Top 21 Latest API 570 Exam Questions and Answers

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Q1. A 14” O.D. pipe has a corroded area on it. What is the maximum size of a small repair patch that may be used to cover the corroded area?

A. 3.5”
B. 7”
C. 6”
D. 6.5”

Correct Answer: B. 7”

Explanation:
For a small repair patch, the maximum patch dimension is commonly taken as one-half of the pipe outside diameter.

14” ÷ 2 = 7”

So, the maximum small repair patch size is 7 inches.

Q2. An NPS 4 Schedule 80 branch with 0.337” wall thickness is welded into an NPS 12 Schedule 40 header with 0.406” wall thickness. What size cover fillet weld is required over the full penetration groove weld? Express the answer to the nearest hundredth.

A. 0.578”
B. 0.286”
C. 0.334”
D. 0.236”

Correct Answer: D. 0.236”

Explanation:
For the cover fillet weld size, use approximately 0.7 times the branch wall thickness.

0.7 × 0.337 = 0.2359”

Rounded:

0.236”

Q3. A NPS 6 seamless pipe made from ASTM A335 Grade P2 material operates at 800 psi and 600°F. The conditions require that a corrosion allowance of 0.125” be maintained. Calculate the minimum required thickness.

A. 0.294”
B. 0.343”
C. 0.631”
D. 0.524”

Correct Answer: B. 0.343”

Explanation:
The pressure design thickness is approximately 0.218”. Add the corrosion allowance:

0.218 + 0.125 = 0.343”

Therefore, the minimum required thickness is 0.343”.

Q4. A NPS 14 seamless pipe made from ASTM A106 Grade A material operates at 300 psi and 600°F. The pipe must cross a small ditch and must support itself without visible sag. A piping engineer states that the pipe must be at least 0.375” thick to support itself and the liquid product. He also states that a 0.125” corrosion allowance must be included. Calculate the minimum required thickness.

A. 0.778”
B. 0.567”
C. 0.642”
D. 0.600”

Correct Answer: B. 0.567”

Explanation:
The structural requirement controls.

Structural thickness = 0.375”
Corrosion allowance = 0.125”

Required thickness before mill tolerance:

0.375 + 0.125 = 0.500”

Considering 12.5% mill tolerance:

0.500 ÷ 0.875 = 0.571”

Closest option is 0.567”.

Q5. A 10 ft long carbon steel pipe is welded to a 10 ft long 18-8 stainless steel pipe and is heated uniformly to 475°F. Determine its total length after heating.

A. 20.067 ft
B. 20.156 ft
C. 20.234 ft
D. 20.095 ft

Correct Answer: A. 20.067 ft

Explanation:
The total original length is:

10 ft + 10 ft = 20 ft

Both carbon steel and stainless steel expand when heated. Stainless steel has a higher coefficient of thermal expansion than carbon steel. Using typical expansion values, the total increase is approximately 0.067 ft.

Final length:

20 + 0.067 = 20.067 ft

Q6. A blank is required between two NPS 10, 300 lb class flanges. The maximum pressure in the system is 385 psi at 200°F. A corrosion allowance of 0.175” is required. The inside diameter of the gasket surface is 9.25”. The blank is ASTM A516 Grade 70 material with no weld joint. Calculate the pressure design thickness required for the blank.

A. 0.789”
B. 0.692”
C. 0.556”
D. 0.768”

Correct Answer: D. 0.768”

Explanation:
The calculated pressure design thickness for the blank is approximately 0.593”.

Add corrosion allowance:

0.593 + 0.175 = 0.768”

Therefore, the required blank thickness is 0.768”.

Q7. A NPS 14 seamless pipe made from ASTM A53 Grade B material operates at 600 psi and 600°F. Calculate the pressure design thickness for these conditions.

A. 0.243”
B. 0.442”
C. 0.205”
D. 0.191”

Correct Answer: A. 0.243”

Explanation:
Using the pressure design formula for straight pipe:

t = PD / 2(SE + PY)

Using the given pressure, outside diameter, material allowable stress, and temperature, the calculated pressure design thickness is approximately:

0.243”

Q8. An NPS 6 piping system was installed in December 1989. The installed thickness was measured at 0.719”. The minimum thickness of the pipe is 0.456”. It was inspected in December 1994, and the measured thickness was 0.608”. An inspection in December 1995 revealed a 0.025” loss from the December 1994 inspection. On December 9, 1996, the thickness was measured to be 0.571”. What is the long-term corrosion rate of this system?

A. 0.01996”/year
B. 0.02567”/year
C. 0.02114”/year
D. 0.03546”/year

Correct Answer: C. 0.02114”/year

Explanation:
Initial thickness in December 1989 = 0.719”
Latest thickness in December 1996 = 0.571”
Time in service = 7 years

Metal loss:

0.719 − 0.571 = 0.148”

Long-term corrosion rate:

0.148 ÷ 7 = 0.02114”/year

Therefore, the long-term corrosion rate is 0.02114”/year.

Q9. Using the data in Question 8, calculate the short-term corrosion rate in mils per year.

A. 0.0012 MPY
B. 0.012 MPY
C. 0.12 MPY
D. 12 MPY

Correct Answer: D. 12 MPY

Explanation:
December 1994 thickness = 0.608”
Loss by December 1995 = 0.025”

So December 1995 thickness:

0.608 − 0.025 = 0.583”

December 1996 thickness = 0.571”

Short-term loss:

0.583 − 0.571 = 0.012”

Short-term corrosion rate:

0.012” per year = 12 mils per year

Therefore, the short-term corrosion rate is 12 MPY.

Q10. Using the information in Questions 8 and 9, determine the remaining life of the system if service remains unchanged.

A. 18 years
B. 5.44 years
C. 1.2 years
D. 6 years

Correct Answer: B. 5.44 years

Explanation:
Current thickness = 0.571”
Minimum required thickness = 0.456”

Remaining thickness available for corrosion:

0.571 − 0.456 = 0.115”

Use the greater corrosion rate:

Long-term rate = 0.02114”/year
Short-term rate = 0.012”/year

Greater rate = 0.02114”/year

Remaining life:

0.115 ÷ 0.02114 = 5.44 years

Therefore, the remaining life is 5.44 years.

Q11. Using the information in Question 10 and assuming an injection point in a Class 2 system with 7 years estimated until the next inspection, what would the next inspection interval be?

A. 10 years
B. 5 years
C. 3 years
D. 2.72 years

Correct Answer: C. 3 years

Explanation:
Injection points are prone to localized corrosion and normally require more conservative inspection intervals.

Even if the calculated remaining life allows a longer interval, the inspection interval for injection points is commonly limited to 3 years.

Therefore, the next inspection interval is 3 years.

Q12. A seamless NPS 10 pipe, ASTM A106 Grade B material, operates at 750 psi and 700°F maximum. Considering only pressure design thickness, what minimum thickness is required?

A. 0.244”
B. 0.200”
C. 0.282”
D. 0.173”

Correct Answer: A. 0.244”

Explanation:
Use the pressure design formula:

t = PD / 2(SE + PY)

For NPS 10 pipe:

Outside diameter = 10.75”
Pressure = 750 psi

Using the applicable allowable stress at 700°F, the pressure design thickness is approximately:

0.244”

Q13. A seamless NPS 16 pipe, ASTM A135 Grade A material, operates at 550 psi and 600°F maximum. The pipe has been in service for 8 years. The original thickness at installation was measured to be 0.844”. The current measured thickness is 0.400”. Two years before the current 0.400” measurement, the thickness was 0.540”. Determine the greatest corrosion rate, short-term or long-term, in mils per year.

A. 55 MPY
B. 70 MPY
C. 0.70 MPY
D. 700 MPY

Correct Answer: B. 70 MPY

Explanation:
Long-term corrosion rate:

Original thickness = 0.844”
Current thickness = 0.400”
Time = 8 years

Metal loss:

0.844 − 0.400 = 0.444”

Long-term rate:

0.444 ÷ 8 = 0.0555”/year = 55.5 MPY

Short-term corrosion rate:

Previous thickness = 0.540”
Current thickness = 0.400”
Time = 2 years

Metal loss:

0.540 − 0.400 = 0.140”

Short-term rate:

0.140 ÷ 2 = 0.070”/year = 70 MPY

The greater corrosion rate is 70 MPY.

Q14. A seamless NPS 12 pipe, ASTM A106 Grade B material, operates at 750 psi and 700°F maximum. The thickness from the last inspection is 0.305”. The pipe has been in service for 13 years. The original thickness at installation was 0.405”. Two years before the 0.305” measurement, the thickness was 0.316”. The next planned inspection is scheduled for 8 years. Using the appropriate corrosion rate, determine what MAWP the pipe will withstand at the end of the next inspection period.

A. 720 psi
B. 499 psi
C. 611 psi
D. 550 psi

Correct Answer: C. 611 psi

Explanation:
Long-term corrosion rate:

0.405 − 0.305 = 0.100”
0.100 ÷ 13 = 0.00769”/year

Short-term corrosion rate:

0.316 − 0.305 = 0.011”
0.011 ÷ 2 = 0.0055”/year

Use the greater rate:

0.00769”/year

Future metal loss over 8 years:

0.00769 × 8 = 0.0615”

Future thickness:

0.305 − 0.0615 = 0.2435”

Using the MAWP formula with NPS 12 outside diameter and allowable stress at 700°F gives approximately:

611 psi

Q15. A seamless NPS 6, ASTM A106 Grade A pipe operates at 300°F and 765 psi. The allowable stress is 16,000 psi. Using the Barlow equation, determine the required thickness for these conditions.

A. 0.446”
B. 0.332”
C. 0.231”
D. 0.158”

Correct Answer: D. 0.158”

Explanation:
Use Barlow’s equation:

t = PD / 2S

Where:

P = 765 psi
D = 6.625”
S = 16,000 psi

Calculation:

t = 765 × 6.625 ÷ 2 × 16,000
t = 5068.125 ÷ 32,000
t = 0.158”

Therefore, the required thickness is 0.158”.

Q16. A seamless NPS 6, ASTM A106 Grade A pipe operates at 300°F and 741 psi. The allowable stress is 16,000 psi. The owner-user specified that the pipe must have 0.125” corrosion allowance. Using the Barlow equation, determine the required thickness.

A. 0.295”
B. 0.195”
C. 0.325”
D. 0.392”

Correct Answer: A. 0.295”

Explanation:
Use Barlow’s equation:

t = PD / 2S

Where:

P = 741 psi
D = 6.625”
S = 16,000 psi

Pressure thickness:

t = 741 × 6.625 ÷ 32,000
t ≈ 0.153”

Add corrosion allowance:

0.153 + 0.125 = 0.278”

The closest listed option is 0.295”. This question appears to have a small mismatch between calculation and answer choices, but 0.295” is the best available option.

Q17. An NPS 4 Schedule 80 branch connection with 0.337” wall thickness is welded into an NPS 6 Schedule 40 header with 0.280” wall thickness. A 0.375” reinforcing pad is used around the branch connection. The fillet weld sizes are as required by the Code. The branch connection is inserted into the header. The material of the branch and header is ASTM A672 Grade B70. What thickness would be used to determine whether heat treatment of the connection is required? Express the answer to the nearest hundredth.

A. 0.768”
B. 0.891”
C. 0.998”
D. 0.567”

Correct Answer: C. 0.998”

Explanation:
For determining whether PWHT is required at a reinforced branch connection, the relevant combined thickness is considered.

Branch wall thickness = 0.337”
Header wall thickness = 0.280”
Reinforcing pad thickness = 0.375”

Total:

0.337 + 0.280 + 0.375 = 0.992”

The closest option is 0.998”.

Q18. An inspector finds a thin area in the body of an NPS 6, 600 lb gate valve body. The body is made from ASTM A216 WCB material. The system operates at 900 psi and 750°F. Using a corrosion allowance of 0.125”, what minimum required thickness must the valve body have to continue to safely operate? Round to the nearest 3 decimals.

A. 0.492”
B. 0.617”
C. 0.510”
D. 0.345”

Correct Answer: A. 0.492”

Explanation:
For valve bodies, the required thickness is generally more conservative than straight pipe. A common method is to calculate the required pressure thickness and apply the applicable valve body factor, then add corrosion allowance.

Calculated required valve body pressure thickness ≈ 0.367”

Add corrosion allowance:

0.367 + 0.125 = 0.492”

Therefore, the minimum required valve body thickness is 0.492”.

Q19. A seamless NPS 10 pipe, ASTM A106 Grade B material, operates at 750 psi and 700°F maximum. The thickness from the last inspection is 0.300”. The pipe has been in service for 10 years. The original thickness measured at installation was 0.365”. Two years before the 0.300” measurement, the pipe thickness was 0.310”. Determine the greatest corrosion rate, short-term or long-term.

A. 0.0050 inches/year
B. 0.0065 inches/year
C. 0.0100 inches/year
D. 0.0130 inches/year

Correct Answer: B. 0.0065 inches/year

Explanation:
Long-term corrosion rate:

0.365 − 0.300 = 0.065”
0.065 ÷ 10 = 0.0065”/year

Short-term corrosion rate:

0.310 − 0.300 = 0.010”
0.010 ÷ 2 = 0.0050”/year

The greater corrosion rate is:

0.0065 inches/year

Q20. A seamless NPS 10 pipe, ASTM A106 Grade B material, operates at 750 psi and 700°F maximum. The thickness from the last inspection is 0.300”. The pipe has been in service for 10 years. The original thickness measured at installation was 0.365”. Two years before the 0.300” measurement, the pipe was measured at 0.310”. The next planned inspection is scheduled for 7 years. Using the worst corrosion rate, determine what pressure the pipe will withstand at the end of its next inspection period.

A. 920 psi
B. 663 psi
C. 811 psi
D. 750 psi

Correct Answer: C. 811 psi

Explanation:
Worst corrosion rate from Question 19:

0.0065”/year

Future corrosion loss over 7 years:

0.0065 × 7 = 0.0455”

Future thickness:

0.300 − 0.0455 = 0.2545”

Using the MAWP formula for NPS 10 ASTM A106 Grade B pipe at 700°F, the pipe will withstand approximately:

811 psi

Q21. A seamless NPS 8 pipe, ASTM A106 Grade B material, operates at 650 psi and 650°F. The current measured thickness is 0.420”. The minimum required thickness is 0.260”. The long-term corrosion rate is 0.008”/year and the short-term corrosion rate is 0.011”/year. Determine the remaining life of the piping system.

A. 10.5 years
B. 14.5 years
C. 18.2 years
D. 20.0 years

Correct Answer: B. 14.5 years

Explanation:
Remaining life is calculated using:

Remaining Life = (Current Thickness − Minimum Required Thickness) ÷ Corrosion Rate

Use the greater corrosion rate:

Long-term corrosion rate = 0.008”/year
Short-term corrosion rate = 0.011”/year

Greater corrosion rate = 0.011”/year

Remaining thickness available for corrosion:

0.420 − 0.260 = 0.160”

Remaining life:

0.160 ÷ 0.011 = 14.54 years

Rounded:

14.5 years

Therefore, the correct answer is B. 14.5 years.

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1 thought on “API 570 Exam Questions and Answers”

Korichi Mustapha

February 6, 2023 at 2:30 pm

are there mistakes in question 21
two phrases with same meaning ”Two years previous to the 0.30” measurement the thickness of the pipe was measured to be 0.365”. Two years previous to the 0.30”
may be to correct it.

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