Let's carefully work through the problem step-by-step. --- # Problem Restatement Given the function: \[ f = \frac{dP}{16 \mu L} g(d^2), \] where: - \( P \) is pressure, - \( d \) is the tube diameter, - \( g(d^2) \) is a general function of \( d^2 \), - \( \mu \) (mu) is the dynamic viscosity, - \( L \) is the length of the tube. --- # Goal Find the **sievert (SI) unit** of \( f \), i.e., derive the SI units of \( f \). --- # Step 1: Identify the units of each component Let's identify the SI units of each variable involved: - \( dP \): Pressure difference - SI unit: Pascal (Pa) - \( \mu \): Dynamic viscosity - SI unit: Pascal-seconds (Pa·s) - \( L \): Length - SI unit: meters (m) - \( d^2 \): Square of diameter - SI unit: meters squared (m\(^2\)) - \( g(d^2) \): General function of \( d^2 \) - Since \( g \) is a function of \( d^2 \), it is dimensionless or at least its units depend on \( d^2 \). For the sake of this problem, assume that \( g(d^2) \) is dimensionless, or at least its units cancel out in the ratio. --- # Step 2: Units of the numerator The numerator is: \[ dP \times g(d^2) \] Assuming \( g(d^2) \) is dimensionless: - Units of numerator: \( \text{Pa} \times 1 = \text{Pa} \) --- # Step 3: Units of the denominator The denominator is: \[ 16 \mu L \] Constants like 16 are dimensionless, so focus on the units: - \( \mu \): Pa·s - \( L \): m Thus, \[ \text{Units of denominator} = \text{Pa·s} \times \text{m} = \text{Pa} \times \text{s} \times \text{m} \] --- # Step 4: Combine units Putting it all together: \[ f = \frac{\text{Pa}}{\text{Pa} \times \text{s} \times \text{m}} \times g(d^2) \] Since \( g(d^2) \) is dimensionless, the units of \( f \) are: \[ \frac{\text{Pa}}{\text{Pa} \times \text{s} \times \text{m}} = \frac{1}{\text{s} \times \text{m}} \] --- # **Final SI units of \( f \):** \[ \boxed{ \text{Units of } f = \frac{1}{\text{s} \times \text{m}} = \text{m}^{-1} \text{s}^{-1} } \] --- # **Summary:** The SI unit of \( f \) is **inverse meters per second** (\( \text{m}^{-1} \text{s}^{-1} \)).