HST.542J Solutions to Quiz 1

MASSACHUSETTS INSTITUTE OF TECHNOLOGY Departments of Electrical Engineering, Mechanical Engineering, and the Harvard-MIT Division of Health Sciences and Technology 6.022J/2.792J/BEH.371J/HST542J: Quantitative Physiology: Organ Transport Systems QUIZ 1 SOLUTIONS Problem 1 A. Draw normal P(t) waveforms for the left ventricle, left atrium, and aorta. Show two complete cardiac cycles, and use typical normal values for the pressures. Use the time axis provided in Figure 1.1a, and assume a heart rate of 60 bpm. C.O. = V̇O2 A − CV CO O2 2 = 363 ml O2/min 363 = 6.6 L/min = (200 − 14.5) ml O2/liter blood 55 So SV = 6.6 L/min = 110 cc/beat 60 beat/min (See Figure.) B. The cardiac output was measured using the Fick method. Oxygen uptake 363 ml O2 per minute Arterial oxygen content 200 ml O2 per liter of blood Mixed venous oxygen content 145 ml O2 per liter of blood Using this data together with your P(t) waveforms, draw the corresponding P-V loop for the LV. Assume an end-diastolic LV volume of 170 cc., and a LV “dead” volume of 15 cc for both systole and diastole. Draw linear systolic and diastolic P-V curves, and use the axes provided. See Figure. C. Correlate the following landmarks on the P-V loop with the appropriate points on the P(t) curves using the numeric labels below: a: begin LV contraction b: peak LV pressure c: begin LV filling d: end ejection e: begin LV ejection See Figure. D. “Ejection fraction” (EF) is defined as the percentage of the end-diastolic volume that is ejected during systole. What is the EF in this case? (Normal > 55%.) EF = 110 = 64.7% 170 2 E. A papillary muscle in the LV ruptures. (Assume that there are no functioning controls, and that the system has reached a new steady state.) The new arterial BP (systolic, diastolic, and mean) drops to 60% of its original value. New BP: 120 × 0.6 = 72 80 × 0.6 = 48 110 × 0.6 = 66 (i) Sketch two cardiac cycles showing the new P(t) waveforms, using the axes supplied in Figure 1.1b. Pay particular attention to the new amplitudes of the LV and LA pressures. Assume no change in the left ventricular end-diastolic pressure and volume. See sketches for P(t), P-V loop. Note high pressure in atrium at end-systole due to blood leaking past mitral valve. (ii) Sketch the new P-V loop on the same axes as part (B) above. Estimate the new stroke volume. New stroke volume ≈ 170 − 42 = 128 cc (iii) What is the new ejection fraction (using the definition in part D)? EF = 128 = 75.3% 170 (iv) Crudely approximate the stroke volume delivered to the aorta by making use of the Windkessel approximation. The pulse pressure had decreased from 120 − 80 = 40 mmHg to 72 − 48 = 24 mmHg (60% of prior) So, if we assume SV is proportional to pulse pressure, SV = 0.6 × 110 cc = 66 cc ↑ original SV (v) What is the “forward ejection fraction” (the percentage of the end-diastolic LV volume that is ejected into the aorta)? Forward EF = 66 = 38.8% 170 3 (vi) As a result of the papillary muscle rupture, a murmur appears. Indicate its temporal location on the time axis provided in Figure 1.1. It is heard throughout systole as a regurgitant jet enters the atrium. See Figure. Figure 1.1: a. 160 160 140 140 120 A0 Pressure (mmHg) 120 100 80 60 V 100 40 20 80 60 40 LA 20 0 0 1 2 1 Time (sec) 2 Time (sec) Heart Sound Amplitude Pressure (mmHg) b. 4 Figure 1.2: 200 180 LV Pressure (mmHg) 160 140 b 120 d 100 e 80 60 110 cc 40 128 cc 20 c a 0 20 40 60 80 100 120 140 160 180 200 LV Volume (cc) 5 Problem 2 We have used the lumped parameter model of the cardiovascular system that is shown in Figure 2.1. The following relationship was derived to relate cardiac output to the various model parameters (in operating region I): C.O. = � Cr � 0 − Pth C rS Pms − Pth − PPA D Rv + Ra CaC+aCv + 1 f C rD Figure 2.1: Lumped Parameter Model Part 1 Using this expression and/or graphical analysis explain the expected changes in: (a) cardiac output, (b) arterial blood pressure, and (c) pulse pressure that would result from the following interventions, assuming an uncontrolled CV system and a heart rate of 60 bpm. A. Increasing the peripheral resistance, Ra . B. Decreasing total blood volume. C. Increasing left ventricular contractility. D. Decreasing arterial capacitance, Ca , by a factor of two. E. Increasing the intra-thoracic pressure by 10 mmHg, and Pms by 8 mmHg by blowing into a balloon. Part 2 For each intervention above, sketch the expected qualitative changes in the CO/VR curves using the graphs below. 6 A. Increasing the peripheral resistance, Ra . ) 10 5 0 -5 Cardia c outp ut (n orm al Cardiac Output and Venous Return (L/min.) 15 Equilibrium point Ve no us ret urn ( no rm a l) 0 5 10 Right Atrial Pressure (mmHg) ABP = CO × Ra . There will be a slight decrease in CO, a large increase in ABP, and a change in the τ of ABP decay, but little change (a slight decrease) in pulse pressure. 7 B. Decreasing total blood volume. ) 10 5 0 -5 Cardia c outp ut (n orm al Cardiac Output and Venous Return (L/min.) 15 Equilibrium point Ve no us ret urn ( no rm a l) 0 5 10 Right Atrial Pressure (mmHg) Pms drops and the VR curve shifts to the left. CO drops. ABP drops proportionately because ABP = C.O. × R. Since HR does not change, pulse pressure also drops proportionately. 8 C. Increasing left ventricular contractility. ) 10 5 0 -5 Cardia c outp ut (n orm al Cardiac Output and Venous Return (L/min.) 15 Equilibrium point Ve no us ret urn ( no rm a l) 0 5 10 Right Atrial Pressure (mmHg) CO is not a function of C SL , so there is no change in CO, ABP, or PP. 9 D. Decreasing arterial capacitance, Ca , by a factor of two. ) 10 5 0 -5 Cardia c outp ut (n orm al Cardiac Output and Venous Return (L/min.) 15 Equilibrium point Ve no us ret urn ( no rm a l) 0 5 10 Right Atrial Pressure (mmHg) There is a slight increase in CO from 5.44 L/min to 5.96 L/min. Ca = 2 : VT − V0 4000 − 3200 800 = = = 7.84 102 Ca − Cv 2 + 100 2 7.84 − (−5) − [15 − (−5)] 20 10.84 7.84 + 5 − 2 = = 90.63 cc/sec = CO = � 2 � 1 .1196 .05 + .0196 + .05 .05 + 1 · 102 + 1×20 = 5.44 L/min Pms = Ca = 1 : 800 = 7.92 101 10.92 7.92 + 5 − 2 = CO = = 99.36 cc/sec = 5.96 L/min � 1 � 0.1099 .05 + 1 · 101 + .05 Pms = Mean ABP will rise proportionately. The pulse pressure, however, will double since Pulse Pressure ≈ SV/Ca 10 E. Increasing the intra-thoracic pressure by 10 mmHg, and Pms by 8 mmHg by blowing into a balloon. ) Cardia c outp ut (n orm al Cardiac Output and Venous Return (L/min.) 15 10 Equilibrium point 5 Ve no 0 -5 us ret urn ( no rm a l) 0 5 10 Right Atrial Pressure (mmHg) 15 Original CO by equation: 2 (7.8 + 5) − (15 + 5) 20 12.8 − 2 = = 5.4 L/min CO ≈ .05 + .02 + .05 .12 With new Pth , Pms : CO = 10.8 − 1 (15.8 − 5) − (15 − 5)0.1 = = 81.67 cc/sec = 4.90 L/min .12 .12 11 Table 1: Glossary of Symbols and Nominal Value for Model Parameters Symbol Definition Normal Value V stroke volume 96 cc heart rate 60/min. = 1/sec. T = TS + T D duration of heart cycle 1 sec. TS duration of systole .3 sec. TD duration of diastole .7 sec. C rD diastolic capacitance of RV 20 ml/mmHg C lD diastolic capacitance of LV 10 ml/mmHg C rS minimum systolic capacitance of RV 2 ml/mmHg C lS minimum systolic capacitance of LV .4 ml/mmHg r , Vl Vmax max “maximum” volumes, RV, LV 200 cc VT = V + V0 total volume of blood in peripheral vasculature 4000 ml V0 volume needed to fill peripheral vasculature without increasing pressure 3200 ml Ca arterial capacitance 2 ml/mmHg Cv venous capacitance 100 ml/mmHg Ra arterial resistance 1 mlHg/(ml/sec) Rv resistance to venous return .05 mmHg/(ml/sec) Pth mean intrathoracic pressure -5 mmHg PA0 pulmonary artery pressure (end-systolic) referenced to mean intrathoracic pressure 15 mmHg Pms mean systemic filling pressure (see text) 7.8 mmHg Pv peripheral venous pressure 6.1 mmHg f = 1 T 12 Figure 3.3: sphere is depolarized before repolarization begins. Assume the heart to be in the center of the spherical torso, and that all the assumptions underlying the dipole ECG theory are valid. A. Sketch the three orthogonal scalar waveforms Vx (t), Vy (t), and Vz (t) as defined in Figure 3.4 for one depolarization sequence. Label the time axis in terms of the radius of the spherical heart, a, and the velocity of propagation, v. [Note: try to be as quantitative as possible, but partial credit will be given for a qualitative answer.] 13 Figure 3.4: z x y 14 Symmetry leads to cancellation of the y and z components of the heart vector. So only Vx is non-zero.  0 . We know it is in the x-direction. Its magni­ First, calculate the equivalent heart vector, M  tude is the projection on i x of the individual components. Figure 3.5: h a Let h be the thickness of the shell. At the interface of depolarized and polarized tissue there is a circular boundary of radius a sin θ . An elemental area, d A, may be defined as d A = ha sin θ dα where α is the angle of rotation around the x-axis. Let m  be the elemental current dipole per unit area, and the dipole moment associated with d A will be �π m  = mha sin θ dα � − θ with the x-axis for all α. The projection of m  on i x will therefore m  makes an angle 2 be m x = mha sin2 θ dα The total net x-projection at a given θ would be � 2π Mx (t) = mha sin2 θ (t)dα = 2π mha sin2 θ 0 15 But v t a θ = ωt = So � Mx (t) = 2π mha sin 2 vt a � It is plotted in the figure below. 16 Vx(t) t 0 Vy(t) Vz(t) 17