Pressure gradients at two different sized holes separating 2 chambers at different pressures Why is the pressure gradient by the simplified Bernoulli equation higher at the smaller hole?
 A: (well as far as i could understand your question)
according to the law of continuity which says $A\times v = k$ where k is a constant, $A$ is the area of cross section of the pipe through which the fluid is flowing and the $v$ velocity of the fluid through that cross section. so let is consider two holes, hole 1 (h1) and hole 2 (h2) of $A$ as $A_1$ and $A_2$ and $A_1>A_2$  hence from the above law it is clear that the velocity at $A_2$ is greater than $A_1$ because $A\times v$ has to be constant but this only tells us that velocity is greater at the smaller area of cross section
now according to the Bernoulli principle $P+\frac 12 \rho v^2+\rho gh=constant$ and the pipe at same level has $P+\frac 12 \rho v^2=constant$. now at H1 has the pressure $P_1$ and velocity $v_1$ and H2 has $P_2$ and $v_2$ and we know that $v_2>v_1$. So, according to the Bernoulli equation $P_1+\frac 12 \rho v_1^2=P_2+\frac 12 \rho v_2^2$ and with all the information we see that $P_1$ is greater than $P_2$ 
So from this explanation i want to say that a hole which has a very small cross section area increases the velocity of the fluid very rapidly and hence the pressure changes very rapidly is small interval of time hence pressure gradient (at what rate the pressure changes) is very large at the small holes       
