Respuesta :
Answer:
Pressure at P point is given as
[tex]P_p = 1.042 \times 10^5 Pa[/tex]
Explanation:
As we know by poiseuille's equation of viscous flow of liquid through cylindrical pipe
[tex]Q = \frac{\Delta P \pi r^4}{8\eta L}[/tex]
here we know that
L = 4 cm
[tex]\eta = 1 \times 10^{-3} Pa s[/tex]
r = 1 mm
Q = flow rate
[tex]\pi r^2 v = \frac{\Delta P \pi r^4}{8\eta L}[/tex]
[tex]v = \frac{\Delta P r^2}{8 \eta L}[/tex]
[tex]10 = \frac{\Delta P (10^{-3})^2}{8(1 \times 10^[-3})(0.04)}[/tex]
[tex]3.2 \times 10^{-3} = \Delta P (10^{-6})[/tex]
[tex]\Delta P = 3.2 \times 10^3 Pa[/tex]
now we have
[tex]P_p - P_{atm} = 3.2 \times 10^3[/tex]
[tex]P_p = 1.01 \times 10^5 + 3.2 \times 10^3[/tex]
[tex]P_p = 1.042 \times 10^5 Pa[/tex]
The gauge pressure of the water at point P is;
P_p = 104525 Pa
We are given;
Length of needle; L = 4 cm = 0.04 m
Viscosity of water; η = 1 × 10^(-3) Pa.s
Radius; r = 1 mm = 0.001 m
Speed; v = 10 m/s
From poiseuille's equation of viscous flow of liquid the flow rate through a cylindrical pipe is;
Q = (ΔP•πr⁴)/(8ηL)
Now, flow rate is also expressed as;
Q = πr²v
Thus;
πr²v = (ΔP•πr⁴)/(8ηL)
This reduces to;
v = (ΔP•r²)/(8ηL)
Plugging in the relevant values;
10 = (ΔP × 0.001²)/(8 × 10^(-3) × 0.04)
ΔP = (10 × 8 × 10^(-3) × 0.04)/(0.001²)
ΔP = 3200 Pa
Now, this means that;
P_p - P_atm = ΔP
Where;
P_p is gauge pressure of the water at point P
P_atm is atmospheric pressure = 101325 Pa
Thus;
P_p - 101325 = 3200
P_p = 101325 + 3200
P_p = 104525 Pa
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