Chapter 15, Problem 2CAQ2

Summary Introduction

To review:

The ${\text{DO}}_{\text{2}}$, if a patient receiving 5 cm ${\text{H}}_{\text{2}}\text{O}$ would have received 10 cm ${\text{H}}_{\text{2}}\text{O}$.

Introduction:

The partial pressure of oxygen in the sea is about 100 mm Hg while that in a solution is equal to 3 mL O₂/L of blood. Oxygen is carried by the blood in two forms, that is, the dissolved form and bound to Hb (hemoglobin).

Please refer to Clinical Application Case-2 for further information.

Explanation of Solution

The total oxygen content of arterial blood is denoted by ${\text{CaO}}_{\text{2}}$. It can be calculated by using the following formula:

${\text{CaO}}_{\text{2}}\text{= }\left(\text{Hb × 1}{\text{.34 × SaO}}_{\text{2}}\right)\text{ + }\left({\text{PaO}}_{\text{2}}\text{× 0}\text{.003}\right)$

Putting the values of Hb as 15 g/dL, ${\text{SaO}}_{\text{2}}$ as 91%, and ${\text{PaO}}_{\text{2}}$ as 57% for calculating the ${\text{CaO}}_{\text{2}}$ of Profile 1, we get:

$\begin{array}{l}\begin{array}{l}{\text{CaO}}_{\text{2}}\text{= }\left(\text{15 × 1}\text{.34 × 0}\text{.91}\right)\text{ + }\left(\text{57× 0}\text{.003}\right)\hfill \\ {\text{CaO}}_{\text{2}}\text{= }\left(\text{18}\text{.29}\right)\text{ + }\left(\text{0}\text{.17}\right)\hfill \end{array}\\ {\text{CaO}}_{\text{2}}\text{= 18}\text{.46}\end{array}$

Putting the values of Hb as 15 g/dL, ${\text{SaO}}_{\text{2}}$ as 90%, and ${\text{PaO}}_{\text{2}}$ as 61% for calculating the ${\text{CaO}}_{\text{2}}$ of Profile 2, we get:

$\begin{array}{l}{\text{CaO}}_{\text{2}}\text{= }\left(\text{15 × 1}\text{.34 × 0}\text{.90}\right)\text{ + }\left({\text{61}}_{}\text{× 0}\text{.003}\right)\hfill \\ {\text{CaO}}_{\text{2}}\text{= }\left(\text{18}\text{.09}\right)\text{ + }\left(\text{0}\text{.18}\right)\hfill \\ {\text{CaO}}_{\text{2}}\text{= 18}\text{.27}\hfill \end{array}$

Total oxygen delivery can be calculated by using the following formula:

${\text{DO}}_{\text{2}}{\text{= Q}}_{\text{T}}{\text{× CaO}}_{\text{2}}\text{× 10}$

Putting the values of Q_T as 3.83 and ${\text{CaO}}_{\text{2}}$ as 18.46 for calculating the ${\text{DO}}_{\text{2}}$ of profile 1 in the above formula:

$\begin{array}{l}{\text{DO}}_{\text{2}}\text{= 3}\text{.83× 18}\text{.46× 10}\hfill \\ {\text{DO}}_{\text{2}}\text{= 707}{\text{.018 mL O}}_{\text{2}}\text{/min}\hfill \\ \hfill \end{array}$

Putting the values of Q_T as 3.1 and ${\text{CaO}}_{\text{2}}$ as 18.27 for calculating the ${\text{DO}}_{\text{2}}$ of profile 2:

$\begin{array}{l}{\text{DO}}_{\text{2}}\text{= 3}\text{.1× 18}\text{.27× 10}\hfill \\ {\text{DO}}_{\text{2}}\text{= 566}{\text{.37 mL O}}_{\text{2}}\text{/min}\hfill \end{array}$

The difference between ${\text{DO}}_{\text{2}}$ of profile 1 and profile 2 is:

$\begin{array}{l}{\text{DO}}_{\text{2}}\left(\text{ profile 1}\right)─{\text{DO}}_{\text{2}}\left(\text{profile 2}\right)\hfill \\ \text{= 707}{\text{.018 mL O}}_{\text{2}}\text{/min}─\text{566}{\text{.37 mL O}}_{\text{2}}\text{/min}\hfill \\ \text{=140}{\text{.648 mL O}}_{\text{2}}\text{/min}\hfill \end{array}$

The best PEEP was 5 cm ${\text{H}}_{\text{2}}\text{O}$, which causes 140.648 mL O₂/min more than a PEEP of 10 cm ${\text{H}}_{\text{2}}\text{O}$.

Conclusion

Therefore, it can be concluded that the best PEEP was 5 cm ${\text{H}}_{\text{2}}\text{O}$, which causes 140.648 mL O₂/min more than a PEEP of 10 cm ${\text{H}}_{\text{2}}\text{O}$.