Given:
size of scale model = 4(size of pump)
power ratio of pump and scale model = 5:1
Solution:
Let the diameter of scale model and pump be [tex]d_{s}[/tex] and [tex]d_{p}[/tex] respectively
and head be [tex]H_{s}[/tex] and [tex]H_{p}[/tex] respectively
Now, power, P is given as a function of head(H) and dischagre(Q)
P = [tex]\rho gQH[/tex] (1)
From eqn (1):
[tex]P \propto QH[/tex]
and
[tex]QH \propto \sqrt{H}D^{2}[/tex]
So,
[tex]P \propto H^{\frac{3}{2}} D^{2}[/tex]
Therefore,
[tex]\frac{P_{s}}{P_{p}}[/tex] = [tex]\frac{D_{s}^{2} H_{s}^{\frac{3}{2}}}{D_{p}^{2} H_{p}^{\frac{3}{2}}}[/tex]
[tex]\frac{P_{s}}{P_{p}}[/tex] = [tex]\frac{D_{s}^{2} H_{s}^{\frac{3}{2}}}{D_{p}^{2} H_{p}^{\frac{3}{2}}}[/tex]
[tex]\frac{P_{s}}{P_{p}}[/tex] = [tex]\frac{1^{2}\times 5^{\frac{3}{2}}}{4^{2}\times 1^{\frac{3}{2}}}[/tex]
[tex]\frac{P_{s}}{P_{p}}[/tex] = [tex]\frac{5\sqrt{5}}{16}[/tex]
[tex]{P_{s}}:{P_{p}}[/tex] = [tex]{5\sqrt{5}}:{16}[/tex]
