The system consists of a float B connected to a valve plate A.
Let $y$ be the height of the water level above the valve A.
Opening Condition ($y_1$):
Upward Forces = Downward Forces
$F_B + P_{atm}S_A = mg + (P_{atm} + \rho g y_1)S_A$
$\rho S_B (y_1 – l) g = mg + \rho g y_1 S_A$
Solving for $y_1$:
$$ y_1 = \frac{m + \rho l S_B}{\rho(S_B – S_A)} $$
Closing Condition ($y_2$):
Weight = Buoyant Force
$mg = \rho S_B (y_2 – l) g$
Solving for $y_2$:
$$ y_2 = l + \frac{m}{\rho S_B} $$
The change in level is: $$ \Delta y = y_1 – y_2 = \frac{m + \rho l S_B}{\rho(S_B – S_A)} – \frac{m + \rho l S_B}{\rho S_B} $$ Simplifying: $$ \Delta y = (m + \rho l S_B) \left[ \frac{S_A}{\rho S_B (S_B – S_A)} \right] $$
Filling Time ($t_1$): Rate is $r$. Effective Area is $(S – S_B)$. $$ t_1 = \frac{(S – S_B) \Delta y}{r} $$ Draining Time ($t_2$): Rate is $\eta r – r = (\eta – 1)r$. $$ t_2 = \frac{(S – S_B) \Delta y}{(\eta – 1)r} $$ Total Time ($T$): $$ T = t_1 + t_2 = \frac{(S – S_B) \Delta y}{r} \left( 1 + \frac{1}{\eta – 1} \right) = \frac{\eta (S – S_B) \Delta y}{r(\eta – 1)} $$
Substituting $\Delta y$: $$ T = \frac{\eta}{(\eta – 1)r} \left[ \frac{S_A (S – S_B)}{S_B (S_B – S_A)} \right] \left( l S_B + \frac{m}{\rho} \right) $$
Given: $S=25, S_A=5, S_B=10$ (cm²), $l=10$ cm, $m=100$ g, $r=10$ cc/s, $\eta=1.5$, $\rho=1$ g/cc.
$$ T = \frac{1.5}{(0.5)(10)} \left[ \frac{5 (25 – 10)}{10 (10 – 5)} \right] \left( 10(10) + \frac{100}{1} \right) $$ $$ T = 0.3 \times \left[ \frac{5 \times 15}{50} \right] \times (100 + 100) $$ $$ T = 0.3 \times 1.5 \times 200 $$ $$ T = 90 \text{ s} $$
