Question

During typical urination, a man releases about 400 mL of urine in about 30 seconds through the urethra, which we can model as a tube 4.0 mm in diameter and 20 long. Assume that urine has the same density as water, and that viscosity can be ignored for this flow. 

Part A 

What is the flow speed in the urethra? 

Part B 

If we assume that the fluid is released at the same height as the bladder and that the fluid is at rest in the bladder (a reasonable approximation), what bladder pressure would be necessary to produce this flow? (In fact, there are additional factors that require additional pressure; the actual pressure is higher than this) 

Answer 1

If you have any query about part A, let me know.

Part B.

Using Bernoulli's equation:

P1 + (1/2)*rho*v1^2 + rho*g*h1 = P2 + (1/2)*rho*v2^2 + rho*g*h2

Given that

h2 - h1 = 0 m (from assumption that fluid is released at same height)

rho = density of urine = 1000 kg/m^3 (given that density of urine is same as density of water)

v1 = 0 (Given that initially fluid is at rest in bladder)

v2 = 1.1 m/sec (from part A)

dP = P1 - P2 = Pressure required to produce the flow

So,

P1 - P2 = (1/2)*rho*(v2^2 - v1^2) + rho*g*(h2 - h1)

Using known values:

dP = (1/2)*1000*(1.1^2 - 0^2) + 1000*9.81*(0)

dP = (1/2)*1000*1.1^2

dP = 605 Pa

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