Question

A circular reservoir with a well in the centre is producing oil at a constant rate under pseudo steady state. The rate of pressure change measured at the wellbore is 1.0 psi/day. Pressure at the external radius of the reservoir at a certain time was 3500 psi. After 500 days, the pressure (in psi) at the external radius will be __________.

Hint:

Under pseudo steady state, the pressure decline rate is the same at every point in the reservoir.

Answer 1

Correct Answer: 3000

Step 1: Recall the defining property of pseudo steady state (PSS) flow:
In a closed, bounded circular reservoir producing at a constant rate, flow eventually reaches pseudo steady state once the pressure transient has reached the outer boundary. In this regime, the shape of the pressure profile between the wellbore and the external radius becomes fixed with time, and the entire pressure profile simply shifts downward at the same uniform rate everywhere in the reservoir, from the wellbore up to the external radius. This is the key distinguishing feature of PSS flow compared with the earlier transient flow period, where the depletion rate varies with radial position.
Step 2: Apply this property to the given data:
Since the rate of pressure decline is uniform throughout the reservoir during PSS, the rate measured at the wellbore, 1.0 psi/day, is the same rate of pressure decline that applies at the external radius as well.
Step 3: Compute the total pressure drop over 500 days:
\[ \Delta p = \left(\frac{dp}{dt}\right) \times t = 1.0 \ \text{psi/day} \times 500 \ \text{days} = 500 \ \text{psi} \]
Step 4: Compute the final pressure at the external radius:
The pressure at the external radius at the starting time was 3500 psi. After 500 days of depletion at the uniform PSS rate, \[ p_e(\text{after 500 days}) = 3500 - 500 = 3000 \ \text{psi} \]
Final Answer:
\[ \boxed{3000 \ \text{psi}} \]