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Helga
2 months ago
12

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?
Physics
1 answer:
ValentinkaMS [3.4K]2 months ago
8 0

Answer:

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

Explanation:

1) We will apply the fluid continuity equation

       Q = A v

The area of a circle is

      A = π r² = π d²/4

     

     v = Q / A = Q 4 / π d²

     v = 0.006 4/π 0.08²

     v =  1.19 m/s

2) Apply Bernoulli's equation, considering point 1 as the bladder and point 2 as where the urine exits

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

The problem states that

P₂ = 1.0013 x 10⁵ Pa

v₁ = 0

y₁ = 1 m

y₂ = 0

Density of water (ρ) = 1000 kg/m³

      P₁ + ρ y₁ = P₂ + ½ ρ v₂²

      P₁ = P₂ + ½ ρ v₂² - ρ g y₁

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

      P₁ = 1.013 x 10⁵ + 708.5  - 9800

      P₁ =  92208.5 Pa

      P₁ = 0.922 x 10⁵ Pa

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