Answer:
W = π*g*(0.365625*p + 0.393375)
Explanation:
Solution:
We will take infinitesimal strips of Volume ΔV along the height Δx:
Δ V = pi*r^2 * Height
Δ V = pi*0.5^2 * Δx
Now, we calculate the infinitesimal mass ΔM
ΔM = Density * Volume
ΔM = (1 + p*h)*pi*0.5^2 *Δx
Where, p is measured from water free surface h = x -0.3
ΔM = (1 + (x - 0.3)*p)*pi*0.5^2 *Δx
Force due to gravity is as follows:
ΔF_g = ΔM*g
ΔF_g = (1 + (x - 0.3)*p)*pi*0.5^2 *Δx *g
Work done in moving water distance x:
ΔW = ΔF_g * x
ΔW = (1 + (x - 0.3)*p)*pi*0.5^2 *Δx *g * x
Integrating ΔW over the height b = 0.3 < x < a =0.8 m:
[tex]W = \int\limits^a_b { (1 + (x - 0.3)*p)*pi*0.5^2 *g * x} \, dx \\\\W = pi*0.5^2 *g\int\limits^a_b ({x - 0.3p*x +px^2}) \, dx \\\\W = \frac{pi*g}{4} (0.5x^2 -0.15p*x +\frac{p*x^3}{3} )\\\\W = \frac{pi*g}{4} *(1.62 +1.485p) - \frac{pi*g}{4} *(0.045 - 0.0045p)\\\\W = pi*g*(0.365625p +0.393375)[/tex]
The work done required by the pump is:
W = π*g*(0.365625*p + 0.393375)
