Question

When you urinate, you increase pressure in your bladder to produce the flow. For an elephant, gravity does the work. An elephant urinates at a remarkable rate of 0.0060 m3 (a bit over a gallon and a half) per second. Assume that the urine exits 1.0 m below the bladder and passes through the urethra, which we can model as a tube of diameter 8.0 cm and length 1.2 m. Assume that urine has the same density as water, and that viscosity can be ignored for this flow.

1) What is the speed of the flow? It is 1.2 m/s
2) If we assume that the liquid is at rest in the bladder (a reasonable assumption) and that the pressure where the urine exits is equal to atmospheric pressure, what does Bernoulli's equation give for the pressure in the bladder? (In fact, the pressure is higher than this; other factors are at work. But you can see that no increase in bladder pressure is needed!)

Answer 1

1) v = 1.19 m / s , 2) P₁ = 9.308 10⁴ Pa

Step-by-step explanation:

In this exercise we will simulate the emission of urine as a fluid mechanics system

1) they indicate the urine flow rate Q = 0.0060 m³ / s, they also give the diameter of the tube 8.0 cm, they ask us the speed.

Let's use the continuity equation

Q = v A

The area of ​​a cycling tube is

A = π r² = π d² / 4

we substitute

Q = v π d² / 4

v = 4Q / π d²

let's calculate

v = 4 0.006 / (π 0.08²)

v = 1.19 m / s

2) they ask to find the pressure in the bladder, for this we use the Bernoulli equation, where the index is for the bladder and the index 2 is for the exit point

P₁ + ½ ρ v₁² + ρ g y₁ = P₂ + ½ ρ v₂² + ρ g y₂

in the exercise it indicates that the outlet pressure is equal to the atmospheric pressure P₂ = 1,013 10⁵ Pa, the velocity of the liquid in the bladder is v₁ = 0 and the height difference 1.0 m

P₁ = P₂ + ½ ρ v₂² + ρ g (y₂-y₁)

let's calculate

P₁ = 1.013₁ 10⁵⁵ + ½ 1000 1.19 + 1000 9.8 (0-1)

P₁ = 1.013 105 + 595 - 9800

P₁ = 9.308 10⁴ Pa