UTILITIES | Plumbing Systems | 10 Questions (MODERATE)

PLUMBING SYSTEMS

(10 Questions, Difficulty Level: MODERATE)

by Raison John J. Bassig

Q#1: Given a 3-storey building with a water closet on the top floor located at a vertical distance of 38-1/2' from the water meter at ground level. The fixture requires 8 psi to operate. If the friction head loss is estimated at 15', what pressure is required to raise the water to the top floor and put the minimum psi for the fixture to be used satisfactorily?

a. 17 psi
b. 8.4 psi
c. 45.5 psi
d. 31 psi

Recall the specific weight of water, i.e., 1 cu.ft. of water at a temperature of 40°F (where water is at its maximum density) weighs approx. 62.4 lbs. Since Pressure = Force / Area, a 1 foot deep water on an area of 1 sq.ft. (or 144 sq.in.), will have a pressure of 62.4 lbs / 144 sq.in. = [0.433 psi per foot of water]. So,

Minimum Pressure to raise water at given height = 38.5 ft. x [0.433psi/ft] = 16.67 psi
Minimum Pressure to counteract friction losses = 15.0 ft. x [0.433psi/ft] = 6.49 psi
Minimum Pressure to operate given WC fixture = 8 psi

Total Minimum Pressure Required for the system = 16.67 psi + 6.49 psi + 8 psi
Total Minimum Pressure Required for the system = 31.16 psi or Approx. 31 psi

Therefore, the correct answer is d. 31 psi.

Q#2: In the gymnasium you are designing, the client requires that all the showers be supplied with hot water lines. What should be the maximum temperature of the hot water as delivered?
a. 48.88 degrees Celsius
b. 104 degrees Fahrenheit
c. 37.77 degrees Celsius
d. 140 degrees Fahrenheit

According to the 1999 Revised National Plumbing Code of the Philippines (RNPCP),

"In the absence of local regulations, shower occupancies OTHER THAN DWELLING UNITS SERVED BY INDIVIDUAL WATER HEATERS shall be provided with individual shower control valves of the pressure balance or the thermostatic mixing valve type. Multiple or gang showers may be controlled by a master thermostatic mixing valve in lieu of individually controlled pressure balance or thermostatic mixing valves. Limit stops shall be provided on such valves and shall be ADJUSTED TO DELIVER HOT WATER with a MAXIMUM TEMPERATURE OF 48.88°C."
(Sec. 409.7 of the 1999 RNPCP)

Note that this provision applies to shower occupancies EXCEPT for dwelling units with individual water heaters. Since the question pertains to a Gymnasium (i.e., a PUBLIC building and NOT a dwelling unit), then, the above provision governs.

For historical reference, this is also based on the 1997 UPC (Uniform Plumbing Code) where the 1999 RNPCP was derived,

(Sec. 409.7 of the 1999 RNPCP and Sec. 420.0 of the 1997 UPC)

The choices I gave above are equivalent as follows:

a. 48.88° Celsius = 120° Fahrenheit
b. 104° Fahrenheit = 40° Celsius
c. 37.77° Celsius = 100° Fahrenheit
d. 140° Fahrenheit = 60° Celsius

Therefore, the correct answer is a. 48.88 degrees Celsius (or 120 degrees Fahrenheit).

Q#3: Where the local water pressure is in excess of _____, an approved-type pressure regulator preceded by an adequately-sized strainer shall be installed to reduce the pressure on the building side of the regulator to the required supply pressure.
a. 551 kPa
b. 621 kPa
c. 690 kPa
d. 827 kPa

According to the 1999 Revised National Plumbing Code of the Philippines (RNPCP),

"Where the local water pressure is in excess of 551 kPa, an an approved-type pressure regulator preceded by an adequately-sized strainer shall be installed to reduce the pressure on the building side of the regulator to the required supply pressure."
(Sec. 607.2 of the 1999 RNPCP)

Note that 1 psi = 6.895 kPa. So, 551 kPa is equivalent to 80 psi.

For historical reference, this is also based on the 1997 UPC (Uniform Plumbing Code) where the 1999 RNPCP was derived,

(Sec. 607.2 of the 1999 RNPCP and Sec. 608.2 of the 1997 UPC)

Therefore, the correct answer is a. 551 kPa (or approx. equivalent to 80psi).

Q#4: You are designing a Duplex Residential Building having 4 bedrooms in each unit and a common septic tank that would serve all units. As per the Revised National Plumbing Code of the Philippines, what would be the minimum capacity of this Septic Tank?
a. 2,100 Gallons
b. 1,800 Gallons
c. 11,355 Liters
d. 4,542 Liters

Note that my question states that the building is a DUPLEX Residence which means it is a type of building with MULTIPLE DWELLING UNITS (i.e., two dwelling units to be exact). The question also states that EACH UNIT has 4 bedrooms, which means that there is a total of 8 bedrooms for the whole building served by a single septic tank.

According to the 1999 Revised National Plumbing Code of the Philippines (RNPCP), the "2 Units" of "MULTIPLE DWELLING UNITS - ONE BEDROOM EACH" has a specified minimum septic tank capacity of "1,200 GALLONS or 4,542 LITERS". The table also states that for "EXTRA BEDROOM, 567.8 LITERS EACH."

(Table B-2 of the 1999 RNPCP)

Analyzing the question, since we already counted 1 bedroom in each unit based on Table B-2 (for a total of 2 bedrooms) requiring a capacity of 4,542 L, there are still 3 bedrooms left in each unit (for a total of 6 bedrooms) that are unaccounted for. These 6 remaining bedrooms must then be counted as extra bedrooms at 567.8 L each. So,

Minimum Capacity of Septic Tank = 4,542 L + (6 Extra Bedrooms x 567.8 L)
Minimum Capacity of Septic Tank = 4,542 L + (3,406.8 L)
Minimum Capacity of Septic Tank = 7,948.8 L

Converting this to gallons, we have,

Minimum Capacity of Septic Tank = 7,948.8 L x (1 Gal. / 3.785 L) = 2,100 Gallons.

For historical reference, this is also based on the 1997 UPC (Uniform Plumbing Code) where the 1999 RNPCP was derived,

(Table K-2 of the 1997 UPC)

Therefore, the correct answer is a. 2,100 Gallons.

Q#5: Given a 36"-diameter tank at 78" tall and filled with water at a temperature of about 4°C. Allowing 44.44 lbs. for the tank when it is empty, what is its combined weight when filled?
a. Approx. 1 ton
b. Approx. 11,516 pounds
c. Approx. 1,317 pounds
d. Approx. 2,912 pounds

First, we calculate the volume of the cylindrical tank (for ease of computing, we converted inches to feet, i.e., 36" = 3' and 78" = 6.5'),

Volume of tank = [π d² / 4] x [h]
Volume of tank = [π x (3 ft)² / 4] x [6.5 ft]
Volume of tank = 45.94579 cu.ft.

Now, recall that weight of water at 4°C (or 40°F) is approx. 62.42 lbs. per cu.ft., so,

Weight of water inside filled tank = [Volume of tank] x [Weight of water at given temp.]
Weight of water inside filled tank = [45.94579 cu.ft.] x [62.42 lbs./cu.ft.]
Weight of water inside filled tank = ~2,868 lbs.

Finally, since weight of empty tank is given at [44.44 lbs.], then,

Combined Weight of Water and Tank = [Weight of water inside filled tank] x [Weight of empty tank]
Combined Weight of Water and Tank = [2,868 lbs.] + [44.44 lbs.]
Combined Weight of Water and Tank = ~2,912.44 lbs.

Therefore, the correct answer is d. Approx. 2,912 pounds.

Q#6: Minimum number of lavatories for office or public buildings with 201-400 persons.
a. 2 Lavatories
b. 3 Lavatories
c. 4 Lavatories
d. 5 Lavatories

According to the 1999 Revised National Plumbing Code of the Philippines (RNPCP),

(Table 4-1 of the 1999 RNPCP)

For historical reference, this is also based on the 1997 UPC (Uniform Plumbing Code) where the 1999 RNPCP was derived,

(Table 4-1 of the 1997 UPC)

Therefore, the correct answer is a. 2 Lavatories.

Q#7: The result of a minus pressure in the drainage system. When a large amount of water flows rapidly through the trap, this is automatically developed and the water content of the trap is discharged. When the trap seal is lost, back flow of gases from the sewer line or septic tank will pass into the trap, finds its way to the fixture drain outlet and spread into the room.
a. Backpressure Backflow
b. Backflow
c. Siphonage
d. Back-siphonage

As per 1999 Revised National Plumbing Code of the Philippines (RNPCP), under definition of terms,

"BACKPRESSURE BACKFLOW - occurs due to an increased reverse pressure above the supply pressure. This may be due to pumps, boilers, gravity or other sources of pressure."
(Sec. 203.3 of the 1999 RNPCP)

"BACKFLOW - the flow of water or other liquids, mixtures or substances into the distributing pipes of a potable supply of water from any source other than from its intended source."
(Sec. 203.1 of the 1999 RNPCP)

"SIPHONAGE - a suction created by the flow of liquids in pipes. A pressure less than atmospheric."
(Sec. 220.14 of the 1999 RNPCP)

"BACK-SIPHONAGE - the flowing back of used, contaminated, or polluted water from a plumbing fixture or vessel into a water supply pipe due to a NEGATIVE PRESSURE in such pipe."
(Sec. 203.5 of the 1999 RNPCP)

Note that the definitions in the code above generally pertains to supply of water, but the same concepts/definitions can also be applied to drainage.

The key differences in the above 'seemingly confusing' definitions are:

Siphonage is just a "general term" describing the SUCTION of liquid.
Backflow is also just a "general term" for the REVERSAL OF FLOW of liquid.

There are 2 types of backflow: Backpressure and Back-Siphonage

Backpressure occurs due to POSITIVE PRESSURE.
Back-Siphonage occurs due to NEGATIVE PRESSURE.

To give you an idea, let's say you are drinking a certain beverage on a cup using a straw. By sipping through the straw as you drink, the act can be described as suction (i.e., SIPHONAGE), at the same time, you are then creating NEGATIVE pressure (i.e., BACK-SIPHONAGE) because the beverage is BACKFLOWING from the cup, through the straw, and into your mouth. On the other hand, if you blow through the straw while you are drinking, you are then creating POSITIVE pressure (i.e., BACKPRESSURE) as the beverage from your mouth reverses flow in the straw and goes back to the cup. Both of these aformentioned scenarios describe a REVERSAL of flow (i.e., BACKFLOW).

In my question, the situation describes the removal/siphoning (i.e., suction) of the trap seal and eventual backflow due to a minus pressure (i.e., negative pressure) in the drainage system.

Therefore, the correct answer is d. Back-Siphonage.

Q#8: A pipe connecting upward from a soil or waste stack below the floor and below the horizontal connection to an adjacent vent stack at a point above the floor and higher than the highest spill level of fixtures for preventing pressure changes in the stacks.
a. Circuit Vent
b. Relief Vent
c. Yoke Vent
d. Loop Vent

According to the 1999 Revised National Plumbing Code of the Philippines (RNPCP), under definition of terms,

"CIRCUIT VENT - a GROUP VENT pipe which STARTS IN FRONT of the EXTREME FIXTURE connection on a HORIZONTAL BRANCH and CONNECTS to the VENT STACK. See loop vent, also."
(Sec. 204.8 of the 1999 RNPCP)

"RELIEF VENT - a VERTICAL vent line, the PRIMARY function of which is to provide ADDITIONAL circulation of AIR between the drainage and vent systems or to ACT as an AUXILIARY VENT on a specially designed system such as a "yoke vent" connection BETWEEN the SOIL and VENT STACKS."
(Sec. 219.3 of the 1999 RNPCP)

"YOKE VENT - a pipe connecting UPWARD from a SOIL or WASTE STACK BELOW THE FLOOR and BELOW horizontal connection to an ADJACENT VENT STACK at a point ABOVE THE FLOOR and HIGHER THAN THE HIGHEST SPILL LEVEL of fixtures for PREVENTING PRESSURE CHANGES in the STACKS."
(Sec. 226.2 of the 1999 RNPCP)

"LOOP VENT - a VERTICAL vent connection on a HORIZONTAL SOIL or WASTE PIPE BRANCH at a point DOWNSTREAM of the LAST FIXTURE CONNECTION and TURNING to a HORIZONTAL line ABOVE the HIGHEST OVERFLOW LEVEL of the highest fixture connected thereat; the TERMINUS connected to the STACK VENT in the case of LOOP VENTING or to the VENT STACK nearby in the case of CIRCUIT VENTING."
(Sec. 213.12 of the 1999 RNPCP)

Below is my intepretation and illustration of the definitions of different vent systems mentioned above,

Therefore, the correct answer is c. Yoke Vent.

Q#9: In standard horizontal sanitary drain installations, what is the most recommended maximum slope of the pipes?
a. 0.05 degree
b. 1.15 degree
c. 1 degree
d. 2 degrees

In most, if not all, plumbing codes (such as the 1999 Revised National Plumbing Code of the Philippines), the ideal grade of horizontal excreta drainage piping is prescribed at 20mm/m (i.e., 20mm drop/rise for every 1m of run) or 2% slope.

Given a slope expressed in PERCENTAGE, say 2%, this simply means that the RATIO [rise : run] is equal to 2%, or,

Rise / Run = 0.02

Now, in my question, all of the choices are NOT expressed in ratios NOR percentages of ratios. All choices are expressed in DEGREES (which is a unit for an ANGLE). So, considering rise and run, we must use the Trigonometric Function: [Tangent(Angle) = Opposite / Adjacent], wherein,

Tan(Angle in Degrees) = [Opposite (or Drop/Rise)] / [Adjacent (or Run)]
Tan(Angle in Degrees) = [20mm] / [1000mm]
Tan(Angle in Degrees) = 0.02 (or 2%)

Now, we can compute the Angle (in DEGREES),

Angle in Degrees = ArcTan(0.02)

Which gives us,

Angle = 1.145762838 degrees, or, ~1.15 degrees.

Therefore, the correct answer is b. 1.15 degrees (or 2% slope or 20mm:1m)

Q#10: Which of the following criteria is not factored in the computation for sizing grease interceptors according to the National Plumbing Code of the Philippines?
a. Waste Flow Rate
b. Number of Vehicles per hour
c. Storage Factors
d. Aggregate Kitchen Drain Size

According to the 1999 Revised National Plumbing Code of the Philippines (RNPCP),

"Interceptor Size for Grease/Garbage in Commercial Kitchens:
= [No. of Meals per Peak hour] x [Waste flow rate] x [Retention time] x [Storage factor]

Interceptor Size for Sand-Silt/Oil in Auto Washers:
= [No. of Vehicles per hour] x [Waste flow rate] x [Retention time] x [Storage factor]

Interceptor Size for Silt-Lint/Grease in Laundries/Laundromats:
= [No. of Machines] x [2 Cycles per hour] x [Waste flow rate] x [Retention time] x [Storage factor]"

For historical reference, this is also based on the 1997 UPC (Uniform Plumbing Code) where the 1999 RNPCP was derived,

(Annex B, Sec. B.9 of the 1999 RNPCP and Appendix H, Table H-1 of the 1997 UPC)

Therefore, the correct answer is d. Aggregate Kitchen Drain Size.