The above image represents the reservoir fluid.

To compute for the reservoir fluid, two essential parameters are needed and these parameters are **External pressure (P _{e}) **and

**Flowing Bottom-Hole Pressure (P**.

_{wf})The formula for calculating the average pressure of the reservoir fluid:

P^{*} = √[^{Pe² + Pwf²} / _{2}]

Where;

P^{*} = Average Pressure

P_{e} = External Pressure

P_{wf} = Flowing Bottom-Hole Pressure

Let’s solve an example;

Find the average pressure of a reservoir fluid with an external pressure of 19 and Flowing bottom-hole pressure is 15.

This implies that;

P_{e} = External Pressure = 19

P_{wf} = Flowing Bottom-Hole Pressure = 15

P^{*} = √[^{Pe² + Pwf²} / _{2}]

P^{*} = √[^{19² + 15²} / _{2}]

P^{*} = √[^{361 + 225} / _{2}]

P^{*} = √[^{586} / _{2}]

P^{*} = √[293]

P^{*} = 17.117

Therefore, the **average pressure** is **17.117 psi;.**

**Calculating the External Pressure when the Average Pressure and the Flowing Button-Hole Pressure is Given.**

P_{e} = √[[P^{*} x 2]² – P_{wf}²]

Where;

P_{e} = External Pressure

P^{*} = Average Pressure

P_{wf} = Flowing Bottom-Hole Pressure

Let’s solve an example;

Find the external pressure of a reservoir fluid with an average pressure of 24 and Flowing bottom-hole pressure is 16.

This implies that;

P^{*} = Average Pressure = 24

P_{wf} = Flowing Bottom-Hole Pressure = 16

P_{e} = √[[P^{*} x 2]² – P_{wf}²]

P_{e} = √[[24 x 2]² – 16²]

P_{e} = √[2048]

P_{e} = 45.25

Therefore, the **external pressure **is **45.25****.**