Kat asked in Science & MathematicsPhysics Β· 2 months ago

When your urinary bladder is full, the bladder pressure can reach up to 60 π‘šπ‘š 𝐻2𝑂.?

When your urinary bladder is full, the bladder pressure can reach up to 60 π‘šπ‘š 𝐻2𝑂.

a. What is this pressure in units of Pascal?

b. Assuming that there is no height difference between your urinary bladder and where your urine comes out, calculate the speed at which your urine comes out. The density of urine is 1030 π‘˜π‘”/π‘š3.

c. If the diameter of a urethra is 0.6 π‘šπ‘š, estimate the volume flow rate of urine as it comes out in units of liters per second.

d. If a full bladder constitutes 500 π‘šπΏ of urine, how long will it take you to remove all of the urine from your bladder?

e. The answer in part (d) is not a realistic time for peeing? What should we factor in in order to make it more realistic?

Update:

The answer for letter e confused me a little because of the 0.5 x10^3 part / 21. I hope you can reply to this. Thanks.

2 Answers

Relevance
  • ?
    Lv 7
    2 months ago
    Favourite answer

    When your urinary bladder is full, the bladder pressure can reach up to 60 π‘šπ‘š of fresh water .

    a. What is this pressure p in units of Pascal?

    p = 60/1000 m * 1000 kg/m^3 * 9.806 N/kg = 588 Pa

    b. Assuming that there is no height difference between your urinary bladder and where your urine comes out, calculate the speed at which your urine comes out. The density of urine is 1030 π‘˜π‘”/π‘š3.

    588 = 1030/2*V^2

    V = √ 588/515 = 1.07 m/sec 

    c. If the diameter of a urethra is 0.6 π‘šπ‘š, estimate the volume flow rate of urine as it comes out in units of liters per second.

    Flow = 1.07*10Β  dm/sec * 0.7854*0.006^2 dm^2 = 0.00030 dm^3/sec ( lps)Β 

    d. If a full bladder constitutes 500 π‘šπΏ of urine, how long t will it take you to remove all of the urine from your bladder?

    tΒ  = 0.5*10^4/3 = 1.660 secΒ 

    e. The answer in part (d) is not a realistic time for peeing? What should we factor in in order to make it more realistic?

    Totally unrealistic ; the real time should in the range ofΒ  half minuteΒ 

    the real average diameter of a urethra is 5 π‘šπ‘š, therefore :

    Flow = 1.07*10 dm/sec * 0.7854*0.05^2 dm^2 = 0.021 dm^3/sec ( lps)Β 

    t = 0.5*10^3/21 = 25 secΒ 

    @ kat : 0.7854 = PI/4Β 

  • 2 months ago

    @oubaas Thanks for the answer but im not sure though where did you get 0.7854

Still have questions? Get answers by asking now.