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.

Part A

What is the speed of the flow?

Express your answer with the appropriate units.

v =

Part B

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!)

Express your answer with the appropriate units.

pbladder−patm=

Answer 1

A) d = 8 cm = 0.08 m

Area of the tube, A = pi*d^2/4

= pi*0.08^2/4

= 0.005036 m^2

volume flow rate, dV/dt = A*v

v = (dV/dt)/A

= 0.006/0.005036

= 1.19 m/s <<<<<<<<<<-------------------Answer

B) use Bernoulli's theorem,

P_baldder + rho*g*h1 + (1/2)*rho*v1^2 = P_atm + rho*g*h2 + (1/2)*rho*v^2

P_baldder + rho*g*h + 0 = P_atm + 0 + (1/2)*rho*v^2

P_baldder - P_atm = (1/2)*rho*v^2 - rho*g*h

= (1/2)*1000*1.19^2 - 1000*9.8*1

= -9092 Pa <<<<<<<<<<-------------------Answer