Textbook Question
Intravenous transfusions are often made under gravity, as shown in Fig. 13–59. Assuming the fluid has a density of 1.00 g/cm3, at what height h should the bottle be placed so the liquid pressure is 52 mm-Hg?
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19. Fluid Mechanics
Intro to Pressure
3:50 minutesProblem 89a
Textbook Question
When a person drives or hikes to a higher altitude, and even more during descent, volume changes of trapped air in the middle ear can cause ear discomfort until the middle-ear pressure and exterior pressure are equalized. If a rapid descent at a rate of 7.0 m/s or faster commonly causes ear discomfort, what is the maximum rate of increase in atmospheric pressure (that is, dP/dt) tolerable to most people?
Verified step by step guidance1
Understand the relationship between pressure and altitude: Atmospheric pressure decreases with increasing altitude and increases with decreasing altitude. The rate of change of pressure with respect to time (dP/dt) depends on the rate of descent and the pressure gradient with altitude.
Use the barometric formula to express the relationship between pressure and altitude: \( P = P_0 e^{-h/H} \), where \( P_0 \) is the pressure at sea level, \( h \) is the altitude, and \( H \) is the scale height of the atmosphere (approximately 8,400 m).
Differentiate the barometric formula with respect to time to find \( \frac{dP}{dt} \): \( \frac{dP}{dt} = \frac{dP}{dh} \cdot \frac{dh}{dt} \). Here, \( \frac{dh}{dt} \) is the rate of descent (given as 7.0 m/s), and \( \frac{dP}{dh} \) can be derived from the barometric formula.
Calculate \( \frac{dP}{dh} \) using the derivative of the barometric formula: \( \frac{dP}{dh} = -\frac{P}{H} \). Substitute this into the expression for \( \frac{dP}{dt} \): \( \frac{dP}{dt} = -\frac{P}{H} \cdot \frac{dh}{dt} \).
Substitute the known values: \( H = 8400 \ \text{m} \), \( \frac{dh}{dt} = -7.0 \ \text{m/s} \) (negative because it is a descent), and \( P \) is the atmospheric pressure at the current altitude. Simplify the expression to find the maximum tolerable \( \frac{dP}{dt} \).
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Atmospheric Pressure
Atmospheric pressure is the force exerted by the weight of air above a given point, typically measured in pascals (Pa) or millibars (mb). As altitude decreases, atmospheric pressure increases due to the greater density of air at lower elevations. Understanding how pressure changes with altitude is crucial for comprehending the discomfort experienced during rapid descents.
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Pressure Equalization
Pressure equalization refers to the process by which the pressure in the middle ear is balanced with the external atmospheric pressure. This is typically achieved through the Eustachian tube, which connects the middle ear to the throat. When there is a rapid change in altitude, the inability to equalize pressure quickly can lead to discomfort or pain in the ears.
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Rate of Change of Pressure (dP/dt)
The rate of change of pressure, denoted as dP/dt, quantifies how quickly the atmospheric pressure changes over time. In the context of rapid descents, a high dP/dt can lead to discomfort as the body struggles to equalize the pressure in the middle ear. Understanding this rate helps in determining safe descent speeds to minimize ear discomfort.
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