Calculate the time constant for lung emptying given the ...

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Calculate the time constant for lung emptying given the information below.   Pplateau: 35 cm H2O   Raw: 1.2 cm H2O/L/sec   Cl: 50 ml/cm H2O
 
  A. 0.06 second
  B. 0.384 second
  C. 1.75 seconds
  D. 6.0 seconds

Question 2

The therapist is mechanically ventilating a lung parenchyma injury patient with ARDS at an I:E ratio of 1:1 with a respiratory rate of 18 breaths/minute. What condition is likely to occur from this situation?
 
  A. alveolar overdistention
  B. dynamic hyperinflation
  C. parenchymal lung injury
  D. ventilator induced diaphragmatic dysfunction

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Answer to Question 1

ANS: A
A. Correct response: The ventilation time constant is the product of the airway resistance and the lung compliance, that is, (50 ml/cm H2O = 0.05 L/cm H2O)
Raw  Cl = time constant
(1.2 /sec) (0.05 ) = 0.06 second
The time constant represents the time required for either lung filling or lung emptying.
B. Incorrect response: See explanation A.
C. Incorrect response: See explanation A.
D. Incorrect response: See explanation A.

Answer to Question 2

ANS: B
A. Incorrect response: See explanation B.
B. Correct response: The ventilatory rate controls the Paco2. A reasonable starting point for a mechanically ventilated patient is a frequency between 12 and 20 breaths/minute. Increasing the frequency increases the minute ventilation and generally increases CO2 clearance (hyperventilation). If the respiratory rate is too high, at some point; however, air trapping or dynamic hyperinflation may develop because of inadequate expiratory time. Generally, hyperinflation begins to happen in parenchymal lung injuries at breathing rates exceeding 30 to 35 breaths/minute. Nonetheless, it can occur at much lower frequencies if the I:E ratio is high or when the time constant for lung emptying is high.
C. Incorrect response: See explanation B.
D. Incorrect response: See explanation B.

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