HELP I can\'t get these last two HW problems Glucose solution is administered to
ID: 2032032 • Letter: H
Question
HELP I can't get these last two HW problems
Glucose solution is administered to a patient in a hospital. The density of the solution is 1.138 kg/l. If the blood pressure in the vein is 27.8 mmHg, then what is the minimum necessary height of the IV bag above the position of the needle?
Find the minimum diameter of an l = 18.7 m long brass wire that will stretch no more than 9.53 mm when a mass of 336 kg is hung on the lower end. (Hint: The Young's modulus of brass is 91.0 GPa.)
Glucose solution is administered to a patient in a hospital. The density of the solution is 1.138 kg/l. If the blood pressure in the vein is 27.8 mmHg, then what is the minimum necessary height of the IV bag above the position of the needle?
Explanation / Answer
1)
given
rho_solution = 1.138 kg/L
= 1138 kg/m^3
P = 27.8 mmHg
= rho_mercury*g*h
= 13600*9.8*27.8*10^-3 pa
let H is the height required,
P = rho_solution*g*H
H = P/(rho_solution*g)
= 13600*9.8*27.8*10^-3/(1138*9.8)
= 0.332 m or 33.2 cm <<<<<<=-------------Answer
2)
L = 18.7 m
delta_L = 9.53 mm = 9.53*10^-3 m
F = m*g
= 336*9.8 N
let d is the minimum diameter.
we know, Y = stress/starin
Y = (F/A)/(delta_L/L)
Y*delta_L/L = F/A
Y*delta_L/L = F/(pi*d^2/4)
pi*d^2/4 = F*L/(Y*delta_L)
d = sqrt(4*m*g*L/(Y*delta_L*pi))
= sqrt(4*336*9.8*18.7/(91*10^9*9.53*10^-3*pi))
= 9.51*10^-3 m or 9.51 mm
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