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Modeling Blood Flow in the Cardiovascular System

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Presentation on theme: "Modeling Blood Flow in the Cardiovascular System"β€” Presentation transcript:

1 Modeling Blood Flow in the Cardiovascular System
Mette S Olufsen MA 432 Spring 2017 Department of Mathematics, North Carolina State University NC STATE University

2 Compartment Model

3 Compartment Model π‘ž= 𝑝 1 βˆ’π‘ 𝑅 1 π‘‰βˆ’ 𝑉 𝑑 =𝐢 π‘βˆ’ 𝑝 𝑒π‘₯𝑑 =𝐢𝑝 q V p p1 pext
π‘ž= 𝑝 1 βˆ’π‘ 𝑅 1 π‘‰βˆ’ 𝑉 𝑑 =𝐢 π‘βˆ’ 𝑝 𝑒π‘₯𝑑 =𝐢𝑝 Assuming that 𝑝 𝑒π‘₯𝑑 =0 Dead space volume: 𝑉 𝑑

4 Solution Total volume Flow where
𝑉 π‘‘π‘œπ‘‘ = 𝑉 𝑆𝐴 + 𝑉 𝑆𝑉 + 𝑉 𝑃𝐴 + 𝑉 𝑃𝑉 = 𝑉 0 (constant) 𝑄= 𝑉 0 𝑇 𝑆𝐴 + 𝑇 𝑆𝑉 + 𝑇 𝑃𝐴 + 𝑇 𝑃𝑉 𝑇 𝑆𝐴 = 𝐢 𝑆𝐴 𝐾 𝑅 + 𝐢 𝑆𝐴 𝑅 𝑆 𝑇 𝑆𝑉 = 𝐢 𝑆𝑉 𝐾 𝑅 𝑇 𝑃𝐴 = 𝐢 𝑃𝐴 𝐾 𝑅 + 𝐢 𝑃𝐴 𝑅 𝑃 𝑇 𝑃𝑉 = 𝐢 𝑃𝑉 𝐾 𝐿

5 Solution Flow Volume 𝑄= 𝑉 0 𝑇 𝑆𝐴 + 𝑇 𝑆𝑉 + 𝑇 𝑃𝐴 + 𝑇 𝑃𝑉
𝑄= 𝑉 0 𝑇 𝑆𝐴 + 𝑇 𝑆𝑉 + 𝑇 𝑃𝐴 + 𝑇 𝑃𝑉 𝑉 𝑖 = 𝑇 𝑖 𝑄 = 𝑇 𝑖 𝑉 0 𝑇 𝑆𝐴 + 𝑇 𝑆𝑉 + 𝑇 𝑃𝐴 + 𝑇 𝑃𝑉 𝑇 𝑆𝐴 = 𝐢 𝑆𝐴 𝐾 𝑅 + 𝐢 𝑆𝐴 𝑅 𝑆 , 𝑇 𝑃𝐴 = 𝐢 𝑃𝐴 𝐾 𝑅 + 𝐢 𝑃𝐴 𝑅 𝑃 , 𝑇 𝑃𝑉 = 𝐢 𝑃𝑉 𝐾 𝐿 , 𝑇 𝑆𝑉 = 𝐢 𝑆𝑉 𝐾 𝑅

6 Solution Flow Pressure 𝑄= 𝑉 0 𝑇 𝑆𝐴 + 𝑇 𝑆𝑉 + 𝑇 𝑃𝐴 + 𝑇 𝑃𝑉
𝑄= 𝑉 0 𝑇 𝑆𝐴 + 𝑇 𝑆𝑉 + 𝑇 𝑃𝐴 + 𝑇 𝑃𝑉 𝑃 𝑖 = 𝑉 𝑖 𝐢 𝑖 = 𝑇 𝑖 𝑉 0 𝐢 𝑖 𝑇 𝑆𝐴 + 𝑇 𝑆𝑉 + 𝑇 𝑃𝐴 + 𝑇 𝑃𝑉 𝑇 𝑆𝐴 = 𝐢 𝑆𝐴 𝐾 𝑅 + 𝐢 𝑆𝐴 𝑅 𝑆 , 𝑇 𝑃𝐴 = 𝐢 𝑃𝐴 𝐾 𝑅 + 𝐢 𝑃𝐴 𝑅 𝑃 , 𝑇 𝑃𝑉 = 𝐢 𝑃𝑉 𝐾 𝐿 , 𝑇 𝑆𝑉 = 𝐢 𝑆𝑉 𝐾 𝑅

7 Normal Parameter Values
Systemic Pulmonary Units R Rs 17.5 Rp 1.79 mmHg min/liter C Csa 0.01 Cpa .00667 liter/mmHg Csv 1.75 Cpv 0.08 Right Left Unit K KR 2.8 KL 1.2 liter/min/mmHg V V0 5.0 liter

8 Values of Variables SA: Systemic Arteries SV: Systemic Veins
PA: Pulmonary Arteries PV: Pulmonary Veins P (mmHg) V (liter) SA 100 1.0 SV 2 3.5 PA 15 0.1 PV 5 0.4 Stroke volume 70 ml/beat Heart rate 60 beats/min Cardiac output 5.6 l/min

9 Balancing the two sides of the heart
The flow is conserved Systemic arterial pressure is higher 𝑃 𝑆𝐴 > 𝑃 𝑃𝐴 The left heart pumps harder (bigger muscle) 𝐾 𝐿 < 𝐾 𝑅 The systemic resistance is higher 𝑅 𝑆 > 𝑅 𝑃 Systemic compliance is higher 𝐢 𝑆𝐴 < 𝐢 𝑃𝐴 20 5 2 100

10 Balancing the two sides of the heart
If 𝐾 𝐿 = 𝐾 𝑅 , 𝑅 𝑆 = 𝑅 𝑃 , … then 𝑃 𝑆𝐴 = 𝑃 𝑃𝐴 𝑉 𝑆𝐴 = 𝑉 𝑃𝐴 ….

11 Balancing the two sides of the heart
What happens if 𝐾 𝑅 is suddenly reduced (right heart failure)? Temporarily 𝑄 𝑅 < 𝑄 𝐿 Net transfer of volume to the systemic circulation Raise systemic venous pressure Lower pulmonary venous pressure Driver flow back to equality

12 Ratio pulmonary to systemic volume
Recall that the volume giving Showing that decreasing 𝐾 𝑅 increases the denominator reducing the ratio of the volume in the pulmonary circuit to the systemic circuit. 𝑉 𝑖 = 𝑇 𝑖 𝑄 𝑉 𝑃 𝑉 𝑆 = 𝑉 𝑃𝐴 + 𝑉 𝑃𝑉 𝑉 𝑆𝐴 + 𝑉 𝑆𝑉 = 𝑇 𝑃𝐴 𝑄+ 𝑇 𝑃𝑉 𝑄 𝑇 𝑆𝐴 𝑄+ 𝑇 𝑆𝑉 𝑄 = 𝑇 𝑃𝐴 + 𝑇 𝑃𝑉 𝑇 𝑆𝐴 + 𝑇 𝑆𝑉 = 𝐢 𝑆𝐴 + 𝐢 𝑃𝑉 𝐾 𝐿 + 𝐢 𝑃𝐴 𝑅 𝑃 / 𝐢 𝑆𝐴 + 𝐢 𝑆𝑉 𝐾 𝑅 + 𝐢 𝑆𝐴 𝑅 𝑆

13 Why does this work? Dependence of cardiac output on venous pressure 𝑄=𝐾 𝑃 𝑉

14 What if 𝑸 was a parameter?
Lose equations for the heart Lose 2 equations But only 1 variable Necessary to add one relationship, e.g. replace 𝑉 π‘‘π‘œπ‘‘ =𝑉 𝑆𝐴 + 𝑉 𝑆𝑉 + 𝑉 𝑃𝐴 + 𝑉 𝑃𝑉 with 𝑉 𝑆𝐴 + 𝑉 𝑆𝑉 = 𝑉 𝑆 𝑉 𝑃𝐴 + 𝑉 𝑃𝑉 = 𝑉 𝑃 The two sides become independent

15 Cardiovascular Control
What happens when you start running? Heart rate goes up More blood to the muscles in the legs More blood is pumped out from the heart

16 Cardiovascular control
Arteries Veins BR afferent BR nerves sympathetic nerves para-sympathetic

17 Cardiovascular control

18 Effects of changing the parameters
𝑄= 𝑉 0 𝑇 𝑆𝐴 + 𝑇 𝑆𝑉 + 𝑇 𝑃𝐴 + 𝑇 𝑃𝑉 𝑃 𝑆𝐴 = 𝑇 𝑆𝐴 𝑉 0 𝐢 𝑖 𝑇 𝑆𝐴 + 𝑇 𝑆𝑉 + 𝑇 𝑃𝐴 + 𝑇 𝑃𝑉 What is the effect on of reducing 𝑅 𝑆 by a factor of 2? Normal RS halved Change % Q 5.6 6.2 0.6 11% PSA 100 57 -43 -43%

19 Changing Parameters (HW: 1.3)
Norm KS KL RS RP V0 CSA CSV CPA CPV Q 5.6 6.2 PSA 100 57 PSV 2 PPA 15 PPV 5 VSA 1 VSV 3.5 VPA 0.1 VPV 0.4

20 Sensitivity The sensitivity of π‘Œ with respect to 𝑋
𝜎 π‘Œπ‘‹ = Ξ” log π‘Œ Ξ” log 𝑋 𝜎 π‘Œπ‘‹ = log π‘Œ β€² βˆ’log⁑(π‘Œ) log X β€² βˆ’log(X) 𝜎 π‘Œπ‘‹ = log( π‘Œ β€² /π‘Œ) log( X β€² /X)

21 Sensitivity Example (1) 𝜎 𝑄 𝑅 𝑆 = log( 𝑄 β€² /𝑄) log( 𝑅 𝑆 β€²/ 𝑅 𝑆 )

22 Sensitivity Example (2) 𝜎 𝑃 𝑆𝐴 𝑅 𝑆 = log( 𝑃 β€² /𝑃) log( 𝑅 𝑆 β€²/ 𝑅 𝑆 )

23 Sensitivity In the limit of 𝑋 and π‘Œ small 𝜎 π‘Œπ‘‹ = π‘‘π‘Œ π‘Œ / 𝑑𝑋 𝑋
𝜎 π‘Œπ‘‹ = π‘‘π‘Œ 𝑑𝑋 𝑋 π‘Œ

24 Sensitivity It is possible to show (homework 1.10) that -( 𝜎 𝑄 𝑅 𝑆 )+ 𝜎 𝑃 𝑆𝐴 𝑅 𝑆 β‰ˆ1 Because 𝑃 𝑆𝐴 β‰ˆπ‘„ 𝑅 𝑆 The latter holds because 𝑃 𝑆𝐴 ≫ 𝑃 𝑆𝑉

25 Sensitivities (use parameter change table) HW 1.4
KR KL RS RP V0 CSA CSV CPA CPV Q 𝜎 𝑄 𝐾 𝑅 PSA 𝜎 𝑃 𝑆𝐴 𝐾 𝑅 PSV PPA PPV VSA VSV VPA VPV


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