Question

Pressure gauge at the top of a vertical oil well registers 140 bars. The oil well is 6000 m deep and filled with natural gas down to a depth of 4700 m and filled with oil (density=700 kg/m3) the rest of the way to the bottom of the well at 15°C. The compressibility factor Z=0.80 for natural gas and its molecular weight is 18.9.
Determine the pressure at (a) the natural gas-oil interface and at (b) the bottom of the well at 15°C.

Answer 1

Explanation:

(a)  The given data is as follows.

              Pressure on top ($P_{o}$) = 140 bar = $1.4 \times 10^{7} Pa$       (as 1 bar = $10^{5}$)

              Temperature = $15^{o}C$ = (15 + 273) K = 288 K

         Density of gas = $\frac{PM}{ZRT}$

                $\frac{dP}{dZ} = \rho \times g$

               $\frac{dP}{dZ} = \int \frac{PM}{ZRT}$

                $\int_{P_{o}}^{P_{1}} \frac{dP}{dZ} = \frac{Mg}{ZRT} \int_{0}^{4700} dZ$

           $ln (\frac{P_{1}}{P_{o}}) = \frac{18.9 \times 10^{-3} \times 9.81 \times 4700 m}{0.80 \times 8.314 J/mol K \times 288 K}$

                              = 0.4548

                     $P_{1} = P_{o} \times e^{0.4548}$

                                 = $1.4 \times 10^{7} Pa \times 1.5797$

                                 = $2.206 \times 10^{7} Pa$

Hence, pressure at the natural gas-oil interface is $2.206 \times 10^{7} Pa$.

(b)   At the bottom of the tank,

                 $P_{2} = P_{1} + \rho \times g \times h$

                             = 2.206 \times 10^{7} Pa + 700 \times 9.81 \times (6000 - 4700)[/tex]

                             = $309.8 \times 10^{5} Pa$

                             = 309.8 bar

Hence, at the bottom of the well at $15^{o}C$ pressure is 309.8 bar.