Measurement of cardiac output by Alveolar gas exchange - Inert gas rebreathing Carlo Capelli SS.MM. Università degli Studi di Verona
Soluble Inert gas 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 (C 2 H 2 ), N 2 O 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 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 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 O 2 and diffusion capacity if CO is also added in the mixture It can be applied to healthy adults and children
Model of rebreathing Single compartment of the total volume V s,tot Complete mixing of all the gases in volume composed by lung (V L ), dead (subject s and apparatus, V ds, V ds,rb ) and full bag (Vr b ) 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
Theory.1 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: δf s,tot V s,tot δt = F s,tot α B Q c P B 47 760 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
Mono exponential decay y 1.0 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 Some basic Math! First order differential equation describes the fall of partial pressure/fraction of the gas during rebreathing dy dt = - y τ y = e t/τ 0.1 0.0 0 20 40 60 80 100 120 Time (sec.) y = e kt
F s,tot (t) = F s,tot0 e ( Theory.2 Assuming the total volume (V s,tot ) constant and integrating: Q c α B (P B 47) ) t V s,tot 760 Where F s,tot0 is the concentration at the beginning of the rebreathing If Q L and V s,tot are constant, the fall of F s,tot with time is described by a mono-exponential decay
Teoria.3 If we describe the fall of F s,tot in semi-logarithmic form: ln (F s,tot (t)) = int - Q c α B (P B - 47) V s,tot 760 t The slope depends on Q L By estimating the slope β, we can also obtain Q L Q c = - β V s,tot 760 α B (P B - 47)
Semi-logarithmic transformation 1.00 Time (sec) 0 20 40 60 80 100 120 ln( y) = t /τ 0.10 ln y 0.01 0.00
Decay of N 2 O [N 2 O] as a function of time during rebreathing 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.
Theory.4 We have at least 4 problems, and we need to solve them 1. V s,tot is not constant at all during rebreathing (V st shrinks) 2. We need V s,tot at time 0 (t 0 ). t 0 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 V s,tot 4. Mixing is not instantaneous and is incomplete
The old, faithful respiratory physiology still at work 1. Helium dilution. [He] 1 V s = [He] 2 (V s + CFR) CFR = V s ([He] 1 / [He] 2-1)
V s,tot at t 0 - inert insoluble gas V = F 0 i s,tot F i,eq V rb F i0 ; initial fraction of inert insoluble gas in the bag (SF 6 ) F i,eq : equilibrium fraction of the same gas backextrapolated to t = 0 V rb : bag volume
V rb = 2 l P B = 763 mmhg t a = 23 C RU = 24% F i 0 = 0.114% F i,eq = 0.053% Saturated PH 2 O at 23 C = 21.1 mm Hg Example 1 V s,tot (ATPS) = F 0 i V F rb (ATPS) i,eq V s,tot (ATPS) = 0.114 0.053 2 l = 4.30 l V s,tot (STPD) = 4.30 l 273 763 273+23 24 100 21,1 760 = 3.96 l
Tissue volume V t 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 ΔF s,tot Δt V s,tot + α t V t P B -47 760 = F s,tot α B Q c P B 47 760 Riarranging: F s,tot (t) = F s,tot0 e Q c α B t V s,tot 760 (P B 47) +V t α t
Inclusion of V t in the calcuations F s,tot (t) = F s,tot0 e Q c α B t V s,tot 760 (P B 47) +V t α t α t : Bunsen coefficient of gas solubility in the tissues V t is assumed equal to 600 ml (we can estimate using C 18 O) Q c is calculated as: Q c = - β V s,tot 760 + α t V t α B (P B - 47)
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) β is the slope computed analysing the expiratory values of: ln(f' s (t)) = ln F (t) F 0 s i 0 F i (t) F s F s (t): Normalised fraction of the soluble gas vs. t F s (t): Fraction of the soluble gas vs. t F i0 : Initial fraction of the insoluble gas in the bag F i (t): Fraction of the insoluble gas vs. t F s0 : Initial fraction of the soluble gas in the bag
Example 2 Q c = - β = ln(f ' s,2 β V s,tot 760 + α t V t α B (P B - 47) ' ) ln(f s,1 ) t 2 - t 1 ln(f s,2 ) = -0.223 ln(f s,1 ) = -0.164 t 2 = 14.2 sec t 1 = 9.1 sec β = 0.223 (0.164) (14.2-9.1)/60 = -0.694 / min
Example 2 (cont.) V s,tot = -3.96 l C 1 = 760/ (P B -47) = 1.06 C 2 = V t α = 0.600 0.407 = 0.244 l α b = 0.412 Qc = - (-0.694) 3.96 1.06 + 0.244 0.412 l/min = 7.45 l/min
Determination of V L Alveolar volume at the end of expiration (FRC) (actually it is between FRC and RV) V L BTPS = [V s,tot STPD - (V rb ATPS +V ds,rb ATPS ) C 1 ] C 2 +V ds BTPS C 1 : converting factor for ATPS --->STPD = 0.916 C 2 : converting factor for STPD ---> BTPS = 1.205 V L BTPS = [3.96 - (2 + 0.013) 0.916] 1.205-0.102 = 2.45 l
Determination of V O 2 O 2 uptake remains constant during rebreathing We assume that P A O 2 during rebreathing is > 100 mm Hg If V O 2 is constant: VO 2 = - β V s,tot (STPD)
Determination of V O 2 We have to take into account of: 1. Changes in V rb 2. Incomplete mixing The normalised slope β of the expiratory values of FO 2 is used for the calculations F ' O 2 (t) = ( FO 2 (t)-fo 2,ET (t)) FO I,eq F I (t) + FO (t) 2,ET
Example 4 - V O 2 F ' O 2 (t) = ( FO 2 (t)-fo 2,ET (t)) FO I,eq F I (t) + FO (t) 2,ET F O 2 (t) = Normalised FO 2 FO 2 (t) = FO 2 vs. t FO 2,ET = End tidal FO 2 before rebreathing F I,eq = Equilibrium fraction of the insoluble gas back extrapolated to t 0 F I (t) = Fraction of the insoluble gas vs. t
Determination of V O 2 Example 4 (cont.): β = F O2,2 ' ' F O2,1 t 1 t 2 F O 2,2 = 0.162 F O 2,1 = 0.181 t 2 = 14.2 sec t 1 = 9.1 sec β = 0.162 0.181 = 0.224 / min ( 14.2 9.1) / 60 VO 2 = -( 0.224) / min 3.96 l = 0.89 l/min
Open circuit C 2 H 2 uptake A non rebreathing technique that involves breathing a C 2 H 2 normoxic mixture and measuring C 2 H 2 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 % C 2 H 2, 5 % He, 21 % O 2, N 2 as balance for 20-24 breaths, expiratory fractions of He and C 2 H 2 are continuously measured (mass spectrometer) Advantages: maximal exercise, hypoxia, absence of stimulus on chemoreceptors
Open circuit C 2 H 2 uptake ln(f ' (t)) = ln C 2 H 2 F C 2 H 2 (t) F 0 He F He (t) F 0 C 2 H 2 This normalised C 2 H 2 fraction (corrected for mixing) is back estrapolated to breath 1 Q L = V E P ECO2 (P IC2 P H AC2 ) 2 H 2 λ P ACO2 P AC2 H 2 A= ET; λ: C 2 H 2 partition coefficient (BTPS)
Open circuit C 2 H 2 uptake The method works if the individual λ is determined λ varied from 0.55 to 0.95
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.
Determination of V t Back extrapolation of ln di F s (t) to t 0 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 F s,int at t 0 of ln F (t) F 0 s i F 0 s F i,eq V t = l V STPD s,tot ( 100 α t F s, int - 1) 760 P B - 47