Respuesta :
Answer:
[tex]\Delta P=1060184.8946\ Pa[/tex]
[tex]P_1=124651.2383\ Pa[/tex]
Explanation:
Given:
- density of liquid, [tex]\rho=1290\ kg.m^{-3}[/tex]
- speed of flow at location 1, [tex]v_1=9.83\ m.s^{-1}[/tex]
- diameter of pipe at location 1, [tex]d_1=0.121\ m[/tex]
- diameter of pipe at location 2, [tex]d_2=0.177\ m[/tex]
- height of pipe at location 1, [tex]z_1=8.35\ m[/tex]
We know the Bernoulli's equation of in-compressible flow:
[tex]\frac{P}{\rho.g} +\frac{v^2}{2g} + z=constant[/tex] ........................(1)
Cross sectional area of pipe at location 2:
[tex]A_2=\pi \frac{d_2^2}{4}[/tex]
[tex]A_2=\pi\times \frac{0.177^2}{4}[/tex]
[tex]A_2=0.0246\ m^2[/tex]
Cross sectional area of pipe at location 1:
[tex]A_1=\pi \frac{d_1^2}{4}[/tex]
[tex]A_1=\pi\times \frac{0.121^2}{4}[/tex]
[tex]A_1=0.0115\ m^2[/tex]
Using continuity equation:
[tex]A_1.v_1=A_2.v_2[/tex]
[tex]0.0115\times 9.83=0.0246\times v_2[/tex]
[tex]v_2=4.5953\ m.s^{-1}[/tex]
Now apply continuity eq. on both the locations:
[tex]\frac{P_1}{\rho.g} +\frac{v_1^2}{2g} + z_1= \frac{P_2}{\rho.g} +\frac{v_2^2}{2g} + z_2[/tex]
[tex](P_2-P_1) = \rho.g [\frac{v_1^2}{2g} + z_1-\frac{v_2^2}{2g} ][/tex]
[tex]\Delta P=1290\times 9.8 [\frac{9.83^2}{19.6} + 8.35-\frac{4.5953^2}{19.6} ][/tex]
[tex]\Delta P=154266.016\ Pa[/tex]...................................Ans (a)
Now the mass flow rate at location 1:
[tex]\dot{m_1}=\rho\times \dot{V}[/tex]
[tex]\dot{m_1}=1290\times (0.0115\times 9.83)[/tex]
[tex]\dot{m_1}=145.828\ kg.s^{-1}[/tex]
Now pressure at location 1:
[tex]P_1=\frac{\dot{m_1}\times v_1}{A_1}[/tex]
[tex]P_1=\frac{145.828\times 9.83}{0.0115}[/tex]
[tex]P_1=124651.2383\ Pa[/tex] ...................................Ans (b)
The difference between fluid pressure at location 2 and fluid pressure at location 1 is mathematically given as
dP = 114 kPa
What is the difference between fluid pressure at location 2 and fluid pressure at location 1.?
Question Parameter(s):
Generally, the Bernoulli's equation is mathematically given as
P + ρ*g*y + v² =pipe constant
Where
A1*v1 = A2*v2
π*(0.105/2)²*9.91 = π*(0.167/2)²*v2
v2 = 3.9 m/s
Therefore
P1 + ρ*g*y1 + v1² = P2 + ρ*g*y2 + v2²
dP = 1290*9.8*9.01 + 9.91² - 3.9²
dP = 114 kPa
In conclusion, difference between fluid pressure is
dP = 114 kPa
Read more about Pressure
https://brainly.com/question/25688500
