  5 cm H2O) = 0.054 L/cm H2O
Raw = (45  18 cm H2O) ÷ (60 L ÷ 60 seconds) = 27 cm H2O/L/sec
Time constant = 0.054 L/cm H2O × 27 cm H2O/L/sec = 1.46 seconds.
54.	Figure A represents a normal lung unit. Figure B represents a lung unit with increased airway resistance. Given the same inspiratory time, the unit with increased airway resistance will receive less volume because it has a longer time constant.
55.	Figure A represents a normal lung unit. Figure B represents a lung unit with low compliance. The low compliance lung unit has a shorter time constant; it therefore will fill faster, but with less volume, than the normal lung unit.
Critical Thinking Questions
1.	PIP represents the amount of pressure required to overcome both the resistance of the airways and the elastic resistance of the alveoli and ventilator circuitry. Therefore, PIP minus Pplateau equals the amount of pressure required to overcome airway resistance; in this case: 43  18 = 25 cm H2O.
2.	Acute respiratory distress syndrome causes lung compliance to decrease, which shortens time constants.
3.	Emphysema is characterized by dilation and destruction of lung units from the terminal bronchioles to the alveoli. This tissue destruction causes a loss of elastic recoil, thereby increasing lung compliance. This lengthens the fill and empty time, which is represented by a long time constant.
4.	At 30 weeks of gestation, there is a high risk for developing respiratory distress syndrome because of pulmonary immaturity. Insufficient pulmonary surfactant production leads to low lung compliance and thus short time constants.
5.	If you have a patient actively breathing during the plateau pressure measurement, the reading will be inaccurate. If the patient is actively trying to exhale against the machines breath hold, coughing, or gagging, the manometer reading may be falsely elevated.
Case Studies
Case Study 1

1.	PIP  Pplateau = Pressure required to overcome airway resistance (PTA)

Time	PIP	Pplateau 	PTA	
0800	18 cm H2O	10 cm H2O	8 cm H2O	
1000	24 cm H2O	12 cm H2O	12 cm H2O	
1200	35 cm H2O	11 cm H2O	24 cm H2O	

2.	Raw = PTA ÷ Flow (L/sec)

Time	PTA	Flow (L/sec)	PTA ÷ Flow = Raw	
0800	8 cm H2O	45 L ÷ 60 sec = 0.75 L/sec	8 ÷ 0.75 = 10.7 cm H2O/L/sec	
1000	12 cm H2O	45 L ÷ 60 sec = 0.75 L/sec	12 ÷ 0.75 = 16 cm H2O/L/sec	
1200	24 cm H2O	45 L ÷ 60 sec = 0.75 L/sec	24 ÷ 0.75 = 32 cm H2O/L/sec	

3.	CSTAT = VT ÷ (Pplateau  PEEP)

Time	VT	Pplateau  PEEP	CSTAT	
0800	600 mL	10 cm H2O  5 cm H2O
= 5 cm H2O	600 mL ÷ 5 cm H2O
= 120 mL/cm H2O	
1000	600 mL	12 cm H2O  5 cm H2O
= 7 cm H2O 	600 mL ÷ 7 cm H2O
= 85.7 mL/cm H2O	
1200	600 mL	11 cm H2O  5 cm H2O
= 6 cm H2O	600 mL ÷ 6 cm H2O
= 100 mL/cm H2O	

4.	Time constant = Raw × CSTAT

Time	Raw	CSTAT	Time Constant	
0800	10.7 cm H2O/L/sec	120 mL/cm H2O	1.28 seconds	
1000	16 cm H2O/L/sec	85.7 mL/cm H2O	1.37 seconds	
1200	32 cm H2O/L/sec	100 mL/cm H2O	3.20 seconds	

5.	There is a significant rise in both the transairway pressure and the airway resistance, from 0800 to 1200, along with an increased time constant. This is a clinical manifestation and indication of bronchospasm, increased secretions, or inflammation. Because a single time constant has increased to 3.2 seconds, there will be inadequate time for inspiration and expiration to occur. Therefore, until this situation is corrected, air trapping or auto-PEEP is very likely.

Case Study 2

1.	40 L/min = 40 L ÷ 60 seconds = 0.67 L/sec; 60 L/min = 1 L/sec

Time	PTA	Raw	CSTAT	Time Constant	
1000	40  28
= 12 cm H2O	12 cm H2O ÷ 0.67 L/sec
= 17.9 cm H2O/L/sec	0.550 L ÷ (28  0)
= 0.0196 L/cm H2O	17.9 × 0.0196
 = 0.35 seconds	
1200	47  37
= 10 cm H2O	10 cm H2O ÷ 0.67 L/sec
= 14.9 cm H2O/L/sec	0.550 L ÷ (37  5)
= 0.0172 L/cm H2O	14.9 × 0.0172
= 0.26 seconds	
1400	54  43
= 11 cm H2O	11 cm H2O ÷ 0.67 L/sec
= 16.4 cm H2O/L/sec	0.550 L ÷ (43  7)
= 0.0153 L/cm H2O	16.4 × 0.0153
= 0.25 seconds	
1600	55  33
= 12 cm H2O	12 cm H2O ÷ 1 L/ sec
= 12 cm H2O/L/sec	0.550 L ÷ (33  12)
= 0.0214 L/cm H2O	12 × 0.0214
= 0.26 seconds	

2.	The rising PIP between 1000 and 1400 hours is due to a decrease in static compliance. The decrease in compliance is due to a rise in plateau pressure from 28 cm H2O at 1000 hours to 37 cm H2O at 1200 hours to 43 cm H2O at 1400 hours. This rise demonstrates stiffening of the lungs. The plateau pressure, and therefore static compliance, dropped at 1600 hours.
3.	Minimum inspiratory time should be three time constants, which correspond to 95% filling of the lungs. Therefore, at 1600 hours the inspiratory time should be 0.26 seconds × 3 = 0.78 seconds.
NBRC-Style Questions
1.	C
2.	C
3.	D
4.	B
5.	A
6.	A
7.	D
8.	C
9.	C
10.	B