utilitas (pak sudarminto)

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Utilitas

Dr. Ir. Sudarminto Setyo Yuwono, M.App.Sc

UTILITAS

• Utilitas merupakan unit pokok dalam prasarana produksi.

• Air

• Steam

• Listrik

• Bahan bakar

Air

• Sumber air

• Air tanah : sanitasi pekerja, air ketel, air cadangan

• Air PDAM : proses produksi, sanitasi peralatan

• Persyaratan air

Kebutuhan daya pompa

• Digunakan untuk:

– Menggerak air (energi kinetik) = (v2/2) x faktor kehilangan energi pada tikungan dan sambungan

– Mengangkat air (energi potensial) = gh

– Mengatasi hambatan (gesekan) = (4ƒv2/2) x (L/D)

TABLE 3.1RELATIVE ROUGHNESS FACTORS FOR PIPES

MaterialRoughness factor

(e) Material

Roughness factor

(e)

Riveted steel 0.001- 0.01Galvanized

iron0.0002

Concrete 0.0003 - 0.003Asphalted

cast iron0.001

Wood staves 0.0002 - 0.003Commercial

steel0.00005

Cast iron 0.0003 Drawn tubing Smooth

Friction factors in pipe

TABLE 3.2FRICTION LOSS FACTORS IN FITTINGS

k

Valves, fully open:

gate 0.13

globe 6.0

angle 3.0

Elbows:

90° standard 0.74

medium sweep 0.5

long radius 0.25

square 1.5

Tee, used as elbow 1.5

Tee, straight through 0.5

Entrance, large tank to pipe:

sharp 0.5

rounded 0.05

Contoh

• Pompa untuk menaikkan air setinggi 22 m dengan debit 1,2 m3/menit. Pipa dari galvanis berdiameter 15 cm dengan panjang 120m. Jika ada 8 tikungan berapa kebutuhan daya pompa.

Reynold number

• Assume properties of water at 20°C are density 998 kg m-3, and viscosity 0.001 N s m-2

• Cross-sectional area of pipe A = (π/4)D2

= π /4 x (0.15)2= 0.0177 m-2

Volume of flow V = 1.2 m3 min-1= 1.2/60 m3 s-1= 0.02 m3 s-1.

• Velocity in the pipe = V/A= (0.02)/(0.0177) = 1.13 ms-1

• Now (Re) = Dvρ/µ• = (0.15 x 1.13 x 998)/0.001

= 1.7 x 105

so the flow is clearly turbulent.

friction loss of energy

From Table 3.1, the roughness factor ε is 0.0002 for

galvanized iron

and so

roughness ratio ε /D = 0.0002/0.15 = 0.001

So from Fig. 3.8,

ƒ = 0.0053

Therefore the friction loss of energy

= (4ƒv2/2) x (L/D)

= [4ƒv2L/2D]

= [4 x 0.0053 x (1.13)2 x 120]/(2 x 0.15)

= 10.8 J.

TABLE 3.1RELATIVE ROUGHNESS FACTORS FOR PIPES

MaterialRoughness factor

(e) Material

Roughness factor

(e)

Riveted steel 0.001- 0.01Galvanized

iron0.0002

Concrete 0.0003 - 0.003Asphalted

cast iron0.001

Wood staves 0.0002 - 0.003Commercial

steel0.00005

Cast iron 0.0003 Drawn tubing Smooth

Friction factors in pipe

TABLE 3.2FRICTION LOSS FACTORS IN FITTINGS

k

Valves, fully open:

gate 0.13

globe 6.0

angle 3.0

Elbows:

90° standard 0.74

medium sweep 0.5

long radius 0.25

square 1.5

Tee, used as elbow 1.5

Tee, straight through 0.5

Entrance, large tank to pipe:

sharp 0.5

rounded 0.05

• For the eight right-angled bends, from Table 3.2 we would expect a loss of 0.74 velocity energies at each, making (8 x 0.74) = 6 in all.

velocity energy = v2/2= (1.13)2/2= 0.64 J

• So total loss from bends and discharge energy= (6 + 1) x 0.64 = 4.5 J

There would be one additional velocity energy loss because of the unrecovered flow energy discharged into the reservoir.

Energy loss from bends and discharge

Energy to move 1 kg water

• Energy to move 1 kg water against a head of 22 m of water is

E = Zg= 22 x 9.81= 215.8 J.

• Total energy requirement per kg:Etot = 10.8 + 4.5 + 215.8

= 231.1 J

energy requirement of pump

• and theoretical power requirement= Energy x volume flow x density= (Energy/kg) x kgs-1= 231.1 x 0.02 x 998= 4613 J s-1.

• Now the head equivalent to the energy requirement= Etot/g= 231.1/9.81= 23.5 m of water,

TABLE 3.2FRICTION LOSS FACTORS IN FITTINGS

k

Valves, fully open:

gate 0.13

globe 6.0

angle 3.0

Elbows:

90° standard 0.74

medium sweep 0.5

long radius 0.25

square 1.5

Tee, used as elbow 1.5

Tee, straight through 0.5

Entrance, large tank to pipe:

sharp 0.5

rounded 0.05

TABLE 3.1RELATIVE ROUGHNESS FACTORS FOR PIPES

MaterialRoughness factor

(e) Material

Roughness factor

(e)

Riveted steel 0.001- 0.01Galvanized

iron0.0002

Concrete 0.0003 - 0.003Asphalted

cast iron0.001

Wood staves 0.0002 - 0.003Commercial

steel0.00005

Cast iron 0.0003 Drawn tubing Smooth

Friction factors in pipe

Listrik

LOKASI LUAS (m2) CANDL

E

LUMEN

Taman

Kantin

Mushola

Laboratorium

Perkantoran dan TU

Toilet

Area penyimpanan

produk

Area Proses

Area Penerimaan dan

Pengeluaran Bahan Baku

30,4

13,57

50,53

20

129,50

10

30

130

36

5

6

9

15

34

2

25

36

10

152

81,42

454,77

300

4403

20

750

4680

360

PERHITUNGAN JUMLAH LAMPU

• N= (E.A)/Qlampu.Cu.LLF

• KETERANGAN:

• N = JUMLAH LAMPU

• E =KEKUATAN PENERANGAN(LUX)

• A =LUAS AREA

• QLAMPU =BESAR LUMEN DARI LAMPU

• Cu =KOEFESIEN OF UTILITYè 0.5(STANDART)

• LLF =LIGHT LOST FILTER è 0.7(STANDAR)

• Untuk sistem penerangan langsung dengan warna plafon dan dinding terang, CU ( coeffesien of utilization ) –nya 50-65 %. Light loss factor ( LLF ) = 0,7-0,8. LLF tergantung ; kebersihan sumber cahaya, tipe kap lampu, penyusutan cahaya dari permukaan lampu, dll.

• Suatu ruang laboratorium di pabrik berukuran 20 x 20 m. Direncanakan dipasang lampu TL 36 watt. Berapa buah lampu TL yang dbutuhkan?

• E= 500 LUX

• A=20 x 20 METER= 400 METER2

• QLAMPUè 1 WATT = 75 LUMEN

• 36 WATT = 2700 LUMEN

• Cu = 0.5 (STANDART UNTUK Cu)

• LLF = 0.7

• N= (E.A)/Qlampu.Cu.LLF

• N= (500.400)/(2700.0.5.0.7)

• = 211,64

• DIBUTUHKAN 212 LAMPU PADA RUANGAN TERSEBUT

• Contoh perhitungan : ruangan kantor berukuran 20 x 10 x 3 m direncanakan memakai TL 4 x 40 watt dengan penerangan E = 300 lux. Hitung, jumlah lampu dan daya listrik yang dibutuhkan.

• Penyelesaian : dari tabel,

• Untuk 1 bh TL 40 watt, jumlah lumen = 40 x 75 = 3000 lumen. Untuk 4 TL 40 watt, jumlah lumen = 4 x 3000 = 12.000 lumen.

• Dipilih CU 60 % dan LLF 0,8

• Jumlah lampu yang dibutuhkan ( N ) = E x A dibagi lampu x CU x LLF = 300 x 200 dibagi 12000 x 0,6 x 0,8 = 10,4

• Jadi N = 11 buah 4 x TL 40 watt. Pemakaian watt untuk lampu TL 40 watt termasuk ballast = 50 watt. Jumlah beban dari lampu = 11 x 4 x 50 watt = 2200 watt. Untuk stop kontak peralatan kantor diperhitungkan 20 % dari beban lampu = 20 % x 2200 watt = 440 watt. Total kebutuhan watt = 2640 watt, atau watt/m2 = 13, 4.

RUANG FC LUAS RUANG(ft²)

LUMEN

(FCxLUAS)

KANTOR 30 1184,07 35522,1

PROSES 20 10764,26 215285,2

TOILET 10 775,03 7750,3

LABORATORIUM 30 322,93 9687,9

GUDANG 10

BAHAN BAKU 1614,64 16146,4

BAHAN TAMBAHAN 645,86 6458,6

COOL STORAGE 4305,71 43057,1

MUSHOLA 10 258,34 2853,4

PARKIR 10

RODA 2 430,57 4305,7

MOBIL 645,86 6458,6

RUANG GANTI 10 322,93 3229,3

JALAN+HALAMAN 5 3229,28 16146,4

PEMBUANGAN LIMBAH 5 322,93 1614,65

TOTAL 368245,65

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