1.1 D= 1.1e-5 cm2/s (given) fluid filtration velocity = 1 μm/s (given)

1.1 D= 1.1e-5 cm2/s (given) fluid filtration velocity = 1 μm/s (given)
A) how large of a distance between capillaries would be needed for convection to influence oxygen transport to tissues? B) Based upon reported values for the distance between capillaries, do you think that convection is an important mechanism for oxygen transport in tissues?

Problems at the end of Chapter 1: ICPS
1.1: 𝐷 𝑡𝑖𝑠𝑠𝑢𝑒 =1.1𝑒−5 𝑐 𝑚 2 𝑠 ; fluid filration = 1 um/s A. how large of a distance between capillaries would be needed for convection to influence oxygen transport?

1.1: 𝐷 𝑡𝑖𝑠𝑠𝑢𝑒 =1.1𝑒−5 𝑐 𝑚 2 𝑠 ; fluid filration = 1 um/s A. how large of a distance between capillaries would be needed for convection to influence oxygen transport? 𝑃𝑒= 𝑣𝑒𝑙∗𝐿 𝐷 𝑖𝑗 =1 𝑛𝑜 𝑢𝑛𝑖𝑡𝑠 = 1𝑒−4 𝑐𝑚 𝑠 ∗𝐿 1.1𝑒−5 𝑐 𝑚 2 𝑠 ; L = 0.11 cm

1.1: 𝐷 𝑡𝑖𝑠𝑠𝑢𝑒 =1.1𝑒−5 𝑐 𝑚 2 𝑠 ; fluid filration = 1 um/s B. Quantitatively show importance of convection of oxygen transport in tissues?

1.1: 𝐷 𝑡𝑖𝑠𝑠𝑢𝑒 =1.1𝑒−5 𝑐 𝑚 2 𝑠 ; fluid filration = 1 um/s B. Quantitatively show importance of convection of oxygen transport in tissues? Distance between capillaries = 1e-2 cm and oxygen needs to travel half of this: so… 𝑃𝑒= 𝑣𝑒𝑙∗𝐿 𝐷 𝑖𝑗 =1 𝑛𝑜 𝑢𝑛𝑖𝑡𝑠 = 1𝑒−4 𝑐𝑚 𝑠 ∗0.5𝑒−2 𝑐𝑚 1.1𝑒−5 𝑐 𝑚 2 𝑠 = … so convection is negligible…

Solubility of oxygen in plasma at 37oC is 1.4e-6 mol/L/mmHg (Given)
Heme concentration in RBCs is mol/L = 4CHb (Given) Hct male = 0.45; Hct female = 0.4 (Given) Assume solubility of oxygen in RBCs =solubility in serum (Given) HO2 = oxygen solubility in plasma HHb = oxygen solubility in hemoglobin CO2 = oxygen concentration in blood PO2 = oxygen partial pressure = 95 mmHg arterial and 38 mmHg venous CHb = hemoglobin concentration in blood Sbar = average fractional saturation of hemoglobin =95% art and 70% for venous

Solubility of oxygen in plasma at 37oC is 1.4e-6 mol/L/mmHg (Given)
Heme concentration in RBCs is mol/L = 4CHb (Given) Hct male = 0.45; Hct female = 0.4 (Given) Assume solubility of oxygen in RBCs =solubility in serum (Given) 𝐶 𝑂2 = 𝐻 𝑂2 𝑃 𝑂2 1−𝐻𝑐𝑡 + 4 𝐶 𝐻𝑏 𝑆 𝑏𝑎𝑟 + 𝐻 𝐻𝑏 𝑃 𝑂2 𝐻𝑐𝑡 ? ? HO2 = oxygen solubility in plasma HHb = oxygen solubility in hemoglobin CO2 = oxygen concentration in blood PO2 = oxygen partial pressure = 95 mmHg arterial and 38 mmHg venous CHb = hemoglobin concentration in blood Sbar = average fractional saturation of hemoglobin =95% art and 70% for venous

Heme concentration in RBCs is 0.0203 mol/L = 4CHb (Given)
Solubility of oxygen in plasma at 37oC is 1.4e-6 mol/L/mmHg (Given: Ho2/Po2) Heme concentration in RBCs is mol/L = 4CHb (Given) Hct male = 0.45; Hct female = 0.4 (Given) Assume solubility of oxygen in RBCs =solubility in serum (Given) 𝐶 𝑂2 = 𝐻 𝑂2 𝑃 𝑂2 1−𝐻𝑐𝑡 + 4 𝐶 𝐻𝑏 𝑆 𝑏𝑎𝑟 + 𝐻 𝐻𝑏 𝑃 𝑂2 𝐻𝑐𝑡 ? ? CO2 = HO2*(PO2 )+ 4CHbSbar*Hct HO2 = oxygen solubility in plasma HHb = oxygen solubility in hemoglobin CO2 = oxygen concentration in blood PO2 = oxygen partial pressure = 95 mmHg arterial and 38 mmHg venous CHb = hemoglobin concentration in blood Sbar = average fractional saturation of hemoglobin =95% art and 70% for venous

Heme concentration in RBCs is 0.0203 mol/L = 4CHb (Given)
Solubility of oxygen in plasma at 37oC is 1.4e-6 mol/L/mmHg (Given: Ho2/Po2) Heme concentration in RBCs is mol/L = 4CHb (Given) Hct male = 0.45; Hct female = 0.4 (Given) Assume solubility of oxygen in RBCs =solubility in serum (Given) 𝐶 𝑂2 = 𝐻 𝑂2 𝑃 𝑂2 1−𝐻𝑐𝑡 + 4 𝐶 𝐻𝑏 𝑆 𝑏𝑎𝑟 + 𝐻 𝐻𝑏 𝑃 𝑂2 𝐻𝑐𝑡 ? ? CO2 = HO2*(PO2 )+ 4CHbSbar*Hct How do you calculate the fraction of solubility from this info? HO2 = oxygen solubility in plasma HHb = oxygen solubility in hemoglobin CO2 = oxygen concentration in blood PO2 = oxygen partial pressure = 95 mmHg arterial and 38 mmHg venous CHb = hemoglobin concentration in blood Sbar = average fractional saturation of hemoglobin =95% art and 70% for venous

Gender where Co2 (M) = Ho2*PO2 (M) P02 (mmHg) + 4CHb (M) Sbar (fr) Hct (f) Female art 95 0.0203 0.95 0.4 female ven 38 0.70 male 0.45 PLASMA BOUND TO HEMOGLOBIN 1.724% 98.276% 0.936% 99.064% 1.532% 98.468% 0.832% 99.168%

1.3 Difference in CO2 in serum is 2.27 mL (pre- vs post-lungs) per 100 mL (70%) (Given) And released 1.98 mL Co2 from RBCs (pre- vs post-lungs) per 100 mL (Given) Actual total lost:?

1.3 Difference in CO2 in serum is 2.27 mL (pre- vs post-lungs) per 100 mL (70%) (Given) And released 1.98 mL Co2 from RBCs (pre- vs post-lungs) per 100 mL (Given) Actual total lost = 2.18 mL of Co2/100 mL O2 goes from X to Y mmHg?

1.3 Difference in CO2 in serum is 2.27 mL (pre- vs post-lungs) per 100 mL (70%) (Given) And released 1.98 mL Co2 from RBCs (pre- vs post-lungs) per 100 mL (Given) Actual total lost = 2.18 mL of Co2/100 mL O2 goes from 38 to 100 mmHg Arterial oxygen concentration is XXXXX M vs venous at YYYYY M

1.3 Difference in CO2 in serum is 2.27 mL (pre- vs post-lungs) per 100 mL (70%) (Given) And released 1.98 mL Co2 from RBCs (pre- vs post-lungs) per 100 mL (Given) Actual total lost = 2.18 mL of Co2/100 mL O2 goes from 38 to 100 mmHg Arterial oxygen concentration is M vs venous at M

1.3 Difference in CO2 in serum is 2.27 mL (pre- vs post-lungs) per 100 mL (70%) (Given) And released 1.98 mL Co2 from RBCs (pre- vs post-lungs) per 100 mL (Given) Actual total lost = 2.18 mL of Co2/100 mL O2 goes from 38 to 100 mmHg Arterial oxygen concentration is M vs venous at M (2.4 mM) PV=nRT; 1 mol of gas = 22.4 L at standard thus per 100 mL we gained 5.58 mL of oxygen… Greater propensity to gain oxygen than to lose co2.

1.4 L = 1um thick (Given); s.s. = 0.33 seconds (Given)
Is oxygen diffusion across the alveolus a significant factor in the time required for hemoglobin to oxygenate as it traverses the capillary? A.k.a is this diffusion-limited? L2/Dij = diffusion time = (1e-4)2 cm2/(2e-5cm2/s) = s (vs 0.33 seconds); Not diffusion limited.

1.5 LSA = lateral surface area… order Order Diam (mm) Length #
Circ. (mm) LSA/length *mm^2) total surface area (cm^2) area of each vol of each tot vol (cm^3) cum sa cum vol 11 1 0.024 0.116 300358 E-05 10 2 0.044 0.262 97519 9 3 0.073 0.433 31662 8 4 0.122 0.81 9736 7 5 0.192 1.51 2925 6 0.352 2.72 774 0.533 4.6 202 0.875 8.19 49 1.519 14.26 12 2.486 11.87 5.08 25

1.6 Essentially same as 1.5

Renal artery flow rate = 1.25 L/min
A) Water fraction across glomerulus? B) renal vein flow rate? C) [Na+] leaving glomerulus and in renal vein 1.7 Mass balance (C1*V1=C2*V2 or C1Volrate1*C2Volrate2 RPF = renal plasma flow rate CarteryRPFartery –CveinRPFvein = CurineQbladder In – out = what was diverted RPFartery =RPFvein + Qbladder Water content for male and female = 55% and 60% (could look this up) Per day: CO = 4-8 L/min; vs 990 L/day male and 1080 L/day for female for kidney filtration A) So Males = % and females % B) from table excretion rate of h20 = 1.44 L/day So 1.25 L/min (from problem) –1.44 L/day/1440 min/day=1.249L/min C) CarteryRPFartery –CveinRPFvein = CurineQbladder (Equation 1.6.7) [Na+ vein]/time = [Na+ artery]/time – [Na+ excreted]/time From table: Filtered: 25,200 mmol/day *180 L/day = 140 mM = Cartery From table: excreted: 150 mmol/day CarteryRPFartery –CveinRPFvein = CurineQbladder 140mM*1.25 L/min – Cvein* L/min = Curine* L/min Curine = 150 mmol/day / 1.44 L/day (table) = mM Cvein = mM; slight increase in [Na+] due to loss of water volume

1.8 Permeability = 5e-9 cm/s = km (Given)
Deff in the tissue= 1e-10 cm2/s (Given) Thickness of membrane = 150 um = 150 e-4 cm = 1.5 e-2 cm (Given) A) Biot # = ratio of relative resistances from membrane to tissue = kmL/Deff If Biot >> 1, then tissue-limited or tissue is providing the greatest resistance to transport If Biot << 1, then membrane-limited or membrane is providing the greatest resistance to transport Biot # = (5e-9 cm/s)*(1.5e-6 cm)/(1e-10 cm2/s) = 0.75 Similar transport resistances in membrane compared to tissue.

1.9 VdotOxygen=Q(Cv-Ca) = oxygen consumption rate
Q = pulmonary blood flow Cv or a = venous or arterial oxygen concentrations HO2 = 1.33e-6 M/mmHg HHb = 1.5e-6 M/mmHg 𝐶 𝑂2 = 𝐻 𝑂2 𝑃 𝑂2 1−𝐻𝑐𝑡 + 4 𝐶 𝐻𝑏 𝑆 𝑏𝑎𝑟 + 𝐻 𝐻𝑏 𝑃 𝑂2 𝐻𝑐𝑡 Negative! M/mmHg mmHg Fraction Fraction Fraction mmHg M/mmHg 𝑃 𝑜2 𝑃 50 =26 𝑚𝑚𝐻𝑔 𝑃 𝑜2 𝑃 50 =26 𝑚𝑚𝐻𝑔 = 𝑆 𝑏𝑎𝑟 HO2 = oxygen solubility in plasma HHb = oxygen solubility in hemoglobin CO2 = oxygen concentration in blood PO2 = oxygen partial pressure = 95 mmHg arterial and 38 mmHg venous CHb = hemoglobin concentration in blood Sbar = average fractional saturation of hemoglobin =95% art and 70% for venous

𝑃 𝑜2 𝑃 50 =26 𝑚𝑚𝐻𝑔 2.6 1+ 𝑃 𝑜2 𝑃 50 =26 𝑚𝑚𝐻𝑔 2.6 = 𝑆 𝑏𝑎𝑟

1.9 continued 𝑆 𝑏𝑎𝑟 𝑓𝑜𝑟 𝑣𝑒𝑛𝑜𝑢𝑠 𝑓𝑟𝑎𝑐𝑡𝑖𝑜𝑛 (𝑠𝑎𝑚𝑒 𝑓𝑜𝑟 𝑣𝑒𝑛𝑜𝑢𝑠 𝑑𝑢𝑟𝑖𝑛𝑔 𝑟𝑒𝑠𝑡 𝑜𝑟 𝑒𝑥𝑒𝑟𝑐𝑖𝑠𝑒) = 𝑃 𝑜2 𝑃 50 =26 𝑚𝑚𝐻𝑔 𝑃 𝑜2 𝑃 50 =26 𝑚𝑚𝐻𝑔 = assuming P02 is 100 mmHg, 0.971 𝑆 𝑏𝑎𝑟 𝑓𝑜𝑟 𝑎𝑟𝑡𝑒𝑟𝑖𝑎𝑙 𝑓𝑟𝑎𝑐𝑡𝑖𝑜𝑛 𝑎𝑡 𝑟𝑒𝑠𝑡 = 𝑃 𝑜2 =40 𝑚𝑚𝐻𝑔 𝑃 𝑃 𝑜2 𝑃 = 𝑆 𝑏𝑎𝑟 𝑓𝑜𝑟 𝑎𝑟𝑡𝑒𝑟𝑖𝑎𝑙 𝑓𝑟𝑎𝑐𝑡𝑖𝑜𝑛 𝑑𝑢𝑟𝑖𝑛𝑔 𝑒𝑥𝑒𝑟𝑐𝑖𝑠𝑒 = 𝑃 𝑜2 =15 𝑚𝑚𝐻𝑔 𝑃 𝑃 𝑜2 𝑃 =

Gender where state Co2 (M) = Ho2*PO2 (M) P02 (mmHg) (1-Hct) + Hct (f) (4CHb (M) Sbar Hhb*PO2 Po2) Female art rest 5.32E-05 40.00 0.6 0.4 0.0203 0.754 6.00E-05 exercise 2.00E-05 15.00 0.193 2.25E-05 female ven 1.33E-04 100.00 0.971 1.50E-04 male 0.55 0.45 Ho2 =1.33e-6 M/mmHg Hhb =1.5e-6 M/mmHg VdotOxygen Q* (Cv -- Ca) Rest 5.8 Exercise 25 Male moles/min L/min M

1.10 A) determine rate of oxygen removal from lung venules given:
Inspired air is 21% oxygen; alveolar oxygen partial pressure is 105 mmHg; blood flow rate of 5 L/min; respiration of 10 breaths per minutes/ and 0.56 L per breath; alveolar volume and dead volume is 0.15 L and T=37C; R = L*atm/mol/K and 1atm = 760 mmHg 𝑉 𝑑𝑜𝑡 𝑖𝑛ℎ𝑎𝑙𝑒𝑑 𝐶 𝑖𝑛ℎ𝑎𝑙𝑒𝑑 − 𝐶 𝑎𝑙𝑣𝑒𝑜𝑙𝑎𝑟 =𝑄( 𝐶 𝑣𝑒𝑛𝑜𝑢𝑠 − 𝐶 𝑎𝑟𝑡𝑒𝑟𝑖𝑎𝑙 )

Cinhaled=Pinhaled/RT and Calveolar=Palveolar/RT
𝑉 𝑑𝑜𝑡 𝑖𝑛ℎ𝑎𝑙𝑒𝑑 𝐶 𝑖𝑛ℎ𝑎𝑙𝑒𝑑 − 𝐶 𝑎𝑙𝑣𝑒𝑜𝑙𝑎𝑟 =𝑄( 𝐶 𝑣𝑒𝑛𝑜𝑢𝑠 − 𝐶 𝑎𝑟𝑡𝑒𝑟𝑖𝑎𝑙 ) PART S PV = nRT; P/RT = n/V = C Cinhaled=Pinhaled/RT and Calveolar=Palveolar/RT Cinhaled=(Pinhaled=0.21*1 atm)/(RT= L*atm/mol/K*310K)= M Calveolar=(Palveolar=105/760 atm)/RT = M Males: 0.56 L/breath and females: 0.45 L/breath Dead volume males: 0.19L and female: 0.14 L Males: 10 breaths/min (Inhaled vol – dead volume) Males = 3.7 L/min and females = 3.1 L/min 𝑉 𝑑𝑜𝑡 𝑖𝑛ℎ𝑎𝑙𝑒𝑑 𝐶 𝑖𝑛ℎ𝑎𝑙𝑒𝑑 − 𝐶 𝑎𝑙𝑣𝑒𝑜𝑙𝑎𝑟 = moles of oxygen/min for males and moles of oxygen/min for females Knowing , 22,400 L of oxygen per mole then 236 mL of oxygen/min for males and 197 mL of oxygen/min for females

1.10B What is 𝑉 𝑑𝑜𝑡 𝑖𝑛ℎ𝑎𝑙𝑒𝑑 𝐶 𝑖𝑛ℎ𝑎𝑙𝑒𝑑 − 𝐶 𝑎𝑙𝑣𝑒𝑜𝑙𝑎𝑟 =𝑄( 𝐶 𝑣𝑒𝑛𝑜𝑢𝑠 − 𝐶 𝑎𝑟𝑡𝑒𝑟𝑖𝑎𝑙 ) given that you know: during exercise, blood flow rate can be 25 L/min, and rises to 30 breaths per minute and oxygen demand increases to 4 L/min. Partial pressure of oxygen in the pulmonary artery declines to 20 mmHg but remains at 100 mmHg in the pulmonary vein. Given that 4CHb is M; Hct for females is 0.4 and for males it is HO2 = 1.33e-6 M/mmHg; HHb = 1.5e-6 M/mmHg 𝐶 𝑂2 = 𝐻 𝑂2 𝑃 𝑂2 1−𝐻𝑐𝑡 + 4 𝐶 𝐻𝑏 𝑆 𝑏𝑎𝑟 + 𝐻 𝐻𝑏 𝑃 𝑂2 𝐻𝑐𝑡 𝑃 𝑜2 𝑃 50 =26 𝑚𝑚𝐻𝑔 𝑃 𝑜2 𝑃 50 =26 𝑚𝑚𝐻𝑔 = 𝑆 𝑏𝑎𝑟 = for f/m vein (0.971) artery (0.336) note the pulmonary artery during rest is ~0.754. Hence 𝐶 𝑂2 is: Gender where exc Co2 (M) = Ho2 (M/mmHg)*PO2 P02 (mmHg) (1-Hct) + Hct (f) (4CHb (M) Sbar Hhb*PO2 Po2) Female art Exc 2.66E-05 20.00 0.6 0.4 0.0203 3.00E-05 female ven 1.33E-04 100.00 1.50E-04 male 0.55 0.45 Ho2 =1.33e-6 M/mmHg Hhb =1.5e-6 M/mmHg 𝑉 𝑑𝑜 𝑡 𝑖𝑛ℎ𝑎𝑙𝑒𝑑 = 𝑄 𝐶 𝑣𝑒𝑛𝑜𝑢𝑠 − 𝐶 𝑎𝑟𝑡𝑒𝑟𝑖𝑎𝑙 𝐶 𝑖𝑛ℎ𝑎𝑙𝑒𝑑 − 𝐶 𝑎𝑙𝑣𝑒𝑜𝑙𝑎𝑟 = 4 𝐿/𝑚𝑖𝑛 𝐶 𝑣𝑒𝑛𝑜𝑢𝑠 − 𝐶 𝑎𝑟𝑡𝑒𝑟𝑖𝑎𝑙 𝐶 𝑖𝑛ℎ𝑎𝑙𝑒𝑑 − 𝐶 𝑎𝑙𝑣𝑒𝑜𝑙𝑎𝑟 ; V dot oxygen = V dot inhalation (𝐶 𝑖𝑛ℎ𝑎𝑙𝑒𝑑 − 𝐶 𝑎𝑙𝑣𝑒𝑜𝑙𝑎𝑟 ) Vdotinhaled = Q* (Cv -- Ca)/(Cinhaled-Calveolar) C=n/V = P/RT Female 25 / Male L/min M

1.11 ans 𝑃𝑜𝑤𝑒𝑟=𝑤𝑜𝑟𝑘∗𝑏𝑝𝑠=𝑏𝑝𝑠 𝑃 𝑎 𝑑𝑉 =𝑏𝑝𝑠 𝑃 𝑎 𝑑𝑒𝑙𝑡𝑎𝑉 = [J/s]
Stroke volume Peripheral Resistance Powder of Left side of heart Athlete Rest and exercise Sedentary Person ans Athlete: Raise from 5 L/min to 25 L/min Raise from 60 bpm to 105 bpm Raise arterial pressure from 100 to 130 mmHg Sedentary Person: Raise from 72 bpm to 125 bpm Raise arterial pressure from 100 to 150 mmHg exercise CO= HR* SV Athlete 25 105 SP 125 0.2 rest 5 60 72 Periph resistance = mean art pressure/ CO 5.2 130 6 150 20 100 𝑃𝑜𝑤𝑒𝑟=𝑤𝑜𝑟𝑘∗𝑏𝑝𝑠=𝑏𝑝𝑠 𝑃 𝑎 𝑑𝑉 =𝑏𝑝𝑠 𝑃 𝑎 𝑑𝑒𝑙𝑡𝑎𝑉 = [J/s] bps = beats/sec Mean arterial pressure is assumed to be constant And in Pa, not mmHg V = ventricular volume exercise CO= HR* SV SV m^3 arterial pressure Power (use m^3 not L and use Pa not mmHg) Athlete 25 105 SP 125 0.2 0.0002 rest 5 60 E-05 72 E-05 WHAT?!?!?! Problem 1.11 is 1.11?! That is very exciting!

1.12 V dot oxygen = V dot inhalation (𝐶 𝑖𝑛ℎ𝑎𝑙𝑒𝑑 − 𝐶 𝑎𝑙𝑣𝑒𝑜𝑙𝑎𝑟 )
Body adapts to reduced barometric pressure to extract sufficient oxygen and do work. @3650 m the barometric pressure = 485 mmHg For an oxygen pressure drop of 30 mmHg, determine oxygen uptake rate for a respiration of 20 breaths per min. Typically the partial pressure of oxygen is mmHg so mmHg *485/760 = mmHg and the alveolar pressure would thus be mmHg with a pressure drop of 30 mmHg. C = P/RT = C= 101.85 mmHg 1/760 mmHg/atm = M L atm/(mol*K) 310 K 71.85 V dot oxygen = V dot inhalation (𝐶 𝑖𝑛ℎ𝑎𝑙𝑒𝑑 − 𝐶 𝑎𝑙𝑣𝑒𝑜𝑙𝑎𝑟 ) V dot inhalation =f(Vinhalation −Vdead) = 20 bpm*(0.56L-0.19L) = 7.4L/min Vdot oxygen = 7.4 L/min* (Cinhaled - Calveolar) ( M M) mol/min Assumed to be the same Estimate oxygen saturation in venous blood if hct rises to 0.6 and the partial pressure of oxygen blood is at a partial pressure equal to 98% of the alveolar level The venous blood oxygen saturation will be 0.98 of the alveolar so mmHg *0.98 = mmHg and so oxygen saturation is 𝑃 𝑜2 = 𝑃 50 =26 𝑚𝑚𝐻𝑔 𝑃 𝑜2 = 𝑃 50 =26 𝑚𝑚𝐻𝑔 = 𝑆 𝑏𝑎𝑟 =0.9302

Basal metabolic rate of resting individual = 1650 kcal/day
Fraction of metabolic energy used to pump blood: recalling problem 1.11 (ans: 1.11 J/s ) resting indivudal expends 1650 kcal/day how is kcal = Joule 4.184 kJ/kcal 6903.6 kJ/day 287.65 kJ/hr kJ/min kJ/s J/s at rest 1.11 J/s expended for heart so of energy is expended on pumping heart

Filtration rate mM Excretion rate []urine/[]plasma water 180 L/day 1.44 Na+ 25200 mmol/day 140 150 transport out of urine K+ 720 4 100 Into urine glucose 800 0.5 Out of urine Urea 933 467

1.15 Low mw sugar for assessing Kidney function (i.v. and
Check blood and urine conc In time. Inulin is not reabsorbed 𝐶 𝑖𝑛𝑢𝑙𝑖𝑛 𝑖𝑛 𝑝𝑙𝑎𝑠𝑚𝑎 𝐺𝐹𝑅= 𝐶 𝑖𝑛𝑢𝑙𝑖𝑛𝑒 𝑖𝑛 𝑢𝑟𝑖𝑛𝑒 𝑄 𝑢𝑟𝑖𝑛𝑒 Plasma 0.001 g/mL GFR = 0.125 1.44 L/day 180 1 mL/min 7.5 L/hr L/min 125

1.1 D= 1.1e-5 cm2/s (given) fluid filtration velocity = 1 μm/s (given)

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1.1 D= 1.1e-5 cm2/s (given) fluid filtration velocity = 1 μm/s (given)

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Presentation on theme: "1.1 D= 1.1e-5 cm2/s (given) fluid filtration velocity = 1 μm/s (given)"— Presentation transcript:

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