An intravenous (IV) system is supplying saline solution to a patient at the rate of 0.14 cm³/s through a needle of radius 0.185 mm and length 2.3 cm. The gauge pressure of the blood in the patient's vein is 8.00 mm Hg. (Assume the viscosity of saline solution is the same as water and the temperature is 20°C) p = 1025kg

A) Calculate the gauge pressure created at a depth of 1.75 m in a saline solution, assuming its density to be that of sea water.

B)  Calculate the new flow rate if the height of the saline solution is decreased to 1.05 m in cm³/s. The viscosity of water is 0.001002 Pa • s.

C)  At what height would the direction of flow be reversed in cm?

C) At what height would the direction of flow be reversed in cm?

Part A.

Gauge pressure at depth of 'h' is given by:

P = rho*g*h

rho = density of saline water = 1025 kg/m^3

g = 9.81 m/sec^2

h = depth = 1.75 m

So,

P = 1025*9.81*1.75 = 17596.7 N/m^2 = 1.76*10^4 N/m^2

Part B.

Using Poiseuelle's law:

Q = dP*pi*r^4/(8*n*L)

r = radius of needle = 0.185 mm = 0.185*10^-3 m

n = Viscosity of water = 0.001002 Pa.sec

L = length of needle = 2.3 cm = 0.023 m

dP = Pressure difference = P2 - P1

P2 = Pressure at height of 1.05 m = rho*g*h1 = 1025*9.81*1.05

P1 = 8.00 mmHg = 1066.58 N/m^2 (since 1 mmHg = 133.32 N/m^2)

So,

Q = (1025*9.81*1.05 - 1066.58)*pi*(0.185*10^-3)^4/(8*0.001002*0.023)

Q = 1.89*10^-7 m^3/sec

Since 1 cm^3 = 10^-6 m^3

Q = (1.89*10^-7 m^3/sec)*(1 cm^3/10^-6 m^3)

Q = 1.89*10^-1 cm^3/sec

Q = 0.189 cm^3/sec

Part C.

Given that gauge pressure of blood in the vein is 8.00 mmHg, So

P = rho*g*h2

h2 = P/(rho*g)

P = 8.00 mmHg = 1066.58 N/m^2

h2 = 1066.58/(1025*9.81)

h2 = 0.106 m = 10.6 cm

So direction of flow will be reversed at the height of 10.6 cm