8 Practice problems

The sections these problems draw on: the heart (the pump and its pressures), waves (\(v = f\lambda\) and the echo), ultrasound (pulse–echo depth and Doppler), conservation of mass and conservation of energy (the physics behind Bernoulli), Bernoulli (\(\Delta p \approx 4v^2\)), and closing the loop (the combined diagnosis).

Review

The problems in this section are built from the following relations, using blood density \(\rho \approx 1060~\text{kg/m}^3\), tissue sound speed \(c \approx 1540~\text{m/s}\), and \(1~\text{mmHg} = 133.3~\text{Pa}\).

Flow rate: \(Q = A\,v\) (area \(\times\) speed), in m³/s.
Continuity (incompressible, fixed \(Q\)): \(A_1 v_1 = A_2 v_2\). Circular orifice: \(A \propto d^2\).
Simplified Bernoulli (jet through a valve): \(\Delta p \approx 4\,v_{\max}^2\) (\(\Delta p\) in mmHg, \(v_{\max}\) in m/s); invert as \(v_{\max} = \sqrt{\Delta p/4}\).
Wave equation: \(v = f\lambda\); in fixed tissue \(\lambda = c/f\).
Pulse–echo depth: \(d = \tfrac{1}{2} c\,t\) (the \(\tfrac12\) is the round trip).
Doppler velocity: \(v = \dfrac{c\, f_D}{2 f_0 \cos\theta}\), and forward \(f_D = \dfrac{2 f_0 v\cos\theta}{c}\).
Resolution & penetration: finest detail \(\approx \lambda\); imaging reach \(\approx 500\,\lambda\).
Cardiac output: \(\text{CO} = \text{HR}\times\text{SV}\) (heart rate \(\times\) stroke volume).

Clinical note. \(\Delta p \approx 4v^2\) is the peak (maximum-instantaneous) gradient. The guideline criteria for severe aortic stenosis are a peak jet velocity \(\geq 4~\text{m/s}\), an aortic valve area \(\leq 1.0~\text{cm}^2\), or a mean gradient \(\geq 40~\text{mmHg}\) (Baumgartner et al. 2017).

Continuity and Bernoulli

Problem 1

A healthy mitral valve passes its peak diastolic inflow at \(v_{\text{healthy}} = 0.9~\text{m/s}\). Disease narrows the valve so that its opening diameter is halved, while the heart keeps driving the same volumetric flow \(Q\) through it. Treat the orifice as a circle.

By what factor does the opening area change?
Find the peak jet velocity through the stenotic valve.
Estimate the pressure gradient across the healthy valve and across the stenotic valve, and state the factor by which stenosis has raised it.

Answer 1

A circle’s area is \(A = \pi d^2/4\), so \(A \propto d^2\). Halving the diameter multiplies the area by \((1/2)^2 = \mathbf{1/4}\) — a quarter, not a half. (This is the trap worth remembering: a “half” narrowing in diameter is really a quarter in area.)
Continuity at fixed flow gives \(v \propto 1/A\), so quartering the area multiplies the velocity by 4: \(v_{\text{stenotic}} = 4 \times 0.9 = \mathbf{3.6~m/s}\).
With \(\Delta p \approx 4v^2\) (mmHg): healthy \(\Delta p_{\text{healthy}} \approx 4(0.9)^2 = \mathbf{3.24~mmHg}\); stenotic \(\Delta p_{\text{stenotic}} \approx 4(3.6)^2 = \mathbf{51.8~mmHg}\). The gradient grew by \(51.8/3.24 = \mathbf{16\times}\). Because velocity scales as \(1/A\) and \(\Delta p\) as velocity squared, \(\Delta p\) scales as \((A_{\text{healthy}}/A_{\text{stenotic}})^2 = 4^2 = 16\) — the quadratic-on-top-of-quadratic penalty that makes even a modest narrowing so punishing for the heart.

Problem 2

Some patients have two stenosed valves at once. Consider a heart in which blood must cross a narrowed mitral valve (area \(A_m = 1.5~\text{cm}^2\)) to fill the left ventricle and then a narrowed aortic valve (area \(A_a = 1.0~\text{cm}^2\)) to leave it. When each valve was healthy it passed blood at a peak velocity of \(1.0~\text{m/s}\) through its normal opening area — \(A_{m,0} = 5.0~\text{cm}^2\) for the mitral, \(A_{a,0} = 3.5~\text{cm}^2\) for the aortic. Assume the heart drives the same flow through each valve now as it did when that valve was healthy.

Find the peak jet velocity across each stenotic valve.
Estimate the pressure gradient across each valve.
Which valve imposes the larger gradient, and why is that not simply the one with the smaller opening?

Answer 2

Continuity says the flow through a valve is fixed at its healthy value, \(Q = v_0 A_0\), so the new jet speed is \(v = v_0\,(A_0 / A)\).

Mitral: \(v_m = 1.0 \times (5.0/1.5) = \mathbf{3.33~m/s}\).
Aortic: \(v_a = 1.0 \times (3.5/1.0) = \mathbf{3.5~m/s}\).

With \(\Delta p \approx 4v^2\):

Mitral: \(\Delta p_m \approx 4(3.33)^2 = \mathbf{44.4~mmHg}\).
Aortic: \(\Delta p_a \approx 4(3.5)^2 = \mathbf{49~mmHg}\).

The aortic valve imposes the larger gradient (49 vs 44 mmHg), even though its \(1.0~\text{cm}^2\) opening is not dramatically smaller than the mitral \(1.5~\text{cm}^2\). What matters for the velocity — and hence, squared, for the gradient — is the ratio of healthy to diseased area, \(A_0/A\). That ratio is \(3.5/1.0 = 3.5\) for the aortic valve but only \(5.0/1.5 = 3.33\) for the mitral, so the aortic jet is slightly faster and its gradient slightly higher. What determines the load is the fractional narrowing, not the absolute orifice size.

Problem 3

A coronary artery is partly blocked by plaque, narrowing its diameter to one-third of the healthy value over a short segment. Upstream of the plaque the blood moves at \(v_1 = 0.20~\text{m/s}\), and the heart maintains the same volumetric flow \(Q\) through the artery.

By what factor does the cross-sectional area at the plaque fall, and by what factor does the blood speed up as it passes through?
Find the blood’s speed in the narrowed segment.
Estimate the pressure drop across the plaque with the simplified Bernoulli relation. Is neglecting the upstream velocity justified here?

Answer 3

Area scales as diameter squared, \(A \propto d^2\), so shrinking the diameter to \(1/3\) shrinks the area to \((1/3)^2 = \mathbf{1/9}\). Continuity at fixed flow, \(v \propto 1/A\), then speeds the blood up by \(\mathbf{9\times}\).
\(v_2 = 9 \times 0.20 = \mathbf{1.8~m/s}\).
Simplified Bernoulli: \(\Delta p \approx 4 v_2^2 = 4(1.8)^2 = \mathbf{13~mmHg}\) (more precisely 12.96). Neglecting \(v_1\) is well justified: the full form \(\Delta p = \tfrac12\rho(v_2^2 - v_1^2)/133.3 = 530(1.8^2 - 0.2^2)/133.3 = 12.7~\text{mmHg}\) differs by under 2%, because \(v_1^2 = 0.04\) is tiny beside \(v_2^2 = 3.24\). This is exactly the “narrow jet fed by a slow, wide vessel” condition the simplified formula was built for. (Note this is a continuity question — we held the flow \(Q\) fixed and asked how fast the blood must move. That is the pressure-submodule reading of the plaque; how a long, thin, viscous channel sets its own flow is a different story for a later section.)

From a frequency shift to a velocity to a pressure

Problem 4

A catheter study measures a peak pressure gradient of \(\Delta p = 64~\text{mmHg}\) across a stenotic aortic valve (Heys et al. 2010). A cardiologist wants to know what a non-invasive Doppler scan should have seen instead.

Invert the simplified Bernoulli equation to find the peak jet velocity implied by this gradient.
If a cardiac probe transmitting at \(f_0 = 2.0~\text{MHz}\) were aimed straight down that jet (\(\cos\theta = 1\)) in tissue (\(c = 1540~\text{m/s}\)), what Doppler frequency shift \(f_D\) would its echo carry?
Is that shift audible?

Answer 4

\(v_{\max} = \sqrt{\Delta p / 4} = \sqrt{64/4} = \sqrt{16} = \mathbf{4~m/s}\).
Use the forward Doppler relation \(f_D = 2 f_0 v \cos\theta / c\): \[f_D = \frac{2(2.0\times10^{6})(4)(1)}{1540} = \frac{1.6\times10^{7}}{1540} = 1.04\times10^{4}~\text{Hz} \approx \mathbf{10.4~kHz}.\]
Yes — \(10.4~\text{kHz}\) sits inside the human hearing range (20 Hz – 20 kHz).

Problem 5

Before locating anything, a scanner has to choose a probe. A cardiologist selects a \(f = 5~\text{MHz}\) probe and sends pulses into tissue (\(c = 1540~\text{m/s}\)).

What is the wavelength of this pulse in tissue, and roughly what is the finest detail it can resolve?
One echo returns \(t = 91~\mu\text{s}\) after the pulse is sent (\(1~\mu\text{s} = 10^{-6}~\text{s}\)). How deep is the structure that produced it?
A rule of thumb puts the effective imaging depth at about \(500\) wavelengths. Does this probe reach the structure in (b)?

Answer 5

\(\lambda = c/f = 1540/(5\times10^{6}) = 3.08\times10^{-4}~\text{m} = \mathbf{0.308~mm}\). Since a wave resolves detail about as fine as its wavelength, this probe can distinguish features down to roughly \(\mathbf{0.3~mm}\).
Pulse–echo depth: \(d = \tfrac12 c\,t = \tfrac12 (1540)(91\times10^{-6}) = 0.0701~\text{m} \approx \mathbf{7.0~cm}\) — the factor of one-half because the echo travels out and back.
Effective depth \(\approx 500\lambda = 500 \times 0.308~\text{mm} = 154~\text{mm} = \mathbf{15.4~cm}\), comfortably beyond the 7.0 cm structure. So yes — this probe both resolves fine detail and reaches the target. (Push the frequency higher for even finer detail and the reach would shrink. There is a resolution–penetration trade-off.)

Problem 6

A cardiac probe at \(f_0 = 2.0~\text{MHz}\) is aligned with an aortic-valve jet (\(\cos\theta = 1\)), tissue speed \(c = 1540~\text{m/s}\), and the returning echo is shifted by \(f_D = 7.8~\text{kHz}\).

Find the blood-jet velocity.
Convert it to the pressure gradient across the valve.
A peak jet velocity of about \(4~\text{m/s}\) (gradient \(\approx 64~\text{mmHg}\)) is the usual flag for severe aortic stenosis (Baumgartner et al. 2017). How would you grade this valve?

Answer 6

\(v = \dfrac{c\, f_D}{2 f_0 \cos\theta} = \dfrac{(1540)(7.8\times10^{3})} {2(2.0\times10^{6})(1)} = \dfrac{1.201\times10^{7}}{4.0\times10^{6}} = \mathbf{3.0~m/s}\).
\(\Delta p \approx 4v^2 = 4(3.0)^2 = \mathbf{36~mmHg}\).
A jet of \(3.0~\text{m/s}\) (peak gradient near \(36~\text{mmHg}\)) falls short of the severe cutoff — a peak velocity \(\geq 4~\text{m/s}\) (see the Toolkit note) — so this is a moderate stenosis, worth watching but not yet critical (Baumgartner et al. 2017). The whole grading came from the pitch of an echo: Doppler gave the velocity, and Bernoulli translated it into pressure, without requiring a catheter.

Problem 7

Trace the entire instrument on a suspected aortic stenosis. A \(f_0 = 2.5~\text{MHz}\) cardiac probe is pressed to the chest (tissue \(c = 1540~\text{m/s}\)). The aortic valve returns its echo \(t = 130~\mu\text{s}\) after the pulse, and when the beam is turned to Doppler mode (aligned with the jet, \(\cos\theta = 1\)) the echo comes back shifted by \(f_D = 11.4~\text{kHz}\).

How deep is the valve beneath the probe?
What is the peak velocity of the jet crossing it?
What pressure gradient does that imply?
The patient’s heart rate is \(70~\text{bpm}\) and stroke volume \(70~\text{mL}\). What cardiac output is the left ventricle sustaining while generating this extra gradient on every beat? Name the sections each of the four steps drew on.

Answer 7

Depth from timing: \(d = \tfrac12 c\,t = \tfrac12 (1540)(130\times10^{-6}) = 0.100~\text{m} = \mathbf{10.0~cm}\) (pulse–echo, ultrasound).
Velocity from the Doppler shift: \(v = \dfrac{c\,f_D}{2 f_0 \cos\theta} = \dfrac{(1540)(11.4\times10^{3})} {2(2.5\times10^{6})(1)} = \dfrac{1.756\times10^{7}}{5.0\times10^{6}} = \mathbf{3.5~m/s}\) (Doppler, ultrasound).
Gradient from Bernoulli: \(\Delta p \approx 4(3.5)^2 = \mathbf{49~mmHg}\) (Bernoulli).
Cardiac output: \(\text{CO} = \text{HR}\times\text{SV} = 70 \times 70 = 4900~\text{mL/min} = \mathbf{4.9~L/min}\) (the heart) — a normal resting output, but bought at the cost of an extra \(\approx 49~\text{mmHg}\) the ventricle must raise on every stroke (a moderate stenosis, edging toward severe).

Problem 8

Doppler and continuity together can even measure a valve’s opening area. All the blood the ventricle ejects must pass first through the left-ventricular outflow tract (LVOT), the smooth channel just below the aortic valve, and then through the valve orifice itself; by continuity, the flow \(Q = A v\) is the same at both. A scan measures the LVOT diameter as \(d_{\text{LVOT}} = 2.0~\text{cm}\) and, with Doppler, the blood speed there as \(v_{\text{LVOT}} = 1.0~\text{m/s}\); across the valve the Doppler jet reaches \(v_{\text{AV}} = 4.0~\text{m/s}\).

Find the LVOT cross-sectional area (treat it as a circle).
Use continuity to find the aortic valve area \(A_{\text{AV}}\).

Answer 8

\(A_{\text{LVOT}} = \pi (d_{\text{LVOT}}/2)^2 = \pi (1.0)^2 = \mathbf{3.14~cm^2}\).
Continuity, \(A_{\text{LVOT}} v_{\text{LVOT}} = A_{\text{AV}} v_{\text{AV}}\), so \[A_{\text{AV}} = A_{\text{LVOT}}\,\frac{v_{\text{LVOT}}}{v_{\text{AV}}} = 3.14 \times \frac{1.0}{4.0} = \mathbf{0.785~cm^2}.\] The fast jet (\(4\) vs \(1~\text{m/s}\)) betrays an orifice four times smaller than the outflow tract feeding it.

References

Baumgartner, Helmut, Judy Hung, Javier Bermejo, et al. 2017. “Recommendations on the Echocardiographic Assessment of Aortic Valve Stenosis: A Focused Update from the European Association of Cardiovascular Imaging and the American Society of Echocardiography.” European Heart Journal – Cardiovascular Imaging 18 (3): 254–75. https://doi.org/10.1093/ehjci/jew335.

Heys, Jeffrey J., Nicole Holyoak, Anna M. Calleja, Marek Belohlavek, and Hari P. Chaliki. 2010. “Revisiting the Simplified Bernoulli Equation.” The Open Biomedical Engineering Journal 4: 123–28. https://doi.org/10.2174/1874120701004010123.