Question

A milkshake has viscosity of 0.60Pa⋅s. To drink this shake through a straw of diameter 0.56 cm and length 22 cm, you need to reduce the pressure at the top of the straw to less than atmospheric pressure.
If you want to drain a 440mL shake in 2.0 minutes, what pressure difference is needed? You can ignore the height difference between the top and the bottom of the straw.

Express your answer with the appropriate units.

Answer 1

Answer:

The pressure difference is $2.01X10^{4} Pa$.

Explanation:

To calculate the pressure difference, we use the fluid-kinetic equation:

Δp = (8μLQ)/(πR⁴)

Step 1: Convert the units of the given terms to SI units:

Diameter = 0.56 cm X ($\frac{1 m}{100 cm}$ = $5.6X10^{-3} m$

∴ Radius, R = $5.6X10^{-3} /2=2.8X10^{-3} m$

Length, L = 22 cm X ($\frac{1 m}{100 cm}$ = 0.22 m

Volumetric flow rate, Q = $440X10^{-3} L$ ÷ 2.0 minutes = $220X10^{-3} \frac{L}{min}$

$220X10^{-3} \frac{L}{min}$ X $\frac{1m^{3} }{1000L}$ X $\frac{1 min}{60 s}$ = $3.67X10^{-6} \frac{m^{3} }{s}$

Step 2: Input the terms into the fluid-kinetic equation:

Δp = (8 X 0.60 X 0.22 X 3.67$X10^{-6}$) / (3.142 X ($(2.8X10^{-3} )^{4}$)

Δp = 20067.43 Pa = $2.01X10^{4} Pa$.

N.B: The pressure difference is lower than the atmospheric pressure (101300 Pa), which means you can drink the milkshake through the straw.