Question

A kill mud of appropriate density is required to be injected in a well such that the shut-in pressure is $6.8 \times 10^{6}\ \mathrm{Pa}$ at a depth of $3500\ \mathrm{m}$. Here, the shut-in pressure is the quantity by which the bottom-hole pressure exceeds the hydrostatic pressure of the original mud at the given depth. The density of the original mud is $1100\ \mathrm{kg/m^3}$. The density of the kill mud is _________ kg/m$^3$ (rounded to two decimal places).

Hint:

For kill-mud calculations at a given depth, add the shut-in pressure head \(\left(\dfrac{P_{\text{si}}}{gh}\right)\) to the original mud density.

Answer 1

Step 1: Write pressure balance at depth $h=3500$ m.
Bottom-hole pressure with kill mud must balance the original hydrostatic plus shut-in:
\[ \rho_k g h = \rho_o g h + P_{\text{si}} \;\Rightarrow\; \rho_k = \rho_o + \frac{P_{\text{si}}}{g h}. \] Step 2: Substitute values.
$\rho_o=1100\ \mathrm{kg/m^3}$,\ $P_{\text{si}}=6.8\times10^{6}\ \mathrm{Pa}$,\ $g=9.81\ \mathrm{m/s^2}$,\ $h=3500\ \mathrm{m}$.
\[ \frac{P_{\text{si}}}{g h}= \frac{6.8\times10^{6}}{9.81\times3500}=198.0486\ \mathrm{kg/m^3}. \] Step 3: Compute kill-mud density.
\[ \rho_k = 1100 + 198.0486 = \mathbf{1298.0486\ \mathrm{kg/m^3}} \Rightarrow \boxed{\text{to two decimals: }1298.05\ \mathrm{kg/m^3}}. \]