Charge Discharge System

We have neither received nor given any help while we prepare the project

except team members.

Betül BULUT Serhenk ÇELİK

A POOL DISCHARGE/CHARGE SYSTEM

FOR A SUMMER RESORT

Course Name: KMU 224-21 FLUID MECHANICS

Group Members;

Member 1:

Name: Serhenk

Family Name: ÇELK

ID Number: 20519613

Member 2:

Name: Betül

Family Name: BULUT

ID Number: 20519593

Dear Prof. Dr. Abdurrahman TANYOLAÇ

A project involving the design of a pool charge-discharge piping and pump system for

a summer resort having 200 bungalows was made before by a contractor. This system was not

builded proper enough to service the villagers. The pool was being filled by the system in 28

hours and evacuated in 32 hours. This means that villagers will not be able to swim during

approximately three days. So the problem is filling and evacuating the pool as soon as

possible with a cheap design.

Now it must be considered that the system present should be completely removed or

some parts may be used.

The object of the new project is filling and evacuating the pool in 8 hours (4 hours for

each). The existing system’s pipes have very small diameters so this makes difficult to fill and

evacuate the pool in 8 hours. The existing pipes’ diameters shoul be broadened.

There is one thing that slowers the system. The collection well.

At the beginning, evacuation of the pool to the collection well is faster than the

evacuation of the collection well to the sea. This causes overflowing. After a while evacuation

of the pool getting slower than the evacuation of the collection well. And this causes the pump

to work without water in it, so it breaks down. This shows that the collection well must be

removed. Instead it a filter should be used.

Finally the material of the pipes must be considered. More enduring and expensive

polyethylen pipe or cheap but having low resistance of outside strikings, PVC pipe. PE pipe

will be more efficient because of endurance.

INTRODUCTION OF THE DESIGN:

Design and the renovation of a pool charge-discharge piping and pump system for a

summer resort place of 200 bungalows is done in this Project. A new configuration of pipeline

and pumps speed up the existing system. Different system for filling-discharge is used to

think about the minimum cost of the system and reasonable prices are chosen for the new

system. The pool is discharged an refilled in 4 hours. That is the minimum cost of the whole

system or construct new pool system.

The old evacuating system’s pipes to the collection well was under the pool, so it

would be difficult to broaden the pipes path under the pool, so the pipes are placed near the

wall of the pool.

The evacuating pipe double is 11 + 2 x 0.2 = 11.4 m to the pump. It is 2.05 m under

the surface of the sea water level. Then it goes 320 m to the sea for evacuating. So the total

length is 331.4 m.

The filling pipe is 8 m from the sea to the pump. After that there is 332 m to go to the

pool. Total path is 340 m

There are six gate valves, for discharging system, there are two of them for pool exit

pipe system, one for discharging pipe system after the discharging pump and one for just

before the sea exit. For filling system, there is one of them after the filling system’s pump,

and one before the pool entrance.

At the beginning of the filling pipe there is a filter. The friction loss in filter is

neglected. Because of the higher values of friction losses.

ASSUMPTIONS :

1-) The density and viscosity of seawater (Aegean Sea) at 20 0C are used in all calculations.

2-) The same canal will be used for each piping design.

3-) In friction loss calculations the ratio of A1/A2 is neglected because the area of the pool is

too big compared to pipe area at the pool exit.

4-) In the filter the friction loss is neglected, because of choosing higher motor pump from the

pump list.

5-) The distance between pool and pump is taken as 11 m.

6-) The distances between at both pump exits, the gate valves and the tees is taken to be very

short and neglected.

7-) It was assumed that there was no elbows on the piping at the hillside.

8-) Friction losses for both filters and pumps are neglected.

9-) At discharge system, the distance between tee at the end of the hillside and sea is

neglected.

10-) It is assumed that the pool is open for 4 months in a year.

CALCULATIONS :

Seawater at 20°C;

ρ = 1038 kg/m³ µ = 1.005 x 10ˉ ³ Pa.s

volumetric flowrate = Q = 1000 m³/ 4 h = 250 m³/h x 1h/3600s = 0.0694 m³/s

mass flowrate = m = 0.0694 m³/s x 1038 kg/m³ = 72.04 kg/s

DISCHARGE SYSTEM 1 :

Pool exit :

ID1 = 100 mm

A1 = 0.00785 m2

V1 = Q / A1 = 0.0347 m³/s / 0.00785 m2

= 4.42 m/s

( volumetric flow rate is 0.0347 m³/s, because there are two exits at the pool exit, so

volumetric flow rate should be divided two equal values)

Nre1 = D1V1ρ / µ = 456513 ( Turbulent flow)

Pump1 entrance :

D2 = 160 mm

ID2 = 160 – (2 x 9.50) = 0.14 m

A2 = 0.015 m2

V2 = Q / A2 = 0.0694 m³/s / 0.015 m2

= 4.63 m/s


Pump1 exit :

D3 = 125 mm

ID3 = 125 – (2 x 7.40) = 0.11 m

A3 = 0.0095 m2

V3 = Q / A3 = 0.0694 m³/s / 0.0095 m2 = 7.30 m/s


320 m long pipe :

D4 = 200 mm

ID4 = 200 – (2 x 11.90) = 0.176 m

A4 = 0.024 m2

V4 = Q / A4 = 0.0694 m³/s / 0.024 m2 = 2.89 m/s


FRICTION LOSSES :

• Friction loss from the pool exit to pump1 exit :

Friction loss in 90° elbow

h f = 4 x K f V2

/ 2 = 4 x (0.75) x (4.42) 2

/ 2 = 29.3 J/kg

Friction loss in tee

h f = K f V2

/ 2 = 1 x (4.42) 2

/ 2 = 9.7 J/kg

Friction loss in gate valve

h f = K f V2

/ 2 = 2 x (0.17) x (4.42) 2

/ 2 = 3.32 J/kg

Friction loss in couplings at pump

Coupling at pump entrance h f = K f V2

/ 2 = (0.04) x (4.63) 2

/ 2 = 0.43 J/kg

Coupling at pump exit h f = K f V2

/ 2 = (0.04) x (7.3) 2

/ 2 = 1.06 J/kg

Coupling at pool exit h f = K f V2

/ 2 = (0.04) x (4.42) 2

/ 2 = 0.78 J/kg

Friction loss in 11.4 m polyethylene pipe

ε = 4.6 x 10-5

m ε / D = 3.3 x 10-4

Nre = 669484 f = 0.0030

Ff = 4f ( ∆L / D ) ( v22

/ 2 ) = 4 x 0.0030 x (11.4 / 0.14) ( 4.632

/ 2) = 10.5 J/kg

Σ F = 55.1 J/kg

• Friction loss from pump1 exit to sea :

Friction losses in expansion

h ex = (1 – A3/A4)

2 V3

2/2 = (1 – 0.0095/0.024)

2 (7.3)

2 / 2 = 9.73 J/kg

Friction losses in 90° elbow

h f = 2 x K f V

2 / 2 = 2 x (0.75) x (2.89)

2 / 2 = 6.3 J/kg


h f = 2 x K f V2

/ 2 = 2 x (0.35) x (2.89) 2

/ 2 = 2.9 J/kg


h f = 2 x K f V2

/ 2 = 2 x 1 x (2.89) 2

/ 2 = 8.35 J/kg

Friction loss in couplings

h f = 2 x K f V2

/ 2 = 2 x (0.04) x (2.89) 2

/ 2 = 0.33 J/kg

Friction loss in 320 m polyethylene pipe

ε = 4.6 x 10-5

m ε / D = 2.6 x 10-4

Nre = 525342 f = 0.0029

Ff = 4f ( ∆L / D ) ( v22

/ 2 ) = 4 x 0.0029 x ( 320 / 0.176 ) x (2.892 / 2) = 88.1 J/kg

Σ F = 115.7 J/kg

TOTAL FRICTION LOSS IN DISCHARGE SYSTEM :

Σ F = 55.1 + 115.7 = 170.8 J/kg

MECHANICAL ENERGY BALANCE FOR DISCHARGE SYSTEM :

1 / 2 ( v42 – v1

2 ) + g ( z4 – z1 ) + ( P4 – P1) / ρ + Σ F + Ws = 0

1 / 2 ( 2.892

– 4.422

) + 9.81(1.85) + (– 18838) / 1038 + 170.8 + Ws = 0

Ws = - 166 J/kg x 72.04 kg/s = - 11958 J/s = -11.9 kW

Wp = - Ws / η = - (-11.58 kW / 0.65)

Wp = 18.3 kW

PRESSURE DROP CALCULATION FROM TOP OF THE POOL TO PUMP

EXIT :

The maximum pressure in the system is on the pump exit, so we can use mechanical

balance from pool exit to pump exit.

V1 = 4.42 m/s

V3 = 7.3 m/s

z3 – z1 = - 0.2 m

P1 = hρg + Patm = 120163 Pa

1 / 2 ( v32 – v1

2 ) + g ( z3 – z1 ) + ( P3 – P1) / ρ + Σ F + Ws = 0

1 / 2 ( 7.32

– 4.422

) + 9.81 ( -0.2 ) + (P3 – 120163) / 1038 + 55.1 – 217.6 = 0

P 3 = 273332 Pa = 2.73 bar

We choose the pipe of maximum pressure endurance 10 bar. 40% more of calculation

value of pressure must be less than 10 bar.

2.73 x 1.40 = 3.8 bar


Pool exit :

ID1 = 100 mm

A1 = 0.00785 m2

V1 = Q / A1 = 0.0347 m³/s / 0.00785 m2

= 4.42 m/s




Pump1 entrance :

D2 = 200 mm

ID2 = 200 – (2 x 11.90) = 0.176 m

A2 = 0.024 m2

V2 = Q / A2 = 0.0694 m³/s / 0.024 m2

= 2.89 m/s


Pump1 exit :

D3 = 160 mm

ID3 = 125 – (2 x 9.50) = 0.14 m

A3 = 0.0015 m2

V3 = Q / A3 = 0.0694 m³/s / 0.0015 m2 = 4.63 m/s


320 m long pipe :

D4 = 225 mm

ID4 = 225 – (2 x 13.40) = 0.198 m

A4 = 0.0308 m2

V4 = Q / A4 = 0.0694 m³/s / 0.0308 m2 = 2.25 m/s


FRICTION LOSSES :



h f = 4 x K f V2

/ 2 = 4 x (0.75) x (4.42) 2

/ 2 = 29.3 J/kg


h f = K f V2

/ 2 = 1 x (4.42) 2

/ 2 = 9.7 J/kg


h f = K f V2

/ 2 = 2 x (0.17) x (4.42) 2

/ 2 = 3.32 J/kg



/ 2 = (0.04) x (2.89) 2

/ 2 = 0.17 J/kg


/ 2 = (0.04) x (4.63) 2

/ 2 = 0.43 J/kg


/ 2 = (0.04) x (4.42) 2

/ 2 = 0.78 J/kg


ε = 4.6 x 10-5

m ε / D = 2.6 x 10-4

Nre = 525342 f = 0.0029

Ff = 4f ( ∆L / D ) ( v22

/ 2 ) = 4 x 0.0029 x (11.4 / 0.176) ( 2.892

/ 2) = 3.14 J/kg

Σ F = 46.8 J/kg



h ex = (1 – A3/A4)

2 V3

2/2 = (1 – 0.0015/0.0308)

2 (4.63)

2 / 2 = 2.82 J/kg


h f = 2 x K f V

2 / 2 = 2 x (0.75) x (2.25)

2 / 2 = 3.8 J/kg


h f = 2 x K f V2

/ 2 = 2 x (0.35) x (2.25) 2

/ 2 = 1.8 J/kg


h f = 2 x K f V2

/ 2 = 2 x 1 x (2.25) 2

/ 2 = 5.1 J/kg


h f = 2 x K f V2

/ 2 = 2 x (0.04) x (2.25) 2

/ 2 = 0.2 J/kg


ε = 4.6 x 10-5

m ε / D = 2.3 x 10-4

Nre = 418298 f = 0.0027

Ff = 4f ( ∆L / D ) ( v22

/ 2 ) = 4 x 0.0027 x ( 320 / 0.198 ) x (2.252 / 2) = 44.2 J/kg

Σ F = 58.0 J/kg


Σ F = 46.8 + 58.0 = 104.8 J/kg


1 / 2 ( v42 – v1

2 ) + g ( z4 – z1 ) + ( P4 – P1) / ρ + Σ F + Ws = 0

1 / 2 ( 2.252

– 4.422

) + 9.81(1.85) + (– 18838) / 1038 + 104.8 + Ws = 0

Ws = - 98.5 J/kg x 72.04 kg/s = - 7095 J/s = -7.1 kW

Wp = - Ws / η = - (-7.1 kW / 0.65)

Wp = 10.9 kW


EXIT :



V1 = 4.42 m/s

V3 = 4.63 m/s

z3 – z1 = - 0.2 m

P1 = hρg + Patm = 120163 Pa

1 / 2 ( v32 – v1

2 ) + g ( z3 – z1 ) + ( P3 – P1) / ρ + Σ F + Ws = 0

1 / 2 ( 4.632

– 4.422

) + 9.81 ( -0.2 ) + (P3 – 120163) / 1038 + 46.8 – 98.5 = 0

P 3 = 174878 Pa = 1.75 bar



1.75 x 1.40 = 2.45 bar


Pool exit :

ID1 = 100 mm

A1 = 0.00785 m2

V1 = Q / A1 = 0.0347 m³/s / 0.00785 m2

= 4.42 m/s




Pump1 entrance :

D2 = 250 mm

ID2 = 250 – (2 x 14.80) = 0.22 m

A2 = 0.038 m2

V2 = Q / A2 = 0.0694 m³/s / 0.038 m2

= 1.83 m/s


Pump1 exit :

D3 = 180 mm

ID3 = 180 – (2 x 10.7) = 0.158 m

A3 = 0.00196 m2

V3 = Q / A3 = 0.0694 m³/s / 0.00196 m2 = 3.54 m/s


320 m long pipe :

D4 = 225 mm

ID4 = 225 – (2 x 13.40) = 0.198 m

A4 = 0.0308 m2

V4 = Q / A4 = 0.0694 m³/s / 0.0308 m2 = 2.25 m/s


FRICTION LOSSES :



h f = 4 x K f V2

/ 2 = 4 x (0.75) x (4.42) 2

/ 2 = 29.3 J/kg


h f = K f V2

/ 2 = 1 x (4.42) 2

/ 2 = 9.7 J/kg


h f = K f V2

/ 2 = 2 x (0.17) x (4.42) 2

/ 2 = 3.32 J/kg



/ 2 = (0.04) x (1.83) 2

/ 2 = 0.07 J/kg


/ 2 = (0.04) x (3.54) 2

/ 2 = 0.25 J/kg


/ 2 = (0.04) x (4.42) 2

/ 2 = 0.78 J/kg


ε = 4.6 x 10-5

m ε / D = 2.1 x 10-4

Nre = 578057 f = 0.0026

Ff = 4f ( ∆L / D ) ( v22

/ 2 ) = 4 x 0.0026 x (11.4 / 0.22) ( 1.832

/ 2) = 0.9 J/kg

Σ F = 45.1 J/kg



h ex = (1 – A3/A4)

2 V3

2/2 = (1 – 0.00196/0.0308)

2 (3.54)

2 / 2 = 0.83 J/kg


h f = 2 x K f V

2 / 2 = 2 x (0.75) x (2.25)

2 / 2 = 3.8 J/kg


h f = 2 x K f V2

/ 2 = 2 x (0.35) x (2.25) 2

/ 2 = 1.8 J/kg


h f = 2 x K f V2

/ 2 = 2 x 1 x (2.25) 2

/ 2 = 5.1 J/kg


h f = 2 x K f V2

/ 2 = 2 x (0.04) x (2.25) 2

/ 2 = 0.2 J/kg


ε = 4.6 x 10-5

m ε / D = 2.3 x 10-4

Nre = 415820 f = 0.0027

Ff = 4f ( ∆L / D ) ( v22

/ 2 ) = 4 x 0.0027 x ( 320 / 0.22 ) x (2.252 / 2) = 44.2 J/kg

Σ F = 55.9 J/kg


Σ F = 45.1 + 55.9 = 101 J/kg


1 / 2 ( v42 – v1

2 ) + g ( z4 – z1 ) + ( P4 – P1) / ρ + Σ F + Ws = 0

1 / 2 ( 2.252

– 4.422

) + 9.81(1.85) + (– 18838) / 1038 + 101 + Ws = 0

Ws = - 93.9 J/kg x 72.04 kg/s = - 6764 J/s = -6.76 kW

Wp = - Ws / η = - (-6.76 kW / 0.65)

Wp = 10.4 kW So the pump type is SNM 150/200


EXIT :



V1 = 4.42 m/s

V3 = 3.54 m/s

z3 – z1 = - 0.2 m

P1 = hρg + Patm = 120163 Pa

1 / 2 ( v32 – v1

2 ) + g ( z3 – z1 ) + ( P3 – P1) / ρ + Σ F + Ws = 0

1 / 2 ( 3.542

– 4.422

) + 9.81 ( -0.2 ) + (P3 – 120163) / 1038 + 44.3 – 72.05 = 0

P 3 = 154635 Pa = 1.5 bar



1.5 x 1.40 = 2.1 bar < 10 bar

FILLING SYSTEM 1 :

Sea exit :

ID5 = 15 cm

A5 = 0.0177 m2

V5 = Q / A5 = 0.0694 m³/s / 0.0177 m2

= 3.92 m/s


Pump2 entrance :

D6 = 225 mm

ID6 = 225 - ( 2 x 13.40) = 0.198 m

A6 = 0.0308 m2

V6 = Q / A6 = 0.0694 m³/s / 0.0308 m2 = 2.25 m/s


Pump2 exit :

D7 = 180 mm

ID7 = 180 – (2 x 10.7) = 0.158 m

A7 = 0.0196 m2

V7 = Q / A7 = 0.0694 m³/s / 0.0196 m2 = 3.54 m/s


320 m long pipe :

D4 = 225 mm

ID4 = 225 – (2 x 13.40) = 0.198 m

A4 = 0.0308 m2

V4 = Q / A4 = 0.0694 m³/s / 0.0308 m2 = 2.25 m/s


Pool entrance :

D4 = 225 mm

ID4 = 225 – (2 x 13.40) = 0.198 m

A4 = 0.0308 m2

V4 = Q / A4 = 0.0694 m³/s / 0.0308 m2 = 2.25 m/s


FRICTION LOSSES :

• Friction loss from sea to pool entrance :

Friction loss in 8 m galvanized pipe

ε = 4.6 x 10-5

m ε / D = 1.0 X 10-3

Nre5 = 607307 f = 0.0036

Ff = 4f ( ∆L / D ) ( v22

/ 2 ) = 4 x 0.0036 ( 8 / 0.15 ) x (3.92 / 2) = 5.9 J/kg

Friction loss in coupling

Pump entrance h f = K f V2

/ 2 = (0.04) x (2.25) 2 / 2 = 0.1 J/kg

Pump exit h f = K f V2 / 2 = (0.04) x (3.54)

2 / 2 = 0.25 J/kg

320 m pipe h f = 2 x K f V2

/ 2 = 2 x (0.04) x (2.25)2 / 2 = 0.2 J/kg

Friction loss in expansion

h ex = (1 – A7/A4) 2

V72/2 = (1 – 0.0196/0.0308)

2 (3.54)

2 / 2 = 1.47 J/kg


h f = 2 x K f V2

/ 2 = 2 x (0.75) x (2.25) 2

/ 2 = 3.8 J/kg


h f = 2 x K f V2

/ 2 = 2 x (0.35) x (2.25) 2

/ 2 = 1.8 J/kg

Friction loss in 45° elbow (galvanized pipe)

h f = K f V2

/ 2 = (0.35) x (3.92) 2

/ 2 = 2.7 J/kg


h f = 2 x K f V2

/ 2 = 2 x 1 x (2.25) 2

/ 2 = 5.06 J/kg


h f = 2 x K f V2

/ 2 = 2 x (0.17) x (2.25)2 / 2 = 0.86 J/kg


ε = 4.6 x 10-5

m ε / D = 2.3 x 10-4

Nre = 415820 f = 0.0027

Ff = 4f ( ∆L / D ) ( v22

/ 2 ) = 4 x 0.0027 x ( 320 / 0.198 ) x (2.252 / 2) = 44.2 J/kg


ε = 4.6 x 10-5

m ε / D = 2.3 x 10-4

Nre = 415820 f = 0.0027

Ff = 4f ( ∆L / D ) ( v22

/ 2 ) = 4 x 0.0027 x ( 12.85 / 0.198 ) x (1.832 / 2) = 1.8 J/kg


h ex = (1 – A5/A6) 2

V52/2 = (1 – 0.0177/0.0308)

2 (3.92)

2 / 2 = 1.4 J/kg

TOTAL FRICTION LOSS IN CHARGE SYSTEM :

Σ F = 68.8 J/kg

MECHANICAL ENERGY BALANCE FOR CHARGE SYSTEM :

1 / 2 ( v42 – v5

2 ) + g ( z4 – z5 ) + ( P4 – P5 ) / ρ + Σ F + Ws = 0

1 / 2 ( 2.252

– 3.922

) + 9.81 (1.5) + (101325-116599) / 1038 + 68.8 + Ws = 0

Ws = -63.7 J/kg x 72.04 kg/s = -4588 J/s = -4.6 kW

Wp = - Ws / η = - ( -4.6 kW) / 0.65

Wp = 7.1 kW

PRESSURE DROP CALCULATION FROM BOTTOM OF THE SEA TO

PUMP EXIT :



1 / 2 ( v72 – v5

2 ) + g ( z7 – z5 ) + ( P7 – P5 ) / ρ + Σ F + Ws = 0

1 / 2 ( 3.542

– 3.922

) + 9.81( 2.5) + ( P7 - 116599 ) / 1038 + 6.38 – 63.7 = 0

P7 = 151834 Pa = 1.5 bar



1.5 x 1.40 = 2.1 bar

FILLING SYSTEM 2 :

Sea exit :

ID5 = 15 cm

A5 = 0.0177 m2

V5 = Q / A5 = 0.0694 m³/s / 0.0177 m2

= 3.92 m/s


Pump2 entrance :

D6 = 200 mm

ID6 = 200 - ( 2 x 11.90) = 0.176 m

A6 = 0.024 m2

V6 = Q / A6 = 0.0694 m³/s / 0.024 m2 = 2.89 m/s


Pump2 exit :

D7 = 160 mm

ID7 = 160 – (2 x 9.5) = 0.14 m

A7 = 0.015 m2

V7 = Q / A7 = 0.0694 m³/s / 0.015 m2 = 4.63 m/s


320 m long pipe :

D4 = 225 mm

ID4 = 225 – (2 x 13.40) = 0.198 m

A4 = 0.0308 m2

V4 = Q / A4 = 0.0694 m³/s / 0.0308 m2 = 2.25 m/s


Pool entrance :

D4 = 225 mm

ID4 = 225 – (2 x 13.40) = 0.198 m

A4 = 0.0308 m2

V4 = Q / A4 = 0.0694 m³/s / 0.0308 m2 = 2.25 m/s


FRICTION LOSSES :



ε = 4.6 x 10-5

m ε / D = 1.0 X 10-3

Nre5 = 607307 f = 0.0036

Ff = 4f ( ∆L / D ) ( v22

/ 2 ) = 4 x 0.0036 ( 8 / 0.15 ) x (3.92 / 2) = 5.9 J/kg



/ 2 = (0.04) x (2.89) 2 / 2 = 0.17 J/kg

Pump exit h f = K f V2 / 2 = (0.04) x (4.63)

2 / 2 = 0.43 J/kg


/ 2 = 2 x (0.04) x (2.25)2 / 2 = 0.2 J/kg


h ex = (1 – A7/A4) 2

V72/2 = (1 – 0.015/0.0308)

2 (4.63)

2 / 2 = 3.93 J/kg


h f = 2 x K f V2

/ 2 = 2 x (0.75) x (2.25) 2

/ 2 = 3.8 J/kg


h f = 2 x K f V2

/ 2 = 2 x (0.35) x (2.25) 2

/ 2 = 1.8 J/kg


h f = K f V2

/ 2 = (0.35) x (3.92) 2

/ 2 = 2.7 J/kg


h f = 2 x K f V2

/ 2 = 2 x 1 x (2.25) 2

/ 2 = 5.06 J/kg


h f = 2 x K f V2

/ 2 = 2 x (0.17) x (2.25)2 / 2 = 0.86 J/kg


ε = 4.6 x 10-5

m ε / D = 2.3 x 10-4

Nre = 415820 f = 0.0027

Ff = 4f ( ∆L / D ) ( v22

/ 2 ) = 4 x 0.0027 x ( 320 / 0.198 ) x (2.252 / 2) = 44.2 J/kg


ε = 4.6 x 10-5

m ε / D = 2.3 x 10-4

Nre = 415820 f = 0.0027

Ff = 4f ( ∆L / D ) ( v22

/ 2 ) = 4 x 0.0027 x ( 12.85 / 0.198 ) x (1.832 / 2) = 1.8 J/kg


h ex = (1 – A5/A6) 2

V52/2 = (1 – 0.0177/0.024)

2 (3.92)

2 / 2 = 0.53 J/kg


Σ F = 71.18 J/kg


1 / 2 ( v42 – v5

2 ) + g ( z4 – z5 ) + ( P4 – P5 ) / ρ + Σ F + Ws = 0

1 / 2 ( 2.252

– 3.922

) + 9.81 (1.5) + (101325-116599) / 1038 + 71.18 + Ws = 0

Ws = -66.1 J/kg x 72.04 kg/s = -4761 J/s = -4.8 kW

Wp = - Ws / η = - ( -4.8 kW) / 0.65

Wp = 7.3 kW


PUMP EXIT :



1 / 2 ( v72 – v5

2 ) + g ( z7 – z5 ) + ( P7 – P5 ) / ρ + Σ F + Ws = 0

1 / 2 ( 4.632

– 3.922

) + 9.81( 2.5) + ( P7 - 116599 ) / 1038 + 6.82 – 66.1 = 0

P7 = 149727 Pa = 1.5 bar



1.5 x 1.40 = 2.1 bar

FILLING SYSTEM 3 :

Sea exit :

ID5 = 15 cm

A5 = 0.0177 m2

V5 = Q / A5 = 0.0694 m³/s / 0.0177 m2

= 3.92 m/s


Pump2 entrance :

D6 = 180 mm

ID6 = 180 - ( 2 x 10.7) = 0.158 m

A6 = 0.0196 m2

V6 = Q / A6 = 0.0694 m³/s / 0.0196 m2 = 3.54 m/s


Pump2 exit :

D7 = 140 mm

ID7 = 140 – (2 x 8.3) = 0.123 m

A7 = 0.012 m2

V7 = Q / A7 = 0.0694 m³/s / 0.012 m2 = 5.8 m/s


320 m long pipe :

D4 = 225 mm

ID4 = 225 – (2 x 13.40) = 0.198 m

A4 = 0.0308 m2

V4 = Q / A4 = 0.0694 m³/s / 0.0308 m2 = 2.25 m/s


Pool entrance :

D4 = 225 mm

ID4 = 225 – (2 x 13.40) = 0.198 m

A4 = 0.0308 m2

V4 = Q / A4 = 0.0694 m³/s / 0.0308 m2 = 2.25 m/s


FRICTION LOSSES :



ε = 4.6 x 10-5

m ε / D = 1.0 X 10-3

Nre5 = 607307 f = 0.0036

Ff = 4f ( ∆L / D ) ( v22

/ 2 ) = 4 x 0.0036 ( 8 / 0.15 ) x (3.92 / 2) = 5.9 J/kg



/ 2 = (0.04) x (3.54) 2 / 2 = 0.25 J/kg

Pump exit h f = K f V2 / 2 = (0.04) x (5.8)

2 / 2 = 0.67 J/kg


/ 2 = 2 x (0.04) x (2.25)2 / 2 = 0.2 J/kg


h ex = (1 – A7/A4) 2

V72/2 = (1 – 0.012/0.0308)

2 (5.8)

2 / 2 = 7.9 J/kg


h f = 2 x K f V2

/ 2 = 2 x (0.75) x (2.25) 2

/ 2 = 3.8 J/kg


h f = 2 x K f V2

/ 2 = 2 x (0.35) x (2.25) 2

/ 2 = 1.8 J/kg


h f = K f V2

/ 2 = (0.35) x (3.92) 2

/ 2 = 2.7 J/kg


h f = 2 x K f V2

/ 2 = 2 x 1 x (2.25) 2

/ 2 = 5.06 J/kg


h f = 2 x K f V2

/ 2 = 2 x (0.17) x (2.25)2 / 2 = 0.86 J/kg


ε = 4.6 x 10-5

m ε / D = 2.3 x 10-4

Nre = 415820 f = 0.0027

Ff = 4f ( ∆L / D ) ( v22

/ 2 ) = 4 x 0.0027 x ( 320 / 0.198 ) x (2.252 / 2) = 44.2 J/kg


ε = 4.6 x 10-5

m ε / D = 2.3 x 10-4

Nre = 415820 f = 0.0027

Ff = 4f ( ∆L / D ) ( v22

/ 2 ) = 4 x 0.0027 x ( 12.85 / 0.198 ) x (1.832 / 2) = 1.8 J/kg


h ex = (1 – A5/A6) 2

V52/2 = (1 – 0.0177/0.012)

2 (5.8)

2 / 2 = 3.79 J/kg


Σ F = 75.1 J/kg


1 / 2 ( v42 – v5

2 ) + g ( z4 – z5 ) + ( P4 – P5 ) / ρ + Σ F + Ws = 0

1 / 2 ( 2.252

– 3.922

) + 9.81 (1.5) + (101325-116599) / 1038 + 75.1 + Ws = 0

Ws = -70.3 J/kg x 72.04 kg/s = -5064 J/s = -5.1 kW

Wp = - Ws / η = - ( -5.1 kW) / 0.65

Wp = 8 kW So the pump type is SNM 125/200


PUMP EXIT :



1 / 2 ( v72 – v5

2 ) + g ( z7 – z5 ) + ( P7 – P5 ) / ρ + Σ F + Ws = 0

1 / 2 ( 5.82

– 3.922

) + 9.81( 2.5) + ( P7 - 116599 ) / 1038 + 6.95 – 70.3 = 0

P7 = 147415 Pa = 1.47 bar



1.47 x 1.40 = 2.05 bar < 10 bar

COST CALCULATIONS :

• Annual electricity cost :

- One pump 20 kW

- Two pumps 20 kW x 2 = 40 kW

Cost = (2 x 20 kW x 4h /days x 120 days / year) x 0.09 $ / kW.h = 1728 $ / year

• Pump cost :

- One pump ( SNM 150 / 200 ) = 1500 YTL x 0.8 ( 20% discount) = 1200 YTL

- Two pumps 2 x 1200 YTL = 2400 YTL x 1 $ / 1.240 YTL = 1935 $

Annual pump cost = 1935 $ / 15 years = 129.03 $ / year

• Pipe cost :

- 8 m, 6’’( 15 cm ) Galvanized pipe 8 m x 62 YTL/m = 496 YTL

- 11.40 m, ( D = 250 mm ) PE pipe 11.40 m x 30,19 € / m x 1.95 YTL / 1 € x

0.60( 40% discount)

= 402 YTL

- 342.40 m, ( D = 225 mm ) PE pipe 320 m x 24.66 € / m x 1.95 YTL / 1 € x

0.60( 40% discount)

= 9232 YTL

Σ pipe cost = 496 + 402 + 9232 = 10130 YTL x 1$ / 1.240 YTL = 8169 $

Annual galvanized pipe cost = 375.76 $ / 25 years = 15 $ / year

Annual PE pipe cost = 7445 $ / 50 years = 148.9 $ / year

Σ annual pipe cost = 163.9 $ / year

• Fitting cost :

90° elbow cost

- 4 units x ( 125 mm ) 4 x 18,34 € = 73.3 € x 0.65 ( 35% discount) = 47.6 €

- 3 units x ( 225 mm ) 4 x 42.92 € = 171.6 € x 0.65 ( 35% discount) = 111.5 €

- 1 unit x (180 mm) 1 x 30.06 € = 30.06 € x 0.65 ( 35%discount) = 19.5 €

45° elbow cost

- 1 unit x (140 mm) 1 x 14.86 € = 14.86 € x 0.65 ( 35%discount) = 9.65 €

- 2 units x (180 mm) 2 x 21.30 € = 42.6 € x 0.65 ( 35%discount) = 27.70 €

- 2 units x (225 mm) 2 x 29.89 € = 59.8 € x 0.65 (35%discount) = 38.87 €

Tee cost

- 2 units x ( 225 mm) 2 x 56.21 € = 112.4 € x 0.65 ( 35%discount) = 73.06 €

- 1 unit x ( 125 mm) 1 x 23.35 € = 23.35 € x 0.65 ( 35%discount) = 15.17 €

Gate valve cost

- 2 units x ( 100 mm ) 2 x 302 YTL = 604 YTL x 0.75 (25% discount ) = 453 YTL

- 1 unit x ( 125 mm ) 1 x 430 YTL = 430 YTL x 0.75 (25% discount ) = 322 YTL

- 1 unit x ( 150 mm ) 1 x 550 YTL = 550 YTL x 0.75 (25% discount ) = 412 YTL

- 2 units x ( 200 mm ) 2 x 796 YTL = 1592 YTL x 0.75 (25% discount ) = 1194

YTL

Coupling

- 2 units x ( 100 mm ) 2 x 13.90 YTL = 27.8 YTL x 0.85 ( 15% discount) = 23.63

YTL

- 1 unit x ( 125 mm ) 1 x 18.40 YTL = 18.40 YTL x 0.85 ( 15 % discount) = 15.64

YTL

- 2 units ( 150 mm ) 2 x 22.00 YTL = 44.00 x 0.85 (15% discount) = 37.4 YTL

- 3 units ( 200 mm) 2 x 33.00 YTL = 66.00 x 0.85 (15% discount) = 56.1 YTL

= 203.5 €

Filter

- 1 unit ( 150 mm ) 1 x 313.5 YTL = 313.5 YTL x 0.80 ( 20% discount ) = 250.8

YTL

- 1 unit x ( 125 mm ) 1 x 243.0 YTL = 243 YTL x 0.80 ( 20% discount ) = 194.4

YTL

Σ fitting cost = ( 343.05 € x 1.57 $ / 1 € ) + ( 4708 YTL x 1 $ / 1.24 YTL) = 4334.5 $

Annual fitting cost = 4334.5 $ / 15 years = 288.97 $ / year

Annual maintenance :

- For two pump 1935 $ x 3 / 100 = 58.05 $

- For galvanized pipe 380 $

- For PE pipe 300 $

Σ annual maintenance = 738.05 $ / year

Cost of construction / handling :

- For PE pipe 4334.5 $ x 10 / 100 = 433.45 $

- For galvanized pipe 375.76 $ x 15 / 100 = 56.36 $

Σ cost of construction/handling = 489.81 $

Cost of removing/filling soil :

- For 8 m galvanized pipe the galvanized pipe part of the old system has not been

removed, so there is no cost for that part.

- For 342.40 m PE pipe = 342.40 m x 0.5 m x 1 m = 171.2 m3 x 15 $/m

3 = 2568 $

Σ cost of removing/filling soil = 2568 $

TOTAL COST :

1935 $ + 8169 $ + 4334.5 $ + 489.81 $ + 2568 $ = 16299 $

With 18% KDV

16299 $ x 1.18 = 19233 $

The cost of electricity and the cost for maintenance is not included to total cost;

because they are operational costs, which will arise when the system is in operation.

TOTAL ANNUAL COST:

738.05 $ + 288.97 $ + 163.9 $ + 1728 $ + 129.03 $ = 3047.9 $/year

The costs for construction/handling and removing/filling is not included because, they

are done for once and do not change anymore.

With 18% KDV

1.18 x (738.05 $ + 288.97 $ + 163.9 $ + 129.03 $) + 1728 $ = 3284.42 $/year

DISCUSSION:

In this Project , a new design of a pool for a summer resort place of 200 bungalows

that charges-discharge the pool at 4 hours, and a new system with minimum cost is

constructed. The existing pipeline system discharges and refills of the pool very slow;

discharges at 28 hours, fills at 32 hours. The existing system is repaired with new equipments

and some changes, so that people can use the pool 16 hours a day. In the new design,

collection well is cancelled and pump is connected to the pool with a pipeline. A filter is put

entrance of the pump, the pump is protected from undesired materials and expensive

ornament things are taken back from the filter. Three different ID pipe are used to calculate

three different pump powers for both filling and discharge system. Total costs are calculated

with using three different calculations, and the best ones are used in the new design.

In the new design Project, polyethylene pipes are used; more expensive but this type of

pipeline will be sturdier and not required the handling cost for connection of units. For

example polyethylene pipes are sold with 200m (rulo) and they are flexible, they don’t need

elbows for a long path; but PVC pipes are sold with 8m constant long so they need 39

coupling and 3 elbows with 45 degree to construct. These extra fittings are passed over with

using polyethylene pipes. In the filling system water at initial conditions; when globe valve is

fully closed, remaining water is protected inside the pipe by using check valve. Friction factor

for polyethylene pipe can be taken as 75% of that commercial steel tubing given in

corresponding figure reference to term Project explanation .The salt percent of the Aegean

Sea is assumed that 4% of the water, sea water and pool water temperature is assumed at

20˚C. The density and viscosity is taken at these conditions from the datas at Transport

Process and Unit Operations book. The whole sale market prices for polyethylene pipes for

different ID and maximum pressure endurance are taken from EGE YILDIZ Rüzgarlı, ULUS

Prices do not include KDV If we get equipment from retail store with a cash, they make 40%

discount to the equipments. Difference between KDV and discount is found and calculated

are made at these price lowering. The exit is taken for maximum pressure endurance of the

different ID pipes are found .After that the accuracy of pipes in the new design are shown in

the calculations. Different parameters for pump choosing to get the volumetric flow rate of the

initial conditions and pump height of the hill. Pumps are chosen that is tied to these

parameters. The flow rate capacity of the pump can be enhanced to the desired volumetric

flow rate by an adaptor. The minimum cost of the new design is done after all assumptions

and calculations.

Documents

Charge Discharge System