4. (20 pts) A human bladder can hold 15.5 ounces of urine from the kidney. A pat

Question

4. (20 pts) A human bladder can hold 15.5 ounces of urine from the kidney. A patient has a chronic bladder leak due to insufficient closure on the urethra, with a leak rate constant, ki. The patient has a constant input rate from the kidneys, R. Solve a single compartment model of the patient's bladder: input-output + dt (5 pts) First, write a change equation for the liquid volume ( generation - consumption) and put it into our standard differential equation form. (10 pts) Now, you should find the integration factor and solve the differential equation. Solve for the integration constant by using an initial volume within the bladder of V, (5 pts) Given k,-0.015/min, V,-0.65 ounces, and R = 0.275 ounce/minute, determine the amount of time until the patient has a full bladder and will need to relieve themselves (i.e when there is 15.5 ounces in the bladder). a. b. c.

Explanation / Answer

Let V(t) be the volume of liquid in th ebladder at any time t

There is no consumption within the bladder Consumption =0

There is no input   input =0

output is kV, constant leak rate k

generation rate is R - const

the rate of change of liquid volume in the bladder

dV/dt = R- kV

This the differential eq.

making it variable seperable form

dV/(R-kV) = dt

-ln(R-kV)/k = t + c , c is integration const.

re-write it in exponntial form

R-kV = C exp (-kt) , C is the new integration const. ( kec )

intial condition V = Vo when t=0

C = R-kVo

V(t) = R/k -(R/k -Vo) exp (-kt)

given Vo = 0.65 ounces, R = 0.275 oun/min, k = 0.015 /min

bladder size = 15.5 oun.

15.5 = 0.275/0.015 -( 0.275/0.015 - 0.65) exp(-0.015 t)

t = 122.14 min. the bladder will be full.

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