Respuesta :
Answer:
P = 98052.64 Pa or 98.05 kPa
Explanation:
Using Bernoulli's equation;
P+rho*v^2/2+ rho*g*z = constant
at high end;
dia= 1.2 m
flow, Q = 5400 L/min = 0.09 m^3/s
therefore, velocity at the high end, v1 = Q/A =0.09/(pi()*(1.2/2)^2) = 0.08 m/s
pressure, P = 68.67 kPa
Solving for elevation, z
assume lower end is reference line. then higher end will be 'x' m high wrt to lower end.
x= 300*sin(tan^-1(1/100)) = 3 m
that means higher end will be 3 m above with respect to lower end
Similarly for lower end;
dia= 0.6 m
flow, Q = 5400 L/min = 0.09 m^3/s
therefore, velocity at the high end, v2 = Q/A =0.09/(pi()*(0.6/2)^2) = 0.318 m/s
assume pressure, P
z=0
put all values in the formula, we get;
68.67*10^3+1000*0.08^2/2+ 1000*9.81*3 =P+1000*0.318^2/2+ 0
solving this, we get;
P = 98052.64 Pa or 98.05 kPa
According to Bernoulli's principle, a lower pressure than expected at the
lower end because the velocity of the fluid is higher.
- The pressure at the low end is approximately 97.964 kPa.
Reasons:
The given parameter are;
Length of the pipe, L = 300 m
Slope of the pipe = 1 in 100
Diameter at the high end, d₁ = 1.2 m
Diameter at the low end, d₂ = 0.6 m
The volume flowrate, Q = 5,400 L/min
Pressure at the high end, P = 68.67 kPa
Required:
Pressure at the low end
Solution:
The elevation of the pipe, z₁ = [tex]300 \, m \times \frac{1}{100} = 3 \, m[/tex]
z₂ = 0
The continuity equation is given as follows;
Q = A₁·v₁ = A₂·v₂
[tex]\sqrt[n]{x} \displaystyle Q = 5,400 \, L/min = 5,400 \, \frac{L}{min} \times \frac{1 \, m^3}{1,000 \, L} \times \frac{1 \, min}{60 \, seconds} = \mathbf{ 0.09 \, m^3/s}[/tex]
Therefore;
[tex]\displaystyle 0.09 = \frac{\pi}{4} \times 1.2^2 \times v_1 = \mathbf{\frac{\pi}{4} \times 0.6^2 \times v_2}[/tex]
[tex]\displaystyle v_1 = \frac{0.09}{\frac{\pi}{4} \times 1.2^2 } \approx 0.0796[/tex]
[tex]\displaystyle v_2 =\frac{0.09}{\frac{\pi}{4} \times 0.6^2 } \approx \mathbf{0.318}[/tex]
The Bernoulli's equation is given as follows;
[tex]\displaystyle \frac{p_1}{\rho \cdot g} + \frac{v_1^2}{2 \cdot g} + z_1 = \mathbf{ \frac{p_2}{\rho \cdot g} + \frac{v_2^2}{2 \cdpt g} + z_2}[/tex]
Therefore;
[tex]\displaystyle p_2 = \left(\frac{p_1}{\rho \cdot g} + \frac{v_1^2}{2 \cdot g} + z_1 - \left( \frac{v_2^2}{2 \cdot g} + z_2 \right)\right) \times \rho \cdot g = p_1 + \frac{\rho}{2} \cdot \left(v_1^2} -v_2^2 \right)+ \rho \cdot g \left(z_1 - z_2\right)[/tex]
The density of water, ρ = 997 kg/m³
Which gives;
[tex]\displaystyle p_2 = 68670 + \frac{997 }{2} \times \left(0.0796^2- 0.318^2 \right) + 997 \times 9.81 \times \left( 3 - 0 \right) \approx \mathbf{97,964.46}[/tex]
- The pressure at the low end, p₂ ≈ 97,964.46 Pa ≈ 97.964 kPa
Learn more about Bernoulli's principle here:
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