Respuesta :
Answer:
Pressure applied to the needle is 7528 Pa
Explanation:
As we know by poiseuille's law of flow of liquid through a cylindrical pipe
the rate of flow through the pipe is given as
[tex]Q = \frac{\Delta P \pi r^4}{8\eta L}[/tex]
now we know that
[tex]Q = 0.06 \times 10^{-6} m^3/s[/tex]
radius = 0.2 mm
Length = 6.32 cm
[tex]\eta = 1\times 10^{-3} Pa s[/tex]
now we have
[tex]6 \times 10^{-8} = \frac{\Delta P \pi (0.2 \times 10^{-3})^4}{8(1 \times 10^{-3})6.32 \times 10^{-2}}[/tex]
[tex]3.03 \times 10^{-11} = \Delta P 5.02 \times 10^{-15}[/tex]
[tex]\Delta P = 6028 Pa[/tex]
now we have
[tex]P - 1500 = 6028 Pa[/tex]
[tex]P = 7528 Pa[/tex]
The gauge pressure (in Pa) needed at the entrance of the needle to cause this flow is; 7492 Pa
We are given;
Rate of flow; Q' = 0.06 cm³/s = 0.06 × 10⁻⁶ m³/s
Radius; r = 0.2 mm = 0.0002 m
Length; L = 6.32 cm = 0.0632 m
viscosity of the saline solution; η = 1 × 10⁻³ Pa.s
Gauge Pressure; P = 1500 Pa
According to Hagen–Poiseuille's law, the equation for Flow rate of liquid through a cylindrical pipe is;
Q' = ΔP*πr⁴/(8ηL)
ΔP is the pressure difference between the two ends
Plugging in the relevant values gives;
0.06 × 10⁻⁶ = ΔP(π * 0.0002⁴)/(8 * 1 × 10⁻³ * 0.0632)
ΔP = [0.06 × 10⁻⁶ × (8 * 1 × 10⁻³ * 0.0632)]/(π * 0.0002⁴)
ΔP = 0.030336 × 10⁻⁹/(5.0625 × 10⁻¹⁵)
ΔP = 5992 Pa
Thus;
gauge pressure needed at the entrance of the needle to cause this flow is;
5992 Pa + 1500Pa = 7492 Pa
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