MindTap for Des Jardins' Cardiopulmonary Anatomy & Physiology, 2 terms Printed Access Card
MindTap for Des Jardins' Cardiopulmonary Anatomy & Physiology, 2 terms Printed Access Card
7th Edition
ISBN: 9781337794923
Author: Des Jardins, Terry
Publisher: Cengage Learning
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Chapter 6, Problem 11RQ
Summary Introduction

Case summary: A 37-year-old woman met with an accident and is on a volume-cycled mechanical ventilator.

Characters in the case: A 37-year-old woman met with an accident.

Adequate information: The barometric pressure is 745 mm Hg. She was receiving FlO2 of 0.50.

To determine: The total oxygen delivery, the arterial venous content difference, the intrapulmonary shunting, oxygen consumption, and the oxygen extraction ratio.

Expert Solution & Answer
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Explanation of Solution

Given information:

Cardiac output: 6 L/min

Hb: 11g/dL

PaO2: 60 mm Hg

Pv¯O2: 35 mm Hg

PaCO2: 38 mm Hg

With this information,

PAO2=(BP-PH2O) FlO2-PaCO2(1.25)=(745-47) 0.5 - 38(1.25)=(698)0.5 - 47.5=349 - 47.5=301.5

CcO2=(Hb×1.34)+(PAO2×0.003)=(11×1.34)+(301.5×0.003)=14.74+0.90=15.64

CaO2=(Hb×1.34×SaO2)+(PaO2×0.003)=(11×1.34×0.90)+(60×0.003)=13.266+0.18=13.44

Cv¯O2=(Hb×1.34×Sv¯O2)+(Pv¯O2×0.003)=(11×1.34×0.65)+(35×0.003)=9.581+0.105=9.686

A)

Totaloxygen delivery=QT×CaO2×10=6×13446×10=806.76 ml O2/min.

B)

Arterial-venous oxygen content difference C(av)O2=CaO2Cv¯O2=13.449.68=3.760mL/dL O2

C)

Intrapulmonary shunting (QSQT)=CcO2-CaO2CcO2-CvO2=15.64-13.4415.64-9.68=2.1985.958=0.37 %

D)

Oxygen consumption V˙O2=Q˙T[C(a-v¯)O2×10]=6×3.760×10=225.6mLO2/min

E)

Oxygen extraction ratio=CaO2Cv¯O2CaO2=13.449.6813.44=3.76013.446=0.279

Conclusion

Therefore it can be concluded that 806.76mLO2/min is the total oxygen delivery, 225.6mLO2/min is the oxygen consumption, the intrapulmonary shunting is 36.8%, 3.7mL/dL is the arterial-venous oxygen content difference, and 0.279 is the oxygen extraction ratio.

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Case StudyAUTOMOBILE ACCIDENT VICTIMA 37-year-old woman is on a volume-cycled mechanical ventilator on a day when barometric pressure is 745 mm Hg. The patient is receiving an FIO2 of .50. The following clinical data are obtained:•    Hb: 11 g%•    PaO2: 60 mm Hg•    SaO2 = 90%•    PvO2: 35 mm Hg•    Svo2 = 65•    PaCO2: 38 mm Hg•    Cardiac Output: 6 L/minuteBased on the information above, approximately what is the patient’s PAO2?
5. Calculate the pulmonary shunt fraction for a patient with the following data: Pb = 760 mm Hg; Hb = 10 g/dL; respiratory quotient = 0.8; FIO2 = 0.6; PaO2 = 100 mm Hg; SaO2 = 93%; PaCO2 = 45 mm Hg; = 36 mm Hg; = 70%.a. 13%b. 26%c. 30%d. 41%
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