Safety Systems for Petroleum Processing

Abstract

The processing of petroleum can be an inherently hazardous activity. The processes we use are handling a flammable material. They use strong, often hazardous, chemicals and employ high pressures and temperatures to convert the oil into finished products. All these factors present safety risks for both personnel and those living near a process facility. Still, the industry experiences very few incidents while processing millions of barrels of oil each day. Refineries place a high value on safety. This chapter discusses four key areas which contribute toward safety performance: personal protective equipment and systems, Process Safety Management (PSM), pressure safety, and temperature safety. A relief valve sizing example is presented.

David S. J. Jones: deceased.

Steven A. Treese has retired from Phillips 66.

Appendix: Example Calculation for Sizing a Relief Valve

Problem

A vessel containing naphtha C₅–C₈ range is uninsulated and is not fireproofed. The vessel is vertical and has a skirt 15′ in length. Dimensions of the vessel are I/D 6′0″, T-T 20′0″, and liquid height to HLL 16′0″. Calculate the valve size for fire condition relief. Set pressure is 120 psig.

Solution

Latent heat of naphtha at 200 °F is 136 Btu/lb=H _L

Q = 21,000 FA^0.82

A = Wetted area and is calculated as follows:

$$ \begin{array}{l}\mathrm{Liquid}\ \mathrm{height}\ \mathrm{above}\ \mathrm{grade}=15+16\ \mathrm{ft}\\ {} =31\ \mathrm{ft}\end{array} $$

Therefore wetted surface of vessel need only be taken to 25 f. above grade which is 25–15 = 10 f. of vessel height:

$$ \begin{array}{rl} {\rm Wetted}\,{\rm Surface} &= \pi D \times 10{\rm ft}\,{\rm for walls} \cr &= 188.5 \, \text{sq ft plus} \, 28.3 \, \text {sq ft for bottom} \cr &= 216.8 \, \text {sq ft} \cr Q &= 21,000 \times 1.0 \times (216.8)^{0.82} 1.729 \times 10^6 {\rm Btu/h} \cr Q/H_{\rm L} &= {{1.729 \times 10^6 } \over {136}} = 12,713\,{\rm lbs /h = }W \cr A &= {{W\sqrt T \sqrt Z } \over {CKP_1 K_{\rm b} \sqrt M }}\end{array} $$

where:

• A = effective discharge area in sq in.

• W = flow through valve in lbs/h, 12,713.

• T = absolute temp of inlet vapor, 460 + 200 = 660 R (Bubble point of C₅-C₈ at, say, 10 psig).

• Z = 0.98 (nC₅).

• C = 356.06 (based on C _A /C _V = 1.4).

• K = 0.65 (typical coefficient of discharge).

• K _b = 0.9

• M = 100 (use C₇).

• P ₁ = set pressure of valve, 134.7 psia.

$$ \begin{array}{l}A=\frac{12,713\times 25.7\times 0.99}{356\times 0.65\times 134.7\times 0.9\times 1}=1.153\ \mathrm{in}{.}^2\\ {} \mathrm{nearest}\ \mathrm{orifice}\ \mathrm{size} = `J' \mathrm{at} 1.287\ \mathrm{in}{.}^2\end{array} $$