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Air samples from a process area are continuously drawn
Chapter 10, Problem 10.1(choose chapter or problem)
Air samples from a process area are continuously drawn through a ¼-in diameter tube to an analytical instrument that is located 40 m away. The tubing has an outside diameter of 6.35 mm and a wall thickness of 0.762 mm. The flow rate through the transfer line is \(10\ \mathrm {c m^{3} / s}\) for ambient conditions of \(20^{\circ} \mathrm{C}\) and 1 atm. The pressure drop in the transfer line is negligible. Because chlorine gas is used in the process, a leak can poison workers in the area. It takes the analyser 5 s to respond after chlorine first reaches it. Determine the amount of time that is required to detect a chlorine leak in the processing area. State any assumptions that you make. Would this amount of time be acceptable if the hazardous gas were carbon monoxide, instead of chlorine? (Adapted from: Student for Safety, Health, and Loss Prevention in Chemical Processes, AIChE Centre for Chemical Process Safety, NY, 1990).
Questions & Answers
QUESTION:
Air samples from a process area are continuously drawn through a ¼-in diameter tube to an analytical instrument that is located 40 m away. The tubing has an outside diameter of 6.35 mm and a wall thickness of 0.762 mm. The flow rate through the transfer line is \(10\ \mathrm {c m^{3} / s}\) for ambient conditions of \(20^{\circ} \mathrm{C}\) and 1 atm. The pressure drop in the transfer line is negligible. Because chlorine gas is used in the process, a leak can poison workers in the area. It takes the analyser 5 s to respond after chlorine first reaches it. Determine the amount of time that is required to detect a chlorine leak in the processing area. State any assumptions that you make. Would this amount of time be acceptable if the hazardous gas were carbon monoxide, instead of chlorine? (Adapted from: Student for Safety, Health, and Loss Prevention in Chemical Processes, AIChE Centre for Chemical Process Safety, NY, 1990).
ANSWER:Step 1 of 5
Calculate the value of the inside diameter.
\(D_{i} =\text { outside diameter }-2(\text { thickness }) \)
\(=6.35-2(0.762) \)
\( =4.826 \mathrm{~mm}\)