Mechanisms of control in Cardiovascular circulation

Math 392 - Mathematical Models in Biology

January, 2014

Analysis of equations

Q = KLPpv

Q = KRPsv

Vsa = CsaPsa

Vsv = CsvPsv

Vpa = CpaPpa

Vpv = CpvPpv

Q =1

Rs(Psa − Psv)

Q =1

Rp(Ppa − Ppv)

Vsa + Vsv + Vpa + Vpv = V0.

Cardiac output equilibriumQ: What makes QR = QL?

Q ≈ 5.6 L/mVs ≈ 4.5 L, Vp ≈ 0.5 L.

Even a small drop in QR compared to QL can lead to emptying of thepulmonary system.

If KR is reduced, then the following happensQR < QL

Ps increases, Pp decreasesQR increases, while QL decreases, leading to a balance with new volumeproportion

Vp

Vs=

Tpv + Tpa

Tsa + Tsv=

Cpa/KL + CpaRp + Cpv/KL

Csa/KR + CsaRs + Csv/KR

Cardiac output equilibriumQ: What makes QR = QL?

Q ≈ 5.6 L/mVs ≈ 4.5 L, Vp ≈ 0.5 L.

Even a small drop in QR compared to QL can lead to emptying of thepulmonary system.

If KR is reduced, then the following happensQR < QL

Ps increases, Pp decreasesQR increases, while QL decreases, leading to a balance with new volumeproportion

Vp

Vs=

Tpv + Tpa

Tsa + Tsv=

Cpa/KL + CpaRp + Cpv/KL

Csa/KR + CsaRs + Csv/KR

Cardiac output equilibriumQ: What makes QR = QL?

Q ≈ 5.6 L/mVs ≈ 4.5 L, Vp ≈ 0.5 L.

Even a small drop in QR compared to QL can lead to emptying of thepulmonary system.

If KR is reduced, then the following happensQR < QL

Ps increases, Pp decreasesQR increases, while QL decreases, leading to a balance with new volumeproportion

Vp

Vs=

Tpv + Tpa

Tsa + Tsv=

Cpa/KL + CpaRp + Cpv/KL

Csa/KR + CsaRs + Csv/KR

Cardiac output equilibriumQ: What makes QR = QL?

Q ≈ 5.6 L/mVs ≈ 4.5 L, Vp ≈ 0.5 L.

Even a small drop in QR compared to QL can lead to emptying of thepulmonary system.

If KR is reduced, then the following happensQR < QL

Ps increases, Pp decreasesQR increases, while QL decreases, leading to a balance with new volumeproportion

Vp

Vs=

Tpv + Tpa

Tsa + Tsv=

Cpa/KL + CpaRp + Cpv/KL

Csa/KR + CsaRs + Csv/KR

Cardiac output equilibriumQ: What makes QR = QL?

Q ≈ 5.6 L/mVs ≈ 4.5 L, Vp ≈ 0.5 L.

Even a small drop in QR compared to QL can lead to emptying of thepulmonary system.

If KR is reduced, then the following happensQR < QL

Ps increases, Pp decreasesQR increases, while QL decreases, leading to a balance with new volumeproportion

Vp

Vs=

Tpv + Tpa

Tsa + Tsv=

Cpa/KL + CpaRp + Cpv/KL

Csa/KR + CsaRs + Csv/KR

Dependence of flow on pressure

The previous regulatory mechanism relies on the dependence of Q on P.

Suppose Q does not depend on P, and is thus, a parameter in the equations.The two pump equations are eliminated, leaving 8 unknowns, (Vi ,Pi ) and 6equations (4 compliance, 2 resistance).

Introduce new parameters, Vs, Vp:

Vsa + Vsv = Vs

Vpa + Vpv = Vp.

This leads to 8 equations with 8 unknowns, but no mechanism guaranteeing areasonable relationship between Vs and Vp. So the relationship of Q and P iscrucial for a controlled partitioning of blood into the pulmonary and systemiccirculations.

Dependence of flow on pressure

The previous regulatory mechanism relies on the dependence of Q on P.

Suppose Q does not depend on P, and is thus, a parameter in the equations.The two pump equations are eliminated, leaving 8 unknowns, (Vi ,Pi ) and 6equations (4 compliance, 2 resistance).

Introduce new parameters, Vs, Vp:

Vsa + Vsv = Vs

Vpa + Vpv = Vp.

This leads to 8 equations with 8 unknowns, but no mechanism guaranteeing areasonable relationship between Vs and Vp. So the relationship of Q and P iscrucial for a controlled partitioning of blood into the pulmonary and systemiccirculations.

Dependence of flow on pressure

The previous regulatory mechanism relies on the dependence of Q on P.

Suppose Q does not depend on P, and is thus, a parameter in the equations.The two pump equations are eliminated, leaving 8 unknowns, (Vi ,Pi ) and 6equations (4 compliance, 2 resistance).

Introduce new parameters, Vs, Vp:

Vsa + Vsv = Vs

Vpa + Vpv = Vp.

This leads to 8 equations with 8 unknowns, but no mechanism guaranteeing areasonable relationship between Vs and Vp. So the relationship of Q and P iscrucial for a controlled partitioning of blood into the pulmonary and systemiccirculations.

Dependence of flow on pressure

The previous regulatory mechanism relies on the dependence of Q on P.

Suppose Q does not depend on P, and is thus, a parameter in the equations.The two pump equations are eliminated, leaving 8 unknowns, (Vi ,Pi ) and 6equations (4 compliance, 2 resistance).

Introduce new parameters, Vs, Vp:

Vsa + Vsv = Vs

Vpa + Vpv = Vp.

This leads to 8 equations with 8 unknowns, but no mechanism guaranteeing areasonable relationship between Vs and Vp. So the relationship of Q and P iscrucial for a controlled partitioning of blood into the pulmonary and systemiccirculations.

Need for an external control

During an exercise, the following is observed:

Arterioles dialate→ Rs falls→ QL rises→ Psa is mantained

Here the increase in QL comes from an increase in the heart rate with strokevolume = const.

Q: What are the effects of a drop in Rs?

Q =V0

Tsa + Tsv + Tpa + Tpv, Psa =

V0

Csa

Tsa

Tsa + Tsv + Tpa + Tpv

Normal Rs → Rs/2 Change % ChangeQ 5.6 L/m 6.2 L/m +0.6 L/m +11%Psa 100 mmHg 57 mmHg -43 mmHg -43%

This is inadequate to sustain exercise of any length.

Need for an external control

During an exercise, the following is observed:

Arterioles dialate→ Rs falls→ QL rises→ Psa is mantained

Here the increase in QL comes from an increase in the heart rate with strokevolume = const.

Q: What are the effects of a drop in Rs?

Q =V0

Tsa + Tsv + Tpa + Tpv, Psa =

V0

Csa

Tsa

Tsa + Tsv + Tpa + Tpv

Normal Rs → Rs/2 Change % ChangeQ 5.6 L/m 6.2 L/m +0.6 L/m +11%Psa 100 mmHg 57 mmHg -43 mmHg -43%

This is inadequate to sustain exercise of any length.

Need for an external control

During an exercise, the following is observed:

Arterioles dialate→ Rs falls→ QL rises→ Psa is mantained

Here the increase in QL comes from an increase in the heart rate with strokevolume = const.

Q: What are the effects of a drop in Rs?

Q =V0

Tsa + Tsv + Tpa + Tpv, Psa =

V0

Csa

Tsa

Tsa + Tsv + Tpa + Tpv

Normal Rs → Rs/2 Change % ChangeQ 5.6 L/m 6.2 L/m +0.6 L/m +11%Psa 100 mmHg 57 mmHg -43 mmHg -43%

This is inadequate to sustain exercise of any length.

Need for an external control

During an exercise, the following is observed:

Arterioles dialate→ Rs falls→ QL rises→ Psa is mantained

Here the increase in QL comes from an increase in the heart rate with strokevolume = const.

Q: What are the effects of a drop in Rs?

Q =V0

Tsa + Tsv + Tpa + Tpv, Psa =

V0

Csa

Tsa

Tsa + Tsv + Tpa + Tpv

Normal Rs → Rs/2 Change % ChangeQ 5.6 L/m 6.2 L/m +0.6 L/m +11%Psa 100 mmHg 57 mmHg -43 mmHg -43%

This is inadequate to sustain exercise of any length.

SensitivityQ: How sensitive are Q and Psa to changes in Rs?

Let Y = Y (X ), define the sensitivity of Y to changes in X as

σYX =∆ log Y∆ log X

=log Y (X + ∆X )− log Y (X )

log(X + ∆X )− log X=

log(Y ′/Y )

log(X ′/X )≈ ∆Y

Y

/∆XX

.

σYX measures roughly the rate of relative changes. For example,

Y = aX n =⇒ σYX =log a(X + ∆X )n − log aX n

log(X + ∆X )− log X n = n,

so in particular

Y ∼ X =⇒ σYX = 1, and if Y ∼ 1/X =⇒ σYX = −1.

From the changes in Q and Psa, while Rs → Rs/2,

σQRs =log 6.2− log 5.6

log(1/2)≈ −0.15, σPsaRs =

log 100− log 57log(1/2)

≈ +0.81

Observe that −σQRs + σPsaRs ≈ 1, which follows from Psa ≈ QRs.

SensitivityQ: How sensitive are Q and Psa to changes in Rs?

Let Y = Y (X ), define the sensitivity of Y to changes in X as

σYX =∆ log Y∆ log X

=log Y (X + ∆X )− log Y (X )

log(X + ∆X )− log X=

log(Y ′/Y )

log(X ′/X )≈ ∆Y

Y

/∆XX

.

σYX measures roughly the rate of relative changes. For example,

Y = aX n =⇒ σYX =log a(X + ∆X )n − log aX n

log(X + ∆X )− log X n = n,

so in particular

Y ∼ X =⇒ σYX = 1, and if Y ∼ 1/X =⇒ σYX = −1.

From the changes in Q and Psa, while Rs → Rs/2,

σQRs =log 6.2− log 5.6

log(1/2)≈ −0.15, σPsaRs =

log 100− log 57log(1/2)

≈ +0.81

Observe that −σQRs + σPsaRs ≈ 1, which follows from Psa ≈ QRs.

SensitivityQ: How sensitive are Q and Psa to changes in Rs?

Let Y = Y (X ), define the sensitivity of Y to changes in X as

σYX =∆ log Y∆ log X

=log Y (X + ∆X )− log Y (X )

log(X + ∆X )− log X=

log(Y ′/Y )

log(X ′/X )≈ ∆Y

Y

/∆XX

.

σYX measures roughly the rate of relative changes. For example,

Y = aX n =⇒ σYX =log a(X + ∆X )n − log aX n

log(X + ∆X )− log X n = n,

so in particular

Y ∼ X =⇒ σYX = 1, and if Y ∼ 1/X =⇒ σYX = −1.

From the changes in Q and Psa, while Rs → Rs/2,

σQRs =log 6.2− log 5.6

log(1/2)≈ −0.15, σPsaRs =

log 100− log 57log(1/2)

≈ +0.81

Observe that −σQRs + σPsaRs ≈ 1, which follows from Psa ≈ QRs.

SensitivityQ: How sensitive are Q and Psa to changes in Rs?

Let Y = Y (X ), define the sensitivity of Y to changes in X as

σYX =∆ log Y∆ log X

=log Y (X + ∆X )− log Y (X )

log(X + ∆X )− log X=

log(Y ′/Y )

log(X ′/X )≈ ∆Y

Y

/∆XX

.

σYX measures roughly the rate of relative changes. For example,

Y = aX n =⇒ σYX =log a(X + ∆X )n − log aX n

log(X + ∆X )− log X n = n,

so in particular

Y ∼ X =⇒ σYX = 1, and if Y ∼ 1/X =⇒ σYX = −1.

From the changes in Q and Psa, while Rs → Rs/2,

σQRs =log 6.2− log 5.6

log(1/2)≈ −0.15, σPsaRs =

log 100− log 57log(1/2)

≈ +0.81

Observe that −σQRs + σPsaRs ≈ 1, which follows from Psa ≈ QRs.

SensitivityQ: How sensitive are Q and Psa to changes in Rs?

Let Y = Y (X ), define the sensitivity of Y to changes in X as

σYX =∆ log Y∆ log X

=log Y (X + ∆X )− log Y (X )

log(X + ∆X )− log X=

log(Y ′/Y )

log(X ′/X )≈ ∆Y

Y

/∆XX

.

σYX measures roughly the rate of relative changes. For example,

Y = aX n =⇒ σYX =log a(X + ∆X )n − log aX n

log(X + ∆X )− log X n = n,

so in particular

Y ∼ X =⇒ σYX = 1, and if Y ∼ 1/X =⇒ σYX = −1.

From the changes in Q and Psa, while Rs → Rs/2,

σQRs =log 6.2− log 5.6

log(1/2)≈ −0.15, σPsaRs =

log 100− log 57log(1/2)

≈ +0.81

Observe that −σQRs + σPsaRs ≈ 1, which follows from Psa ≈ QRs.

Neural control: Baroreceptor loopNeed a mechanism to hold Psa constant and to guarantee blood flow tononexercising tissue, i.e.

σPsaRs = 0 and σQRs = −1

This is achieved by the baroreceptor loop:

The loop adjusts the heart rate F to keep Psa = P∗ constant.

Neural control: Baroreceptor loopNeed a mechanism to hold Psa constant and to guarantee blood flow tononexercising tissue, i.e.

σPsaRs = 0 and σQRs = −1

This is achieved by the baroreceptor loop:

The loop adjusts the heart rate F to keep Psa = P∗ constant.

Neural control: Baroreceptor loopNeed a mechanism to hold Psa constant and to guarantee blood flow tononexercising tissue, i.e.

σPsaRs = 0 and σQRs = −1

This is achieved by the baroreceptor loop:

The loop adjusts the heart rate F to keep Psa = P∗ constant.

Effect of the baroreceptor loopIn this adjusted model F becomes a variable, while Psa = P∗ is a parameter.

QR = FCRPsv

QL = FCLPpv

To solve, ignore Psv in the systemic resistance equation (2% error),

QRs = P∗

ignore the volume of pulmonary blood in the total volume equation (10% error)

Vsa + Vsv = V0.

Q =P∗

Rs.

CsaP∗ + CsvPsv = V0 ⇒ Psv =V0 = CsaP∗

Csv.

F =Q

CRPsv=

P∗Csv

RsCR(V0 − CsaP∗).

Effect of the baroreceptor loopIn this adjusted model F becomes a variable, while Psa = P∗ is a parameter.

QR = FCRPsv

QL = FCLPpv

To solve, ignore Psv in the systemic resistance equation (2% error),

QRs = P∗

ignore the volume of pulmonary blood in the total volume equation (10% error)

Vsa + Vsv = V0.

Q =P∗

Rs.

CsaP∗ + CsvPsv = V0 ⇒ Psv =V0 = CsaP∗

Csv.

F =Q

CRPsv=

P∗Csv

RsCR(V0 − CsaP∗).

Effect of the baroreceptor loopIn this adjusted model F becomes a variable, while Psa = P∗ is a parameter.

QR = FCRPsv

QL = FCLPpv

To solve, ignore Psv in the systemic resistance equation (2% error),

QRs = P∗

ignore the volume of pulmonary blood in the total volume equation (10% error)

Vsa + Vsv = V0.

Q =P∗

Rs.

CsaP∗ + CsvPsv = V0 ⇒ Psv =V0 = CsaP∗

Csv.

F =Q

CRPsv=

P∗Csv

RsCR(V0 − CsaP∗).

Effect of the baroreceptor loopIn this adjusted model F becomes a variable, while Psa = P∗ is a parameter.

QR = FCRPsv

QL = FCLPpv

To solve, ignore Psv in the systemic resistance equation (2% error),

QRs = P∗

ignore the volume of pulmonary blood in the total volume equation (10% error)

Vsa + Vsv = V0.

Q =P∗

Rs.

CsaP∗ + CsvPsv = V0 ⇒ Psv =V0 = CsaP∗

Csv.

F =Q

CRPsv=

P∗Csv

RsCR(V0 − CsaP∗).

Effect of the baroreceptor loop

Q =P∗

Rs, Psa = P∗, F =

P∗Csv

RsCR(V0 − CsaP∗).

σQRs = −1, σPsaRs = 0

In the new model, Q = Q(P∗,Rs)In particular, σQV0 = 0, while in the old model σQV0 = 1.

Blood loss leads to increased heart rate (F ), which compensates for decreasein the stroke volume.

F breaks down when V0 = CsaP∗, i.e. when V0 = Vsa, and Vsv = 0.

Effect of the baroreceptor loop

Q =P∗

Rs, Psa = P∗, F =

P∗Csv

RsCR(V0 − CsaP∗).

σQRs = −1, σPsaRs = 0

In the new model, Q = Q(P∗,Rs)In particular, σQV0 = 0, while in the old model σQV0 = 1.

Blood loss leads to increased heart rate (F ), which compensates for decreasein the stroke volume.

F breaks down when V0 = CsaP∗, i.e. when V0 = Vsa, and Vsv = 0.

Effect of the baroreceptor loop

Q =P∗

Rs, Psa = P∗, F =

P∗Csv

RsCR(V0 − CsaP∗).

σQRs = −1, σPsaRs = 0

In the new model, Q = Q(P∗,Rs)In particular, σQV0 = 0, while in the old model σQV0 = 1.

Blood loss leads to increased heart rate (F ), which compensates for decreasein the stroke volume.

F breaks down when V0 = CsaP∗, i.e. when V0 = Vsa, and Vsv = 0.

Effect of the baroreceptor loop

Q =P∗

Rs, Psa = P∗, F =

P∗Csv

RsCR(V0 − CsaP∗).

σQRs = −1, σPsaRs = 0

In the new model, Q = Q(P∗,Rs)In particular, σQV0 = 0, while in the old model σQV0 = 1.

Blood loss leads to increased heart rate (F ), which compensates for decreasein the stroke volume.

F breaks down when V0 = CsaP∗, i.e. when V0 = Vsa, and Vsv = 0.

Effect of the baroreceptor loop

Q =P∗

Rs, Psa = P∗, F =

P∗Csv

RsCR(V0 − CsaP∗).

σQRs = −1, σPsaRs = 0

In the new model, Q = Q(P∗,Rs)In particular, σQV0 = 0, while in the old model σQV0 = 1.

Blood loss leads to increased heart rate (F ), which compensates for decreasein the stroke volume.

F breaks down when V0 = CsaP∗, i.e. when V0 = Vsa, and Vsv = 0.

AutoregulationQ: What controls Rs on the local level?

1 There is a range of pressures, for which Q is insensitive to ∆P.2 When ∆P = const, Q depends on the rate of O2 consumption.

Define:

[O2]a − arterial oxygen concentration, in (liters of O2)/(liters of blood)[O2]v − venous oxygen concentration, in (liters of O2)/(liters of blood)

M − metabolic rate of the tissue (O2 consumption rate), in liters/minute

Fick’s principle:

Q[O2]a −Q[O2]v = M ⇒ [O2]v = [O2]a −M/Q.

When [O2]v = 0, thenQ = Q∗ = M/[O2]a

which is the minimum blood flow to sustain the metabolic rate M.

AutoregulationQ: What controls Rs on the local level?

1 There is a range of pressures, for which Q is insensitive to ∆P.2 When ∆P = const, Q depends on the rate of O2 consumption.

Define:

[O2]a − arterial oxygen concentration, in (liters of O2)/(liters of blood)[O2]v − venous oxygen concentration, in (liters of O2)/(liters of blood)

M − metabolic rate of the tissue (O2 consumption rate), in liters/minute

Fick’s principle:

Q[O2]a −Q[O2]v = M ⇒ [O2]v = [O2]a −M/Q.

When [O2]v = 0, thenQ = Q∗ = M/[O2]a

which is the minimum blood flow to sustain the metabolic rate M.

AutoregulationQ: What controls Rs on the local level?

1 There is a range of pressures, for which Q is insensitive to ∆P.2 When ∆P = const, Q depends on the rate of O2 consumption.

Define:

[O2]a − arterial oxygen concentration, in (liters of O2)/(liters of blood)[O2]v − venous oxygen concentration, in (liters of O2)/(liters of blood)

M − metabolic rate of the tissue (O2 consumption rate), in liters/minute

Fick’s principle:

Q[O2]a −Q[O2]v = M ⇒ [O2]v = [O2]a −M/Q.

When [O2]v = 0, thenQ = Q∗ = M/[O2]a

which is the minimum blood flow to sustain the metabolic rate M.

AutoregulationQ: What controls Rs on the local level?

1 There is a range of pressures, for which Q is insensitive to ∆P.2 When ∆P = const, Q depends on the rate of O2 consumption.

Define:

[O2]a − arterial oxygen concentration, in (liters of O2)/(liters of blood)[O2]v − venous oxygen concentration, in (liters of O2)/(liters of blood)

M − metabolic rate of the tissue (O2 consumption rate), in liters/minute

Fick’s principle:

Q[O2]a −Q[O2]v = M ⇒ [O2]v = [O2]a −M/Q.

When [O2]v = 0, thenQ = Q∗ = M/[O2]a

which is the minimum blood flow to sustain the metabolic rate M.

AutoregulationQ: What controls Rs on the local level?

1 There is a range of pressures, for which Q is insensitive to ∆P.2 When ∆P = const, Q depends on the rate of O2 consumption.

Define:

[O2]a − arterial oxygen concentration, in (liters of O2)/(liters of blood)[O2]v − venous oxygen concentration, in (liters of O2)/(liters of blood)

M − metabolic rate of the tissue (O2 consumption rate), in liters/minute

Fick’s principle:

Q[O2]a −Q[O2]v = M ⇒ [O2]v = [O2]a −M/Q.

When [O2]v = 0, thenQ = Q∗ = M/[O2]a

which is the minimum blood flow to sustain the metabolic rate M.

Model for Rs

[O2]v can be used as a (regulating) indicator for Rs

For example, suppose

R = R0[O2]v, where R = ∆P/Q

combining equations we have

M = Q([O2]a − [O2]v) = Q[O2]a −∆PR0

,

hence

Q =M

[O2]a+

∆PR0[O2]a

= Q∗ +∆P

R0[O2]a.

Model for Rs

[O2]v can be used as a (regulating) indicator for Rs

For example, suppose

R = R0[O2]v, where R = ∆P/Q

combining equations we have

M = Q([O2]a − [O2]v) = Q[O2]a −∆PR0

,

hence

Q =M

[O2]a+

∆PR0[O2]a

= Q∗ +∆P

R0[O2]a.

Model for Rs

[O2]v can be used as a (regulating) indicator for Rs

For example, suppose

R = R0[O2]v, where R = ∆P/Q

combining equations we have

M = Q([O2]a − [O2]v) = Q[O2]a −∆PR0

,

hence

Q =M

[O2]a+

∆PR0[O2]a

= Q∗ +∆P

R0[O2]a.

Model for Rs

σQP is less in the new modelQ ≥ Q∗, min flow guaranteedWhen ∆P = const, ∆Q = ∆M/[O2]a = ∆Q∗.When ∆P = const, an increase in M results in a decrease in R.When [O2]a changes, Q adjusts, so that Q[O2]a = const.