The pressure gradient in a pipe carrying water is $600\mathrm{Pa}/\mathrm{m}.$ What pressure gradient would be required to maintain the same flow rate in the same pipe if it were carrying olive oil? Treat the flow as the flow of a viscous fluid. The viscosity of water is $1.0\times {10}^{-3}\mathrm{Pa}\cdot \mathrm{s}$ and the viscosity of olive oil is $8.2\times {10}^{-2}\mathrm{Pa}\cdot \mathrm{s}$.

• A. $23\text{kPa}/\mathrm{m}$
• B. $36\text{kPa}/\mathrm{m}$
• C. $58\text{kPa}/\mathrm{m}$
• D. There is not enough information to answer the question.
• E. $49\text{kPa}/\mathrm{m}$

Answer: (E)

Verified step by step guidance

1. Understand that the flow of a viscous fluid in a pipe is described by Poiseuille's Law, which relates the flow rate to the pressure gradient, the viscosity of the fluid, and the dimensions of the pipe.
2. Recall Poiseuille's Law: \( Q = \frac{\pi r^4 \Delta P}{8 \eta L} \), where \( Q \) is the flow rate, \( r \) is the radius of the pipe, \( \Delta P \) is the pressure difference, \( \eta \) is the dynamic viscosity, and \( L \) is the length of the pipe.
3. Since the flow rate \( Q \) and the pipe dimensions \( r \) and \( L \) remain constant, the pressure gradient \( \frac{\Delta P}{L} \) is directly proportional to the viscosity \( \eta \).
4. Set up the proportionality: \( \frac{\Delta P_{water}}{L} \times \eta_{water} = \frac{\Delta P_{oil}}{L} \times \eta_{oil} \).
5. Solve for the new pressure gradient for olive oil: \( \frac{\Delta P_{oil}}{L} = \frac{\Delta P_{water}}{L} \times \frac{\eta_{oil}}{\eta_{water}} \). Substitute the given values to find the required pressure gradient for olive oil.