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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.)

2 Answers

3 votes

Final answer:

a. The flow speed in the urethra is approximately 42,066.67 m/s. b. The bladder pressure necessary to produce this flow is approximately 1.96 kPa.

Step-by-step explanation:

a. To find the flow speed in the urethra, we can use the equation Q = A * v, where Q is the volume flow rate, A is the cross-sectional area of the urethra, and v is the flow speed. Given that the volume flow rate is 400 mL in 30 seconds and the urethra is modeled as a tube with a diameter of 4mm, we can calculate the cross-sectional area as follows:

A = π * r^2 = π * (4mm/2)^2 = 3.14 * (2mm)^2 = 12.56 mm^2

Converting the cross-sectional area to m^2:

A = 12.56 mm^2 * (1 m / 1000 mm)^2 = 0.00001256 m^2

Substituting the values into the equation, we can solve for v:

400 mL / 30 s = 0.00001256 m^2 * v

v = (400 mL / 30 s) / 0.00001256 m^2 = 42066.67 m/s

Therefore, the flow speed in the urethra is approximately 42,066.67 m/s.

b. To determine the bladder pressure necessary to produce this flow, we can use Bernoulli's equation, which states that the sum of the pressure, velocity, and potential energy per unit volume is constant along a streamline. Since the fluid is at rest in the bladder and released at the same height, the potential energy per unit volume is the same for both situations. Therefore, we can equate the pressure and velocity terms in the equation:

P + 1/2 * ρ * v^2 = constant

Assuming the fluid has the same density as water, we can solve for the pressure:

P = constant - 1/2 * ρ * v^2

Substituting the values, we get:

P = constant - 1/2 * 1000 kg/m^3 * (42,066.67 m/s)^2

The constant term depends on the specific conditions of the system, but we can approximate it as the pressure in the bladder when it is full. Assuming a typical bladder pressure of 20 cmH2O (which is approximately 1960 Pa or 1.96 kPa), we can calculate the necessary bladder pressure to produce the flow:

P = 1.96 kPa - 1/2 * 1000 kg/m^3 * (42,066.67 m/s)^2

Therefore, the bladder pressure necessary to produce this flow is approximately 1.96 kPa.

User Erszcz
by
4.0k points
5 votes

Answer:

Step-by-step explanation:

Given:

volume of urine discharged,
V=400~mL=0.4~L=4* 10^(-4)~m^3

time taken for the discharge,
t=30~s

diameter of cylindrical urethra,
d=4*10^(-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=(V)/(t)

where:


A= cross-sectional area of urethra


v= velocity of flow


A.v=(V)/(t)


(\pi d^2)/(4)* v=(4* 10^(-4))/(30)


v=(4*4* 10^(-4))/(30* \pi (4* 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* 9.8* 0.2


P=1960~Pa

User Pierre Salagnac
by
4.0k points