Problem 6 - Biofluid Mechanics: An Introduction to Fluid Mechanics, Macrocirculation, and Microcirculation

Question

The variation of air viscosity as a function of temperature can be represented by the following equation $$ \mu=\frac{\left(1.458 \times 10^{-6} \frac{\mathrm{kg}}{\mathrm{msK} K^{\frac{1}{2}}}\right) T^{\frac{3}{3}}}{1+\frac{110.4 K}{T}} $$ Estimate changes in air viscosity as a function of altitude in Earth's atmosphere. Assume that the temperature is approximately \(-60^{\circ} \mathrm{C},-40^{\circ} \mathrm{C}\), and \(-80^{\circ} \mathrm{C}\) at an altitude of \(10 \mathrm{~km}, 30 \mathrm{~km}\), and \(75 \mathrm{~km}\), respectively.

Step-by-Step Solution

Answer

At 10 km, viscosity is approximately 3.08 x 10^-5 kg/(m.s); at 30 km, 3.62 x 10^-5 kg/(m.s); at 75 km, 2.54 x 10^-5 kg/(m.s).

Step 1: Convert Temperatures to Kelvin

The temperature given is -60°C for 10 km,-40°C for 30 km, and-80°C for 75 km.Convert these temperatures to Kelvin using the formula: \[ T(K) = T(°C) + 273.15 \] Thus, we have - for 10 km, \( T = -60 + 273.15 = 213.15 \) K,- for 30 km, \( T = -40 + 273.15 = 233.15 \) K,- for 75 km, \( T = -80 + 273.15 = 193.15 \) K.

Step 2: Calculate Viscosity at 10 km

Using the provided formula: \[ \mu = \frac{1.458 \times 10^{-6} \times (213.15)^{3/2}}{1 + \frac{110.4}{213.15}} \]- Calculate the numerator: \( 1.458 \times 10^{-6} \times (213.15)^{3/2} = 4.684 \times 10^{-5} \)- The denominator: \( 1 + \frac{110.4}{213.15} = 1.518 \)- Viscosity: \( \mu \approx \frac{4.684 \times 10^{-5}}{1.518} \approx 3.08 \times 10^{-5} \) kg/(m.s).

Step 3: Calculate Viscosity at 30 km

Using the formula: \[ \mu = \frac{1.458 \times 10^{-6} \times (233.15)^{3/2}}{1 + \frac{110.4}{233.15}} \]- Calculate the numerator: \( 1.458 \times 10^{-6} \times (233.15)^{3/2} = 5.337 \times 10^{-5} \)- The denominator: \( 1 + \frac{110.4}{233.15} = 1.474 \)- Viscosity: \( \mu \approx \frac{5.337 \times 10^{-5}}{1.474} \approx 3.62 \times 10^{-5} \) kg/(m.s).

Step 4: Calculate Viscosity at 75 km

Using the formula: \[ \mu = \frac{1.458 \times 10^{-6} \times (193.15)^{3/2}}{1 + \frac{110.4}{193.15}} \]- Calculate the numerator: \( 1.458 \times 10^{-6} \times (193.15)^{3/2} = 4.001 \times 10^{-5} \)- The denominator: \( 1 + \frac{110.4}{193.15} = 1.572 \)- Viscosity: \( \mu \approx \frac{4.001 \times 10^{-5}}{1.572} \approx 2.54 \times 10^{-5} \) kg/(m.s).

Key Concepts

Temperature Conversion

Often, when studying scientific problems, especially those involving physical properties, temperatures are given in Celsius. However, many scientific formulas, like the one used to calculate air viscosity, require temperatures in Kelvin.
The Kelvin scale is a thermodynamic temperature scale where absolute zero, the point at which there is a complete absence of heat energy, is 0 K. In contrast, Celsius starts at 0 °C, which is the freezing point of water. To convert Celsius to Kelvin, you add 273.15 to the Celsius temperature:

• -60 °C becomes 213.15 K
• -40 °C becomes 233.15 K
• -80 °C becomes 193.15 K

This conversion is crucial for calculating various physical properties, not only viscosity. Hence, always remember: a quick addition of 273.15 transforms Celsius to Kelvin.

Altitude Effect on Air Properties

Air properties change with altitude due to variations in pressure, temperature, and density in different layers of the atmosphere. Essentially, as altitude increases, the atmosphere becomes thinner, which affects these properties. Let’s explore these effects on air viscosity.
When you ascend from ground level to higher altitudes, the pressure significantly drops. A drop in pressure typically results in a decrease in air density. However, the temperature also plays a vital role in influencing air viscosity at these elevations. At higher altitudes, the temperature is generally lower, which directly affects viscosity.

Specific Examples:

• At 10 kilometers, the temperature might be around -60 °C, leading to a specific viscosity value.
• At 30 kilometers, with further lower temperatures at approximately -40 °C, the viscosity changes again.
• Reaching 75 kilometers with a chilling -80 °C temperature, results in the lightest and least viscous air due to its reduced kinetic energy.

In these examples, as you go higher, since air temperature decreases, it could counterbalance the thinning effect of lower pressure, contributing to changes in viscosity. Understanding this relationship is critical when studying aircraft performance, meteorology, and even climate science.

Calculating Viscosity Based on Temperature

Viscosity measures a fluid's resistance to deformation or flow. In air, viscosity is a dynamic property influenced by temperature. The formula used to calculate air viscosity, as derived from Sutherland’s formula, evaluates this relationship.
To estimate air viscosity at different temperatures, the formula provided is:\[\mu = \frac{1.458 \times 10^{-6} \times T^{3/2}}{1 + \frac{110.4}{T}}\]Where:

• \( T \) is the temperature in Kelvin.
• The constant \( 1.458 \times 10^{-6} \) \(\frac{\text{kg}}{\text{msK}^{\frac{1}{2}}}\) scales the viscosity.

Practical Calculations:

For example, at 10 km with a temperature of 213.15 K:

• Calculate the numerator: \( 1.458 \times 10^{-6} \times (213.15)^{3/2} = 4.684 \times 10^{-5} \)
• Calculate the denominator: \( 1 + \frac{110.4}{213.15} = 1.518 \)
• Thus, viscosity \( \mu \approx 3.08 \times 10^{-5} \) kg/(m.s)

Repeat similar calculations for a temperature of 233.15 K at 30 km and 193.15 K at 75 km to see how viscosity varies based on temperature at different altitudes. This method helps in understanding how temperature modifies viscosity, offering insights into air resistance or fluid flow changes under varying atmospheric conditions.