Code Boundary Reminder: While the calculation baseline is always $\frac{\text{RL}}{2}$, API 653 imposes rigid maximum caps (e.g., 5 years for external visual, 10 years for UT thickness, or 20 years for internal inspections unless utilizing an approved RBI program).
If corrosion is accelerating, use the short-term (worst-case) rate. If it’s stable, use the long-term trend.
🧮 SECTION 1 — CORROSION RATE (Q1–Q10)
Q.1
0.600 → 0.450 in over 5 years
Solution:
CR=0.600−0.4505=0.030 in/yrCR = \frac{0.600 – 0.450}{5} = 0.030\ \text{in/yr}CR=50.600−0.450=0.030 in/yr
✔ Answer: 0.030 in/yr
Q.2
0.500 → 0.400 in over 4 years
CR=0.500−0.4004=0.025CR = \frac{0.500 – 0.400}{4} = 0.025CR=40.500−0.400=0.025
✔ Answer: 0.025 in/yr
Q.3
0.450 → 0.300 in over 6 years
CR=0.450−0.3006=0.025CR = \frac{0.450 – 0.300}{6} = 0.025CR=60.450−0.300=0.025
✔ Answer: 0.025 in/yr
Q.4
0.800 → 0.600 in over 10 years
✔ CR = 0.020 in/yr
Q.5
Loss = 0.150 in over 3 years
✔ CR = 0.050 in/yr
Q.6
0.700 → 0.560 in over 7 years
✔ CR = 0.020 in/yr
Q.7
0.900 → 0.720 in over 9 years
✔ CR = 0.020 in/yr
Q.8
Loss = 0.120 in over 6 years
✔ CR = 0.020 in/yr
Q.9
0.650 → 0.500 in over 5 years
✔ CR = 0.030 in/yr
Q.10
Loss = 0.100 in over 2 years
✔ CR = 0.050 in/yr
⏳ SECTION 2 — REMAINING LIFE (Q11–Q20)
Q.11
t = 0.300, tmin = 0.200, CR = 0.025
RL=0.300−0.2000.025=4RL = \frac{0.300 – 0.200}{0.025} = 4RL=0.0250.300−0.200=4
✔ Answer: 4 years
Q.12
t = 0.400, tmin = 0.250, CR = 0.030
✔ RL = 5 years
Q.13
t = 0.350, tmin = 0.200, CR = 0.025
✔ RL = 6 years
Q.14
t = 0.500, tmin = 0.300, CR = 0.040
✔ RL = 5 years
Q.15
t = 0.280, tmin = 0.200, CR = 0.020
✔ RL = 4 years
Q.16
t = 0.450, tmin = 0.250, CR = 0.050
✔ RL = 4 years
Q.17
t = 0.600, tmin = 0.300, CR = 0.030
✔ RL = 10 years
Q.18
t = 0.420, tmin = 0.220, CR = 0.020
✔ RL = 10 years
Q.19
t = 0.380, tmin = 0.200, CR = 0.030
✔ RL = 6 years
Q.20
t = 0.500, tmin = 0.350, CR = 0.025
✔ RL = 6 years
📅 SECTION 3 — INSPECTION INTERVAL (Q21–Q25)
Q.21
RL = 6 years
Interval=62=3Interval = \frac{6}{2} = 3Interval=26=3
✔ Answer: 3 years
Q.22
RL = 10 → ✔ 5 years
Q.23
RL = 8 → ✔ 4 years
Q.24
RL = 4 → ✔ 2 years
Q.25
RL = 12 → ✔ 6 years
🌊 SECTION 4 — HYDROSTATIC PRESSURE (Q26–Q30)
Q.26
Height = 10 ft
P=0.433×10=4.33P =0.433×10 = 4.33
✔ Answer: 4.33 psi
Q.27
Height = 20 ft → ✔ 8.66 psi
Q.28
Height = 15 ft → ✔ 6.50 psi
Q.29
Height = 30 ft → ✔ 12.99 psi
Q.30
Height = 25 ft → ✔ 10.83 psi
🧠 FINAL EXAM PATTERN TAKEAWAYS
🧮 Most repeated calculations:
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Corrosion rate
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Remaining life
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Inspection interval
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Hydrostatic pressure
⚠️ Common mistakes:
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Forgetting to divide RL by 2
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Unit conversion errors
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Wrong subtraction order in CR
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Overthinking simple math
want to know more? study our free lessons and quiz at:
API 653 Storage Tank Inspector Online Training Course
