1
Lung gas composition and transfer analysis: O2 and CO2 diffusion coefficients and metabolic rates D. N. Ghista1, K. M. Loh2 & D. Ng3 1School of Mechanical and Production Engineering, Nanyang Technological University, Singapore 639798; email:mdnghista@ntu,edu.sg 2School of Engineering (Electronics), Nanyang Polytechnic, Singapore 3Department of Nuclear Medicine and PET, Singapore General Hospital, Singapore Abstract The primary function of the lung is to (i) oxygenate the blood and thereby provide oxygen to the cells for metabolization purposes, and (ii) to remove the collected CO2 from the pulmonary blood. Herein, we will analyze the compositions of the inspired and expired air per breath, and from there compute the O2 consumption and CO2 production rates. Next, we derive expressions for diffusion coefficients
2OD and 2COD in terms of the evaluated cardiac output,
O2 and CO2 concentrations in arterial and venous blood, alveolar and blood O2 and CO2 partial pressures. We then take up a typical case study, and demonstrate the computation of
2OD and2COD , to represent the lung-performance capability
to oxygenate the blood.
2
1 Introduction The lung functional performance is characterized by (i) its ventilatory capacity, to bring air (and hence O2) into the alveoli, and (ii) its capacity to transfer O2 and CO2 into and from the pulmonary capillary bed. Hence, the O2 and CO2 diffusion coefficients as well as the O2 consumption rate and the CO2 production rate represent the lung-performance indices. 2. Lung air composition analysis (and 2O consumption and
2CO production rates) We carry out a mass balance analysis, involving:
(i) compositions of air breathed in and out (ii) consumption or losses of O2, CO2 and H2O.
The Table 1 below provides clinical data on partial pressures and volumes of
2N , 2O , 2CO and 2H O of atmospheric air breathed in and expired in one breath cycle. The monitored breathing rate (BR) = 12 breaths/min, and we assume PH2O at 37οC = 47 mmHg.
Table 1: Inspired air composition and partial pressures.
Atmospheric Air Expired Air Respiratory gases mmHg ml/% mmHg ml/%
N2 597 393.1 78.55%
566 393.1 74.5%
O2 159 104.2 20.84%
120 80.6 15.7%
CO2 0.3 0.2 0.04%
27 19.1 3.6%
H2O 3.7 2.5 0.49%
47 32.6 6.2%
Total 760 500 100%
760 525.3 100%
It can be noted that the expired air volume exceeds the inspired air volume for this particular breath cycle. The H2O loss of 30.1ml (=32.6 ml – 2.5 ml) contributes the major portion of this difference.
3
2.1 Calculation of 2O consumption-rate and CO2 production-rate We now determine the O2 consumption rate and CO2 production rates from the inspired and expired gases Assuming the patient breathes at 12 times per minute we have
2O Consumption Rate = (Inspired 2O – Expired 2O ) × 12 = (104.2–80.6) × 12
= 283.2 ml/min
2CO Production Rate = (Expired 2CO – Inspired 2CO ) × 12 = (19.1–0.2) × 12 = 226.8 ml/min
The amount of water vapor in the humidified expired air amounts to 6.2% of the expired air (compared to 0.49% of the dry inspired air) corresponding to the partial-pressure ratio of water vapor in the expired air (= 47/760). The volume of the dry expired air =(525.3–32.6) ml=492.7 ml. Now, assume that out of 500 ml of inspired air, the dead space air volume (not taking part in gas transfer process) is 150 ml and the alveolar air volume is 350 ml. We next compute the dead space air volume composition.
Fig. 1: Dead-space volume
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2.2 Dead space air composition The clinical data of expired air composition is:
2
2
2
2
N = 393.1 ml
O = 83.36 ml
CO = 16.87 ml
H O = 34.15 ml
Total 527.49 ml
Now, the dead-space air will be made up of (i) a dry air portion from the inspired air (assumed to be =141 ml), plus (ii) the water vapor taken up by the dry air (estimated to be =9 ml) since the expired air portion of 141 ml will not have under gone O2 and CO2 transfer, it’s composition is the same as that of inspired air:
2N = 111 ml (78.55%), 2O = 29.40 ml (20.84%), 2CO = 0.06 ml (0.04%),
2H O = 0.69 ml (0.49%). When this inspired air (in the dead space) of 141 ml is fully humidified, it will take up a further X ml of 2H O vapor, in the ratio of the partial-pressures, as:
47
0.0659141 713
X= =
X = 0.0659×141 = 9.29 ml∴ of H O2 vapor (which is close to our estimate).
So, by adding 9.29 ml of H O2 vapor to 0.69 ml of water vapor in the inspired air volume of 141 ml, the total water vapor in the dead-space air is 9.98 ml. The humified dead-space air composition will be:
N = 111.00 ml (= 73.78%)2O = 29.40 ml (= 19.55%)2CO = 0.06 ml (= 0.04%)2H O = 9.98 ml (= 6.63%)2Total = 150.44 ml
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2.3 Alveolar air composition and partial pressures We can now compute the alveolar air composition, by subtracting the dead-space air from the expired air. These values are tabulated in column 4 of the Table below. Expired Air
( ml ) Dead Space Air ( ml )
Alveolar Air ( ml )
Alveolar Air Partial Pressure
(mmHg) N2 393.1 111.00 282.1 569.41 O2 80.53 29.40 51.13 103.21
CO2 19.12 0.06 19.06 38.47 H2O 34.21 9.98 24.23 48.91 Total 526.96 150.44 376.52 760
Finally, we compute the partial pressure of O2 and CO2 (as well as of N2 and H2O), so that we can determine next the diffusion coefficients of O2 and CO2 based on the monitoring of arterial and venous blood concentrations. These values are tabulated in column 5 of the above table.
Fig. 2: Schematic of blood-gas concentration in the Pulmonary Capillary.
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3. Lung gas-exchange model & parametric analysis 3.1 Expressions for
2OD and 2COD
The gas exchange between the alveolar air and pulmonary capillary blood is represented by the following 2O and 2CO conservation equations. (Fig. 2):
22 2VE VE AE AE
OO OQ C Q C V°
= + (from the alveolar air to capillary blood)
2
2 2 2 2( ) ;O
avcapAE AE AE
O O O OQ C P D P P= + ∆ = , (1)
in which 2
capOP =
2
PRBOP ( 2O concentration of the pre oxygenated blood)
22 2
VE VE AE AECOCO COQ C Q C V
°= −
2
2 2 2 2( ) ;CO
avcapAE AE VE
CO CO CO COQ C P D P P= − ∆ = , (2)
in which
2
capCOP =
2
PRBCOP ( 2CO concentration of the pre oxygenated blood),
wherein (i) ABQ and VBQ are arterial and venous blood flow-rates;
(at venous end) (at arterial end) ,AB VE VB AEQ Q Q Q= =
(ii) 2
alOP and
2
capOP are the alveolar and capillary 2O partial pressures
(iii) 2 2
2and are the alveolar and capillary partial pressure al capCO COP P CO
(iv) 2OD and
2COD are the 2O and 2CO diffusion coefficients
(v) 2
2 2 average of ( ) over the capillary length;O
avcapal
O OP P P= − 2
2 2average over the capillary length of CO al cap
av CO OP P P= − Now we can equate the arterial and venous blood flow rates, as
ABQ = VBQ =Q=(SV)/(EP) / 60CO SV, EP and CO being the stroke volume, ejection period and cardiac output, respectively. Hence the above equations can be rewritten as:
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(vi) 2OV is the 2O transfer rate from alveolar air to capillary blood
(= 2O consumption rate), 2COV is the 2CO transfer rate from capillary blood to
alveolar air. From eqn. (1):
2
2
2 2 2 2 2 2
2 2 2
( ) ;
( )
O AEav
Oav
capVE VE AB ABO O O O O O
VB ABO O O
PQ C Q C P D P P
QC QC P D
== + =
= +
2 2
2 2 2 2( ) ( )
( ) ( )
AB VB
O Oav av
VE AEO O O OQ C C Q C C
P P=
− −=
2OD . (3)
From eqn. (2):
2
2 2( )
( )
VE
COav
ABCO COQ C C
P
−=
2COD , (4)
wherein (i) Q,
2
VEOC and
2
AEOC ,
2
VECOC and
2
AECOC can be monitored because
2 2 2 2 2 2 2 2 and = and and and = and AB AB VB VBVE VE AE AE
O CO O CO O CO O COC C C C C C C C
(ii) 2OD and
2COD (eqns. (3) & (4)) represent the lung gas-exchange
parameters. Now from eqns. (3) and (4), if we want to evaluate the diffusion coefficients
2OD and2COD , we need to also express
2
alOP ,
2
capOP and
2
alCOP ,
2
capCOP in terms of
monitorable quantities. In this regard,
2
2
2 2 2 2 2 2
2 2 2
( ) ;
( )
CO AEav
COav
capVE VE AE AE VBCO CO CO CO CO CO
VE AEO CO CO
Q C Q C P D P P P
QC QC P D
= − = =
= −
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(i) Alveolar 2
alOP can be expressed in terms of V
°(the ventilation rate) and
2OV°
(the 2O consumption rate) as Fig. 3 :
O2
m
1V V V
e
⎡ ⎤⎛ ⎞⎢ ⎥⎜ ⎟⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦
° ° °−
= −
⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦
2
2alO 1
k
P k , (5)
where mV is the maximum ventilation rate and 2OV (the 2O consumption rate or
absorption rate from the alveoli)= 2 2O O )( AB VBQ C C− . Eqn. (5) implies that as
mV V⎛ ⎞⎜ ⎟⎝ ⎠
increases, (the exponential term decreases, and) 2O
alP increases (as in
Fig. 3), and as 2OV increases
2OalP decreases (as in Fig. 3).
(ii) Alveolar 2
alCOP can be expressed in terms of V
° and 2OV
° as in Fig. 4.
CO2
V
V mV
e
°
°
⎡ ⎤⎛ ⎞⎢ ⎥⎜ ⎟⎢ ⎥⎜ ⎟⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦
°−
=4
alCO 32
k
P k , (6)
where2CO
°V (the 2CO production rate or excretion rate from the blood)
=2 2CO CO )( VB ABQ C C− . This equation implies that as mV V increases,
2COalP
decreases; also, as 2COV increases (the exponential term decreases, and hence)
2COalP increases.
(iii) Blood
2OP can be obtained in terms of blood 2CO , from the 2O
disassociation curve (providing concentrations in arterial and venous blood), is represented in Fig. 5, as:
O2
O22 2
5
O O(1 )
mP
kPmC C e
−
= − , or *
O22
*O 1
PC e
−= − 5k
, (7)
9
where
• 2O
mC and 2O
mP are the maximum values of blood 2O partial pressure
• *2 2 2CO CO COm=
• 2 2
*O O O2
mP P P=
(iv) Blood
2COP can be obtained in terms of 2COC , from the 2CO
disassociation curve or 2CO concentration in arterial and venous blood can be represented as per Fig. 6 as:
( )CO CO2 22 2
CO O(1 )
mP6k PmC C e−
= −
( )CO CO2 22
*2*or, 1 1
mCO
k Pk P P 66COC e e
−−== − − (8)
3.2 Alveolar 2 2O and CO partial pressure expressions
Now, let us refer eqn. (4) for the 2
alOP partial pressure curve (Fig. 3), represented
by the eqn.:
2
2O 12 O
m
Vk VVal
1P k e
°
°
⎡ ⎤⎛ ⎞⎢ ⎥⎜ ⎟⎢ ⎥⎜ ⎟⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦
°−
= −
⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦
*
21
2 Ok V V
1k e
° °⎡ ⎤⎢ ⎥⎢ ⎥⎣ ⎦
−
= −
⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦
, where *
V°
= m
V
V
°
° (9)
wherein V is the alveolar ventilation rate (in liters/min), mV°
is the maximum
ventilation rate (= 50 liters/min) and 2O
V is the 2O consumption rate (in
liters/min). Herein, the coefficients 1k and 2k can be determined by having this
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equation match the Fig. 3 data. Note, in this equation, when 0V°= ,
2O 0alP =
from the equation ,which satisfies the data.
Now for 2OV =0.25 liters/min, when *
V°
= m
V
V
°
° =0.5, 2O
alP =140 mmHg . Hence ,
0.520.25140 1 (1 )
22
k k1 1k e k e
⎡ ⎤⎢ ⎥⎣ ⎦
−−= − = −
⎡ ⎤⎢ ⎥⎢ ⎥⎣ ⎦
. (10)
Also, when 2OV =1 / minl , *
V°
=0.3 / minl , 2
100alOP = mmHg. Hence
0.30.31100 1 (1 )
22
k k1 1k e k e
⎡ ⎤⎢ ⎥⎣ ⎦
−−= − = −
⎡ ⎤⎢ ⎥⎢ ⎥⎣ ⎦
. (11)
From eqns. (10) and (11), we get:
2 2
0.3 0.3(1 )140 1
100 (1 ) 1
2 2
2 2
k k1
k k1
k e e
k e e
− −
− −− −
= =− −
0.3 2
2 0.3, 40 , so that
140 140 100 100
or 100 140 4.18
2 2
2 2
k k
k k2k
e e
e e min/l
− −
− −
∴ − = −
= + = (12)
Upon substituting 4.18 min/l2k = into eqn. (10) we obtain:
(2 4.18) that140 (1 ), so 140 mmHg1 1k e k− ×= − ≈ (13) Hence, the
2
alOP curve can be represented by:
*
2
2
4.18
O ,140 1OV V
alP e
° °⎡ ⎤⎢ ⎥⎢ ⎥⎣ ⎦
−
= −
⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦
(14)
11
wherein 2OV =Q2 2
)( AB VBO OC C− and
*
50 / minV V l°
=
Now, let us look at the
2COalP expression:
2 2
2
*
CO
CO COVk V k V V4 4V mal
3 3P k e k e
⎡ ⎤⎛ ⎞⎡ ⎤⎢ ⎥⎜ ⎟⎢ ⎥⎢ ⎥⎜ ⎟⎢ ⎥⎢ ⎥⎜ ⎟
⎝ ⎠⎣ ⎦ ⎣ ⎦
° ° °− −°
= = ,
We note from Fig. 4 that for 2 0.2 / minCOV l°
= and 0.2mV°
= , 2CO
alP =12. Hence, from the above eqn., we get: 12 4-k
3k e= (15)
Also, for 2 0.8 /OV l min= and 0.2mV°
= , 2CO
alP = 62 mmHg. Hence
40.200.80
36244
kk
3k e k e⎡ ⎤⎢ ⎥⎣ ⎦
−−= = (16)
From eqn.s (15) and (16), we get:
4
23
, so that
12
62
12 2ln 2.46
62 3
44
4
k k
k
4
ee
e
k k4
−−= =−
= − =⎛ ⎞⎜ ⎟⎝ ⎠
(17)
Substituting 2.464k = into eqn. (16), we obtain :
2.464 ,62 114.683 3k e k
−
∴= = (18) Hence, the
2
alCOP curve can be represented as
22.46
CO2114.68
COV V
V malP e
°°
°
⎡ ⎤⎛ ⎞⎢ ⎥⎜ ⎟⎢ ⎥⎜ ⎟⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦
−
= (19)
wherein *
50 l/minV V°
= and 2COV =Q2 2
)( VB ABCO COC C−
12
Fig. 3: Effect on alveolar 2OP of (i) alveolar ventilation, and (ii) rate of Oxygen
absorption from alveolar 2OP or 2O consumption rate [from Guyton
(1971), p. 476].
Fig. 4: Effect on alveolar 2COP of alveolar ventilation and rate of carbon dioxide
excretion from the blood or 2CO production rate [from Guyton (1971), p. 476].
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3.3 Arterial and venous 2 2O and CO partial pressure expressions
We now need to express 2O
ABP and 2CO
VBP in terms of 2O
ABC and 2CO
VBC .
So that let us look at the 2O disassociation curve, as shown in Fig. 5.
|max2 22 2O |max 1 5 O Ok P P
OC C e⎡ ⎤⎣ ⎦= −⎡ ⎤
⎢ ⎥⎣ ⎦-
,
*2*
2or, 1 5 Ok P
OC e−
= − , (20)
where 2
O* 2O O |max2
CC C= , 2*
2|max2
OO
O
PP
P= .
From Fig. 5, at 2
*OP =
40 mmHg0.29
140 mmHg= (for normal venous blood), and
2
*OC
150.75
20= = .
Hence from eqn. (20):
0.290.75 1
4.78.
5k
5
e
k
−= −
∴ = (21)
Also, 2
* 95 mmHg0.68
140 mmHgOP = = (for normal arterial blood), and
*2
190.95
20OC = = .
Hence from (20):
0.68 , or0.95 1 4.45ke k5−= − = . (22)
So, we take the average value of 5k :
(4.78+4.4)= 4.59
25k∴ = . (23)
Then the 2O disassociation curve is given by:
14
2
2 2
O4.59140
O O 0.2 1
P
BC C e
⎡ ⎤⎢ ⎥⎢ ⎥⎣ ⎦
−
= = −
⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦
, (24)
and
22 2
OO O
140 0.2 0.2ln 30.5 ln
4.59 0.2 0.2B BPC C
= =− −
⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦
. (25)
Finally, we look at the 2CO disassociation curve
2
|max22
2*
|max2 22
/ max
*
(1 ),
or, 1 1 .
CO
CO
CO
CO CO
PP
CO
PP P
CO
C e
C e e
−
−−
= −
= − = −
6
2
66
k
CO
kk
C (26)
Based on Fig. 6, when 2
* 20 mmHg0.14
140 mmHgCOP = = , 2
* 380.475
80COC = = , so that
,0.140.475 1 4.6066
ke k−= − = (27)
when 2
*CO
70 mmHg0.5
140 mmHgP = = ,
2
* 600.75
80COC = = , so that
0.5 ,0.75 1 2.7766
ke k−= − = . (28) So we take the average value of 6k :
(4.60+2.77)= 3.69
26k = (29)
Then the CO2 concentration is given (from eqns. 26-29) by:
CO2
2 2
3.69140
CO CO 0.8 1
P
BC C e
⎡ ⎤⎢ ⎥⎢ ⎥⎣ ⎦
−
= = −⎡ ⎤⎢ ⎥⎢ ⎥⎣ ⎦
(30)
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Figure 5: O2 dissociation curves, showing the total oxygen in each 100 ml of normal blood, the portion dissolved in the water of the blood [from Guyton (1971), p. 485].
Figure 6: The carbon dioxide dissociation curve [from Guyton (1971), p. 491].
17
3.4 Sequential procedure to compute 2OD and
2COD
(1) We first monitor: ( ), ( )V t V t , SV(stroke volume), EP(cardiac ejection
period), 2O
VBC , 2O
ABC , 2CO
VBC and 2CO
ABC ( 2O and 2CO concentrations
in pre oxygenated and post oxygenated blood). (2) We substitute the values of
2OABC (=
2OVEC ) and
2OVBC (=
2OAEC ) into eqn.
(3), and the values of 2CO
ABC (=2CO
VEC ) and 2CO
VBC (=2CO
AEC ) into eqn. (4).
(3) We next determine:
Q= SV/ejection period, (33)
2O ( )V t =Q2 2O O( )VBABC C− , (34)
2CO ( )V t =Q2 2CO CO )( ABVBC C− . (35)
(4) We then substitute the expressions for 2O ( )V t and
2CO ( )V t into the
eqns. for 2O
alP (eqn. (14)) and 2CO
alP (eqn. (19)).
(5) We substitute the monitored values of
2OVBC (=
2OAEC ) and
2COVBC
(=2CO
AEC ) into eqns. (25) and (32), to obtain the values of 2O
AEP and
2COAEP .
(6) Now, in order to determine the values of the lung gas-exchange
parameters 2OD and
2COD , we substitute into eqns. (3) and (4) for Q
from eqn. (33), 2O
alP from (14), 2CO
alP from eqn. (19), 2O
VBP from eqn.
(26), and 2CO
VBP from eqn. (32) .
3.5 Determining 2OD and
2COD Fig. 7 illustrates the variation of 2OP (=
2 2 2 2O O O Ocapal al ABP P P P− = − ) along the
length (l) of the capillary bed.
18
2 2
2
.
*( ) . max
*Let
Then we can express:
(36)OOOf l
l l lm
P P
=
=
( )2 2 22 2
1
max max0
* *( ) F .
Then,
(37)O OOO Of l dlP P Pav
⎛ ⎞⎜ ⎟ =∫⎜ ⎟⎝ ⎠
=
Based on data [3], since 2 12 mmHgOavP = for 2O
max 65 mmHgP = , we have
2F 0.185.O =
We can similarly determine the average value of 2CO
avP from Fig. 8 as:
2 2
2
.
*( ). max
*Let
Then, we can represent figure 8 as:
(38)COCOOf l
l l lm
P P
=
=
( )2 2 22 2
1
max max0
* *( ) F .
Then,
(39)O COCOCO COf l dlP P Pav
⎛ ⎞⎜ ⎟ =∫⎜ ⎟⎝ ⎠
=
Based on data [3], since 2CO 0.5 mmHgavP = for 2COmax 5 mmHgP = , we have
2F 0.1.CO =
19
From the 2OavP and 2CO
avP expressions, we can determine the O2 consumption and the CO2 production rates, as follows:
2 2222 2 2
Total consumed2AB VBO OO
OO O Oav avav
Q C C
P P
O VD
P
−⎛ ⎞⎜ ⎟⎝ ⎠= = = (40)
2 2222 2 2
Total C produced2VB ABCO COCO
COCO CO COav avav
Q C C
P P
O VD
P
−⎛ ⎞⎜ ⎟⎝ ⎠= = = . (41)
4. Case Studies (A) We monitor the partial pressures blood concentrations of O2 and CO2 as:
2 20.13AE VB
O OC C= = , 2 2
0.18VE ABO OC C= = ,
2 20.525,AE VB
CO COC C= =
2 20.485VE AB
CO COC C= =
From eqn. (26), we obtain:
22
0.2 0.230.5 ln 30.5 ln
0.2 0.2 0.13
32.02 mmHg (42)
VBO VB
OP
C= =
− −
=
⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥⎣ ⎦⎢ ⎥⎣ ⎦
From eqn. (32), we obtain:
We now also monitor Q=5 l/min, *
0.1V°
= and 5 /minV l= : Then from eqn. (34):
2( )OV t =Q ( )2 2
AB VBO OC C− , so that from the above data,
2
0.8 0.837.94 ln 37.94 ln
0.8 0.8 0.5252
40.51 mmHg (43)
VBCO
VBPCO C= =
− −
=
⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥⎣ ⎦⎢ ⎥⎣ ⎦
20
2( )OV t =5000 × 0.05=250 mlO2/min consumption rate (44)
From eqn. (35):
2( )COV t =Q ( )2 2
VB ABCO COC C− =5000(0.04)
= 200 mlCO2 /min production rate. (45)
Now, from eqn. (14):
For *°
= 0.1V and 2 2O O :0.25 , we obtain alV l P
°=
2
2
*4.18
4.18 0.1 0.25
(46)140 1
140 1 113.7 mmHg
V V Oal
OP e
e
⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦
⎡ ⎤⎣ ⎦
° °−
−
= −
= − =
⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦
⎡ ⎤⎣ ⎦
From eqn. (19), for *°
= 0.1V and 2 2CO CO :0.20 l, we obtain alV P
°=
[ ]
*
CO2
2
2.192.19 0.1 0.2
CO 107.18 107.18
35.86 mmHg (47)
V V
alP e e
° °⎡ ⎤⎢ ⎥−⎢ ⎥ −⎣ ⎦= =
=
Now, we can evaluate the diffusion coefficients: From eqns. (3, 37, 42 and 46):
2 2
2
( )
5000(0.18 0.13)16.84 mlO /min/mmHg. (48)2(113.7 31.2) 0.18
AB VBO O
Oav
Q C C
P
−=
−= =
− ×
2OD
21
From eqn. (4):
2 22 2
CO COCO CO
( )
5000(0.04)430.11 m CO /min/mmHg. (49)2(40.51 35.86) 0.1
VB AB
av
Q C C
P
l
−=
= =− ×
D
(B) Alternately, we derive data from:
(i) the inspired and expired air analysis (such as that carried out in section 2.3):
O2 consumption rate = 283.2 ml/min, CO2 production rate = 226.8 ml/min,
2O 103.03 mmHgalP = and2CO 38.41 mmHgalP =
and (ii) venous blood gas analysis:
2O 0.13VBC = ,2CO 0.548VBC =
Then, as per eqn. (42),
2O 31.2 mmHgVBP = (50)
corresponding to 2
0.13VBOC = and, as per eqn. (43):
2
0.8 0.837.94 ln 37.94 ln
0.8 0.8 0.5482
43.84 mmHg (51)
VBCO
VBPCO C= =
− −
=
⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥⎣ ⎦⎢ ⎥⎣ ⎦
We obtain, from air composition analysis, that 2
283.3 ml/min( )OV t = (52)
and 2CO 226.8 ml/min.( )V t = (53)
22
Hence,
2
2 2O
283.221.90 mlO /min/mmHg, (54)2(103.03 31.2) 0.18
OOav
V
P=
= =− ×
D
and
2
2 2
COCO CO
226.8417.68 mlCO /min/mmHg. (55)2(43.84 38.41) 0.1
av
V
P=
= =− ×
D
The advantage of this method (B) over (A) is that it does not require monitoring of the cardiac output, and is hence simpler to implement clinically.
23
Figure. 7: Uptake of oxygen by the pulmonary capillary blood. (The curve in this figure was constructed from data in Mihorn and Pulley:Biophys. J., 8:337,1968). [from Guyton (1971), p. 434.]
Figure. 8: Diffusion of carbon dioxide from the pulmonary blood into the alveolus. (The curve in this figure was constructed from data in Mihorn and Pulley:Biophys. J., 8:337,1968). [From Guyton (1971), p. 435.]
24
References [1] A. C. Guyton, M.D., Text Book of Medical Physiology, 8th edition, HBJ I.E.
Saunders, ISBN 0-7216-3994-1, pp.422-442, 1991. [2] J. H. Comroe, Physiology of Respiration, 2nd edition, Year Book Medical
Publishers Incorporated, ISBN 0-8251-1827-9, pp. 268-269, 1977. [3] D. O. Conney, Biomedical Engineering Principles, Marcel Dekker Inc.,
ISBN 0-8247-6347-5, pp. 248-251 342-397, 1976.
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