NACE CP 2 Quiz 2 100% Correctly Solved 2024

NACE CP 2 Quiz 2 100% Correctly Solved 2024 A structure-to-electrolyte potential measures the potential difference between A. Two different structures acting as electrodes B. The structure and a standard reference electrode C. The structure and the anodes D. Two different points on the structure - Answer-B. The structure and a standard reference electrode If 5.0 A of battery current passes through the outside leads of a 4-wire current span spaced at 30 m (100 ft), resulting in a voltage drop of 5.0 mV with the test current applied and of 1 mV without the test current, what is a the calibration factor of the current span? A. 0.00025 A/mV B. 4.0 A/mV C. 1.25 A/mV D. 5.0 A/mV - Answer-C. 1.25 A/mV I = 5 A EC = 5 mV ENC = 1 mV K = I/(EC - ENC) = 5 A/(5 mV - 1 mV) = 1.25 A/mV Polarized potentials are measured A. when the rectifiers are ON B. just after the rectifiers are turned OFF C. after the rectifiers are left off for several days D. just after the Rectifiers are turned ON - Answer-B. just after the rectifiers are turned OFF Polarization is determined by A. Subtracting the ON from the instant OFF potential B. Measuring the instant OFF potential C. Subtracting the instant OFF from the Native (free corroding) potential D. Substracting the IR drop from the ON potential - Answer-C. Subtracting the instant OFF from the Native (free corroding) potential Current pickup on a pipeline is indicated when A. The structure-to-electrolyte potential over the pipe is more electronegative than each side B. The structure-to-electrolyte potential over the pipe is more electropositive than each side C. The two electrode method (side drain) shows the electrode over the pipe is more electropositive D. A CIS shows a more electropositive potential with the rectifier ON than OFF - Answer-B. The structure-to-electrolyte potential over the pipe is more electropositive than each side A 61 cm diameter (24 in) pipe has a current span at each end of a 2 km (1.24 miles) section with the following information determined: EONts1 = -1200 mVcse EONts2 = -1140 mVcse EOFFts1 = -1000 mVcse EOFFts2 = -980 mVcse IONts1 = 3 A IONts2 = 2.8 A IOFFts1 = 0.2 A IOFFts2 - = 0.2 A What is the pipe-to-earth resistance (ohm)? A. 0.9 ohm B. 9 ohm C. 66.7 ohm D. 900 ohm - Answer-A. 0.9 ohm Rc = DeltaE/Ic Ic = Delta1ts1 - DeltaIts2 = (3 - 0.2) - (2.8 - 0.2) Ic = 0.2 A DeltaE = (DeltaEts1 - DeltaEts2)/2 DeltaE = (()+(1140-980))/2 = 180 mV Rc = 180 mV/0.2 A = 0.9 ohm A 61 cm diameter (24 in) pipe has a current span at each end of a 2 km (1.24 miles) section with the following information determined: EONts1 = -1200 mVcse EONts2 = -1140 mVcse EOFFts1 = -1000 mVcse EOFFts2 = -980 mVcse IONts1 = 3 A IONts2 = 2.8 A IOFFts1 = 0.2 A IOFFts2 - = 0.2 A What is the specific coating resistance (ohm-m2)? A. 0.00023 ohm-m2 B. 0.023 ohm-m2 C. 3450 ohm-m2 D. 345000 ohm-m2 - Answer-C. 3450 ohm-m2 D = 61 cm = 0.61 m L= 2 km = 2000 m A = piDL = pi(0.61 m)(2000 m) = 3830.8 m2 rCE = A*Rc = (3830.8 m2)(0.9 ohm) = 3447.72 ohm-m2