Cairo: Pilbeam's Mechanical Ventilation, 6th Edition

Chapter 06: Initial Ventilator Settings

WorkBook Answer Key

Key Terms Crossword Puzzle

Chapter Review Questions
1. Minute ventilation
2. VO2 = 250 mL/min and VCO2 = 200 mL/min  and 
3. Body mass, surface area
4. (a) VT, (b) f, (c) inspiratory gas flow, (d) flow waveform, (e) I:E ratio, (f) pressure limit, (g) inflation hold, and (h) PEEP
5. 2.18 m2
6. 1.74 m2
7. 1.92 m2

	Calculations	Minute Ventilation	
8.	3.5 × 1.6 = 5.6 L/min + (5.6 × 0.18) = 5.6 + 1 =	6.6 L/min	
9.	4 × 2.8 = 11.2 L/min + (11.2 × 0.15) = 11.2 + 1.7 =	12.9 L/min	
10.	4 × 2.3 = 9.2 L/min + (9.2 × 0.1) = 9.2 + 0.9 =	10.1 L/min	
11.	3.5 × 1.9 = 6.65 L/min + (6.65 × 0.2) = 6.65 + 1.33 =	8.0 L/min	
12. 6-8 mL/kg IBW
13. 4-8 mL/kg IBW
14. 300-400 mL
15. 540-720 mL
16. The normal range for spontaneous VT in a human is 5-7 mL/kg.
17. The normal range for spontaneous breathing rate in a human is 12-18 breaths/min.
18. 105 + 5(62  60) = 115 lb ÷ 2.2 = 52 kg. Using 6-8 mL/kg, VT should range between 312 and 416 mL at a rate of 8-12 breaths/min. For example, 400 mL × 10 breaths/min = 4 L/min.
19. 106 + 6(71  60) = 172 lb ÷ 2.2 = 78 kg. Using 6-8 mL/kg, VT should range between 468 and 624 mL at a rate of 8-12 breaths/min. For example, 600 mL × 10 breaths/min = 6 L/min.
20. 106 + 6(69  60) = 160 lb ÷ 2.2 = 73 kg. Using 4-8 mL/kg, VT should range between 292 and 584 mL at a rate of 15-25 breaths/min. For example, 400 mL × 20 breaths/min = 8 L/min.
21.
	Tidal Volume	Respiratory Rate	Minute Ventilation	
a.	750 mL	12 breaths/min	0.75L × 12 = 9 L/min	
b.	580 mL	10 breaths/min	0.58L × 10 = 5.8 L/min	
c.	6.9 ÷ 15 = 0.46 L = 460 mL	15 breaths/min	6.9 L/min	
d.	8.3 ÷ 20 = 0.415 = 415 mL	20 breaths/min	8.3 L/min	
e.	460 mL	5.2 ÷ 0.46 = 11 breaths/min	5.2 L/min	
f.	660 mL	9.5 ÷ 0.66 = 14 breaths/min	9.5 L/min	
22. (a) 30 breaths/min: 60 seconds ÷ 30 breaths/min = 2 seconds
(b) 15 breaths/min: 60 seconds ÷ 15 breaths/min = 4 seconds
(c) 12 breaths/min: 60 seconds ÷ 12 breaths/min = 5 seconds
(d) 10 breaths/min: 60 seconds ÷ 10 breaths/min = 6 seconds
23. (a) TI = 1 second; f = 30; TE = TCT  TI; 2  1 second = TE = 1 second
(b) TI = 0.75 second; f = 15; TE = TCT  TI; 4  0.75 second = TE = 3.25 seconds
(c) TI = 1.25 seconds; f = 12; TE = TCT  TI; 5  1.25 seconds = TE = 3.75 seconds
(d) TI = 2 seconds; f = 10; TE = TCT  TI; 6  2 seconds = TE = 4 seconds
24. Use this formula: I:E = TI /TI:TE/TI
(a) 1 second/1 second:1 second/1 second = 1:1 ratio
(b) 0.75 second/0.75 second:3.25 seconds/0.75 second = 1:5 ratio
(c) 1.25 second/1.25 second:3.75 seconds/1.25 second = 1:3 ratio
(d) 2 seconds/2 seconds:4 seconds/2 seconds = 1:2 ratio
25. TCT = 60 ÷ rate = 60 ÷ 12 = 5 seconds. Therefore, one breath cycle is 5 seconds long and is represented by the following rectangle.
←5 seconds→	
The I:E ratio shows how much of the rectangle is allotted to inspiratory time and how much to expiratory time. The ratio given is 1:3. That means that 1 part of the rectangle is for inspiration and 3 parts are for expiration. If the two numbers from the ratio are added, the sum shows the number of equal parts into which the TCT is divided.
1.25 second	1.25 second	1.25 second	1.25 second	
TCT ÷ (I + E) = 1 part of the rectangle. 5 seconds ÷ (1 + 3) = 5/4 = 1.25 second. Each part of the rectangle is equal to 1.25 second. Because TI is one part, it is equal to 1.25 second. TE = 5  1.25 = 3.75 seconds.
26. TCT = 60 ÷ 25 = 2.4 seconds
←2.4 seconds→	
I:E = 1:2; I + E = 1 + 2 = 3 equal parts. 2.4 seconds ÷ 3 = 0.8 seconds
0.8 second	0.8 second	0.8 second	
TI = 0.8 second; TE = 1.6 second
27. Flow rate × Inspiratory time = Tidal volume. 50 L/min = 50 L/60 sec = 0.83 L/sec. 0.83 L/sec × 1 second = 0.83 L, or 830 mL.
28. Tidal volume ÷ Flow rate = Inspiratory time. 0.7 L ÷ 0.83 L/sec = 0.84 second.
29. Shorten; increase; decrease.
30. When slower flow rates are used, the TI will be longer and the TE will be shorter, which may result in air trapping and possible cardiovascular side effects. However, PIP may be reduced, and gas exchange may be improved.
31. Descending or decelerating waveform
32. Sine waveform
33. Descending or constant waveforms
34. In patients with hypoxemia and low lung compliance, the descending flow pattern may be beneficial, because it keeps PIP low and Paw high, improving gas distribution.
35. The measurement of plateau pressure, which helps estimate alveolar pressure for the calculation of static compliance
36. The plateau pressure is unattainable because the patient is actively breathing, which results in an unstable reading.
37. PC-CMV, bilevel PAP, and PRVC
38. PSV, PC-CMV, bilevel PAP, Paug, PRVC, and VS, APRV (every vent but Drager), C
39. PSV, bilevel PAP, Paug, and VS
40. PC-CMV and PRVC
41. Advantages include providing flow on demand and limiting pressures to avoid overinflation. Disadvantages include alveolar shearing from rapid initial flows to heterogeneous lungs and the variation in tidal volume as lung characteristics change.
42. Elastic recoil is weakened in a patient with COPD; therefore, small airways tend to collapse during exhalation, because thoracic pressures are greater than intraluminal pressure. Low levels of PEEP can increase the intraluminal pressure and prevent this collapse on exhalation.
43. It is established by the change in pressure between PIP and baseline (PEEP + auto-PEEP).
44. C = ΔV ÷ ΔP; therefore, ΔP = ΔV ÷ C. Compliance for this patient is 400 mL ÷ 18 cm H2O = 22.2 mL/cm H2O. ΔP = 750 mL ÷ 22.2 mL/cm H2O = 33.8 cm H2O. The PIP needs to be increased to 33.8 cm H2O to achieve 750 mL.
45. C = ΔV ÷ ΔP; therefore, ΔP = ΔV ÷ C. Compliance for this patient is 550 mL ÷ 14 cm H2O = 39.3 mL/cm H2O. ΔP = 800 mL ÷ 39.3 mL/cm H2O = 20.4 cm H2O. The PIP needs to be increased to 20.4 cm H2O to achieve 800 mL.
46. The easiest method for establishing initial levels of pressure support is to use the formula for transairway pressure (PIP  Pplateau), the values for which are measured during volume ventilation.
47. PTA = PIP  Pplateau = 28  20 cm H2O = 8 cm H2O PS.
48. The patient is actively exhaling prior to flow cycling, which is causing pressure cycling.
49. The flow cycling variable can be adjusted to a higher flow cycling percentage. This will allow more time for the patient to exhale.
50. Flow cycle percentage × Peak inspiratory flow = Flow rate level for cycling. 0.2 × 50 L/min = 10 L/min.
51. 0.4 × 50 L/min = 20 L/min
52. The 40% flow cycling percentage will have the shortest inspiratory time, because the ventilator will cycle when the peak inspiratory flow drops from its peak at 50 L/min to 20 L/min (rather than the 10 L/min caused by the 20% setting).
53. (a) The initial PIP should be set at the Pplateau level taken during VC-CMV; (b) if Pplateau is not available, use PIP from VC-CMV minus 5 cm H2O as a starting point; and (c) if volume settings are not available, use a setting of 10-15 cm H2O with simultaneous volume measurements and adjust appropriately.
54. (a) IPAP, 5-10 cm H2O; EPAP, 2-5 cm H2O; rate ≤25 breaths/min, and an initial oxygen flow of 2 L/min to ensure adequate oxygenation.
(b) The target tidal volume is 4-8 mL/kg or more.
55. Pressure-regulated volume control (PRVC)
56. The first breath is a test breath and is a volume-targeted breath with an inspiratory hold. This establishes the static compliance and airway resistance. The first pressure breath in PRVC is equal to the plateau pressure measured during the test breath.
57. Ventilators rely on the upper pressure limit to determine how high to go when increasing the pressure to deliver the set VT. The upper pressure limit will not permit pressure to exceed that setting. If the patient coughs or forcibly exhales during inspiration, this upper pressure limit protects the patient’s lungs from barotrauma.
58. (1) Bronchospasm, (2) increased secretions, (3) pneumothorax, and (4) decreased lung compliance. (Anything that will increase airway resistance or decrease lung compliance.)
59. Observation of the flow curve can allow estimation of the Palv during PC-CMV. If flow drops to zero prior to the end of inspiration, the set pressure is reaching the alveolar level by the end of inspiration. If flow does not drop to zero before inspiration ends, Palv is less than the set pressure.
Critical Thinking Questions
1. Factors to consider before placing a patient on mechanical ventilatory support include the patient’s metabolic rate, body surface area, ideal body weight, lung pathology, acid-base balance, and oxygenation status, as well as tubing compliance.
2. Decreasing compliance causes an increase in pressure (Pplateau) during VV and a decrease in VT during PV. Improving compliance decreases pressure during VV and increases VT delivery during PV.
3. The two PSV safety systems are (1) a maximum safety pressure system that pressure cycles the ventilator when there is a sudden rise in pressure (usually occurs with an active exhalation) and (2) a time cycle to prevent increased TI due to a leak in the system.
Case Studies
Case Study 1

1. Intubate and place on mechanical ventilation. (The patient is suffering from a neuromuscular disorder and is already showing signs of breathing difficulties. She is having difficulty swallowing, which necessitates protection of the airway.)
2. BSA = 1.82 m2; 1.82 m2 × 3.5 = 6.4 L/min
3. VT 6-8 mL/kg, because the patient has no underlying lung problems. Her IBW is 105 + 5(68  60) = 145 lb ÷ 2.2 = 66 kg. Therefore, VT range is 396-528 mL. The set rate should be 10-12 breaths/min.
4. Volume ventilation, because the patient has no underlying pulmonary problems.
5. VC-CMV; rate, 12 breaths/min; VT, 500 mL; flow rate needed to achieve this VT when TI = 1 second, FIO2 to maintain saturations.
NBRC-Style Questions
1. A [CSTAT = 600 mL ÷ (47  10) = 16.2 cm H2O. IBW = 106 + 6(72  60) = 178 lb ÷ 2.2 = 81 kg. 6 mL/kg × 81 kg = 486 mL. ΔP = ΔV ÷ C = 486 mL ÷ 16.2 mL/cm H2O = 30 cm H2O]
2. C
3. B (PIP  Pplateau = Initial PSV setting; 31  23 = 8 cm H2O)
4. B
5. B (TCT = 60 ÷ 35 = 1.71; TCT  TI = TE = 1.71  1 = 0.71; TI:TE = 1:0.71, or 1.4:1)
6. A (TCT = 60 ÷ 15 = 4 seconds; I + E = 4; therefore, TI = 1 second and flow = 0.5 L ÷ 1/60 sec = 30 L/min)
7. D [MMV = 4 × BSA = 4 × 2.3 = 9.2 L/min plus 9% for each °C above 37° = 9.2 + (3 × 9%) = 9.2 + (27% × 9.2) = 11.7 L/min]
8. C
9. B
10. A