Biology 8.12 - Heart rate, stroke volume and cardiac output

Cardiac output during exercise

In a person, the circulatory system must deliver oxygen and glucose to respiring cells and carry carbon dioxide away. During exercise, active muscle cells have a greater demand, so the heart must pump a greater volume of blood each minute.

Three quantities describe this pumping:

• Heart rate (HR): the number of heart beats each minute, measured in $\mathrm{beats}\,\mathrm{min}^{-1}$ or bpm.
• Stroke volume (SV): the volume of blood pumped by one ventricle in one beat, often measured in $\mathrm{cm}^{3}\,\mathrm{beat}^{-1}$.
• Cardiac output (CO): the volume of blood pumped by one ventricle each minute, measured in $\mathrm{cm}^{3}\,\mathrm{min}^{-1}$ or $\mathrm{dm}^{3}\,\mathrm{min}^{-1}$.

The units reveal why the relationship works:

$\frac{\mathrm{cm}^{3}}{\mathrm{beat}}\times\frac{\mathrm{beats}}{\mathrm{min}}=\frac{\mathrm{cm}^{3}}{\mathrm{min}}$

The beats cancel, leaving a volume each minute. During exercise, heart rate rises and stroke volume can rise, so multiplying them gives a larger cardiac output. The larger output supplies active muscles with materials for respiration more rapidly and removes carbon dioxide more rapidly.

Worked example: calculating cardiac output

A person's stroke volume during light exercise is $75\,\mathrm{cm}^{3}\,\mathrm{beat}^{-1}$ and heart rate is $80\,\mathrm{beats}\,\mathrm{min}^{-1}$.

1. Select the relationship: $\mathrm{CO} =\,\mathrm{SV} \times\,\mathrm{HR}$.
2. Check units: $\mathrm{cm}^{3}/\mathrm{beat}$ and $\mathrm{beats}/\mathrm{min}$ are compatible.
3. Substitute: $\mathrm{CO} = 75 \times 80$.
4. Calculate: $\mathrm{CO} = 6000\,\mathrm{cm}^{3}\,\mathrm{min}^{-1}$.
5. Convert if required: $1000\,\mathrm{cm}^{3} = 1\,\mathrm{dm}^{3}$, so $6000\,\mathrm{cm}^{3}\,\mathrm{min}^{-1} = 6.0\,\mathrm{dm}^{3}\,\mathrm{min}^{-1}$.
6. Sense-check: $75\,\mathrm{cm}^{3}$ is pumped on each of $80$ beats, so a total of several thousand cubic centimetres per minute is reasonable.

At a harder exercise intensity, a stroke volume of $110\,\mathrm{cm}^{3}\,\mathrm{beat}^{-1}$ and heart rate of $140\,\mathrm{beats}\,\mathrm{min}^{-1}$ would give:

$\mathrm{CO} = 110 \times 140 = 15\,400\,\mathrm{cm}^{3}\,\mathrm{min}^{-1} = 15.4\,\mathrm{dm}^{3}\,\mathrm{min}^{-1}$

This numerical increase is consistent with the increased transport demand of active muscles. It does not mean that heart rate alone is cardiac output: one is beats per minute, while the other is blood volume per minute.

Calculating cardiac quantities

The same equation can find heart rate or stroke volume. Rearrange before substituting:

$\mathrm{HR}=\frac{\mathrm{CO}}{\mathrm{SV}}$

$\mathrm{SV}=\frac{\mathrm{CO}}{\mathrm{HR}}$

Worked example: calculating heart rate

During exercise, cardiac output is $15\,000\,\mathrm{cm}^{3}\,\mathrm{min}^{-1}$ and stroke volume is $100\,\mathrm{cm}^{3}\,\mathrm{beat}^{-1}$.

1. Rearrange: $\mathrm{HR}=\frac{\mathrm{CO}}{\mathrm{SV}}$.
2. Substitute compatible units: $\mathrm{HR} = 15\,000 / 100$.
3. Calculate: $\mathrm{HR} = 150\,\mathrm{beats}\,\mathrm{min}^{-1}$.
4. Check: $100\,\mathrm{cm}^{3}/\mathrm{beat} \times 150\,\mathrm{beats}/\mathrm{min} = 15\,000\,\mathrm{cm}^{3}/\mathrm{min}$, so the answer returns the given cardiac output.

Worked example: calculating stroke volume with conversion

Cardiac output is $10.8\,\mathrm{dm}^{3}\,\mathrm{min}^{-1}$ and heart rate is $120\,\mathrm{beats}\,\mathrm{min}^{-1}$. The answer is required in $\mathrm{cm}^{3}\,\mathrm{beat}^{-1}$.