Draining of a tank with two holes

Im having a problem with equations that im creating for a model that i have built. Using the bernoulli's equation and the continuity equation, I am trying to solve for draining time. The model is a tank that is filled 25 inches with water, the tank then has 2 holes drilled into it with a diameter of 3/4" each, one is 5 inches up from the bottom and the other is 10 inches up from the bottom. The tank diameter is 4". The flow is quasi-steady so when using the bernoulli's equation it can be modeled as steady flow. Pressure at the top of the tank is equal to pressure at the holes so it can simplify out of bernoulli's along with the velocity at the top of the tank which is considered as 0. Im not sure if this is the right place for the question but I was hoping someone could help me out. If not then I hope someone can direct me to somwhere that can help me.

from the looks of it, it don't think you'll get much help here. you may want to try http://www.eng-tips.com/

The tank will never fully drain, since the two oulets for the system are both above 5" from the bottom. Also, I don't think the steady-state assumption is a good one. The flow rate of the water out will vary with the height of the water - at the beginning, the rate will be at its highest; at the end, the water will barely be dripping out.

im sorry, i said that wrong. the system cannot be considered steady-state because water is only flowing out. This idea is completely different from the idea that the flow rate changes depending on the height of the water.

the answer should involve the integral, from t=0s to t=t secs, of the flow rate of the lower hole. When this flow rate equals zero, then the water is at its lowest point and the time to drain has gone by. So, take the integral of the flow of the lower hole, set it equal to zero, and solve for time t.