I have an interesting question to share with you about how to estimate minimum mud weight required to safely TD the well.
The question is shown below.
7” casing shoe was set at 6,500’MD/5,000’ TVD. The geologist team in town expects 2 hydrocarbon reservoirs and information is listed below;
Formation sand A: Expected depth 5,500’ TVD, pressure gradient is 0.48 psi/ft.
Formation sand B: Expected depth 8,800’ TVD, pressure gradient is 0.49 psi/ft.
The planned TD is 9200’MD/9000’TVD and the drilling team requires 250 psi overbalance while drilling.
What is the mud weight required to drill the well with 250 psi overbalance?
First of all, let’s draw a simple diagram like this.
Knowledge required for this example:
Let’s take a look at each point.
Formation sand A:
Formation pressure of sand A = 0.48 x 5500 = 2,640 psi
With over balance of 250 psi, the hydrostatic pressure required is 2,890 (2640 + 250) psi.
Convert pressure into mud weight = 2890 ÷ (0.052 x 5500) = 10.2 ppg.
Formation sand B:
Formation pressure of sand B = 0.49 x 8800 = 4,312 psi
With over balance of 250 psi, the hydrostatic pressure required is 4,562 (4312 + 250) psi.
Convert pressure into mud weight = 4562 ÷ (0.052 x 8800) = 10.0 ppg
Please always use round up number for the mud weight.
For this case, you must use 10.2 ppg mud weight in order to drill to TD safely.
The point that I would like to emphasize on this topic is that you need to calculate how much pressure required for all sands. Please do not use only the sand that has the maximum pressure gradient.