Input information
catheter flow rate | flow rate | 1 mL/s (milliliter per second) catheter length | 15 cm (centimeters) pressure at the end of catheter proximal to site of insertion | 1900 Pa (pascals) pressure at the end of catheter distal to site of insertion | 3100 Pa (pascals) dynamic viscosity | 0.00682 poise
Results
internal radius of catheter | 682.6 µm (micrometers) = 0.02687 inches = 0.6826 mm (millimeters) = 0.06826 cm (centimeters) = 6.826×10^-4 meters
Possible intermediate steps
Calculate the internal radius of catheter using the following information: known variables | | FR | flow rate | 1 mL/s L | catheter length | 15 cm P_1 | pressure at the end of catheter proximal to site of insertion | 1900 Pa P_2 | pressure at the end of catheter distal to site of insertion | 3100 Pa η | dynamic viscosity | 0.00682 P Convert known variables into appropriate units using the following: 1 mL/s = 1×10^-6 m^3/s: 1 cm = 0.01 m: 1 Pa = 1000 g/(m s^2): 1 Pa = 1000 g/(m s^2): 1 P = 100 g/(m s): known variables | | FR | flow rate | 1/1000000 m^3/s L | catheter length | 3/20 m P_1 | pressure at the end of catheter proximal to site of insertion | 1.9×10^6 g/(m s^2) P_2 | pressure at the end of catheter distal to site of insertion | 3.1×10^6 g/(m s^2) η | dynamic viscosity | 0.682 g/(m s) The relevant equation that relates internal radius of catheter (r), flow rate (FR), catheter length (L), pressure at the end of catheter proximal to site of insertion (P_1), pressure at the end of catheter distal to site of insertion (P_2), and dynamic viscosity (η) is: FR = ((P_2 - P_1) (π r^4))/(8 L η) FR = (π r^4 (-P_1 + P_2))/(8 L η) is equivalent to (π r^4 (-P_1 + P_2))/(8 L η) = FR: (π r^4 (-P_1 + P_2))/(8 L η) = FR Divide both sides by (π (P_2 - P_1))/(8 η L): r^4 = -(8 FR L η)/(π (P_1 - P_2)) Take the positive fourth root of both sides: r = 2^(3/4) π^(-1/4) (-(FR L η)/(P_1 - P_2))^(1/4) Substitute known variables and constants into the equation: known variables | | FR | flow rate | 1/1000000 m^3/s L | catheter length | 3/20 m P_1 | pressure at the end of catheter proximal to site of insertion | 1.9×10^6 g/(m s^2) P_2 | pressure at the end of catheter distal to site of insertion | 3.1×10^6 g/(m s^2) η | dynamic viscosity | 0.682 g/(m s) constant | | π | pi | 3.14159 | : r = 2^(3/4) (-(0.682 g/(m s)×1×10^-6 m^3/s×0.15 m)/(1.9×10^6 g/(m s^2) - 3.1×10^6 g/(m s^2)))^(1/4)×3.14159^(-1/4) Separate the numerical part, 2^(3/4) (-(0.682×1×10^-6×0.15)/(1.9×10^6 - 3.1×10^6))^(1/4)×3.14159^(-1/4), from the unit part, ((g/(m s)×m^3/s m)/(g/(m s^2) + g/(m s^2)))^(1/4) = m: r = 2^(3/4) (-(0.682×1×10^-6×0.15)/(1.9×10^6 - 3.1×10^6))^(1/4)×3.14159^(-1/4) m Evaluate 2^(3/4) (-(0.682×1×10^-6×0.15)/(1.9×10^6 - 3.1×10^6))^(1/4)×3.14159^(-1/4): r = 6.8259×10^-4 m Convert 6.8259×10^-4 m into µm (micrometers) using the following: 1 m = 1×10^6 µm: Answer: | | r = 682.6 µm
Equation
FR = ((P_2 - P_1) (π r^4))/(8 L η) | r | internal radius of catheter FR | flow rate L | catheter length P_1 | pressure at the end of catheter proximal to site of insertion P_2 | pressure at the end of catheter distal to site of insertion η | dynamic viscosity