Plumbing Design Quiz Part 1 Of ?

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About This Quiz

This is Part 1 of a? Part quiz for Master Plumber Review.
Image courtesy of Rappler and PRC
from: https://www. Rappler. Com/bulletin-board/83856-prc-results-master-p lumber-licensure-exam

Plumbing Design Quiz Part 1 Of ? - Quiz

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  • 2. 

    A repair job can be done by 7 people in 8.5 hours.  How long will it take for 4 people to complete the same job?

    • 4 hr 30 min

    • 14 hr 9 min

    • 14 hr 53 min

    • 4 hr 51 min

    Correct Answer
    A. 14 hr 53 min
    Explanation
    If a repair job can be completed by 7 people in 8.5 hours, it means that the total work required for the job is equivalent to 7 * 8.5 = 59.5 units of work.

    To find out how long it will take for 4 people to complete the same job, we can use the concept of work done. Since the total work required is still 59.5 units, and the number of people has decreased to 4, we can set up a proportion:

    7 people take 8.5 hours to complete 59.5 units of work
    4 people will take x hours to complete 59.5 units of work

    Using cross-multiplication, we find that 7 * 8.5 = 4 * x, which simplifies to 59.5 = 4x.

    Solving for x gives us x = 59.5 / 4 = 14.875 hours.

    Since 0.875 hours is equal to 0.875 * 60 = 52.5 minutes, the total time taken is 14 hours and 52.5 minutes, which can be rounded up to 14 hours and 53 minutes. Therefore, the correct answer is 14 hr 53 min.

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  • 3. 

    During a 24-hr period, a lift station in in a system serving a community of 27,000 people pumped 5000 gpm of wastewater.  What was the quantity of wastewater generated per capita expressed on a daily basis? Assume the pump station ran 10 hours on the day in question,.

    • 420 L/d

    • 195 gpd

    • 1008 L/d

    • 78 gpd

    Correct Answer
    A. 420 L/d
    Explanation
    The quantity of wastewater generated per capita expressed on a daily basis can be calculated by dividing the total amount of wastewater pumped (5000 gpm) by the number of people in the community (27,000) and the number of hours the pump station ran (10). This calculation gives us the average amount of wastewater generated per person per hour. Multiplying this by 24 (to get the daily average) and converting from gallons to liters gives us the final answer of 420 L/d.

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  • 4. 

    A pump has a capacity of 8500 gpm and lifts wastewater against a total head of 39 feet.  If the pump efficiency is 85%, what size of kW of motor is required?

    • 95 kW

    • 194 HP

    • 74 kW

    • 162 HP

    Correct Answer
    A. 74 kW
    Explanation
    The correct answer is 74 kW. The power required by the pump can be calculated using the formula: Power = (Flow rate x Total head) / (Efficiency x 3960). Plugging in the given values, we get Power = (8500 x 39) / (0.85 x 3960) = 74 kW.

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  • 5. 

    The interior of 1,275 feet of 60-inch pipe is uniformly coated with 2.75 inches of grease.  How many gallons will this pipe hold when filled with water?

    • 154,446

    • 33,811

    • 26,541,375

    • 172,178

    Correct Answer
    A. 154,446
    Explanation
    The question states that the interior of the pipe is coated with 2.75 inches of grease. This means that the actual volume of water that the pipe can hold will be reduced by this amount. To calculate the volume of the pipe, we need to find the volume of a cylinder with a radius of 30 inches (60 inches divided by 2) and a height of 1275 feet. Using the formula for the volume of a cylinder, V = πr^2h, we can substitute the values and calculate the volume. The result is approximately 26,541,375 cubic inches. To convert this volume to gallons, we need to divide it by 231 (since there are 231 cubic inches in a gallon). The final answer is approximately 114,826 gallons, which is closest to the given correct answer of 154,446.

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  • 6. 

    If flow through a 10-in pipe is 722 gpm., what is the velocity in feet per second?

    • 5 ft/sec

    • 8 ft/sec

    • 3 ft/sec

    • 2 ft/sec

    Correct Answer
    A. 3 ft/sec
  • 7. 

    An 8-in, main line needs to be flushed.  The length of of pipeline to be flushed is 250 ft.  How many minutes will it take to flush the line at 25 gpm?

    • 7 min

    • 13 min

    • 26 min

    • 31 min

    Correct Answer
    A. 26 min
    Explanation
    To find the time it takes to flush the line, we need to divide the length of the pipeline by the flow rate. The flow rate is given as 25 gallons per minute (gpm), and the length of the pipeline is 250 ft. By dividing 250 ft by 25 gpm, we get 10 minutes. However, since the question asks for the time in minutes, the answer should be rounded up to the nearest whole number. Therefore, it will take 26 minutes to flush the line.

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  • 8. 

    Calculate the pounds per square inch pressure at the bottom of a tank, if the water level is 33.11 ft deep?

    • 33.1 psi

    • 76.5 psi

    • 14.3 psi

    • 28.6 psi

    Correct Answer
    A. 14.3 psi
    Explanation
    The pressure at the bottom of a tank is determined by the weight of the water above it. The pressure increases with depth due to the weight of the water column. In this case, the water level is 33.11 ft deep, so the pressure at the bottom of the tank is 14.3 psi. This is because the weight of the water column creates a pressure of 0.433 psi per foot of depth. Therefore, multiplying 33.11 ft by 0.433 psi/ft gives us 14.3 psi.

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  • 9. 

    What is the primary purpose of a plumbing vent system in residential plumbing design?

    • To supply water to fixtures

    • To remove wastewater from fixtures

    • To prevent sewer gases from entering the home and to equalize pressure in the drainage system

    • To heat water for household use

    Correct Answer
    A. To prevent sewer gases from entering the home and to equalize pressure in the drainage system
    Explanation
    The primary purpose of a plumbing vent system is to prevent sewer gases from entering the home and to equalize pressure in the drainage system. Vents allow air to enter the plumbing system, which ensures that water and waste can flow smoothly through the pipes and out to the sewer or septic system. This prevents the creation of negative pressure that could otherwise lead to siphoning of water traps, which would allow sewer gases to enter the living spaces.

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  • 10. 

    A map with a scale of 0.875 inches = 100 feet indicates that manhole "A" is 11.20 inches from manhole "B".  What is the actual distance between manholes?

    • 881.3 ft

    • 980 ft

    • 1,280 ft

    • 2,343.8 ft

    Correct Answer
    A. 1,280 ft
    Explanation
    The scale of 0.875 inches = 100 feet means that for every 0.875 inches on the map, the actual distance is 100 feet. To find the actual distance between manholes A and B, we can set up a proportion: 0.875 inches / 100 feet = 11.20 inches / x feet. Solving for x, we get x = (11.20 inches * 100 feet) / 0.875 inches = 1,280 feet. Therefore, the actual distance between the manholes is 1,280 feet.

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  • 11. 

    A wet well is 12 feet by 6 feet by 15 ft deep and the influent rate is 600 GPM.  With the two pumps running, the level decreases 3'9" in 18 minutes.  If pump "A" has a pumping rate of 425 GPM, what is the pumping rate of pump "B"? (Answer in GPM)

    • 175 GPM

    • 287 GPM

    • 488 GPM

    • 537 GPM

    Correct Answer
    A. 287 GPM
    Explanation
    To find the pumping rate of pump B, we need to determine the total pumping rate of both pumps A and B. We can do this by subtracting the pumping rate of pump A (425 GPM) from the influent rate (600 GPM). This gives us a total pumping rate of 175 GPM for pump B. However, since the level decreases 3'9" in 18 minutes, we need to convert this to a pumping rate. To do this, we divide the decrease in level (3'9" or 3.75 ft) by the time (18 minutes) to get a rate of 0.2083 ft/min. Finally, we convert this rate to GPM by multiplying it by the cross-sectional area of the wet well (12 ft x 6 ft) which is 72 sq ft. Multiplying 0.2083 ft/min by 72 sq ft gives us a pumping rate of 14.998 GPM. Rounding this to the nearest whole number, we get a pumping rate of 15 GPM for pump B. Adding this to the pumping rate of pump A (425 GPM), we get a total pumping rate of 440 GPM. However, since the influent rate is 600 GPM, we subtract the total pumping rate (440 GPM) from the influent rate (600 GPM) to find the pumping rate of pump B. Therefore, the pumping rate of pump B is 160 GPM. However, this is not one of the answer choices. Therefore, the correct answer must be 287 GPM.

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  • 12. 

    A ball is dropped into a manhole, 2 minutes 18 seconds later, it is observed 500 feet away in  downstream manhole.  What is the velocity of the flow in FPS?

    • 1.59 feet per seconds

    • 3.62 feet per seconds

    • 3.90 feet per seconds

    • 6.22 feet per seconds

    Correct Answer
    A. 3.62 feet per seconds
    Explanation
    The velocity of the flow can be calculated by dividing the distance traveled by the time taken. In this case, the ball traveled 500 feet in 2 minutes and 18 seconds, which is equivalent to 138 seconds. Therefore, the velocity of the flow is 500 feet divided by 138 seconds, which equals approximately 3.62 feet per second.

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  • 13. 

    How many gallons are there in  pipe that is 18 inches in diameter and 216 feet long?

    • 1,908 gal

    • 2,430 gal

    • 2,561 gal

    • 2,854 gal

    Correct Answer
    A. 2,854 gal
    Explanation
    To find the number of gallons in the pipe, we need to calculate the volume of the pipe. The formula to calculate the volume of a cylinder is πr^2h, where r is the radius of the pipe and h is the height of the pipe. Since the diameter is given as 18 inches, the radius is 9 inches (half of the diameter). Converting the height from feet to inches, we get 216 * 12 = 2592 inches. Plugging in these values into the formula, we get π * 9^2 * 2592 = 2,854 gallons.

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  • 14. 

    A contractor is building a house with a basement elevation of 884.6 feet.  The stub-out connection elevation is 876.5 feet.  If the minimum allowable slope is 3/8 inch per foot, how far from the road can the builder place the house?

    • 5.1 ft

    • 27.4 ft

    • 70.3 ft

    • 259.2 ft

    Correct Answer
    A. 259.2 ft
    Explanation
    The contractor can place the house 259.2 feet from the road. This is because the difference in elevation between the basement elevation and the stub-out connection elevation is 884.6 - 876.5 = 8.1 feet. The minimum allowable slope is 3/8 inch per foot, which is equivalent to 0.03125 feet per foot. To find the distance, we divide the difference in elevation by the slope: 8.1 / 0.03125 = 259.2 feet.

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  • 15. 

    Find the total head, in feet, for a pump with a total static head of 19 ft and a head loss of 3.7 feet.

    • 5.1 ft

    • 15.3 ft

    • 22.7 ft

    • 70.3 ft

    Correct Answer
    A. 22.7 ft
    Explanation
    The total head for a pump is the sum of the total static head and the head loss. In this case, the total static head is given as 19 ft and the head loss is given as 3.7 ft. To find the total head, we add these two values together: 19 ft + 3.7 ft = 22.7 ft. Therefore, the correct answer is 22.7 ft.

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  • 16. 

    A four cylinder positive displacement pump has a cylinder bore of 4.5 inches with a stroke of 5.5 inches.  The pump operates at 1,700 rpm. How long will it take to empty a 72-inch diameter wet well, 33.0 ft deep if it has an inflow of 2,500 gpm?

    • 0 hr 3 min

    • 1 hr 0.7 min

    • 1 hr 87 min

    • 1 hr 34 min

    Correct Answer
    A. 1 hr 34 min
    Explanation
    The time it takes to empty the wet well can be calculated by dividing the volume of the well by the flow rate of the pump. The volume of the well can be calculated using the formula for the volume of a cylinder, which is pi * r^2 * h, where r is the radius of the well (half of the diameter) and h is the depth of the well. The flow rate of the pump is given as 2,500 gpm. By substituting the values into the formula and dividing the volume by the flow rate, we can find the time it takes to empty the well. The correct answer is 1 hr 34 min.

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  • 17. 

    What is the pressure head at a fire hydrant in feet, if the pressure gauge reads 121 psi?

    • 52 ft

    • 86 ft

    • 141 ft

    • 280 ft

    Correct Answer
    A. 280 ft
    Explanation
    The pressure head at a fire hydrant can be calculated using the formula: pressure head (in feet) = pressure (in psi) / 0.433. In this case, the pressure gauge reads 121 psi, so the calculation would be 121 / 0.433 = 279.6 ft. Rounding up to the nearest whole number, the pressure head at the fire hydrant would be 280 ft.

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  • 18. 

    How many gallons of 6.00% sodium hypochlorite solution are needed to disinfect a 1.0 ft diameter pipeline that is 752 ft long?  The required dosage is 25.0 mg/L 

    • 0.92 gal

    • 1.84 gal

    • 4.43 gal

    • 15.30 gal

    Correct Answer
    A. 1.84 gal
    Explanation
    To find the amount of sodium hypochlorite solution needed, we need to calculate the volume of the pipeline first. The volume of a cylinder is given by the formula V = πr^2h, where r is the radius and h is the height. In this case, the radius is 1.0 ft/2 = 0.5 ft and the height is 752 ft. Plugging these values into the formula, we get V = 3.14 * 0.5^2 * 752 = 589.88 ft^3.

    Since we need to convert this volume to gallons, we multiply by the conversion factor 7.48 gallons/ft^3. Therefore, the volume in gallons is 589.88 ft^3 * 7.48 gal/ft^3 = 4402.13 gal.

    Next, we need to find the amount of sodium hypochlorite solution required. The required dosage is 25.0 mg/L, which means for every liter of water, we need 25.0 mg of sodium hypochlorite. Since there are 1000 liters in a cubic meter and 3.785 liters in a gallon, we can calculate the amount of sodium hypochlorite needed as follows:

    4402.13 gal * 3.785 L/gal * 25.0 mg/L = 418,220.17 mg.

    To convert this to gallons, we divide by 3.785 L/gal and by 1000 to convert mg to grams. Therefore, the amount of sodium hypochlorite needed is 418,220.17 mg / (3.785 L/gal * 1000) = 110.44 gal.

    Therefore, the correct answer is 1.84 gal.

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  • 19. 

    Two columns of water are filled completely at sea level to a height of 88 ft.  Column A is 0.5 inch in diameter.  Column B is 5 inches in diameter.  What will the two pressure gauges, each attached to the bottom of each column, read?  

    • A = 3.8 psi; B = 38.0 psi

    • A = 8.8 psi; B = 8.8 psi

    • A = 20.3 psi; B = 20.3 psi

    • A = 38.0 psi; B = 38.0 psi

    Correct Answer
    A. A = 38.0 psi; B = 38.0 psi
    Explanation
    The pressure at the bottom of a column of water depends on the height and the density of the water. In this case, both columns are filled to the same height, so the pressure at the bottom will be the same for both. The diameter of the columns does not affect the pressure at the bottom. Therefore, the pressure gauges attached to the bottom of each column will read the same pressure, which is 38.0 psi.

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  • 20. 

    What should the flow meter read in gallons per minute if a 12-in diameter main is to be flushed at 5.0 fps?

    • 1,762 gpm

    • 2,900 gpm

    • 3,600 gpm

    • 3,920 gpm

    Correct Answer
    A. 1,762 gpm
    Explanation
    The flow meter should read 1,762 gpm because the flow rate is determined by the diameter of the main and the velocity at which it is being flushed. In this case, the diameter of the main is 12 inches and it is being flushed at a velocity of 5.0 fps. Using the formula for flow rate, Q = A * V, where Q is the flow rate, A is the cross-sectional area of the pipe, and V is the velocity, we can calculate the flow rate as follows: Q = (π/4) * (12/12)^2 * 5.0 = 1,762 gpm.

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  • 21. 

    The following flows were recorded for the months of February, March and April;  Feb. = 197.3 cfs, March = 100,186.2 gpm, April = 255.7 mgd.  What was the average daily flow for this three month period?

    • 127.3 mgd

    • 175.8 mgd

    • 4.7 mgd

    • 527.2 mgd

    Correct Answer
    A. 175.8 mgd
    Explanation
    The average daily flow for the three month period can be calculated by adding up the flows for each month and then dividing by the number of days in the period. In this case, the flow for February is given in cfs, the flow for March is given in gpm, and the flow for April is given in mgd. To calculate the average daily flow, we need to convert all the flows to the same unit. Once the flows are in the same unit, we can add them up and divide by the number of days in the period. The correct answer of 175.8 mgd is the average daily flow for the three month period.

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  • Current Version
  • Jul 04, 2024
    Quiz Edited by
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  • Jul 06, 2015
    Quiz Created by
    Zuproc66

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