| 1 |
What is the primary advantage of using cup lump rubber (CLR) in cold mix asphalt (CMA)?
|
Lowers production cost |
|
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7 |
-.50
-.25
+.25
เต็ม
0
-35%
+30%
+35%
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| 2 |
Which property of CMA is significantly improved by the addition of CLR?
|
Nutritional value |
|
การเพิ่ม CLR ลงใน CMA ช่วยปรับปรุงประสิทธิภาพการละลายน้ำแข็งเป็นหลัก ทำให้กลายเป็นสารกำจัดน้ำแข็งที่มีประสิทธิภาพมากขึ้น โดยเฉพาะอย่างยิ่งในอุณหภูมิที่ต่ำกว่า |
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7 |
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| 3 |
If the tensile strength of CMA increases by 26% due to the addition of CLR and the original tensile strength was 5 MPa, what is the new tensile strength?
|
6.3 MPa |
|
5*0.26=1.3+5=6.3 MPa |
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7 |
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| 4 |
Given that the rut depth decreases by 70% when CLR is added to CMA and the original rut depth was 10 mm, what is the new rut depth?
|
3 mm |
|
10*0.7=7
10-7=3 Mm |
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7 |
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| 5 |
If the CMA with CLR has a TSR (Tensile Strength Ratio) value of 104% and the minimum requirement is 80%, by what percentage does the TSR exceed the requirement?
|
30% |
|
104-80=24
24/80*100=30% |
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7 |
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-35%
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| 6 |
What is the potential increase in moisture damage resistance for CMA-CR compared to conventional CMA if the improvement is 12%?
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12% |
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7 |
-.50
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เต็ม
0
-35%
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| 7 |
If the shear resistance of CLR-modified CMA increases due to the membrane effect, which physical property is most directly influenced?
|
Tensile strength |
|
คุณสมบัติทางกายภาพที่ได้รับอิทธิพลโดยตรงมากที่สุดจากการเพิ่มขึ้นของความต้านทานแรงเฉือนเนื่องจากผลกระทบของเมมเบรนคือ ความต้านทานแรงเฉือน |
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7 |
-.50
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| 8 |
The viscosity of CLR-modified bitumen at 135°C is 1.16 Pa·s. If the shear rate is 50 s^-1, what is the shear stress?
|
58 Pa |
|
τ=1.16 Pa\cdotps×50 s −1 𝜏 = 58 Pa τ=58 Pa |
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7 |
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| 9 |
If the mass loss in the Cantabro test for CMA-CR is 14.6% and the maximum accepted limit is 20%, by how much does CMA-CR fall below the limit?
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5.4% |
|
20-14.6=5.4% |
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7 |
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| 10 |
What is the significance of using cup lump rubber in the context of environmental sustainability?
|
It enhances the nutritional content of the asphalt. |
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7 |
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| 11 |
What is one of the emerging trends affecting China's oil and gas pipeline development strategies?
|
Digitization |
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7 |
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| 12 |
What is the proportion of natural gas pipelines in the total length of long-distance oil and gas pipelines in China as of 2022?
|
60% |
|
|
https://www.sciencedirect.com/science/article/pii/S2950276424000035 |
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| 13 |
If the total length of long-distance oil and gas pipelines in China is 180,000 km, how many kilometers are dedicated to natural gas pipelines?
|
108,000 km |
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7 |
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| 14 |
According to the article, if the target length for oil and gas pipelines is 210,000 km by 2025, how many kilometers need to be constructed from the 2022 total?
|
30,000 km |
|
21000-18000=30000 |
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| 15 |
If the failure rate of oil and gas pipelines in Europe is 0.29 per year per hundred kilometers, what is the failure rate per year for a pipeline network of 1,000 kilometers?
|
2.9 failures |
|
Multiply the failure rate per kilometer by the total length of the pipeline network:
Failure rate for 1,000 kilometers
=
0.0029
×
1
,
000
=
2.9
failures per year
Failure rate for 1,000 kilometers=0.0029×1,000=2.9 failures per year |
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7 |
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| 16 |
What is one of the main causes of oil and gas pipeline failures in China according to the article?
|
Oil theft through drilling |
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7 |
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| 17 |
Assuming the failure rate in the United States is 0.14 per year per hundred kilometers, calculate the expected number of failures per year for a 1,500 kilometers pipeline network.
|
2.1 failures |
|
Failure rate per kilometer=
100
0.14
=0.0014 failures per year per kilometer |
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7 |
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| 18 |
If a pipeline defect inspection technology improves detection efficiency by 25% and the current detection efficiency is 80%, what will be the new detection efficiency?
|
100% |
|
0.25*80+80=100 |
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| 19 |
If a vibration signal monitoring system faces a 15% reduction in noise levels due to a new technology and the original noise level was 200 units, what is the new noise level?
|
170 units |
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7 |
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| 20 |
For a hydrogen pipeline with an embrittlement rate of 0.05% per year, calculate the total embrittlement after 10 years.
|
0.5% |
|
Subtract 1 to find the total embrittlement:
1.005
−
1
=
0.005
or
0.5
%
1.005−1=0.005 or 0.5% |
|
7 |
-.50
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-35%
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