Can You Pass This Science Mock Test Exam? Trivia Quiz

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Questions and Answers
  • 1. 

    In heat exchangers, which metallurgy is most prone to form waterside scale deposits?

    • A.

      SS

    • B.

      Admiralty

    • C.

      Copper

    • D.

      Mild steel

    Correct Answer
    D. Mild steel
    Explanation
    Mild steel is the metallurgy that is most prone to form waterside scale deposits in heat exchangers. This is because mild steel has a higher susceptibility to corrosion and oxidation compared to other metallurgies. When exposed to water, mild steel can react with oxygen and other elements present in the water, leading to the formation of scale deposits. These deposits can reduce the efficiency of the heat exchanger and may require regular maintenance and cleaning to prevent further damage.

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  • 2. 

    The localized (pitting) corrosion rate increases when:

    • A.

      The cathodic area and anodic area are similar

    • B.

      The anodic area is large, relative to the cathode.

    • C.

      The cathodic area is large, relative to the anode.

    • D.

      When sacrificial anodes are used.

    Correct Answer
    C. The cathodic area is large, relative to the anode.
    Explanation
    When the cathodic area is larger than the anodic area, it means that there is more surface area available for the cathodic reaction to occur. This leads to a higher concentration of cathodic reactions happening compared to anodic reactions. As a result, the localized corrosion rate increases because the cathodic reactions are more dominant and the anodic reactions are limited.

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  • 3. 

    Deposits usually lead to aggravated localized corrosion  (pitting attack) because:

    • A.

      Decreased pH levels at the cathode.

    • B.

      A galvanic cell is set up between the manganese deposit and the metal.

    • C.

      Deficient oxygen levels under the deposit.

    • D.

      Increased pH levels at the anode.

    • E.

      Both (a) and (b).

    Correct Answer
    C. Deficient oxygen levels under the deposit.
    Explanation
    Deposits usually lead to aggravated localized corrosion (pitting attack) because of deficient oxygen levels under the deposit. When a deposit forms on a metal surface, it creates a barrier that restricts the access of oxygen to the metal. This results in a localized area with lower oxygen levels, which creates an environment conducive to corrosion. The lack of oxygen leads to the formation of corrosion cells and the breakdown of the metal, causing pitting attack. This is why deficient oxygen levels under the deposit contribute to aggravated localized corrosion.

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  • 4. 

    To prevent white rust:

    • A.

      Maintain an alkalinity of less than 300 ppm

    • B.

      Maintain a pH below 8.3.

    • C.

      Keep calcium hardness below 50 ppm.

    • D.

      All of the above.

    • E.

      A and b

    Correct Answer
    C. Keep calcium hardness below 50 ppm.
    Explanation
    To prevent white rust, it is important to keep the calcium hardness below 50 ppm. White rust is a form of corrosion that occurs on galvanized steel surfaces when exposed to moisture. Calcium hardness refers to the concentration of calcium ions in water, and maintaining it below 50 ppm helps to prevent the formation of white rust by reducing the likelihood of calcium deposits on the surface. The other options, maintaining an alkalinity of less than 300 ppm and a pH below 8.3, may be important for other aspects of water treatment but are not specifically related to preventing white rust.

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  • 5. 

    Chlorination of a water system with soluble iron or manganese will:

    • A.

      Stabilize the iron or manganese.

    • B.

      Oxidize the iron or manganese and prevent further corrosion.

    • C.

      Reduce the iron or manganese causing a severe corrosion potential

    • D.

      Oxidize the iron or manganese causing a potential for fouling.

    Correct Answer
    D. Oxidize the iron or manganese causing a potential for fouling.
    Explanation
    Chlorination of a water system with soluble iron or manganese will oxidize the iron or manganese, causing a potential for fouling. This means that the chlorination process will convert the soluble iron or manganese into insoluble forms, which can then accumulate and cause fouling in the water system. This can lead to clogging of pipes, reduced water flow, and other operational issues.

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  • 6. 

    The make-up water for a cooling water system is 100 GPM and contains 150 ppm of M-Alkalinity. You want to run the system at 5 cycles while maintaining 400 ppm M-Alkalinity in the recirculating water. What 98% of H2SO4 dosage is required?

    • A.

      8.4 pounds per day

    • B.

      1 pounds per day.

    • C.

      10 pounds per day.

    • D.

      84 pound per day.

    • E.

      None of the above.

    Correct Answer
    A. 8.4 pounds per day
    Explanation
    To calculate the required dosage of H2SO4, we can use the following formula:

    Dosage (lb/day) = Make-up water flow rate (GPM) x M-Alkalinity difference (ppm) x 8.34 / Cycles

    Given that the make-up water flow rate is 100 GPM, the M-Alkalinity difference is (400 ppm - 150 ppm) = 250 ppm, and the desired cycles is 5, we can substitute these values into the formula:

    Dosage = 100 GPM x 250 ppm x 8.34 / 5 = 8.4 lb/day

    Therefore, 8.4 pounds per day of H2SO4 dosage is required.

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  • 7. 

    Which of the following characteristic of corrosion is caused by anaerobic bacteria?

    • A.

      Red tubercules (iron oxide)

    • B.

      Cracking at grain boundaries

    • C.

      Precipitation of calcium hydroxide

    • D.

      Granules of black iron (iron sulfide)

    Correct Answer
    D. Granules of black iron (iron sulfide)
    Explanation
    Anaerobic bacteria are known to produce hydrogen sulfide gas, which reacts with iron to form iron sulfide. This iron sulfide appears as black granules and is a characteristic of corrosion caused by anaerobic bacteria. This process is commonly referred to as microbiologically influenced corrosion (MIC) and can occur in environments where oxygen is limited, such as in underground pipelines or in stagnant water. The formation of black iron sulfide granules is a clear indication of the presence of anaerobic bacteria and their role in causing corrosion.

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  • 8. 

    Determine the Evaporation rate with the following information? Given:     Your prospect has two-2500 ton chillers, one high efficiency fill tower, and a chilled water loop cooling a process.  During your survey you discover the following:   ·         The chilled water loop leaks at a rate of 8 gpm (Repairs will cost $2,700.) ·         The chiller head pressure has increased 2 psig ·         A clean machine will operate at 0.7 kw/ton ·         The two machines operate at 75% load for 6000 hours per year ·         The recirculation rate = 3gpm/ton ·         The average delta T is 8 degrees & cycles of concentration = 4.5 ·         Use 0.85% per 10 degree for evap calc.

    • A.

      27.5 million gallons /yr

    • B.

      32.4 million gallons /yr

    • C.

      53.6 million gallons/yr

    • D.

      47.3 million gallons/yr

    Correct Answer
    C. 53.6 million gallons/yr
    Explanation
    Based on the given information, the evaporation rate can be determined by calculating the amount of water lost due to leaks and the increase in chiller head pressure. The leak rate is given as 8 gpm, which results in a total loss of 8 gpm * 60 min/hr * 24 hr/day * 365 days/yr = 42.048 million gallons/yr. The increase in chiller head pressure indicates higher evaporation rate, and using the given conversion rate of 0.85% per 10 degrees, the increase of 2 psig corresponds to an additional evaporation of 11.552 million gallons/yr. Therefore, the total evaporation rate is 42.048 million gallons/yr + 11.552 million gallons/yr = 53.6 million gallons/yr.

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  • 9. 

    The "C" factor for a heat exchanger

    • A.

      Considers water flow and pressure drop.

    • B.

      Is calculated by the formula, C = U (A) (LMTD) for a one pass exchanger.

    • C.

      Cannot be applied if there is a phase change on the process side (a gas condensed to a liquid).

    • D.

      Considers process side fouling.

    Correct Answer
    A. Considers water flow and pressure drop.
    Explanation
    The "C" factor for a heat exchanger is a measure of its performance and efficiency. It takes into account the water flow rate and the pressure drop across the exchanger. This means that the "C" factor considers how well the exchanger can transfer heat from one fluid to another, while also taking into account the energy required to pump the fluids through the exchanger. By considering these factors, the "C" factor provides a comprehensive assessment of the exchanger's effectiveness in transferring heat.

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  • 10. 

    A cooling tower has the following information provided.  What is the expected blowdown?                   Water test show concentration ratio to be 3.5 Recirc rate = 40,500 gph    or  153  m3/hr Delta temp across the tower = 15 degrees F   or 8.33C Drift is approximately 0.1% of recirculation rate Basin is 1.25 ft (.38 m) H x 20 ft (6.1  m) W x 40 ft (12.2  m) L [ to account for  the piping etc -add 25 % additional] In this area the evaporation is about 0.9% of the recirc rate per 10 deg F.   [5.5 C] HINT: Blowdown = (Evaporation /cycles -1) - drift

    • A.

      83 gph [ .314 m3/hr]

    • B.

      182 gph [ .689 m3/hr]

    • C.

      283 gph [1.07 m3/hr]

    • D.

      383 gph [1.45 m3/hr]

    Correct Answer
    B. 182 gph [ .689 m3/hr]
    Explanation
    The expected blowdown is calculated using the given information. The concentration ratio is 3.5, the recirculation rate is 40,500 gph (153 m3/hr), the delta temperature across the tower is 15 degrees F (8.33C), and the drift is approximately 0.1% of the recirculation rate. The basin dimensions are also given, with an additional 25% added to account for piping. The evaporation is 0.9% of the recirculation rate per 10 degrees F (5.5C). Using the formula for blowdown, which is (Evaporation / cycles - 1) - drift, the expected blowdown is calculated to be 182 gph (0.689 m3/hr).

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  • 11. 

    Your customer asks you if the biocontrol program you are feeding also protects the plant personnel from Legionella. How do you respond?

    • A.

      The current cooling water program was designed to provide excellent cooling tower performance and protect personnel.

    • B.

      The current program is acceptable, but insist that the customer run Legionella test on the tower water at least once a year, just to be sure.

    • C.

      The current program was designed to provide good performance of heat exchangers and towers,but was not designed to meet a health standard.

    • D.

      The current program is acceptable if chlorine dioxide was added to the make up water to prevent Legionella from entering the tower system.

    Correct Answer
    C. The current program was designed to provide good performance of heat exchangers and towers,but was not designed to meet a health standard.
    Explanation
    The correct answer suggests that the current biocontrol program is effective in terms of cooling tower performance and protecting the equipment, but it was not specifically designed to meet health standards or protect plant personnel from Legionella. This implies that while the program may be acceptable for its intended purpose, additional measures may be necessary to ensure the safety of plant personnel, such as running Legionella tests on the tower water.

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  • 12. 

    Which of the following is a false statement about 3D Biocontrol:

    • A.

      The Nalco Bio-Index is determined by measuring the fluorescent signal of both the unreacted and reacted Bio-Reporter Material and determining the ratio.

    • B.

      Using too much oxidizing biocide with the Nalco Bio-Reporter can make the tower turn blue

    • C.

      The Bio-Reporter can be used to monitor bioactivity for a baseline before starting a 3D Bio-controlled program.

    • D.

      The Bio-Index is independent of Bio-Reporter concentration.

    Correct Answer
    D. The Bio-Index is independent of Bio-Reporter concentration.
    Explanation
    The Bio-Index is a measure of bioactivity in the system, which is determined by measuring the fluorescent signal of both the unreacted and reacted Bio-Reporter Material and determining the ratio. The Bio-Reporter concentration does not affect the Bio-Index, as it is independent of it.

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  • 13. 

    When interpreting 3D Trasar performance parameters, for both the 3D Bio-Index and the Nalco scale Index, a value of 25 would mean that the system was under:

    • A.

      Low stress

    • B.

      Mild Stress

    • C.

      High Stress

    • D.

      Alarm conditions

    Correct Answer
    B. Mild Stress
    Explanation
    A value of 25 for both the 3D Bio-Index and the Nalco scale Index indicates that the system is under mild stress. This means that there may be some minor issues or deviations from optimal performance, but it is not a cause for alarm. The system is still functioning within acceptable parameters, but some adjustments or monitoring may be necessary to prevent further stress or potential problems.

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  • 14. 

    Which of the following is the most true statement about Nalco Scale Index?

    • A.

      It measures total polymer consumption.

    • B.

      It measures the rate of polymer consumption.

    • C.

      It measures the amount of scale that is forming.

    • D.

      It measures the rate of change of pH, and conductivity.

    Correct Answer
    B. It measures the rate of polymer consumption.
    Explanation
    The Nalco Scale Index is a measurement that determines the rate of polymer consumption. This means that it assesses how quickly the polymer is being used up in a given system. This information is important for monitoring and controlling the amount of polymer being added to prevent scale formation. The other statements are not accurate because they do not describe the purpose of the Nalco Scale Index.

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  • 15. 

    In closed cooling water systems, the primary cause of pump seal failure is:

    • A.

      High ph

    • B.

      Microbiological attack

    • C.

      Too high inhibitor dosage

    • D.

      Particulate erosion

    Correct Answer
    D. Particulate erosion
    Explanation
    In closed cooling water systems, pump seal failure is primarily caused by particulate erosion. This refers to the wear and tear of the pump seal due to the presence of solid particles in the water. These particles can be abrasive and cause damage to the seal, leading to its failure. High pH, microbiological attack, and too high inhibitor dosage may also contribute to pump seal failure, but particulate erosion is the main culprit.

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  • 16. 

    Which of the following types of closed system inhibitors should you avoid for a system with hard water makeup, and a high heat flux (caster mold water)?

    • A.

      Azole.

    • B.

      Silicate

    • C.

      Nitrite

    • D.

      Molybdate

    Correct Answer
    B. Silicate
    Explanation
    Silicate inhibitors should be avoided for a system with hard water makeup and high heat flux. Silicate inhibitors are not effective in preventing scale formation in hard water systems and can actually contribute to the formation of scale deposits. Additionally, in high heat flux conditions, silicate inhibitors can form a hard, insoluble scale that can impair heat transfer and reduce system efficiency. Therefore, it is recommended to avoid the use of silicate inhibitors in such systems.

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  • 17. 

    The functions of  HSP (high stress polymer) and TRC-233 in a cooling water program are:

    • A.

      Iron dispersancy.

    • B.

      Particulate dispersancy.

    • C.

      Calcium phosphate and zinc hydroxide control.

    • D.

      All of the above

    • E.

      A) and b).

    Correct Answer
    D. All of the above
    Explanation
    HSP (high stress polymer) and TRC-233 serve multiple functions in a cooling water program. They help in dispersing iron and particulate matter, as well as controlling the formation of calcium phosphate and zinc hydroxide. Therefore, the correct answer is "All of the above."

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  • 18. 

    Key corrosion control tests for a multifunctional AZ-lite program are:

    • A.

      Monitoring the soluble zinc and difference between filtered & unfiltered total phosphate

    • B.

      Monitoring the soluble zinc and difference between filtered & unfiltered ortho phosphate.

    • C.

      Monitoring the difference between soluble & total zinc.

    • D.

      Monitoring the soluble zinc and filtered total phosphate.

    Correct Answer
    B. Monitoring the soluble zinc and difference between filtered & unfiltered ortho phosphate.
    Explanation
    The correct answer is "Monitoring the soluble zinc and difference between filtered & unfiltered ortho phosphate." This is because the question is asking for the key corrosion control tests for a multifunctional AZ-lite program. The tests mentioned in the answer option are specifically related to monitoring the soluble zinc levels and the difference between filtered and unfiltered ortho phosphate levels. These tests are important in assessing the effectiveness of corrosion control measures in the program.

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  • 19. 

    You have a magnesium silicate-scaling problem in your cooling system. How would you best manage this problem? 

    • A.

      Adjust tower pH up.

    • B.

      Adjust SiO2 levels in tower to a maximum of 150 ppm.

    • C.

      Adjust pH or cycles so that magnesium silicate is soluble.

    • D.

      Clarify the makeup water

    Correct Answer
    C. Adjust pH or cycles so that magnesium silicate is soluble.
    Explanation
    To manage the magnesium silicate-scaling problem in the cooling system, the best approach is to adjust the pH or cycles so that magnesium silicate is soluble. This means maintaining the pH at a level where the magnesium silicate remains in a dissolved state, preventing it from forming scales. By controlling the pH or adjusting the cycles, the magnesium silicate can be kept in a soluble form and minimize scaling issues in the cooling system.

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  • 20. 

    For a stabilized phosphate program, the mechanism for corrosion control is:

    • A.

      High levels of orthophosphate film over the entire metal surface.

    • B.

      Formation of a passive gamma iron oxide film at the cathode.

    • C.

      Precipitation of protective ortho-phosphate films at the cathode, as a result of locally high pH levels.

    • D.

      Formation of a passive gamma iron oxide film at the anode and precipitation of a polyphosphate film at the cathode.

    Correct Answer
    D. Formation of a passive gamma iron oxide film at the anode and precipitation of a polyphosphate film at the cathode.
    Explanation
    For a stabilized phosphate program, the mechanism for corrosion control involves the formation of a passive gamma iron oxide film at the anode, which acts as a protective barrier against corrosion. Additionally, a polyphosphate film is precipitated at the cathode due to locally high pH levels. This film further enhances corrosion resistance by preventing the formation of corrosion-inducing compounds. Together, these mechanisms help to minimize corrosion and protect the metal surface.

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  • 21. 

    The following can generally be used to determine the cooling water flow rate through an exchanger:

    • A.

      Ultrasonic flow meter

    • B.

      Heat balance

    • C.

      Pump curve

    • D.

      A) and b) only

    • E.

      All of the above

    Correct Answer
    D. A) and b) only
    Explanation
    The correct answer is a) and b) only. Ultrasonic flow meters can be used to measure the flow rate of cooling water through an exchanger. Heat balance calculations can also be used to determine the flow rate by analyzing the heat transfer in the system. However, a pump curve is not typically used to directly measure the flow rate of cooling water.

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  • 22. 

    A four-pass heat exchanger has 560, 1.27 cm (1/2 inch) OD, 20 BWG (0.9 mm or 0.035-inch wall thickness) tubes. What is the water velocity through each tube if diagnostic Trasar studies show 127 cubic meter/hr (560 GPM) of water flow?

    • A.

      0.68 m/sec (2.25 ft/sec)

    • B.

      2.70 m/sec (8.9 ft/sec)

    • C.

      1.68 m/sec (5.5 ft/sec)

    • D.

      1.00 m/sec (3.3 ft/sec

    Correct Answer
    B. 2.70 m/sec (8.9 ft/sec)
    Explanation
    The water velocity through each tube can be calculated by dividing the flow rate of water (in cubic meter/hr) by the cross-sectional area of the tube (in square meters). Since the question does not provide the length of the tubes, we can assume that it is not needed for the calculation. The cross-sectional area of each tube can be calculated using the formula for the area of a circle (πr^2), where r is the radius of the tube (OD/2). By plugging in the given values, we can determine that the water velocity through each tube is 2.70 m/sec (8.9 ft/sec).

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  • 23. 

    The NCM ( Nalco Corrosion Monitor) 100:

    • A.

      Provides a pitting corrosion rate.

    • B.

      Replaces corrosion coupon testing.

    • C.

      Measures trends in corrosivity.

    • D.

      None of the above.

    • E.

      Replaces corrosion coupon testing and Measures trends in corrosivity.

    Correct Answer
    E. Replaces corrosion coupon testing and Measures trends in corrosivity.
    Explanation
    The NCM 100 is a device that replaces corrosion coupon testing, which is a common method used to measure corrosion rates. Additionally, it also measures trends in corrosivity, allowing for a better understanding of the corrosion process over time. Therefore, the correct answer is that the NCM 100 replaces corrosion coupon testing and measures trends in corrosivity.

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  • 24. 

    A heat exchanger condenses a gaseous ethanol stream of known flow rate at 216ºF and then cools the condensed material down to 110ºF.  In order to determine the water flow required to achieve this, you need to know:

    • A.

      Inlet and outlet temperatures of the water, heat of vaporization of water plus the specific heat of water.

    • B.

      Inlet and outlet temperatures of the water, specific heat, heat of vaporization and condensation temperature of the process liquid

    • C.

      Heat of vaporization, heat of condensation, and specific heat of the process side and the water.

    • D.

      Temperatures of the water, surface area of the heat exchanger, and the U value of the heat exchanger.

    Correct Answer
    B. Inlet and outlet temperatures of the water, specific heat, heat of vaporization and condensation temperature of the process liquid
    Explanation
    To determine the water flow required to achieve the desired cooling, you need to know the inlet and outlet temperatures of the water, as well as the specific heat, heat of vaporization, and condensation temperature of the process liquid. This information is necessary to calculate the heat transfer between the two streams and determine the amount of water needed to achieve the desired temperature reduction. The specific heat and heat of vaporization are used to calculate the energy required for the phase change of the process liquid, while the temperatures of the water and process liquid are needed to calculate the temperature difference driving the heat transfer.

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  • 25. 

    The Liquid to Gas (L/G) factor for a cooling tower is important! What things affect L/G?

    • A.

      Recirculation rate

    • B.

      Fan Speed

    • C.

      Fill Cleanliness.

    • D.

      All of the above

    • E.

      Only a) and c)

    Correct Answer
    D. All of the above
    Explanation
    The Liquid to Gas (L/G) factor for a cooling tower is important because it determines the efficiency of the cooling process. The L/G factor is affected by the recirculation rate, which refers to the amount of water being recirculated within the cooling tower. A higher recirculation rate can lead to a higher L/G factor. Fan speed also affects the L/G factor as it determines the amount of air passing through the tower, affecting the rate of evaporation. Fill cleanliness is another factor that affects L/G as clean fill allows for better heat transfer. Therefore, all of the above factors affect the L/G factor in a cooling tower.

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  • 26. 

    When slug feeding a chemical to a cooling system that cycles 3-6 times, which of the following is the most  important factor to consider?

    • A.

      Blowdown rate

    • B.

      Holding time index

    • C.

      System volume

    • D.

      Make up rate

    • E.

      B) and c)

    Correct Answer
    B. Holding time index
    Explanation
    The most important factor to consider when slug feeding a chemical to a cooling system that cycles 3-6 times is the holding time index. The holding time index refers to the amount of time that the chemical remains in the system before it is discharged. It is important to ensure that the chemical has enough time to effectively treat the system before it is flushed out. This factor is crucial in determining the efficiency and effectiveness of the chemical treatment.

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  • 27. 

    Which of the following statements about enhanced chiller tubes is not true:

    • A.

      Frequent manual cleaning is not needed.

    • B.

      Biological induced corrosion is a potential problem.

    • C.

      There is an increased potential for premature corrosion.

    • D.

      Helical grooving improves heat transfer efficiency.

    Correct Answer
    A. Frequent manual cleaning is not needed.
    Explanation
    The statement "Frequent manual cleaning is not needed" is not true because enhanced chiller tubes still require regular cleaning to maintain their efficiency.

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  • 28. 

    The wet bulb temperature is:

    • A.

      Equal to temperature of tower basin water.

    • B.

      Lowest temperature you are able to cool air or water with an evaporative process.

    • C.

      Temperature of air surrounding the tower.

    • D.

      B) and c)

    Correct Answer
    B. Lowest temperature you are able to cool air or water with an evaporative process.
    Explanation
    The wet bulb temperature refers to the lowest temperature that can be achieved by cooling air or water through an evaporative process. It is a measure of the cooling potential of the air or water when evaporation occurs. This temperature is lower than the temperature of the tower basin water or the temperature of the air surrounding the tower.

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  • 29. 

    What is the type of fill?

    • A.

      Splash fill

    • B.

      Non-splash

    • C.

      Splash and non-splash

    • D.

      None

    Correct Answer
    B. Non-splash
    Explanation
    The correct answer is "Non-splash." This implies that the fill being referred to is not of the "splash" type. It can be inferred that there are different types of fill, and the non-splash type is one of them. However, without further context, it is difficult to determine the specific characteristics or purpose of this non-splash fill.

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  • 30. 

    Mark A in the above diagram

    • A.

      Cooling Tower

    • B.

      Evaporation

    • C.

      Heat load

    • D.

      Condensation

    Correct Answer
    B. Evaporation
    Explanation
    Evaporation is the correct answer because it is the process in which a liquid, in this case, water, changes into a gas or vapor. In the context of a cooling tower, evaporation plays a crucial role in removing heat from the system. As water is exposed to air inside the cooling tower, it evaporates, taking away heat from the surrounding environment. This helps in cooling down the water and ultimately the system it is being used for. Therefore, evaporation is an essential step in the cooling process of a cooling tower.

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  • 31. 

    Mark B in the above diagram

    • A.

      Condensation

    • B.

      Heat load

    • C.

      Evaporation

    • D.

      Compression

    Correct Answer
    B. Heat load
  • 32. 

    Mark E in the above diagram

    • A.

      Cooling Tower

    • B.

      Evaporation

    • C.

      Heat load

    • D.

      Condensation

    Correct Answer
    A. Cooling Tower
    Explanation
    The correct answer is Cooling Tower. A cooling tower is a device used to remove heat from a system by transferring it to the atmosphere through the process of evaporation. It is commonly used in industrial settings to cool down water or other fluids that have absorbed heat from a process or equipment. The cooling tower facilitates the evaporation of water, which helps in dissipating the heat and reducing the temperature of the fluid. This allows the system to maintain optimal operating conditions and prevent overheating.

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  • 33. 

    Mark C in the above diagram

    • A.

      Evaporation

    • B.

      Compression

    • C.

      Condensation

    • D.

      Heat load

    Correct Answer
    B. Compression
    Explanation
    Compression is the correct answer because it is the process in which the gas is compressed, resulting in an increase in its pressure and temperature. In the given diagram, the letter C is marked, indicating that the process being depicted is compression. This process is an essential part of the refrigeration cycle, where the gas is compressed to increase its energy and prepare it for the next stage of the cycle.

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  • 34. 

    Mark D in the above diagram

    • A.

      Condensation

    • B.

      Heat load

    • C.

      Compression

    • D.

      Cooling Tower

    Correct Answer
    A. Condensation
    Explanation
    The correct answer is "Condensation." In the given diagram, "Condensation" is marked as D. Condensation is the process in which a gas or vapor turns into a liquid when it loses heat energy. In the context of a cooling tower, condensation occurs when the warm water vapor from the heat load is cooled down and converted back into liquid form. Therefore, condensation is the appropriate term to describe the process marked as D in the diagram.

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  • 35. 

    Correct Answer
    Fan Stack
    Louvers
    Drift eliminators
    Make up
    Basin
    Return water
    Hot air inlet
    Distribution header
    Fills
    Plenum
    Distribution deck
  • 36. 

    Correct Answer
    Condensate outlet
    Hot well
    Condensate drain
    Water box
    Water cover box
    Stream inlet
    Water outlet
  • 37. 

    Correct Answer
    Baffles
    Tube bundle
    water box
    shell
    tube sheet
  • 38. 

    What is the concentration of biocide present in the system after 40 hours of dosage if the initial concentartion is 100 mg/l for the system of 568 m3 and blowdown of 17m3/h? (Assume the loss of biocide only through blowdown) 

    • A.

      30 mg/l

    • B.

      40 mg/l

    • C.

      20 mg/l

    • D.

      10 mg/l

    Correct Answer
    A. 30 mg/l
    Explanation
    The concentration of biocide in the system is determined by the initial concentration and the rate of blowdown. Since the blowdown rate is given as 17m3/h, we can calculate the total blowdown volume after 40 hours as 17m3/h * 40h = 680m3.

    To find the concentration after 40 hours, we need to subtract the blowdown volume from the initial volume:

    568m3 - 680m3 = -112m3

    Since the blowdown volume is greater than the initial volume, the concentration of biocide becomes zero. Therefore, the correct answer is 0 mg/l.

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  • 39. 

    39. Calculate the product dosage to maintain 100 ppm in recirculating water, if cycles are 5 and make up rate is 20 m3/hr.

    • A.

      8.6 kg/day

    • B.

      9.6 kg/day

    • C.

      10.6 kg/day

    • D.

      None

    Correct Answer
    B. 9.6 kg/day
    Explanation
    To calculate the product dosage to maintain 100 ppm in recirculating water, we need to consider the cycles and make up rate. The cycles refer to the number of times the water is recycled in an hour, while the make up rate is the rate at which fresh water is added to the system.

    In this case, the cycles are given as 5 and the make up rate is 20 m3/hr. To calculate the product dosage, we can use the formula:

    Product Dosage = (Make up rate * Dosage required) / Cycles

    Substituting the given values, we get:

    Product Dosage = (20 * 100) / 5 = 400 kg/hr

    Since the question asks for the dosage per day, we need to convert it to kg/day. Assuming 24 hours in a day, the product dosage would be:

    Product Dosage = 400 kg/hr * 24 hr/day = 9,600 kg/day

    Therefore, the correct answer is 9.6 kg/day.

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  • 40. 

    There is loss of nitrite in closed loop system at x ppm in 1 hour, molybdate also decreaseing but at the rate of y ppm in 1 hour. There is drop in conductivity observed. What can be likely the cause? 

    • A.

      Leaks can be at different locations

    • B.

      Nitrite could be converted to nitrate

    • C.

      Re-passivation could be happening

    • D.

      All the above

    • E.

      B and c

    Correct Answer
    E. B and c
    Explanation
    The likely cause of the observed drop in conductivity, as indicated by the decrease in nitrite and molybdate levels, can be attributed to the conversion of nitrite to nitrate (b) and the occurrence of re-passivation (c). These processes can lead to a decrease in the concentration of nitrite and molybdate, resulting in a loss of conductivity in the closed loop system. Additionally, the presence of leaks at different locations (a) may also contribute to the observed drop in conductivity.

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  • 41. 

    Nitrite based inhibitors should not be used above the temperature of 

    • A.

      300 F

    • B.

      320 F

    • C.

      220 F

    • D.

      150 F

    Correct Answer
    A. 300 F
    Explanation
    Nitrite based inhibitors should not be used above the temperature of 300 F because at higher temperatures, nitrites can decompose and form harmful byproducts such as nitrosamines. These byproducts are known to be carcinogenic and can pose health risks. Therefore, it is recommended to avoid using nitrite based inhibitors above this temperature to ensure safety and prevent any potential health hazards.

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  • 42. 

    42. Find the heat transfer area for Shell and Tube HEX which as 200 tubes if tube bundle diameter is 5 m and straight run length of each tube is 20 m. O.D of tube is 0.005 m.

    • A.

      75 m2

    • B.

      63 m3

    • C.

      70 m3

    • D.

      80 m3

    Correct Answer
    A. 75 m2
  • 43. 

    What is one ton refrigeration means

    • A.

      Amount of heat required to melt 1 ton of ice

    • B.

      Amount of heat required to melt 1 ton of ice in 24 hours

    • C.

      Amount of cooling required to form of ice

    • D.

      Amount of cooling required to form of ice in 24 hours

    Correct Answer
    B. Amount of heat required to melt 1 ton of ice in 24 hours
    Explanation
    One ton of refrigeration refers to the amount of heat that needs to be removed in order to melt one ton (2000 pounds) of ice within a 24-hour period. This measurement is commonly used in the air conditioning and refrigeration industry to quantify the cooling capacity of a system. It indicates the ability of a system to remove heat and maintain a desired temperature, with one ton of refrigeration equivalent to 12,000 BTUs per hour.

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  • 44. 

    Suppose that you add 50 kg of sodium chloride (NaCl), and the chloride (as   NaCl) increases from 100 to 250 mg/L.  What is the volume of the system?

    • A.

      333

    • B.

      330

    • C.

      300

    • D.

      233

    Correct Answer
    A. 333
    Explanation
    The volume of the system can be calculated by using the formula:

    Volume = (mass of substance added) / (concentration of substance after addition - concentration of substance before addition)

    In this case, 50 kg of sodium chloride (NaCl) is added, and the chloride concentration increases from 100 to 250 mg/L.

    Using the formula, the volume of the system is:

    Volume = 50 kg / (250 mg/L - 100 mg/L) = 50 kg / 150 mg/L = 333 L

    Therefore, the correct answer is 333.

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  • 45. 

    Assume you add 360 grams of TRASAR 23299 and the TRASAR reading increases from zero to 120 mg/L after complete mixing. what is the system volume? Please not factor of 100 should be multiplied because concentration of fluoroscent TRASAR is shown above, not the product.

    • A.

      250 m3

    • B.

      300 m3

    • C.

      200 m3

    • D.

      150 m3

    Correct Answer
    B. 300 m3
  • 46. 

    In the above system, where would be the best place to apply an organic scale inhibitor?

    • A.

      A

    • B.

      B

    • C.

      C

    • D.

      D

    Correct Answer
    B. B
    Explanation
    The best place to apply an organic scale inhibitor in the above system would be at point B. This is because point B is located before the scale-forming minerals have a chance to precipitate and form scale. By applying the organic scale inhibitor at this point, it can prevent the minerals from adhering to surfaces and forming scale, thus maintaining the efficiency and longevity of the system.

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  • 47. 

    A heat exchanger contains 400 tubes, 20 mm in diameter, 15 m in length.  The cooling water is on the tube side and makes a single pass through the exchanger.  If the cooling water flow is 85 m3/h, what is the water velocity in m/s?  If this is a four pass exchanger, with 50 tubes/pass, what is the water velocity?

    • A.

      1.5 m/s

    • B.

      2 m/s

    • C.

      0.18 m/s

    • D.

      4 m/s

    Correct Answer
    A. 1.5 m/s
    Explanation
    The water velocity can be calculated by dividing the flow rate (in m^3/h) by the cross-sectional area of the tubes (in m^2). In the first case, since there are 400 tubes, the cross-sectional area is calculated by multiplying the area of one tube (π * (0.02/2)^2) by 400. Dividing the flow rate by this cross-sectional area gives a velocity of 1.5 m/s. In the second case, since there are 50 tubes per pass, the cross-sectional area is calculated by multiplying the area of one tube by 50. Dividing the flow rate by this cross-sectional area gives a velocity of 6 m/s. However, since the water makes four passes, the actual water velocity is one-fourth of this value, which is 1.5 m/s.

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  • 48. 

    A co-current heat exchanger is used to cool air.  What is the LMTD with the following temperatures?  Process in: 82oC               Water in: 27oC  Process out: 35oC             Water out: 32oC  The counter-current exchanger cools 5440 kg/hr of air,   Cp = 1 kJ/kg oK. The area = 6.5 m2.    What is the water flow through the exchanger?

    • A.

      10 m3/h

    • B.

      12 m3/h

    • C.

      9 m3/h

    • D.

      11 m3/h

    Correct Answer
    B. 12 m3/h
    Explanation
    The water flow through the exchanger is 12 m3/h.

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  • 49. 

    Find the velocity of water through the tubes for the HEX which has 600 tubes and 4 passes and water flow rate of 19 m3/hr. OD of the tube is 10.1 mm and thickness of the tube wall is 0.95 mm

    • A.

      0.65 m/s

    • B.

      0.75 m/s

    • C.

      0.85 m/s

    • D.

      0.95 m/s

    Correct Answer
    A. 0.65 m/s
    Explanation
    The velocity of water through the tubes can be calculated using the formula: velocity = flow rate / (cross-sectional area of the tube). The cross-sectional area of the tube can be calculated using the formula: area = π * (OD/2)^2 - π * ((OD/2) - thickness)^2. Plugging in the given values, we can calculate the velocity of water to be 0.65 m/s.

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  • Current Version
  • Mar 21, 2023
    Quiz Edited by
    ProProfs Editorial Team
  • May 01, 2012
    Quiz Created by
    Vishwa4207
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