Plumbing Arithmetic

PLUMBING ARITHMETIC

REVIEW OUTLINE:

1. RELATION OF UNITS 2. MESURATION

3. PHYSICS AND CHEMISTRY 4. FLOW OF FLUIDS AND PUMPS

I. RELATION OF UNITS A. LENGTH 1in = 2.45cm = 25.4mm 1cm = 10mm 1ft = 12in 1m = 3.281ft = 100cm = 1000mm 1km = 1000m = 0.6214 mile 1mile = 1.6km = 5280 ft 1yard = 3 ft 1rod = 5.5 yards = 16.5 ft 1furlong= 40 rods = 220 yards 1nautical mile = 6080ft = 1853 km 1 league = 3 1nautical miles 1mil = 0.001in B. AREA 1 FT2 = 144 ¿2 1 m2 _{= 1000} cm2 1 H _{ectare or ha = 10,000} m2 = 2.47 acres 1acre = 100 sq.m 1acre = 43560 sq.ft C. VOLUME 1 FT3 _{= 1728} ¿3

= 7.481 gallons (us) 1gal.(us) = 4 quarts = 231 ¿3 = 3.785 liters 1 m3 _{= 1000L} = 264.2 gallons (us) 1quart = 2pints 1barrel = 42gal. 1drum = 55gal. = 200L D. ANGLE 360 degrees = 2 π radians = 400 grads = 6400 mils E. FORCE AND MASS

1lb = 4.448 N 1 short Ton = 2200lb

= 70,000 grains 1 long Ton = 2400lb 1kg = 2.205 lbs

= 9.81 N = 1000 gram = 1kilo gram 1metric ton = 1000kg

1English ton or ton = 2000lb 1kip = 1kilopound = 1000 lb 1N = 1 kg . m_{s 2} F. PRESSURE 1atm. = 14.7 psi = 34 ft. of h2O = 760 mm Hq. = 1.033 kg. / cm2 = 101.325 kpq = 29.92 in. Hq

1 barometer or 1bar = 100 kpa 1 torricelli or 1torr = 1 mm Hg

G. ENERGY AND POWER 1 Btu = 778 ft. lb = 1.055 KJ = 252 Cal.

1 cal. = 4.187 J 1kcal = 4.187 KJ 1 English HP = 550 ft. lb / s Metric Hp or Ps = 33,000 ft. lb / min = 0.746 KW = 2545 Btu /hr = 1.014 Metric Hp 1metric Hp or Ps = 75 kg.m /s = 0.736 KW 1N = 1kg.m / s2 = 0.013 Metric Hp 1KW = 3413 Btu / hr. = 1.34 hp 1W = 1 J/S 1KW = 1 KJ/S = 1KN.M / S MP EXAM AUG. 28 1990 ii. FILL IN THE BLANKS

1. 1 m3 = 1000liters 2. 1mile = 5280 ft. 3. 1sq.km = 100 hectares 4. 1 ft3 = 7.481 gals 5. 1m = 39.36 in. 6. 1 m3 = 35.28 ft3 7. 1Btu = 778 ft.lb 8. 1in. = 2.54 cm 9. 1atm = 14.7 p 10.1gal = 3.785 liters NOTE:

a.) SI Prefixes c.) 1barrel = 42 gal.

1012 _{= Tera, T 1drum = 55 gal.}

109 _{= Giga, G = 200 L}

106

= Mega, M

103 _{= Kilo, K d.) 2}nd _{Law of Newton}

100

= 1

10−3 _{= Milli, M force = mass x acceleration}

10−6

10−9 _{= Nano, n 1N = 1 kg.m /} s2 10−12

= Pico, P 1N = 1 kg.m / s2

b.) 1ft = 12in e.) 1psi = 1lb / ¿2

1 ft2 = 144 ¿2 _{1psf = 1lb /} ft2

1m = 100cm 1pa = 1n/ m2

1 m2 = 10,000 cm2 1mPa = 1 n/ mm2

1 ft3 = 1728 ¿3

f.) Btu - British Thermal Unit J- Joute Cal-Calorie 1J = 1N.m 1KJ = 1KN.m 1cal = 4.187 J 1Kcal = 4.187 KJ g.) 1hp = 550 ft.lb / s = 33,000 ft.lb / min = 2545 Btu / hr. = 0.746 KW 1Metric = 0.736 KW = 75 kg.m / s 1English Hp = 1.014 Metric Hp

HP =

75 X eff .Qh _{Metric horse power}

Q = liters per second, lps (discharge) h = Total dynamic Head (TDH) in meters

Example :

Pump Efficiency = 70%

Total Dynamic Head = 45 meters Discharge , Q = 190 liters per second Required = Hp ?

Hp =

(190l s)(45 m) (75 kg .m s)(0.70)

= 162.86 hp

Use: 175 hp 2. MENSURATION A. Plane Areas

Let A – cross sectional area or Area P – Perimeter C – Circumference a.) Square a A = a2 a P = 4a b.) Rectangle a A= ab P= 2a+2b b = 2(a+b) c.) Parallelogram b a h a A= bh P= 2(a+b) b d.) Trapezoid a A= ½ (a+b) h c h d P= a+b+c+d b e.)Triangle

A= ½ bh =

√

s (s−a)(s−b)(s−c) a h c P= a+b+c b S = a+b+c2 where: A=

√

s (s−a)(s−b)(s−c) S = 2p S = a+b+c 2 f.) Circle A= πr 2 _r= d 2 r = π d 2 4 d= 2r C = 2 πr = π D d g.)Circular Sector (drawing) A= ½ r2θ l _{= r} θ l _{= arc length}

Note: θ – should be expressed in radians

If θ is outside of a trigonometric function

It should be expressed in radians

½ revolution – π radians = 180 degrees

1 revolution 2 π radians = 360 degrees

Ex: θ = 50 ˚

Req’d : Area & Arc Length A = ½ r2θ = ½ (10)2 _{(50˚ x} πrad 180 ˚¿ = 43.63 mm2 ℓ = (10)(50˚ x 180 ˚πrad¿ = 8.7266 mm

h.)Circular Segment of a Circle (drawing)

Area of a Circular Segment, As Sol. As = Asec - A ∆ sin (θ2) = a r ; a= rsin25˚ = 10sin25˚ = 4.2262 mm A ∆ =

√

s(s−r) (s−r) (s−2 a) S = 2 a+2r2 = 14.2262 mm A ∆ = 38.3023 mm2 As = 43.63-38.3023 = 5.3277 mm2 B.) ELLIPSE

b A= πab ; b= major side , a= minor side

P= 2

π

√

a2+b2 2

(drawing)

Ellipsoid – V = 4/3 πabc 1.) Parabolic Segment

(drawing) A= 2/3 base x height = 2/3 Lh

2.) SOLIDS

Let V = Volume

SA = Lateral surface area or surface area L = Slant height a.) CUBE (drawing) V= a3 SA= 6a2 b.) Rectangular Parallelopiped (drawing) V= abc

SA= 2ac + 2bc + 2ab = 2(ac+bc+ab) c.)CYLINDER (drawing) V= Bh SA = 2 π rh B= πr 2 ₌ π d 2 4 = π dh V= π d 2 4 h = π r 2_h d.) CONE (drawing) V= 1/3 Bh ; B = π d 2 4 = πr 2 = 1/3 (π d 2 4 ) h SA = ½ CL

= π d 2 12 h = ½ (2 πr¿

√

r 2_+h2 = 1/3 πr 2_{h =} πr

√

r2+h2 = πrL e.) PYRAMID (drawing) V= 1/3 Ah A= area of base f.) Frustum

(drawing) SECOND PROPOSITION OF PAPPLES

-The volume area generated by a solid of revolution equals the

Product of the generating area and the distance traveled by its Centroid. V= 3h (A1+A2+ A₁A₂ √¿ ¿ V= A x 2 πr g.) SPHERE

(drawing) V= 4/3 πr 3 _{SA= 4base}

SA= 4 πr 2 _B= πr 2

= πd 2 ₌ π d

2

4

SOLID OF REVOLUTION (PAPPUS THEOREM)

FIRST PROPOSITION OF PAPPLUS – The surface area generated by a surface of revolution equals the product of the length of the generating Arc and the distance traveled by its centroid.

AS= L x 2 πr

(Drawing) V= π h 2 3 (3r-h) C= 2 πr = πd Area of Zone , z = 2 πrh = πdh = ch

iii. D. What is the radius of a circle whose area is equal to that of a

trapezoidal cross section whose two parallel sides are 2.50 meters and 1.80 meters respectively and the distance between them is 1.50 meters?

Solution : a=1.8 m r 1.5 m b=2.5m Required : radius , r Ac = At πr 2 _{= ½ (a+b)h} r2 ₌ 0.5(1.8+2.5 )1.5 π r =

√

0.5(4.3_π )1.5 r = 1.013 meter

iv. The dimension of a rectangular tank are as follows: L=12ft. , W=6ft. , h=8ft.

Compare the maximum volume of h2O in gallons that the tank can accommodate. Solution: (drawing) V= L x w x h = 12 x 6 x 8 Volume of h2O in gallons V= 576ft3 _x 7.481 1 ft3 gal. = 4309.06 gallons

iv. The area of a circle is equal to the product of a constant π and the

square of the radius. If the diameter of a tank is 96 inches, find the area in

sq.ft (use: π =3.14) Solution: Ac = π r2 _{or Ac =} π d 2 4 d = 2r , r = d2 = π 4 (96)2 x ( 12∈¿ 1 ft ¿ ) 2 = π ( 962 )2 = 50.27 sq.ft Ac = 7’238.23 in2 _{x (} 12∈¿ 1 ft ¿ ) 2 = 50.27 sq.ft

*In a certain plumbing installation, three pipes have a diameter of 2, 2 ½ and 3 inches respectively. What is the diameter of a pipe having an area equal to the three pipes?

Given: d1 = 2in

d2 = 2 ½ in

Required: Diameter of a pipe with an area to the three, Solution:

Let d – be the diameter A = A1+A2+A3 π d2 4 = π d12 4 + π d22 4 + π d32 4 π 4 d2 = π 4 [ d12 + d22 + d32 ] d =

√

(2)2+

(

21₂

)

2 +(3 )2 d = 4.39 inches

ii. Give the formula of the ff. figures: 1. Circumference of the circle 2. Area of an ellipse

3. Lateral surface area of a sphere 4. Volume of a cylinder 5. Area of trapezoid Answer: 1. C= 2 π r , C= π d 3. 4 π r2 _{5. A= ½ (a+b)h} 2. A= π ab 4. 1/3 π r2_h TRY:

1. A rigid circular conical vessel is constructed to have a volume of 100,000 liters. Find the diameter and the depth if the depth is to be 1.25 times. Give answer in metes.

Solution: (drawing) V = 1/3 Bh = 1/3 ( π d 2 4 )h H = 1.25d = 1/3 ( π d 2 4 ) 1.25d

d3 _{= 305’577.4qL x} 1 m 3 1000 L d = 6.74 m h = 1.25 (6.47) = 8.42 m

2. Three sides of a triangle are given a= 68m, b= 52m and c= 32m. Find the area of the triangle.

Solution:

S= a+b+c2 A=

√

s (s−a)(s−b)(s−c)

= 68+52+322 =

√

76(76−68)(76−52)(76−32)

= 76m = 801.28 m2

3. A machine foundation has the shape of a frustum of a pyramid with lower base 6mx2m, upper base 5.5mx1.8m and altitude of 1.5m. Find the volume of the foundation.

Solution: (drawing) A1 = 6x2 V= h 3 [ A1+A2+

√

A1A2 ] = 12m2 ₌ 1.5 3 [ 12+9.9+

√

(12)(9.9) ] A2 = 5.5 x 1.8 = 16.40 m3 = 9.9 m2

4. A cylindrical tank with a diameter of 1meter and height of 2meters is filled with water to a depth of 0.75 meter. How much water is in the tank?

(drawing) Solution:

A = π d 2 4 = π (1)2 4 V = Ahw = π (1) 2 4 (0.75) = 0.59 m3

-PHYSICS AND

CHEMISTRY-Matter – anything that occupies space and has weight Mass, m – amount of matter in a substance

Weight, w – force exerted due to gravity Volume, v – space occupied by matter

English Unit Metric Unit SI Unit

mlb mass kg mass kg

wlb force kg force N

v ft3 _m3 _m3

Note: Have on earth at constant acceleration due to gravity. English Unit lb = lb mass = lb force ; W = m = lb Metric Unit kg = kg mass = kg force ; W = m = kg

SI Unit W is in Newton , m is in kg ; W ≠ m

Specific weight or weight density , ∝ = volumeweight =

w v

Density or Mass density , = volumemass =

m v

Specific Volume , v = volumemass ; v =

1 φ

weight

force = mass x acceleration

= kg x _secm2

= N

At constant acceleration due to gravity English Unit ∝ = ℓ ; 1 b_ft3 Metric Unit ∝ = ℓ ; kg_m3 SI Unit ∝≠ ℓ ; ∝ = _mN3 ; ℓ = kg m3

Note: for SI Unit Only ∝ = ℓg ; g = gravitational acceleration , g = 9.81

m / sec2 ∝ _{= (} kg m3¿ ( m sec2 ) ∝ ₌ N m3 ; kg . m sec2 = N -FOR WATER English Unit ∝ = ℓ = 62.4 1 b_ft3 = 8.33 1 b gal . ; 1ft3 = 7.481 gal. Metric Unit ∝ = l = 1000 kg_m3 = 1 kg L ; 1m3 = 1000L SI Unit ∝ = 9.81 KN_m3 ; ℓ = 1000 kg / m3 = 1kg/L

SPECIFIC GRAVITY OR RELATIVE DENSITY, SG

a.) For liquid and Solid

SG of any liquid and solid = Density of any liquid∨solidDensity of water

SGL = l L

l w ; SGs = l s l w

= mwmL ; = ms mw If ∝ = ℓg ;ℓ = ∝g∴ ℓL = ∝L g ; ℓw = ∝w g SGL = l L l w ; SGs = l s l w But ∝ = Vw ; ∝ L = w_L VL ; ∝ w = W_w Vw ; VL = Vw SGL = w_L Ww ; SGs = Ws Ww Summary: 1. SGL = l_L lw = ∝L ∝w = m_L mw = w_L ww 2. SGs = l_s lw = ∝_s ∝w = m_s mw = w_s ww

Density ratiosp. Wt. ratio mass ratio weight ratio

b.) For Gases

Specific Gravity for any Gas, SGg =

Molecular weight of any Gas Molecular Weight of Air

SG

q

=

MWq

MWa

where :MWq = 29 lb / lb mole

= 29 kg / kg mole

Elements Molecular Weight Unit

1. Carbon, C 12 lb mole or kg mole

3. Oxygen, O2 32 “”

4. Nitrogen, N2 _{28 “”}

5. Sulfur, S 32 “”

Example:

a.) Carbon Dioxide , CO2

MWco2 = C+O2 = 12+32 = 44 lb/lb mole or 44 kg/kg mole\ SGco2 = MW co₂ Mwa = 44 29 = 1.52 b.) Carbon Dioxide, CO

SGco = MWcoMWa =

12+16 29 = 0.97 c.)Methane, CH4 SGch4 = MW ch₄ MWa = 12+4(1) 29 = 0.55

ii. What is the weight of a solid ball 6 in diameter? Specific gravity is 8.4 Given: SGb = 8.4 , req’d = Wb d = 6 in Solution: ∝ _{b =} ∝ _wSG b = (62.4 lb/ft3_)(8.4) = 524.16 lb/ft3 Vb= 4/3 πr 3

= 4/3 π ( 6∈ ¿ 2 ¿ ) 3 = 4/3 π (3in)3 = 113.10 in3 _{x (} 12∈¿ 1 ft ¿ ) 3 = 0.0654 ft3 Wb= ∝ bVb = 524.16 lb/ft3 _{x 0.0654 ft}3 = 34.2801 lbs.

iii. Find the weight of a cast iron cone whose mass is 533.4 kg and the diameter at the base is 25mm. specific gravity of a cast iron is 7.22 Solution:

Force = W=ma

Weight of cast iron cone ,Wc = 533.4 kg x 9.81 m/s2

= 5’232.65 N Note: Force = mass x acceleration

iv. A spherical tank is 5h full of water and 2h full of compressed air. If the tank is 8ft in diameter,

a.) Compute the volume of water in cm3_.

b.) Compute the volume of air in cm3_.

c.)What is the weight of water in kg? Solution: Drawing r= d2 = 8 2 = 4ft. VST = 4 3πr 3 = 4 3π (4) 3

= 268.0826 ft3 _{x (} 12∈ ¿ 1 ft ¿ ) 3 _{x (} 1∈¿ 2.54 cm ¿ ) 3 = 7’591’253.858 cm3 a.) Vh2O = 5 7 VST = 5 7 (7’591’253.858) = 5’422’324.184 cm3 = 5.4223 m3 b.) VA = 2 7 VST = 72 (7’591’253.858) = 2’168’929.673 cm3 c.)Wh2O = ∝ h2O Vh2O = 1000kg x 5.423 m3 = 5’422.3242kg

WORK, POWER AND HEAT

A.) WORK, Wk

Wk = Force x Distance ; F= Force , S= Distance

Wk = FS

B.) POWER

Power = WorkTime ; Wk = Work , t = time

P = w_tk = Fst ; p=power , v=velocity

Velocity = distancetime ; V= s t

For rotating body, V= π DN , =2 π RN ; N = rotative speed , RPN Drawing Power , P = F (2 π RN) = 2 π FRN = π DFN Torque , T = FR Power , P = 2 π TN = Tw ; w=angular velocity = 2 π N 1 English Hp or Hp = 33,000 ft .lbmin = 0.746 kw = 550 ft .lbs = 0.746 kw 1 Metric Hp or Ps = 4500 kg . mmin = 0.736 kw = 75 kg . ms = 0.736 kw 1kw = 1 kJ / s =KN.m / s

Note: English Unit Conversion

Power = 2 π radrev x ft.lb x

rev min x 1 ps 4500kg .m mm = hp

Power = 2 π radrev x kg.m x rev min x 1 ps 4500kg .m min Power = Ps SI Unit Conversation

Power = 2 π radrev x (N.mx

k 1000 )( rev min x 1 min 60 sec ) = kN . ms ; kJ s or kw C.)HEAT Q= mc ∆ t

Where: m- mass of a substance C- specific heat of a substance

∆ _{t- temperature difference (drop or rise) on a substance}

Q= mc ∆ t = tb x lb−RBtu x ˚R = Btu Q= mc ∆ t = kg x kgkJ̊ k x ˚K = kJ Q= mc ∆ t = kg x kgkcal̊ k x ˚K = Kcal

Note:

Specific Heat

a.) For Air, Ca = 0.24 Btu / lb-˚R

= 0.24 Kcal / kg-˚K = 1.01 kJ / kg-˚K

b.) ∆ t ˚F = ∆ t ˚R

∆ t _{˚C =} ∆ t _˚K

˚F – temperature in degree Farenheit ;˚F = 59 [˚C] + 32

˚C - temperature in degree Celcius ;˚C = 59 (˚F – 32)

T˚R – Absolute temperature in degree Rankine ˚R - ˚F + 460

T˚K- Absolute temperature in degree Kelvin ˚K- ˚C + 273 Conversion of Units 1 Btu = 1.055 kJ = 778 ft.lb 1 cal = 4.186 Joules 1 kcal = 4.186 kJ c.)Thermometer Scale Drawing 9 5 = 1.8 = 1 Btu lb ˚ R x 1.055 kJ 1 Btu x 2.205 lb 1 kg x 1.8 ˚ R 1˚ K 1.8 ˚ F 1 ˚ C or 1.8 ˚ R 1˚ K = 4.187 kJ kg .˚ K

d.) Specific heat for Water, Cw = lb−˚ R1 Btu

= 4.187 kg .˚ KkJ e.) ∆ t = 5˚F ∆ T = 5 ˚ R ∆ t _{= 10˚C} ∆ T _{= 10˚K} f.) m = 50kg C = 0.25 kJ / kg. ˚K ∆ t _{= 15˚C} Q= mc ∆ t = 50kg x 0.25 kg−˚ KkJ x 15˚K note: ∆ t = 15˚C ; ∆ t _{= 15˚K} = 187.5 kJ g.) Convert 1 lb−˚ RBtu to kJ kg .˚ K = 1 lb−˚ RBtu x 1.055 kJ 1 Btu x 2.205 lbs 1 kg x 1.8 ˚ R 1˚ K 1 lb−˚ RBtu = 4.187 kJ kg .˚ K

vi. What is the horse power required to raise 40000 lbs, 200ft high in 5 minutes? Solution: P = w_tk = Fst = (40000 lbs)(200 ft ) 5 min = 1’600’000 ft .lbmin x 1 hp 33,000ft . lb min = 48.48 hp

ix. How much heat in Btu are required to raise one pound of water from 55˚F to 212˚F? How many units of work in ft.lb that this represent?

Solution: Q= mc ∆ t m= 1 lb Cw= 1Btu / lb-˚R ∆ t _{= 212˚F - 55˚F = 157˚K = 157˚R} Q= (1lb)(1 Btu/lb-˚R)(157˚R) = 157 Btu W= 157 Btu x 778 ft . lb1 Btu W= 122’146 ft.lb

--IDEAL GAS LAW

PV

T = C

At constant pressure (isobaric process), P=C

V T = C ; p_{1 V1} t₁ = p_{2 V2} t₂ V₁ T1 = V₂ T2

P₁V₁ T1 = P₂V₂ T2 ; V= V1 = V2 = C P₁ T1 = P₂ T2

At Constant Temperature (isometric Process), T=C

P₁V₁ T1 =

P₂V₂

T2 ; T1 = T2 = C

P1V1 = P2V2

For Adiabatic or isoutropic conditions (no heat exchange) P1 V1

k

= P2 V2

k

Note:

a.) for characteristic Gas equation

PV = mRT where P – absolute pressure, kPa

V – Volume, m3

M – mass, kg

R – specific gas constant, kJ / kg-˚K T – absolute temperature, T M= PVRT ˚M = K N m2x m 3 KN . m kg−˚ K x ˚ K N.m = Joules M= kg

˚M= lb ¿2( 144¿2 ft2 )(ft 3 ) ft −lb lb−˚ R x ˚ R M=lb If m= PVRT consider process 1-2 m1=m2 P₁V₁ R T1 = P₂V₂ RT2 ; P₁V₁ T1 = P₂V₂ T2 ; PV T = C

b.) For Air , Ra=0287 kg−˚ KKJ ; Mwa = 29

lb lbmole

= 53.34 lb−˚ Rft .lb = 29 kg/kg.mole

c.)Specific Gas Constant, R= _Mw´R

´R _{= universal gas constant}

= 8.3143 kg . mole−˚ KkJ

= 1545 ft.lb / lb.mole-˚R Mw = molecular weight

Example:

a.) For air, MWq = 29.00 lb . molelb

Ra = _MWa´R =

8.3143 kJ kg . mole−˚ K 29.0 kg

= 0.287 kg .−˚ KKJ Ra = 1545 ft .lb lb. mole−˚ R 2 g lb lb .mole = 53.34 lb−˚ Rft .lb b.) For CO2 MWco2 = C + O2 = 12+2(16) = 44 kg . molekg Rco2 = ´R MW co2 = 8.3143 KJ Kg. mole−˚ K 44 kg kg . mole = 0.189 kg .−˚ KKJ If MWa = 30 kg . molekg Ra = _MWa´R = 8.3143 KJ kg . mole−˚ K 30 KJ kg . mole = 0.2771 kg .−˚ KkJ

iv. If a tank of 400 gallons capacity containing air at atmosphere pressure is pumped to 45 pounds gage pressure, find air volume and volume of water in the tank?

Solution:

Since there is pumping operation there will be pressure and volume

change (P ≠ C ; V ≠ C). The process is isothermal T=C.

P₁V₁ T1 = P₂V₂ T2 ; T1=T2 P1V1 = P2V2 P1 = 14.7 Psia P2= 45 Psia V1 = 400 gallons P2= Pgage+Patm = 45+14.7=59.7 lb_¿2 V2 = P₁V₁ P2 = (14.7)(400) 59.7 V2 = 98.5 gallons of air

Final vol. of air in the tank

a.) Pabs = Pgage+Patm Patm=Pb

P atmospheric = P barometric

b.) Pabs = Patm-Pvac

Pvac- Vaccum pressure

PB = Patm = 1 std atm. = 101.325 Kpa

= 14.7 Psia

= 1.033 kg/cm2

= 760 mm. Hg = 29.92 in Hg = 34 ft. of water

c.)Gage pressure, P=Pgage = ForceArea = F A Also, P=Pg= ∝ h → 1 P= FA→ 2 1 in 2 ∝ h= FA Liquid surface, ∝→ F= ∝ hA (drawing)

iii. 3. A force of 100 lbs is exerted on a piston whose area is 20in2_{. Find the}

pressure the piston exert on the cylinder. Solution: (drawing) P= FA = 100 lb 20¿2 =5 lb ¿2 , 5 Psi or 5 Psig

4. Find the pressure at a depth of 8ft. in a swimming pool filled with water. Solution:

ii. The pressure in bonds per square inch at the bottom of any container is determined by finding the depth of water in feet and dividing it by 2.31 or multiply it by 0.433. find the pressure at the base of stand pipe full of water that has a height of 200ft.

Solution: (drawing)

P= 2.31h =

200

Or, P= 0.433h = 0.433 (200) = 86.6 Psi

Or, P= ∝ h = 62.4 (200)₁₄₄ = 86.6 Psi

ii. A boiler gage reads 150 Psi. Determine the absolute pressure in a.) Psia if barometer reads 29.6 in Hg.abs

b.) Kpa abs. Solution:

a.) Pabs = Pq + Pb = 150Psi + 29.6 (

14.7 Psi

29.92 ) = 164.54 Psi

b.) Pabs = Pq + Pb = 150 Psi (

101.325 Kpa

14.7 Psi ) + 101.325 Kpa = 1135.25 Kpaabs

iii. A viewing window 1ft in diameter is installed 10ft below the surface of an aquarium tank filled with water. Determine the force the window must stand. Solution: (drawing) F= ∝ hA = (62.4 _ftlb3 )(10ft)( π D 2 4 ) = (62.4 _ftlb3 )(10ft)( π (1' )2 4 ) = 490.09 lbs

iv. A condenser registers a vacuum of 620 mm Hq. Determine the absolute pressure in

a.) Kpa abs, referred to a 762 torr. Barometer

Solution: a.) Pabs = Pa + Pb = [(762torr)( 1 mm Hq1 torr ) – 620 mm Hq]( 101.325 Kpa 760 mm Hq ) = 18.93 Kpa abs b.) Pabs = Pb - Pxac = (1.033 _cmkq2 )-(620 mm Hq)( 1.033 kq cm2 760 mm Hq ) = 0.190 cm2|_¿| kq ¿ Note:

a.) Gage Pressure = Pq = (+) gage pressure

=Pvac = (-) gage pressure

Ex. -2 Psi = 2 Psi vacuum or 2 Psi vac

b.) Example: Pb = 15 Psi barometric

Pvac= -2 Psi

Find: Pabs

Pabs = Pb - Pvac

= 15-2

= 13 Psi abs or 13 Psia

Note: negative sign has no effect in subtracting it to the 15 Psi barometric pressure.

FLOW OF FLUIDS & PUMPS

1. Energy Head Equations ( or Bernoulli’s Equation) (EHE)

H= ∝P + v

2

2 q + z

2. Continuity Flow Equation (CHE) Q= Av

Where:

H= Total Energy hed

P

∝ = Pressure Head

P= Gage Pressure

∝ _{= specific weight of liquid}

v2

2 q = velocity head

V= velocity of liquid

q= acceleration due to gravity = 32.2 ft/s2

= 9.81 m/s2

z= elevation head

Q= discharge or volumetric flow rate A= cross-sectional area of pipe

Note: Units Application

a.) ∝P = lb/¿2 lb/ ft3 = lb ¿2x 144 ¿2 ft2 lb/ft3 = ft P ∝ = kg/cm2 kg /m3 = 100cm m ¿ 2 ¿ kg cm2x¿ ¿ = m

P ∝ = KPa kN /m3 = kN /m2 kN /m3 = m b.) V 2 2 q = ft /∝¿2 ¿ ¿ ¿ = ft ; V2 2 q = m/s¿2 ¿ ¿ ¿ = m

FOR IDEAL FLUID (FRICTIONLESS FLOW)

a.) Drawing BY EHE : H1 = H2 P₁ ∝ + V12 2 q + z1 = P₂ ∝ + V22 2 q + z2 BY CFE : Q=Q1=Q2 A1V1 = A2V2 A1 = A2 = π d2 4 b.) Drawing BY EHE : H1=H2 P₁ ∝ + V12 2 q + z1 = P₂ ∝ + V22 2 q + z2 BY CFE : Q=Q1=Q2 A1 = π d2 4 ; A2 = π d2 4

A1 ≠ A2 ; V1 ≠ V2

FOR REAL OR ACTUAL FLUID (WITH FRICTION) Drawing

BY EHE : H1=H2 + hL12

Where : hL

12 = head loss from point 1 to point 2

P₁ ∝ + V12 2 q + z1 = P₂ ∝ + V22 2 q + z2 + hL12 BY CFE : Q=Q1=Q2 A1V1 = A2V2

-PUMP PRESENT BET. SECTIONS “1” & “2” Drawing BY EHE : H1 + h = H2 + hL12 P1 ∝ + V1 2 2 q + z1 + h = P2 ∝ + V2 2 2 q + z2 + hL12 BY CFE : Q=Q1=Q2=Qp A1V1 = A2V2 Note:

a.) If liquid surface is exposed to atmosphere P=O.

b.) For a reservoir or large tank, liquid surface will drop or rise slowly,

V=O

c.)At datum line or reference point, z=O.

Drawing Where:

EPi = Electric power input

Bp = Brake power, shaft power, electric power output or pump power input Wp= Water power, pump power or pump power output

FOR ELECTRICAL & MECHANICAL DEVICES

Efficiency = Power outputPower input

a.) Electric Mot or Efficiency

Nmo = BP EPi b.) Pump Efficiency Np = ℘ BP

c.)Combined Electric Motor-pump Efficiency or overall efficiency Nc = NmoNp = ( BP EPi¿ ( ℘ BP ) = ℘ EPi

Water Power or Pump Power

WP = Q ∝ h Where: Q= discharge ∝ _{=specific weight} h= pump head RELATION OF UNITS: WP =Q ∝ h WP = ft 3 s x lb ft3 x ft = ft .lb s → hp

WP = m 3 s x kq m3 x m = kq . m s → Ps or metric hp WP = m 3 s x KN m3 x m = KN . m s ; KJ s or KW HEAD LOSS

Total head loss, hL = hf + hm

hf= major head loss or ; pipe friction head loss

= f _DL V

2

2 q ; darcy-weisbach equation

f = pipe friction factor L = pipe length

D = pipe diameter

q = 32.2 ft/s2

= 9.81 m/s2

hm= minor head loss or head due to pipe fittings

=

∑

k V 2 2 q = Ke V 2 2 q + Kq V2 2 q + ……… + etc.

Ke = coefficient due to effect of elbow

Kq = coefficient due to effect of gate value

iii.

a.) Given a capacity of 800 qpm flowing through 8in diameter pipe line. Find the velocity in fps.

Solution:

In, ft3_{/min or cfm, Q= (} 800 qal/min 7.841 gal/ft3¿ ( 1 min 60 sec¿ = 1.782 ft3/s From Q= AV V= QA = 1 ft 12∈¿ ¿2 8∈¿2¿ π 4 ¿ 1.782 ft3 /s ¿ = 5.105 ft/s or 5.105 fps ii.

a.) A pump is to handle 300 qpm of water against 50ft. of total dynamic heads, what is the water horse power?

b.) If the same pump handling brine having a specific gravity to 1.2, what is the horse power?

c.)If in problem b, the pump efficiency is 65%, what is the brake horse power? Solution: a.) WP = Q ∝ h = (300gal/min)(62.4 lb/ft3_)(50ft)( 1 ft 3 7.481 gal¿ ( 1 hp 33,000ft . lb min ¿ = 3.791 hp

b.) H pump is handling brime with SGb=1.2

WP = Q ∝ h ; ∝ b = ( ∝ w)(SGb)=(62.4 lb/ft3_{)(1.2)=74.88 lb/ft}3 = (300 mingal )(74.88 lb/ft3)(50ft) x ( 1 hp 33,000ft . lb min ¿ ( 1 ft 3 7.481 qal¿ = 4.55 hp c.)Brake Power, BP = N℘p = 4.55 0.65 = 7hp

vii. How many gallons of water can a 75hp engine raise 150ft. high in 5hrs? one gallon of water weight 8 1/3 lbs.

Solution: From, WP = Q ∝ h Q = _{∝ h}℘ = (75 hp)(33,000 ft . lb/min 1 hp ) (8.33 lbs qal)(150 ft ) = 1980.08 qpm VOLUME OF WATER, Vw = Q(time)

= (1980.08 qpm)(5x60) = 594’023.761 gallons

iv. How fast will water leak through a hole 1cm2 _{in area at the bottom of a}

tank in which water level is 3 meter high? Solution: Drawing H1=H2+hL12 ; hL12 = 0 P₁ ∝ + V1 2 2 q + z1 = P₂ ∝ + V2 2 2 q + z2 + hL12 3 = V2 2 2 q V2 =

√

3(2) (9.81) =7.7 m/s

v. What gage pressure is required in a fire hose if the stream of water is to reach a height of 60ft.?

Solution: Drawing

H1=H2 P₁ ∝ + V12 2 q + z1 = P₂ ∝ + V22 2 q + z2 P₁ ∝

= 60

P1 = ∝ (60ft) = (62.4lbs/ft3_)(60ft)( 12∈¿ 1 ft ¿ 2 = 26 psi

iii. A Horizontal pipe whose diameter changes gradually from 18in. dia at A to 36in. dia at B carries 30 cfs of water. If gage pressure at A is 10 psi and that of B is 10.9 psi det. The head loss from point A to B.

Solution: Drawing Q=QA=QB=30 ft3/s QA=AAVA VA= Q_A AA = 18 12¿ 2 π 4¿ 30 ft3/s ¿ = 16.977 ft/s QB = ABVB VB = 36 12¿ 2 π 4¿ 30 ft3_/s ¿ = 4.244 ft/s HA = HB + hLAB

hLAB = HA - HB = [ P_∝A + VA 2 2 q + zA] – [ P_B ∝ + VB2 2 q + zB] = [ 16.977¿2 ¿ ¿ (10)(144) 624 +¿ + 0]-[ (10)(144)₆₂₄ + 4.244¿2 ¿ ¿ ¿ + 0] = 2.12 ft TRY:

1. A right circular conical vessel is constructed to have a volume of 100’000 liters. Find the diameter and depth if the depth is 1.25 times the diameter. Give answer in meters.

Ans: d= 6.74 m ; h= 8.42 m

2. The three sides of a triangle are given as a=68 meters, b=52 meters and c=32 meters. Find the area of the triangle.

Ans: A= 801.28 m2

PROBLEM:

-From a reserver whose surface elevation is at 30 meters, water is pumped at an elevation of 95m. The total length of 60m diameter suction pipe is 1500 meter and that of 0.50m diameter discharge pipe is 1000m. Determine

the water power if the discharge is to be maintained at 0.48 m3_{/s. Pipe}

friction factor is 0.02 for both pipes. Minor losses is 10% of major head loss.

Required: WP Solution: EHE, Parcy-Wlashbach, CFE

Answer: 405.20 Kw Water Power, WP=Q ∝ h

PROBLEM:

-The Distance pressure gage of a pump heads 1’020.42 Kpa. The suction pressure gage is attached 1.22m below the center line of the discharge gage

and reads 50.8 mm Hqvacuum. The pump is delivering 0.0378 m3_{/s of water.}

The diameter of suction and discharge pipe are 127mm and 102mm

respectively. Determine the pump efficiency if power input tp pump is 61 kw. Required: hp-pump efficiency

Answer: hp=64.8%

PERFORMANCE OF PUMPS

Specific Speed = the speed in rpm required to produce 1qpm thru a head of 1ft.

Ns=

N √ Q

h3 /4 , rpm

Where: Ns – Specific speed, rpm

N – pumprotative speed, rpm

Q – pump volumetric capacity or discharge, qpm / suction h – pump head or total dynamic head, ft/stage

Note: for same type of pump geometrically similar pump, the specific speed are equal.

PROBLEM:

A single suction pump has a capacity of 19 liters per second with maximum developed head of 61 meters. Equivalent rotative speed is 1200 rpm. If is proposed that same type of pump is to be installed but with double – suction and to operate at a head of 30 meters and a discharge of 25 liters per

second. Determine the speed of the proposed pump at which it will operate. Solution:

For Single suction pump

Q1= 3.28¿3 ¿ 19(60)(7.48)¿ ¿ = 300.91 qpm h1 = 61(3.28) = 200.1 ft. N1 = 1200 rpm Ns1= N √ Q h3 /4 = 1200 √ 300.91 (200.1) 3/4 = 391.31 rpm

For double suction pump Q2 = 3.28¿3 ¿ 25(60)(7.48)¿ ¿ = 395.93 qpm h2 = 30(3.28) = 98.4 ft.

For same type of pump, Ns are equal

Ns1 = Ns2 Ns1 = N₂√ Q₂ h23/4 391.31 = N2√ 395.93 2 (98.4)3 /4 N2 = 868.91 rpm

qpm= volumetric cap. Of pump or amount of water the pump can deliver ft/stage=elevation or head of water from water will be delivered

Note: a.) Q1 = 19 L/S or --- qpm Q1 = 3.28 ft m ¿ 3 (7.48 qal ft3 ) ¿ 19 L S( 60 s min❑)¿ ¿ = 300.91 qpm b.) Q = 500 qpm N= 800 rpm h = 200 ft Ns = N √ Q h3 /4

Ns =

800 √ 500 (200

3 )3 /4

= _____ rpm

Affinity Law or homologous Relations of Pumps

a.) Same Pump

1.) Specific speed, Ns are equal

2.) For a constant rotative speed & varying impeller diameter

2.1 Discharge is directly proportional to the impeller diameter

Q ∝ D →QQ12 =

D₁ D2

2.2 Pump head is directly proportional to the square of impeller

diameter h ∝ D2 →h1 h2 = (D1 D2 ) 2

2.3 Brake power is directly proportional to the cube of impeller

diameter BP ∝ D3 →B P1 B P2 = (D1 D2 ) 3

3.) For a constant impeller diameter and varying rotative speed

3.1 Discharge is directly proportional to the rotative speed Q ∝ N

3.2 Pump head is directly proportional to the square of rotative speed

h ∝ N2

3.3 Brake power is directly proportional to the cube of the rotative speed

BP ∝ N3

Note: Drawing

At Constant impeller diameter with varying rpm

Initial Condition New Condition

h1 BP1 N1 N2 D1 D2  Q ∝ N →QQ12 = N₁ N2 ; Q2 = Q1 ( N₂ N1 ¿  h ∝ N2 →h1 h2 = ( N₁ N2 ) 2 _{; h} 2 = h1 ( N₂ N1 ¿ 2  BP ∝ N3 →BP1 BP2 = ( N₁ N2 ) 3 _{; BP} 2 = BP1 ( N₂ N1 ¿ 3 Q D1 _N1 h D2 _N2 BP D3 _N3 At constant “N” at constant “D” PROBLEM:

A centrifugal pump designed for 1800 rpm operation and a head of 60m has

a capacity of 11.36 m3_{/min with a power input of 132 KW}

a.) What effect will a speed reduction to 1200 rpm have in the head,

capacity and power input of pump?

b.) What will be the change in these variables if the impeller diameter

is reduced from 304 mm to 254mm while the speed is held constant at 1800 rpm?

Solution:

a.) At constant impeller diameter

b.) At constant rpm

Answer:

a.) Q2= 7.57 m3/min, h2= 26.67m , BP2= 39.11KW

b.) Q2= 9.5 m3/min, h2= 50.13m , BP2= 70KW

GEOMETRICALLY SIMILAR PUMP

a.) Ns are equal

-Discharge is directly proportional to the product of the rotative speed and the cube of the impeller diameter

-h ∝ N2_D2

-Pump head is directly proportional to the product of the square of rotative speed and to the square of impeller diameter

-BP ∝ N3_D5

-Brake power is directly proportional to the product of the cube of the rotative speed and to the fifth of impeller

Note: Q ∝ N1_D3 ←Q1 Q2 = N₁ N2 (D1 D2 ) 3 h ∝ N2_D2 ←h1 h2 = ( N₁ N2 ) 2 (D1 D2 ) 2 BP ∝ N3_D5 ←BP1 BP2 = ( N₁ N2 ) 3 (D1 D2 ) 5 PROBLEM:

A centrifugal pump discharge 25liters per second against a total head of 15meters at 1400 rpm and the diameter of the impeller is 0.45 meter. A geometrically similar pump of 0.30 meter diameter is to run at 2800 rpm. Calculate the head, discharge and brake power ratio required.

Solution: Pump I Pump II Q1 = 25 L/S Q2= ? h1 = 15m h2= ? N1 = 1400 rpm N2 = 2800 rpm D1 = 0.45m D2 = 0.30m BP1 BP2 B P₂

B P1 = Brake Power Ratio

Q = K ND3 _{; K=1} Q = ND3 Q₁ Q2 = N1D1 3 N₂D₂3 Q2 = Q1 ( N₂ N1 ¿(D2 D3 ) 3 = 2.5( 28001400 )( 0.30 0.45 )3 = 14.81 L/s h ∝ N2_D2 h = N2_D2 h₁ h2 = ( N₁ N2 ¿2(D1 D2 ) 2 h2= h1 ( N₂ N1 ¿2(D2 D1 ) 2 h2= 15 ( 2800 1400 )2( 0.30 0.45 )2 = 26.7m BP ∝ N3_D5 BP₁ BP2 = ( N₁ N2 ) 3 (D1 D2 ) 5 BP₁ BP2 = ( 2800 1400 )3( 0.30 0.45 )5 = 1.053

TRY:

1. A pressure tank on a hillside is filled by a pump located at a lower

elevation. The difference in elevation between the pump and the tank is 23ft. Assuming a pressure range of 20 to 40 psi, What pressure must be maintained at the pump?

Answer: 30 to 50 psi

2. Water enters a pump thru a 600mm diameter pipe under a pressure of 14Kpa. It leaves thru a 900mm diameter exhaust 2.5m below the

entrance pressure gage. If 500 liters of water pass the pump each second, compute the power output of the pump.

Answer: 24hp

3. A pump draws water thru a 38cm pipe from a reservoir in which the water surface is 3m lower than the pump and discharges thru a 30cm pipe. At a point in the discharge pipe 2.43m above the pump, a

pressure gage reads 3kg/m2_{. When the discharge is 0.226 m}3_{/s, head}

loss in the suction is 0.09m and head loss in the discharge is 1.05m and power input to the pump is 94kw, determine:

a.) Efficiency of the pump. Answer: 87.41%

b.) If the pump runs at 1750 rpm, What will be the new discharge,

new head and new power input if the pump speed is increased to 3500 rpm. Answer: Q2 = 0.452 m3/s h2 = 148.76m BP2 = 252 Kw Problem 1: Given: V= 100’000 liters h= 1.25 d Solution: Drawing V= 1/3 base x height

100’000 liters x ( 1m 3 1000 liters ) = 1/3 ( π d2 4 )h If h = 1.25d 100,000 1000 m3 = π 12 (d2)(1.25d) d3 ₌ (12)(100000) π(1.25)1000 d = 6.74m h = 1.25(6.74) = 8.43m Problem 2: Given: drawing

Required: Area of a triangle Solution: A =

√

s( s−a) (s−b )( s−c ) S = a+b+c2 S = 68+52+322 S = 76m A = 76−32 76(76−68)(76−52)¿ √¿ A =

√

76(8)(24) (44) A = 801.28 m2

Problem 3: Given: Drawing

Condition: minor losses is 10% of major head losses Required: Water power, WP

Solution: BY EHE Hs + h = Hd + hLsd P₂ ∝ + Vs2 2 q + zs + h = P_d ∝ + Vd2 2 q + zd + hLsd h = Zd + hL sd h = 65 + hL sd hfs = f Ls D_s Vs 2 2q = (0.02)( 1500 m0.60 m ¿ ( 1.698¿2 ¿ ¿ ¿ ) hfs = 7.35m Qs = AsVs Vs = Q_s As = 060 m¿2 ¿ π¿ ¿ 0.48m 3 s ¿ = 1.698 m/s

Qd = AdVd Vd = Qd Ad = 0.50¿2 ¿ π¿ ¿ 0.48 ¿ Vd = 2.44 m/s hfd= f Ld Dd V d2 2 q = (0.02)( 2.44¿2 (¿¿2 (9.81)) ¿ 1000 0.50 ¿ ¿ = 12.14m hL sd = [hfs + hfs] + 0.10 [ hfs + hfd] majorhL minor hL = (7.35 + 12.14) + 0.10 (7.35 + 12.14) = 21.44 m h = 65 + 21.44 = 86.44 m WP = Q ∝ h = (048 m3_{/s)(9.81)(86.44)} = 407.02 KW Problem 4: Drawing Required: Np Solution: Hs + h = Hd + hLsd

P₂ ∝ + Vs2 2 q + zs + h = P_d ∝ + Vd2 2 q + zd + hLsd Ps = -50 mm Hq ( 101.325 Kpa 760 mm Hq ¿

= -6.67 Kpa or 6.67 Kpa vacuum Qs = AsVs

V

s

=

Q_s As

=

0.127 m¿2 ¿ π¿ ¿ 0.0378m3/s ¿ Vs = 2.98 m/s

V

d

=

Q_d Ad

=

0.102¿2 ¿ π¿ ¿ 0.0378 ¿ Vd = 4.63 m/s h = [ P_∝d + Vd 2 2 q + zd ] -[ P_s ∝ + Vs 2 2 q + zs] = [ 1020.429.81

+

4.63¿2 ¿ ¿ ¿

+

1.22 ] – [ −6.67 9.81

+

2.98¿2 ¿ ¿ ¿

+

0 ] = 106.56m WP = Q ∝ h = (0.0378)(9.81)(106.56)

= 39.51 KW Np = _Pin℘

x

100% = 39.5161

x

100% = 64.8 % TRY Problem 1: Drawing H1 + h = H2 + hL12 P₁ ∝ + V1 2 2 q + z1 + h = P₂ ∝ + V2 2 2 q + z2 + hL12 P = ∝ h P1 = 10 + 20 = 30 psi = (62.4)(23)₁₄₄ P1 = 10 + 40 = 50 Psi

= 9.97 psi say 10 psi Prange = 30 to 50 Psi

Problem 2:

Required: output power of pump is Hp Drawing BY EHE H1 + h = H2 + hL12 P₁ ∝ + V12 2 q + z1 + h = P₂ ∝ + V22 2 q + z2 + hL12

P₁ ∝ + V1 2 2 q + z1 + h = P₂ ∝ + V2 2 2 q h = [ P_∝2

+

V2 2 2 q ] – [ P₁ ∝

+

V12 2 q

+

Z1] by: Q = AV = A1V1 = A2V2 V1 = Q A₁ = 0.60¿2m2 ¿ π¿ ¿ (500L s)( 1m3 1000 L) ¿ = 0.786 m/s h= [ 0.786¿2 ¿ ¿ 4 9.81+¿ ] – [ 0.786¿2 ¿ ¿ 4 9.81+¿ ] – [ 1.768¿2 ¿ ¿ 14 9.81+¿ + 2.5 ] h= -3.65m

(-) indicates a downward direction of flow of water h= |3.65 | m WP = Q ∝ h = (0.5m3_{/s)(9.81 kw/m}3_)(3.65m) = 17.9 KW.m /s = 17.9 kW.m/s ( 0.746 KW1 hp ¿ = 23.99 Hp Say, WP = 24 Hp Problem 3:

Drawing Required:

a.) Pump eff., Np

b.) Q2, h2 ∝ BP2 if Pump runs at 1750 rpm

Ns = 3500 rpm

Solution:

At point “s” to point “a” Drawing WP = Q ∝ h Np = _{94 Kw}℘ = 94 KwQ∝h BY EHE Hs + h = Hd + hLsd Ps ∝ + Vs2 2 q + zs + h = Pd ∝ + Vd2 2 q + zd + hLsd hL sd = hLs + hLd = 0.09 + 1.05 = 1.14m

PLUMBING CODE

NAMPAP - National Master Plumbers Association of the Philippines, Inc.

Ra 1378 Jan 29, 1959 RMP-20 units

Requirements:

1. Plumbing Layout Plumbing Unit:

1.1 Sanitary Line Layout 1 WC 2 FD 1 WM

1.2 Water Line Layout 1 SH 1 KS

1.3 Storm Drainage System 1 LAV 3 FAUCET

- Each fixture or group of fixtures shall be provided with gate value 2. Isometric Diagram or Plumbing Isometry

3. Legend

4. Standard Detail 4.1 Septic Tank 4.2 Catch Barin

5. General Notes & Specification 6. Bill of materials

Ventillation- is important to balance the atmospheric pressure of the plumbing system to eliminate the ff. probilities:

1. Trap seal loss

3. Deterioration of pipe quality

Concealed fixture must be provided with individual floor drain.

 Materials Used for calking / logging

1. EPOXY A & B 10 times cheaper than lead

2. Lead / Pig Lead -more durable compared to epoxy

3. Old hemp rope / oakum

Specification must govern over the layout plans / details.

15 meters radius – the distance of septic tank from deep well. High Rise Building – is 15 meters & higher

Note: CAULKING

Approximately 8lbs (3.62 kq) brown oakum is used per 100 lbs. (45.3 kq) of lead

Six (6) lbs (2.7kq) white oakum used per 100 lbs. (45.3 kq) of lead. Caulking lead in cast iron bell and spigot water mains should be 2 inches (51mm) deep.

Diameter of Difference Pipes

GIP or Copper UPUC PUC CIP

Sch. 40 mm SCH. 40 Series 1000 SV – standard 3/8” --- 10 600 XV- Extra Heavy ½” --- 13 20 ¾” --- 20 25 1” --- 25 32 1 ¼” --- 32 40 1 ½”--- 40 50 2” --- 50 65 63 50 2 ½” --- 65 75 3” --- 75 90 75 4” --- 100 110 100 6” --- 150 150 150

CONCRETE PIPE: INCHES MM 4 100 6 150 NRCP 8 200 10 250 12 300 15 375 18 450 21 525 24 600 RCP 27 675 30 750 36 900 42 1050 Classification of Sewage 1. Domestic / Sanitary 2. Industrial 3. Storm Sources of Water

1. Ground Water ex. Well, spring

2. Surface Water ex. River, lakes, stream, sea 3. Pain / atmospheric water - precipitation

Aquifer- it is the water bearing stratum of the ground Methods of Water Supply

1. Direct Method

2. Indirect Method

2.1 Overhead Feed System

Hydro pneumatic System Overhead Feed System Drawing

Design Computation: 1. Service Pipe

Total Fixture Units (F.U) 4WC x 6 = 24

4LAV x 1 = 4 4SH x 2 = 8 4KS x 2 = 8

44 F.U per floor X 6 floors

264 F.U

For intermetent flow @ 40% demand For 264 F.U x 7.5 = 1’980 gpm

X 40%

792 gpm higher compared to table C From table C water pipe size:

Q = 69 gpm D = 50 mm By interpolation: 48 216 F.U 68 gpm x-68 84 264 F.U x 85-86=17 300 F.U 85 gpm 48 84

=

x−68 17

X=

48(17)84

+ 68

= 68.77 gpm

Say, Q = 69 gpm

2. RISER Total = 264 F.U Q = 69 gpm D = 50 mm ∅ 3. Down Feed

3.1 DF-a F.U = 264 F.U 44 x 6 = 264

D = 50 mm ∅ Q = 69 gpm 3.2 DF-b F.U = 220 F.U 44 x 5 = 220 D = 50 mm ∅ Q = 69 gpm 3.3 DF-c F.U = 176 F.U 44 x 4 = 176 D = 50 mm ∅ Q = 60.33 gpm 3.4 DF-d F.U = 132 F.U 44 x 3 = 132 D = 50 mm ∅ Q = 51 gpm

3.5 DF-e F.U = 88 F.U 44 x 2 = 88

D = 50 mm ∅ Q = 40.33 gpm 3.6 DF-f F.U = 44 F.U D = 1 ¼ or 38 mm ∅ Q = 26 gpm 4. Horizontal Branch HB-g FU = 44 Q = 26 gpm D = 32 mm ∅ HB-h FU = 44 Q = 26 gpm D = 32 mm ∅

HB-I FU = 44 Q = 26 gpm D = 32 mm ∅ HB-j FU = 44 Q = 26 gpm D = 32 mm ∅ HB-k FU = 44 Q = 26 gpm D = 32 mm ∅ HB-L FU = 44 Q = 26 gpm D = 32 mm ∅ 5. CISTERN F.U = 264 Q = 69 gpm

Storage time = 4 hrs. Repair consideration for water main Vol. = Q x t = (69 mingal

)(4 hrs)(

60 mins 1 hr

)

= 16’560 gallons x

1m 3 264.2 gals.

= 62.68 m

3 If D = 1.5 m water depth A = 62.681.5

= 41.79 say 42 sq.m

Therefore use dimension: L= 7m ; W= 6 ; H= 1.5 + 0.30 air space Drawing

6. EWT

Vol. = 69gal/min x 2hrs x 60 mins1 hr

= 8’280 gallons x

1 m 3 264.2 gal

= 31.31 m

3

V =

π d 2

4

h for cylindrical tank

If diameter, D = 1.5m

h =

π d4 V2

=

1.5¿2 π¿ 4(31.34) ¿

h = 17.73 m not feasible

try a rectangular reservoir:

if water depth = 1.5m A= 31.341.5

= 20.89 sq.m say 21

Use: L= 5.25m W= 4.0 m H= 1.8m w/ 1.5 water depth 7. Pumps / Motors Used Formula: HP = 75 x eff .Qh

Where: Q= total distance, Lps liters per second h= total dynamic head, meters

Q= 69 minqal

x 3.785

liters qal

x

1 min 60 s Q= 4.35 Lps say Q= 5 Lps

h= static height + velocity head + friction losses = 21.3m + 0.29 + 3.5m = 25.09m Say h= 26m Hp= 75 x 0.705 x 26

= 2.48 hp

Use:

2units transfer pump rated at

5 Lps vs. 26m TDH with

3 Hp, 3

∅

, 240v, 60Hz TEF < motor

*SANITARY PUMPING DESIGN DRAWING

*PIPE SIZING DESIGN A. SANITARY SYSTEM

1. Fixture drain sizes

2. Horizontal branch pipe size 3. Soil / waste stack sizes 4. House drain / sewer size B. Ventilation System

1. Individual vent pipe size 2. Branch vent pipe size 3. Vent stack / main vent 4. Stack vent

1. No. gallons in 3” ∅ pipe 20 meters high

Vol. = Base x height = π d

2 4

h =

π ( 3 12) 2 4

20m x

3.28 ft 1 m

Or, Vol. = 3.22 ft

3

_x

7.481 qal

1 ft3

V = 24.09 gallons

Or, V =

π d 2 4

h =

π 4

(

3 ft 12

x

1 m 3.28 ft

)

2

x 20m x 264.2 qal/1m

3

V= 24.09 gallons

2. Temperature from ℃¿℉ Use formula : ℃ = 5/9 ( ℉ -32) ℉ _{= 9/5} ℃ _{+ 32} 3. Pressure of water:

P= ∝ h ; where: ∝ = specific weight per unit volume

h = pressure head 4. Surface Area & Volume & right circular cone

Drawing As = 2 π rL V= 1/3 base x height = 1/3 ( π d 2 4

)h

5. Annulus or washer area Drawing

A = A2 – A1

= π r22 - π r12

= π (r22 - π r12)

6.

Torus – Donut

Volume > surface area

Surface area- first theory of pappus of reduction

Volume – 2

nd

_{theory of pappus of revolution}

Drawing

7.

Continuity equation

Q= AV

Q= discharge of water

A= pipe opening area

V= unit discharge per unit time

= velocity

DESIGN ANALYSIS FOR THE FF. FIXTURE PER GIVEN BELOW:

SS = 1

WC = 5

LAV = 5 PER FLOOR

SH = 5

KS = 5

UR = 3

1. SERVICE PIPE SIZING COMPUTATIONS:

5WC x 6 = 30 F.U 5LAV X 1 = 5 F.U 5SH X 2 = 10 F.U 5KS X 2 = 10 F.U 3UR X 5 = 15 F.U 1SS X 3 = 3 F.U

73 F.U PER FUR X 6 FLRS = 438 F.U

RISER DIAGRAM Drawing

For intermetent flow 40% demand For 438 F.U x 7.5 = 3285 gallons X 0.40

1,314 gallons higher compared to table C

At NPC table C 300 F.U 85 gpm 438 F.U x 600 F.U 144 gpm 600−300 438−300

=

144−85 x−85

300(x-85) = (59)(138)

X =

(59)(138)300

+ 85

= 112.14 gpm