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?
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?

a) v =  1.19 m / s , b)   P₁ = 0.922 10⁵ Pa

Explanation:

1) Let's use the fluid continuity equation

       Q = A v

The area of a circle is

      A = π r2 = π d²/4

     v = Q / A = Q 4 / pi d²

     v = 0.006 4/π 0.08²

     v =  1.19 m / s

2) write Bernoulli's equation, where point 1 is the bladder and point 2 is the urine exit point

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

The exercise tell us

P₂ = 1.0013 105 Pa

v₁ = 0

y₁ = 1 m

y₂=0  

Rho (water) = 1000 kg / m³

      P₁ + rho y₁ = P₂ + ½ rho v₂²

      P₁ = P₂ + ½ rho v₂² - rho g y₁

      P₁ = 1.013 10⁵ + ½ 1000 (1.19)² - 1000 9.8 1

      P₁ = 1.013 10⁵ +708.5  - 9800

      P₁ =  92208.5Pa

      P₁ = 0.922 10⁵ Pa

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