Building
Plumbing
Systems
CHAPTER 1
ME432 || ENGINEERING UTILITIES 2
Topics for discussion
Core Concepts
Modern Plumbing Systems
Water: The Substance
Water Supply
Components of a Plumbing System
Plumbing Materials
Fittings and Valves
Plumbing Fixtures
CHAPTER 2
PLUMBING SYSTEM
What?
the art and technique of installing pipes, fixtures, and other
apparatuses in buildings
Why?
bringing in the supply of liquids, study of waste and hazardous
substances and/or ingredients and techniques of removing them thus,
improving health, sanitation, life and property
All pipes and fixtures after installation.
PLUMBING
BUILDING
SYSTEM
TWO PRIMARY FUNCTIONS
• WATER SUPPLY
• WATER DISPOSAL
MODERN PLUMBING SYSTEM
PLUMBING SYSTEM
• WATER SUPPLY SYSTEM
• SANITARY DRAINAGE
CHAPTER 2
• PLUMBING FIXTURES
WATER SUPPLY WATER DISPOSAL
• free from cross-piping connections with • contains many substances that can be
wastewater or unsafe sources of plumbing harmful to human beings and other living
systems creatures
• plumbing material used should meet the • needs proper management and periodic
necessary performance and quality inspection and maintenance
specification
MODERN PLUMBING SYSTEM
CHAPTER 2
WATER (𝑯𝟐 𝑶 ): THE SUBSTANCE
FUNDAMENTAL UNITS
SPECIFIC WEIGHT (γ ) SPECIFIC GRAVITY (sg)
- weight per unit volume - ratio of the specific weight of the
a. Sp. Wt = 9.81 kN/m^3 object to the specific weight of water
b. Sp. Wt = 62.4 lb/ft^3 a. sg<1.0 are less dense than water
b. sg>1.0 are denser than water
𝑊 c. sg of water is 1.0
γ= =ρg ρ𝑜𝑏𝑗𝑒𝑐𝑡 γ𝑜𝑏𝑗𝑒𝑐𝑡
𝑉 sg = =
ρ𝑤𝑎𝑡𝑒𝑟 γ𝑤𝑎𝑡𝑒𝑟
CHAPTER 2
WATER: THE SUBSTANCE
WATER (𝑯𝟐 𝑶 ): THE SUBSTANCE
FUNDAMENTAL UNITS
Volume (V) Volumetric Flow Rate (Q)
- amount of space occupied by a substance - volume of a substance that passes a
a. cubic inches (in3) or cubic feet (ft3) in English system point in a system per unit of time
b. cubic meters (m3) or liters (L) in the SI system
𝑉
Q=
𝑡
Velocity (v) Mass Flow Rate (ṁ)
- rate of linear motion of a substance in one - mass of a substance that passes a point
direction in a system per unit of time
𝑄 ṁ = ρvA
v=
𝐴𝑟𝑒𝑎
CHAPTER 2
WATER: THE SUBSTANCE
SAMPLE PROBLEM
𝑑𝑒𝑛𝑠𝑖𝑡𝑦 𝑜𝑓 𝑜𝑏𝑗𝑒𝑐𝑡
𝑠𝑔 =
𝑑𝑒𝑛𝑠𝑖𝑡𝑦 𝑜𝑓 𝑤𝑎𝑡𝑒𝑟
60.5 𝑙𝑏/𝑓𝑡3
𝑠𝑔 =
62.4 𝑙𝑏/𝑓𝑡3
𝑠𝑔 = 0.9696
CHAPTER 2
WATER: THE SUBSTANCE
SAMPLE PROBLEM
𝑉
Q=
𝑡
25𝑚3 60𝑠 1000 𝐿 1 𝑔𝑎𝑙
Q=
28𝑠 1𝑚𝑖𝑛 1𝑚3 3.785 𝐿
Q= 14153.6139 𝑔𝑝𝑚
CHAPTER 2
WATER: THE SUBSTANCE
SAMPLE PROBLEM
𝑉
𝑄=
𝑡
6 𝑔𝑎𝑙 3.785 𝐿
𝑄= ( )
8𝑠 1 𝑔𝑎𝑙
𝐿
𝑄 = 2.8388
𝑠
CHAPTER 2
WATER: THE SUBSTANCE
SAMPLE PROBLEM
4. Determine the average velocity for water flow in a pipe under the following conditions:
a. A 3⁄4 in diameter, Type L copper tube (0.875 in outside diameter and 0.785 in inside diameter) carrying water at a volumetric flow
rate of 10 gpm.
b. A 2 in diameter, Schedule 40 chlorinated polyvinyl chloride (CPVC) pipe (2.375 in outside diameter and 2.047 in inside diameter)
carrying cold water at a volumetric flow rate of 40 gpm.
Q = 𝐴𝑣
2
𝑔𝑎𝑙 1 𝑚𝑖𝑛 1 𝑓𝑡3 π 1 𝑓𝑡
10 ( )( )= 0.785 𝑖𝑛 ∗ (𝑣)
𝑚𝑖𝑛 60𝑠 7.48 𝑔𝑎𝑙 4 12𝑖𝑛
𝑣 = 6.6294 𝑓𝑡/𝑠
CHAPTER 2
WATER: THE SUBSTANCE
SAMPLE PROBLEM
4. Determine the average velocity for water flow in a pipe under the following conditions:
a. A 3⁄4 in diameter, Type L copper tube (0.875 in outside diameter and 0.785 in inside diameter) carrying water at a volumetric flow
rate of 10 gpm.
b. A 2 in diameter, Schedule 40 chlorinated polyvinyl chloride (CPVC) pipe (2.375 in outside diameter and 2.047 in inside diameter)
carrying cold water at a volumetric flow rate of 40 gpm.
Q = 𝐴𝑣
2
𝑔𝑎𝑙 1 𝑚𝑖𝑛 1 𝑓𝑡3 π 1 𝑓𝑡
40 ( )( )= 2.047 𝑖𝑛 ∗ (𝑣)
𝑚𝑖𝑛 60𝑠 7.48 𝑔𝑎𝑙 4 12𝑖𝑛
𝑣 = 3.998 𝑓𝑡/𝑠
CHAPTER 2
WATER: THE SUBSTANCE
SAMPLE PROBLEM
A liquid is moving through a tube at 18 m/s, the tube has a transverse area of 0.6 m2. The density of the liquid is ρ = 62.4 kg/m3.
What is the amount of mass flowing through the tube?
ṁ = ρvA
𝑘𝑔 𝑚
ṁ = 62.4 3 (18 ) 0.6 𝑚2
𝑚 𝑠
𝑘𝑔
ṁ = 673.92
𝑠
CHAPTER 2
WATER: THE SUBSTANCE
QUESTIONS
SEE YOU NEXT
MEETING!
