Please have patience with me as I have not looked at this sort of theory in many years (I am also very new to StackExchange).
A question asks the following:
A sloping swimming pool two-thirds full of sea water provides the supply for a fixed installation protecting a risk situated above the pool.
The pool measures 12m long, 6m wide and 1m at the shallow end sloping to 2.5m at the deep end. A pipe is placed so that its inlet is at the bottom of the deep end of the pool. A pump imparts pressure to the water in the pipe such that a mercury manometer indicates a pressure difference of 681mm Hg between inlet and outlet. The inlet of the pipe is 80mm diameter; the outlet nozzle is 45mm in diameter and is 5m above the inlet. The inlet velocity of the water is 5m/s, the density of sea water is 1050kgm-3 and the density of mercury is 13600kgm-3.
Use Bernoulli’s theorem to calculate the velocity of the water through the outlet nozzle?
By applying the continuity equation the following is found:
$v_1A_1=v_2A_2$
$v_2=15.8m/s$
However, the solution is given as $v_2=10m/s$
If I apply Bernoulli’s equation:
$$ \frac{P_2 -P_1}{ρ}+\frac{1}{2}(v^2_2-v^2_1)+g(z_2-z_1)=0 $$ $$ \frac{-90792.5}{1050}+\frac{1}{2}(v^2_2-25)+9.81(5)=0 $$
$$ v_2=10m/s $$
This is the answer that the solution gives but continuity is not met as shown above. Is this question ill-posed?
I am very confused why continuity is not met when Bernoulli’s equation is used and why the two methods produce different results. A similar question was asked here but this was for a system without a pump and I am not sure if it is applicable.
My question is: Can Bernoulli’s equation be used when the system has a pump or is it no longer valid due to energy not being conserved?