Cairo: Pilbeam's Mechanical Ventilation, 6th Edition

Chapter 09: Ventilator Graphics

WorkBook Answer Key

Key Terms Crossword Puzzle

Chapter Review Questions
1.	(a) Rectangular, (b) ascending or accelerating ramp, (c) exponential rise, (d) descending or decelerating ramp, (e) sinusoidal or sine, and (f) exponential decay.
2.	The term scalar is used to identify waveforms for pressure, flow, and volume that are graphed relative to time. The term loop is used to describe a graph of two variables other than time, such as pressure-volume and flow-volume loops.
3.	VT = Flow × TI.
4.	The flow of gas into the lungs is dependent on (a) the difference between the pressure from the ventilator (power source) and (b) the pressure inside the lungs.
5.	Decreases (drops); increases (rises).
6.	PTA (transairway pressure) + PA (alveolar pressure)
7.	When the flow is constant.
8.	TI = Volume ÷ Flow. Flow of 60 L/min = 60 L ÷ 60 sec = 1 L/sec. Volume of 600 mL = 0.6 L; therefore, TI = 0.6 L/1 L/sec = 0.6 second.
9.	PA = ΔV ÷ Cs = 800 mL ÷ 25 mL/cm H2O = 32 cm H2O
10.	(a) Inspiration begins, (b) inspiration ends, and (c) maximum or peak expiratory flow
11.	Sinusoidal
12.	45 L/min
13.	50 L/min
14.	1 second
15.	2 seconds
16.	Air trapping and hyperinflation
17.	No. An intrinsic PEEP measurement would not be valid if the patient were actively attempting to take a breath during the maneuver. (An unstable expiratory pause invalidates the measurement.)
18.	Intrinsic PEEP is measured at the end of exhalation.
19.	Active inspiration by the patient in a ventilator that responds to patient flow demand.
20.	1-2 cm H2O
21.	Sensitivity (pressure trigger) and flow settings
22.	Changing the flow pattern during volume ventilation changes the pressure patterns.
23.	45 cm H2O
24.	Descending or decelerating ramp; 40 L/min
25.	600 mL
26.	CSTAT = ΔV ÷ (Pplateau  PEEP) = 600 mL ÷ (20  0) = 30 mL/cm H2O
27.	Raw = (PIP  Pplateau) ÷ Flow = (45  20) ÷ 0.67 L/sec = 37.3 cm H2O/L/sec
28.	There is no intrinsic PEEP.
29.	The flow rate drops to zero before the end of inspiration, because there is an inspiratory pause where there is no flow. After the inspiratory pause the gas is exhaled, giving the negative appearance on the graph.
30.	During decreases in lung compliance, the delivered tidal volume decreases in PC-CMV.
31.	Flow returns to zero when the alveolar pressure becomes equal to the set pressure during PC-CMV.
32.	Descending ramp or exponential decay, which varies with the patient’s lung characteristics.
33.	The largest pressure gradient occurs at the beginning of inspiration.
34.	Pressure, sensitivity, rise time or slope, and flow cycle time.
35.	VC-SIMV (no pressure support). Look closely at the pressure waveform. It has a peak as opposed to a flat top, and the two mechanical breaths do not have the same peak inspiratory pressure. This means that the pressure is variable. In volume control, the volume is set and the pressure is determined by lung characteristics.
36.	PC-SIMV with CPAP and pressure support.
37.	“A” is a pressure-supported breath with a set pressure of 20 cm H2O. “B” is a pressure-controlled breath with a set pressure of 38 cm H2O.
38.	During inspiration, ATC augments flow. During expiration, it provides a slight negative-pressure gradient between the lungs and the upper airways.
39.	Peak flow rate × Flow cycling percentage = Flow rate when cycling occurs. 45 L/min × 0.25 = 11 L/min.
40.	TI = 1 second
41.	45 L/min
42.	10 L/min
43.	(Ending flow ÷ Peak inspiratory flow) × 100 = Flow-cycling percentage; therefore, (10 ÷ 45) × 100 = 22%.
44.	40 L/min
45.	Patients with COPD have increased airway resistance and therefore need a long expiratory time to exhale. During normal inspiration, flow drops toward the end of inspiration. Percent flow cycling sets up a criterion to determine when inspiration will end for a pressure-supported breath. Therefore, when inspiratory flow tapers down to that set percentage of the peak inspiratory flow, inspiration ends. Patients with increased airway resistance require longer times for exhalation. A higher flow cycle percentage ends the inspiration sooner than does a lower flow cycle percentage, giving the patient a longer time for exhalation.
46.	A patient with reduced compliance fills the lungs quickly but with less volume than normal. To allow more volume to enter the lungs for the given pressure, a lower flow cycle percentage should be set.
47.	(18 ÷ 40 L/min) × 100 = 45%
48.	200 mL
49.	(5 ÷ 38 L/min) × 100 = 13%
50.	400 mL
51.	“A” TI = approximately 0.75 second and “B” TI = approximately 1.25 second
52.	The flow cycle percentage was lowered, increasing the inspiratory time.
53.	“A” is PEEP +5 cm H2O; “B” is peak alveolar pressure; and “C” is peak inspiratory pressure.
54.	35 cm H2O
55.	500 mL
56.	There is +5 cm H2O PEEP.
57.	More pressure is needed to deliver the set volume; therefore, the height of the loop (volume) remains the same, but the loop extends farther to the right and tends to flatten out.
58.	Because the pressure remains constant, the loop will reach the same pressure but will be shorter because less volume is being delivered to the patient for each breath.
59.	Airway resistance increases because of bronchospasm, secretions in the airway, or mucosal edema.
60.	Loop “A” represents a spontaneous breath.
61.	Loop “A” has a clockwise direction because a spontaneous breath is negative on inspiration and becomes positive on exhalation.
62.	Loop “B” represents a patient-triggered mandatory breath.
63.	Loop “B” has a counterclockwise direction because after the initial negative deflection caused by the patient, the ventilator delivers the positive-pressure breath (doing all the work).
64.	45 L/min = 0.75 L/sec, Raw = PTA ÷ Flow; 3 cm H2O ÷ 0.75 L/sec = 4 cm H2O/L/sec.
65.	Hysteresis
66.	This loop moves in a clockwise direction.
67.	Constant flow or rectangular waveform
68.	(a) 50 L/min and (b) 55 L/min
69.	425 mL
70.	Either a leak in the patient (bronchopleural fistula) or a leak in the patient circuit.
71.	Air trapping or the presence of intrinsic PEEP.
Critical Thinking Questions
Refer to Figure 9-13 of the Workbook to complete this exercise.
1.	VC-CMV. Constant flow waveforms are associated with volume control. The pressure curves all have peaks, meaning that the pressure is not being maintained.
2.	The patient is not assisting. There are no downward deflections below baseline on the pressure scalar.
3.	Total cycle time = 1.5 second. TI = 1 second. TE = 0.5 second. Set rate = 40 breaths/min.
4.	The peak inspiratory pressure is increasing with each breath.
5.	(1) The exhaled volume is decreasing with every breath and (2) inspiration is beginning before all the volume has been exhaled.
6.	Each breath begins before exhalation is complete. Exhalation should return to zero on the x-axis before the beginning of the next breath.
7.	Rising PIP, dropping exhaled tidal volumes, inspiration beginning before the completion of exhalation, and an inverse I:E ratio all mean that air is being trapped in the lungs. In this case, it is due to the rapid set rate.
8.	Decrease the set rate to allow for adequate expiratory time.
Case Studies
Case Study 1

1.	This loop is wider because of the increased airway resistance associated with COPD. The axis of the loop is more upright, which is a result of an increase in static compliance in the patient’s lungs.
2.	Yes, the patient is triggering the ventilator. The small loop to the left of the volume axis on the pressure scale represents patient effort to trigger the ventilator.
3.	Approximately 650 mL
4.	Approximately 23 cm H2O
5.	The patient has a combined alkalosis from acute alveolar hyperventilation superimposed on chronic hypercapnic respiratory failure. The current tidal volume setting is 12 mL/kg. This is too high; it should be reduced to deliver between 6 and 8 mL/kg, which would give a volume between 330 and 436 mL.

Case Study 2

1.	Spontaneous breath
2.	Inspiration
3.	Expiration
4.	Clockwise
5.	There is no set pressure support. If pressure support were present, the loop would be shifted to the positive side of the x-axis.
6.	VC-CMV (or the mechanical breath during VC-SIMV). A pressure-controlled breath would have a flatter top.
7.	Yes, the patient is triggering the breath. There is a negative deflection along the x-axis, or pressure axis.
8.	Adjust the ventilator sensitivity to be more sensitive to the patient.
9.	The type of flow waveform (constant), the peak inspiratory flow (50 L/min), the peak expiratory flow (50 L/min), and the tidal volume (650 mL).
10.	The current loop has a scooped-out appearance on the flow side.
11.	This is due to increased airway resistance. The most likely cause, in this case, is bronchospasm and/or increased airway secretions.
12.	Aerosolized bronchodilator and tracheobronchial suctioning.
NBRC-Style Questions
1.	A
2.	C
3.	C
4.	B
5.	C
6.	B
7.	A
8.	C
9.	D
10.	B