Pressure Differences Drive Flow in a Lumped Model
A circulation model needs a rule connecting vascular pressure to blood flow. For a segment represented by one effective resistance, the rule is simple: flow is proportional to the pressure difference across that segment. In a network, this relation is paired with conservation of flow at each node. Together, the two rules let us calculate both the pressure at an internal junction and the flows through the beds connected to it.
The earlier material on coupled cardiovascular-renal state variables and feedback topology places circulatory variables within a larger physiological system. Here the focus is narrower: a steady, lumped pressure-flow calculation. It does not represent transient storage of blood in compliant vessels or the time course of feedback.
The sign convention matters. A positive \(Q_{ij}\) means flow from \(i\) to \(j\); a negative result means the actual flow is opposite to the direction chosen for writing the equation. For a positive resistance, flow is always directed from higher pressure toward lower pressure.
Why the Relation Holds, and What It Assumes
Resistance is defined in the linear model as the pressure drop required per unit flow: \(\Delta P=RQ\). Rearranging gives \(Q=\Delta P/R\). This is a constitutive relation for a lumped segment: all the detailed effects of vessel geometry and blood properties are summarized by one effective parameter, \(R\).
A familiar physical basis is Poiseuille flow through a rigid cylindrical tube containing a Newtonian fluid under steady laminar conditions. In that idealization, \(R=8\mu L/(\pi r^4)\), where \(\mu\) is viscosity, \(L\) is tube length, and \(r\) is radius. A vascular bed is not a single rigid tube, so its effective resistance is generally a model parameter rather than a direct description of one vessel. The idealized law nevertheless motivates the proportional relation between pressure drop and flow.
The calculation relies on several assumptions. Pressures at each node are treated as spatially uniform; each segment has a specified, positive resistance; and the resistance is constant over the range of pressures and flows being considered. The calculation is steady, so there is no net accumulation of blood at a node. Finally, the network must be connected to pressure conditions that determine the unknown node pressures. If these assumptions are not reasonable, the algebra can still be solved but may not describe the physiology adequately.
Flow Conservation at a Node
A node is a junction where segments meet. In a steady model with no source, sink, or storage at the junction, total flow entering must equal total flow leaving. For a node \(k\), this can be written as the sum of signed flows equal to zero. Each flow term is calculated from the pressure difference across its own segment.
In this expression, each term is defined as flow from node \(k\) toward a connected node \(j\). Terms are positive when pressure at \(k\) is higher and negative when the connected node has the higher pressure. This signed form handles either direction without needing to guess the direction in advance.
Worked Example: Flow Through One Segment
Suppose a hypothetical vascular segment has an upstream pressure of 80 mmHg, a downstream pressure of 10 mmHg, and an effective resistance of 35 mmHg·min/L. Calculate the flow in the defined upstream-to-downstream direction.
The pressure drop is \(\Delta P=80-10=70\) mmHg. The linear lumped model gives:
The units reduce to L/min, and the positive sign agrees with the higher upstream pressure. This calculation assumes the segment resistance remains 35 mmHg·min/L at the resulting flow and that the pressures represent the steady pressures at the segment ends.
Two Vascular Beds Connected to a Shared Node
Consider a source at pressure \(P_s\) connected through an upstream resistance \(R_0\) to a shared arterial node at unknown pressure \(P_n\). Two vascular beds leave that node and drain to a common venous pressure \(P_v\). Let their resistances be \(R_1\) and \(R_2\), and define their flows as \(Q_1\) and \(Q_2\), directed from the node toward the venous side. Define \(Q_0\) as flow from the source to the node.
Because the node has no storage or additional inflow or outflow in this steady model, its balance is \(Q_0=Q_1+Q_2\). Substitution produces one equation for the unknown node pressure. After solving for \(P_n\), calculate each flow separately and verify the balance numerically. This procedure keeps the pressure and flow for each bed explicit rather than assuming a particular flow split.
Worked Example: Calculating Two Bed Flows From a Known Node Pressure
In a hypothetical network, the shared node pressure is measured or otherwise specified as 60 mmHg, and venous pressure is 5 mmHg. Bed 1 has resistance 55 mmHg·min/L, and bed 2 has resistance 110 mmHg·min/L. Calculate both flows and the total flow leaving the node.
The pressure drop across each bed is \(60-5=55\) mmHg. Applying the pressure-flow relation to each bed gives:
The total outflow is \(Q_1+Q_2=1.000+0.5000=1.500\) L/min. Thus a steady source flow into the node would also need to be 1.500 L/min for this network to satisfy conservation. Both pressure drops are positive in the defined directions, so both computed flows leave the node as specified.
Worked Example: Solving the Node Pressure in a Two-Bed Network
Consider a hypothetical source pressure \(P_s=100\) mmHg, venous pressure \(P_v=5\) mmHg, upstream resistance \(R_0=10\) mmHg·min/L, and bed resistances \(R_1=40\) and \(R_2=20\) mmHg·min/L. Find the shared node pressure and all three flows.
Write the flow expressions and impose the node balance:
Multiply both sides by 40 to clear the denominators. Then collect the terms involving \(P_n\):
Using the unrounded node pressure \(415/7\) mmHg for the flows avoids introducing rounding error into the balance:
The downstream flows sum to \(1.357+2.714=4.071\) L/min at the displayed precision, matching the upstream flow. The node pressure lies between the source and venous pressures, and each flow is positive in its defined direction. These checks are consistent with a steady source-to-vein flow through the network.
Derivation of the Node Equation and Practical Checks
The two-bed equation follows directly from conservation rather than from an assumed pressure at the internal node. In the example, the source flow entering the node is \((P_s-P_n)/R_0\). The two flows leaving are \((P_n-P_v)/R_1\) and \((P_n-P_v)/R_2\). Equating entry and exit yields:
For the stated positive resistances, the left side decreases as \(P_n\) increases, while the right side increases as \(P_n\) increases. Therefore, when the source pressure exceeds the venous pressure, these two sides can intersect at no more than one node pressure. In this connected network, solving the linear equation gives that pressure; calculating the flows afterward supplies an independent conservation check.
Useful diagnostics are straightforward. First check units: dividing mmHg by mmHg·min/L must give L/min. Next check signs: a positive flow should correspond to a pressure drop in the stated direction. Finally, calculate the node residual, \(Q_0-Q_1-Q_2\). A residual other than zero beyond rounding suggests an arithmetic, sign, or equation-setup error. A computed negative flow is not automatically an error; it indicates flow opposite to the direction initially defined and should prompt a check that the model structure and pressure boundary conditions allow that direction.
Choosing and Interpreting the Model
Direct segment calculations are appropriate when both endpoint pressures and the resistance are known. If an internal pressure is unknown, the node-balance equations provide the appropriate extension: write a conservation equation for each unknown internal node and solve the simultaneous system. A dynamic model is a different approach, appropriate when pressures and flows change over time and storage or feedback cannot be ignored. Such a model requires additional state variables and equations; the steady calculation alone cannot predict a transient response.
The lumped linear relation is useful for transparent accounting and for checking whether proposed pressures, resistances, and flows are mutually consistent. It has important limitations. Real vessels can change radius, resistance can depend on flow or physiological state, and pressure and flow can vary over the cardiac cycle. A single resistance and a steady pressure may therefore be inadequate for a question about pulsatility, changing vascular tone, or local flow regulation. In those settings, the values calculated here should be treated as model-dependent approximations, not as universal physiological constants.
A common mistake is to calculate each bed flow from the source pressure rather than from the pressure difference across that bed. For example, using \((P_s-P_v)/R_1\) in the two-bed example incorrectly assigns the full source-to-vein pressure drop to bed 1, ignoring the upstream resistance and the shared node pressure. The correction is to identify each segment’s actual endpoint pressures, write its own pressure difference, and enforce conservation at the junction.
Check Your Understanding
Use the stated steady, linear lumped assumptions. Show the pressure difference and units in each calculation.
- A segment has pressures 72 mmHg and 12 mmHg at its ends and resistance 30 mmHg·min/L. What is flow from the higher-pressure end?
- At a shared node, pressure is 50 mmHg and venous pressure is 5 mmHg. Two beds have resistances 45 and 90 mmHg·min/L. Calculate both bed flows and their sum.
- A node receives 3.0 L/min and has two outgoing flows, 1.2 L/min and 1.1 L/min. What is the flow-balance residual if defined as inflow minus total outflow? Does the node balance?
- In the two-bed network, why is it incorrect to use source pressure as the upstream pressure for both beds when an upstream resistance separates the source from their shared node?
Solutions. (1) The drop is \(72-12=60\) mmHg, so \(Q=60/30=2.000\) L/min. (2) Each bed has a 45 mmHg drop. The flows are \(45/45=1.000\) L/min and \(45/90=0.5000\) L/min, for a total of 1.500 L/min. (3) The residual is \(3.0-(1.2+1.1)=0.7\) L/min. The balance is not satisfied; another outflow, accumulation, or a revised flow value would be needed. (4) The beds begin at the shared node, not at the source. The pressure drop across the upstream resistance means node pressure can differ from source pressure, so bed flows must use node-to-venous pressure differences.