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 mm in diameter and 20 cm long. Assume that urine has the same density as water, and that viscosity can be ignored for this flow.a. What is the flow speed in the urethra?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

Answer:

Explanation:

Given:

volume of urine discharged, $V=400~mL=0.4~L=4\times 10^{-4}~m^3$

time taken for the discharge, $t=30~s$

diameter of cylindrical urethra, $d=4\times10^{-3}~m$

length of cylindrical urethra, $l=0.2~m$

density of urine, $\rho=1000~kg/m^3$

a)

we have volume flow rate Q:

$Q=A.v$ & $Q=\frac{V}{t}$

where:

$A=$ cross-sectional area of urethra

$v=$ velocity of flow

$A.v=\frac{V}{t}$

$\frac{\pi d^2}{4}\times v=\frac{4\times 10^{-4}}{30}$

$v=\frac{4\times4\times 10^{-4}}{30\times \pi (4\times 10^{-3})^2}$

$v=1.06~m/s$

b)

The pressure required when the fluid is released at the same height as the bladder and that the fluid is at rest in the bladder:

$P=\rho.g.l$

$P=1000\times 9.8\times 0.2$

$P=1960~Pa$