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Measurement of cardiac output by

Alveolar gas exchange - Inert gas rebreathing

Carlo Capelli SS.MM.

Università degli Studi di Verona

Soluble Inert gas

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•  They dissolve and do not form bonds with haemoglobyn •  Their alveolar-capillary transfer is perfusion limited •  Therefore, their uptake is proportional to Q’L

•  Acetylene (C2H2), N2O and Freon

•  The most utilized methods are •  Breath holding •  Rebreathing: it is the most suitable method during sub

maximal and maximal exercise

Inert gas-Rebreathing

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•  It is based on •  1) the analysis of the exponential decrease of a soluble inert

gas concentration measured with a fast responding analyser •  2) the estimation of the volume at which the gas is taken up

by the blood. This volume is obtained through the simultaneous measurement of an inert non soluble gas

•  3) the correction for the volume of soluble gas that equilibrates with lung tissue

Why Inert gas-Rebreathing

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•  It easily provides a reliable disappearance curve of the soluble gas

•  It allows at the same time the continuous recording of lung volume during the maneuver by monitoring the dilution of an accompanying inert, not soluble gas

•  The maneuver lasts less than 15 s and it can be repeated after 3-5 minutes of wash out

•  It minimizes V’A/Q’ inequalities •  It allows simultaneous assessment of V’O2 and diffusion

capacity if CO is also added in the mixture •  It can be applied to healthy adults and children

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Model of rebreathing

•  Complete mixing of all the gases in volume composed by lung (VL), dead (subject’s and apparatus, Vds, Vds,rb) and full bag (Vrb)

•  Instantaneous alveolar-to-blood and alveolar-tissue equilibria of the soluble gas •  Constant Q’L during the maneuver •  No soluble gas in the mixed venous blood: no recirculation, sufficient washout •  No significant ventilation-to-perfusion mismatch

•  Single compartment of the total volume Vs,tot

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Theory.1

δFs,tot •Vs,totδt

= Fs,tot • αB • Qc • PB − 47

760

• Partial pressure of a soluble inert in the capillary blood leaving the lung is equal to its alveolar partial pressure

• Therefore, the alveolar-to-capillary uptake can be described as:

• The left hand describes the volume of gas that disappears from the alveoli crossing the barrier

• The right hand is the volume of gas that leaves the lung transported by the blood flow

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Mono exponential decay

0.0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1.0

0 20 40 60 80 100 120

y

Time (sec.)

dydt = - y

τ

y = e−t/τ

•  Some basic Math! •  First order differential equation

describes the fall of partial pressure/fraction of the gas during rebreathing

y = e−kt

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Theory.2

Fs,tot (t) = Fs,tot0

•e(− Qc• αB • (PB−47)

Vs,tot • 760)•t

• Assuming the total volume (Vs,tot) constant and integrating:

• Where Fs,tot0 is the concentration at the beginning of the

rebreathing •  If Q’L and Vs,tot are constant, the fall of Fs,tot with time is

described by a mono-exponential decay

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Teoria.3

ln (Fs,tot (t)) = int - Qc •αB • (PB - 47)

Vs,tot • 760

⎛

⎝⎜⎜

⎞

⎠⎟⎟

• t

•  If we describe the fall of Fs,tot in semi-logarithmic form:

• The slope depends on Q’L • By estimating the slope β, we can also obtain Q’L

Qc = - β •Vs,tot • 760

αB • (PB - 47)

⎛

⎝⎜⎜

⎞

⎠⎟⎟

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Semi-logarithmic transformation

0.00

0.01

0.10

1.00 0 20 40 60 80 100 120

ln y

Time (sec) ln( y)= −t /τ

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Decay of N2O

•  The initial fast drop is due to the rapid uptake from the tissue and blood

•  The gradual slower decay is due to the uptake of the gas form the blood flowing in the capillaries.

[N2O] as a function of time during rebreathing

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Theory.4

•  We have at least 4 problems, and we need to solve them

1. Vs,tot is not constant at all during rebreathing (Vst shrinks)

2. We need Vs,tot at time 0 (t0). t0 is set at half way of the first inspiration during rebreathing

3. A fraction of the soluble gas diffuses in the tissue and disappears from the alveolar volume. Its diffusion, equivalent lung volume (ELV) is larger than Vs,tot

4. Mixing is not instantaneous and is incomplete

The old, faithful respiratory physiology still at work

1.  Helium dilution.

[He]1 Vs = [He]2 (Vs + CFR)

CFR = Vs ([He]1/ [He]2 - 1)

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Vs,tot at t0 - inert insoluble gas

Vs,tot = Fi

0

Fi,eq

• Vrb

•  Fi0; initial fraction of inert

insoluble gas in the bag (SF6)

•  Fi,eq: equilibrium fraction of the same gas backextrapolated to t = 0

•  Vrb: bag volume

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Example 1 •  Vrb = 2 l •  PB = 763 mmHg •  ta = 23 °C •  RU = 24% •  Fi

0 = 0.114% •  Fi,eq = 0.053% •  Saturated PH2O at 23°C = 21.1 mm Hg

Vs,tot (ATPS) = Fi

0

Fi,eq

• Vrb (ATPS)

Vs,tot (ATPS) = 0.114

0.053 • 2 l = 4.30 l

Vs,tot (STPD) = 4.30 l• 273

273+23•

763− 24100

• 21,1

760= 3.96 l

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Tissue volume Vt • The soluble gas diffuses quickly into the lung • ELV is the total diffusing volume, therefore we must tak into

account also the diffusing volume of the tissue

ΔFs,totΔt Vs,tot +

αt•Vt• PB-47⎛

⎝⎜⎞

⎠⎟

760

⎛

⎝

⎜⎜⎜⎜⎜

⎞

⎠

⎟⎟⎟⎟⎟

= Fs,tot • αB • Qc • PB − 47

760

• Riarranging:

Fs,tot (t) = Fs,tot0

•e− Qc• αB

Vs,tot • 760

(PB−47)+Vt •α t

⎛

⎝

⎜⎜⎜⎜⎜

⎞

⎠

⎟⎟⎟⎟⎟

•t

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Inclusion of Vt in the calcuations

• αt: Bunsen coefficient of gas solubility in the tissues • Vt is assumed equal to 600 ml (we can estimate using C18O) • Q’c is calculated as:

Fs,tot (t) = Fs,tot0

•e− Qc• αB

Vs,tot • 760

(PB−47)+Vt •α t

⎛

⎝

⎜⎜⎜⎜⎜

⎞

⎠

⎟⎟⎟⎟⎟

•t

Qc = - β •Vs,tot • 760 + α t •Vt

αB • (PB - 47)

⎛

⎝⎜⎜

⎞

⎠⎟⎟

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Correction for not instantaneous mixing • Soluble gas concentration are corrected to account for the delayed and

incomplete mixing (and the shrinkage of the total volume)

ln(F's (t)) = ln

Fs (t) • Fi0

Fi(t) • Fs0

⎛

⎝⎜⎞

⎠⎟

•  β is the slope computed analysing the expiratory values of:

•  F’s(t): Normalised fraction of the soluble gas vs. t

•  Fs(t): Fraction of the soluble gas vs. t •  Fi

0: Initial fraction of the insoluble gas in the bag

•  Fi(t): Fraction of the insoluble gas vs. t Fs

0: Initial fraction of the soluble gas in the bag

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Example 2

Qc = - β • V

s,tot • 760 + α

t• V

t

αB• (P

B - 47)

⎛

⎝⎜⎞

⎠⎟

β = ln(Fs,2

' ) − ln(Fs,1' )

t2 - t1

ln(F’s,2) = -0.223 ln(F’s,1) = -0.164 t2 = 14.2 sec t1 = 9.1 sec

β = −0.223− (0.164)

(14.2-9.1)/60 = -0.694 / min

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Example 2 (cont.) Vs,tot = -3.96 l C1 = 760/ (PB -47) = 1.06 C2 = Vt • α = 0.600 • 0.407 = 0.244 l αb = 0.412

Q

c = - (-0.694) • 3.96 • 1.06 + 0.244

0.412 l/min = 7.45 l/min

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Determination of VL

• VLBTPS = [Vs,tot

STPD - (Vrb ATPS+Vds,rb

ATPS) • C1] • C2 +VdsBTPS

• Alveolar volume at the end of expiration (FRC) (actually it is between FRC and RV)

C1: converting factor for ATPS --->STPD = 0.916 C2: converting factor for STPD ---> BTPS = 1.205

• VLBTPS = [3.96 - (2 + 0.013) • 0.916] • 1.205 -0.102 = 2.45 l

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Determination of V’O2

•  O2 uptake remains constant during rebreathing

•  We assume that PAO2 during rebreathing is > 100 mm Hg

•  If V’O2 is constant:

VO2 = - β Vs,tot (STPD)

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Determination of V’O2

F'O2(t) = FO2(t)-FO2,ET (t)( )• FOI,eq

FI (t)+ FO2,ET (t)

•  We have to take into account of:

1.  Changes in Vrb 2.  Incomplete mixing •  The normalised slope β of

the expiratory values of FO2 is used for the calculations

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Example 4 - V’O2

F'O2(t) = FO2(t)-FO2,ET (t)( )• FOI,eq

FI (t)+ FO2,ET (t)

•  F’O2(t) = Normalised FO2 •  FO2(t) = FO2 vs. t •  FO2,ET = End tidal FO2 before

rebreathing •  FI,eq = Equilibrium fraction of

the insoluble gas back extrapolated to t0

•  FI(t) = Fraction of the insoluble gas vs. t

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Determination of V’O2 • Example 4 (cont.):

β =

FO2,2

' − FO2,1

'

t1− t

2

•  F’O2,2 = 0.162 •  F’O2,1 = 0.181 •  t2 = 14.2 sec •  t1 = 9.1 sec

β =

0.162− 0.18114.2− 9.1( ) / 60

= −0.224 / min

VO2 = - −0.224( ) / min•3.96 l = 0.89 l/min

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Open circuit C2H2 uptake

•  A non rebreathing technique that involves breathing a C2H2 normoxic mixture and measuring C2H2 open circuit uptake in a short-term, quasi-steady state

•  It acknowledges the problem caused by rapid recirculation and allows for it in the calculations

•  Subjects breaths through a 1-way valve a mixture of 1 % C2H2, 5 % He, 21 % O2, N2 as balance

•  for 20-24 breaths, expiratory fractions of He and C2H2 are continuously measured (mass spectrometer)

•  Advantages: maximal exercise, hypoxia, absence of stimulus on chemoreceptors

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Open circuit C2H2 uptake

ln(F 'C2H2 (t)) = lnFC2H2 (t)× F

0He

FHe (t)× F0C2H2

⎛

⎝⎜

⎞

⎠⎟

•  This normalised C2H2 fraction (corrected for mixing) is back estrapolated to breath 1

QL =VE × PECO2

× (PIC2H2− PAC2H2

)

λ × PACO2× PAC2H2

⎡

⎣⎢⎢

⎤

⎦⎥⎥

A= ET; λ: C2H2 partition coefficient (BTPS)

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Open circuit C2H2 uptake

•  The method works if the individual λ is determined

•  λ varied from 0.55 to 0.95

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References •  Barker RC, Hopkins SR, Kellogg N, Olfert IM, Brutsaert TD,

Gavin TP, Entin PL, Rice AJ, Wgner P. Measurement of cardiac ouptu during exercise by open-circuit acetylene uptake. J Appl Physiol 87: 1506-1512, 1999.

•  Sackner MA, Measurement of cardiac ouptu by alveoalr gas exchange, In: Fahri LE, Tenney SM, editors. Handbook of physiology, sec 3: the respiratory system. vol IV. Bethesda, MD: American Physiological Society; 1987. pp. 233-255.

•  Triebwasser JH, Johnson RL, Burpo RP, Campell JC, Reardon WC, Blomqvist CG Noninvasive detremination of cardiac output by a modifeid acetylene rebreathing procedure utilizing mass spectromete measurements. Aviat Space Environ Med 48: 203-209, 1977.

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Determination of Vt • Back extrapolation of ln di Fs (t) to t0 •  It is given as percent of the normalised initial, theoretical fraction

of the soluble gas before equilibrium with the tissues. • This corrects for impefect mixing and for tissue diffusion

•  Intercept Fs,int at t0 of

ln Fs (t)• Fi

0

Fs0 • Fi,eq

Vt = l

Vs,totSTPD

α t

• (100Fs, int

- 1) • 760

PB - 47