kalkulasi steam trap mfo rev 00

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Asumsi 1. Kebutuhan steam yang dihitung adalah steam untuk meningkat temperature ke 2. Penurunan temperatur sehingga mengharuskan pemanasan adalah 5 C 3. Volume oil yang dipanaskan diasumsikan 155 m3 yang merupakan volume maksim Kalkulasi Kebutuhan Steam per Tangki MFO Initial T 60 C Final T 65 C Heating time 5 hr Maintain T 65 C Wind velocity 10.08 km/h 6.25 mph Ambient T 27.4 C 81.32 F Diameter 6.8 m Height 6.04 m h 1.7 m Tank manhole 20 in 0.508 m Volume oil 155 m3 Delta T 5K Faktor 1 Data Cp, MFO 2.09 kJ/kg K Densitas MFO 988.8 kg/m3 Pressure Steam 8 bar Latent heat 2768.3 kJ/kg 873 Btu/lb Basic Data Calculation Tank capacity 219 m3 153266 kg 62.5 C 144.5 F Tank wall area 129 m2 1387 ft2 Tank roof area 41 m2 437 ft2 Heat up Load Heat up load 320,326 kJ/hr 305,072 Btu/hr Tank wall heat loss 120 Btu/hr/ft2 2 Weight of fluid in tank Average liquid temperature Tank wall heat loss Wind velocity correction

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Page 1: Kalkulasi Steam Trap MFO Rev 00

Asumsi1. Kebutuhan steam yang dihitung adalah steam untuk meningkat temperature ke temperatur maintenance dan steam yang dibutuhkan untuk meningkatkan temperatur mula-mula 155 m3 fuel oil2. Penurunan temperatur sehingga mengharuskan pemanasan adalah 5 C3. Volume oil yang dipanaskan diasumsikan 155 m3 yang merupakan volume maksimal tangki

Kalkulasi Kebutuhan Steam per Tangki MFO

Initial T 60 CFinal T 65 CHeating time 5 hrMaintain T 65 CWind velocity 10.08 km/h

6.25 mphAmbient T 27.4 C

81.32 FDiameter 6.8 mHeight 6.04 mh 1.7 mTank manhole 20 in

0.508 mVolume oil 155 m3Delta T 5 KFaktor 1

DataCp, MFO 2.09 kJ/kg KDensitas MFO 988.8 kg/m3

Pressure Steam 8 barLatent heat 2768.3 kJ/kg

873 Btu/lb

Basic Data CalculationTank capacity 219 m3

153266 kg

62.5 C144.5 F

Tank wall area 129 m21387 ft2

Tank roof area 41 m2437 ft2

Heat up LoadHeat up load 320,326 kJ/hr

305,072 Btu/hr

Tank wall heat loss

120 Btu/hr/ft2

2

1 in

0.20

Weight of fluid in tank

Average liquid temperature

Tank wall heat loss

Wind velocity correction

Insulation thickness

Insulation thickness correction

Page 2: Kalkulasi Steam Trap MFO Rev 00

Wall heat loss 66,181 Btu/hr

Tank roof heat loss

60 Btu/hr/ft2

10,413 Btu/hr

Total heat loadTotal heat load 381,666 Btu/hr

Steam Consumption

437 lb/hr 197 kg/hr

88 lb/hr 39 kg/hr

393.47 kg/hr

78.96 kg/hr

Maximum steam consumption as the function of volume of oil in the tank

Oil volume (m3) Oil weight (kg) Heat up load (Btu/hr) Total heat load (Btu/hr)5 4,944 9,841 86,435 99.01

10 9,888 19,682 96,276 110.28 20 19,776 39,364 115,958 132.83 40 39,552 78,728 155,323 177.92 60 59,329 118,092 194,687 223.01 80 79,105 157,457 234,051 268.10

100 98,881 196,821 273,415 313.19 120 118,657 236,185 312,779 358.28 140 138,434 275,549 352,143 403.37 160 158,210 314,913 391,507 448.46

Tank roof heat loss rate

Tank roof heat loss

Maximum steam consumption

Maintenance steam consumption

Maximum steam consumption for two tanks

Maintenance steam consumption for two tanks

Max steam consumption (lb/hr)

0 30 60 90 120 150 180 -

50

100

150

200

250

300

350

400

450

Kebutuhan steam satu tangki Kebutuhan steam dua tangki

Volume MFO yang akan dipanaskan (m3)

Kons

umsi

stea

m (k

g/hr

)

Asumsi:- Peningkatan temperatur 5 C- Waktu pemanasan 5 jam

Page 3: Kalkulasi Steam Trap MFO Rev 00

0 30 60 90 120 150 180 -

50

100

150

200

250

300

350

400

450

Kebutuhan steam satu tangki Kebutuhan steam dua tangki

Volume MFO yang akan dipanaskan (m3)

Kons

umsi

stea

m (k

g/hr

)

Page 4: Kalkulasi Steam Trap MFO Rev 00

1. Kebutuhan steam yang dihitung adalah steam untuk meningkat temperature ke temperatur maintenance dan steam yang dibutuhkan untuk meningkatkan temperatur mula-mula 155 m3 fuel oil

Page 5: Kalkulasi Steam Trap MFO Rev 00

44.55 89.11 49.63 99.25 59.77 119.54 80.06 160.13 100.35 200.71 120.64 241.29 140.94 281.87 161.23 322.45 181.52 363.03 201.81 403.62

Max steam consumption (kg/hr)

Max steam consumption for two tanks (kg/hr)

Page 6: Kalkulasi Steam Trap MFO Rev 00

0.8 1 1.2 1.4 1.6 1.8 20

0.2

0.4

0.6

0.8

1

1.2f(x) = NaN x + NaNR² = 0

Page 7: Kalkulasi Steam Trap MFO Rev 00
Page 8: Kalkulasi Steam Trap MFO Rev 00

0.8 1 1.2 1.4 1.6 1.8 20

0.2

0.4

0.6

0.8

1

1.2f(x) = NaN x + NaNR² = 0

Page 9: Kalkulasi Steam Trap MFO Rev 00

Basic Sizing Step

I. Inlet and Outlet Pressure Condition of the trap

Inlet Pressure

10 bargInlet pressure to control valve 8.000 bargPressure drop in coil 0.005 barg

7.995 bargPressure drop in coil 0.005 bargInlet pressure to steam trap 7.990 barg

Outlet Pressure

Outlet pressure 0 barg Atmospheric tank system-gravity drainage(see Equipment Type 1, Class A)

Differential Pressure

Differential pressure 7.990 barg115.86 psig

II. Condensate Load Calculation

Asumsi

Pemanasan dalam tangki diasumsikan hanya disebabkan oleh heat loss dari tangki karena pengaruh lingkungan

Heat loss

Tank wall heat loss 66,181.43 Btu/hrTank roof heat loss 10,412.93 Btu/hrTotal tank heat loss 76,594.36 Btu/hr

Condensate load

Condensate load 87.74 lb/hr 39.48 kg/hr

III. Safety factor

Safety factor 3 see matrix

IV. Desired Trap Capacity

Desired trap capacity each tank 118.44 kg/hr

Maximum pressure in steam line supply

Outlet pressure from control valve to coil

Page 10: Kalkulasi Steam Trap MFO Rev 00

V. Available Steam Trap Capacity

See graphic below

Discharge capacity 350 kg/hr (Hot condensate near steam temperature)Discharge capacity 3300 kg/hr (Cold water 21 C)

VI. Selection Criteria

Mechanical Thermostatic Thermodynamic

First level criteria

Safety P P PEfficiency P P PService life P P P

Second level criteria

Ease of checking

7.99 psig

116 psig

Page 11: Kalkulasi Steam Trap MFO Rev 00

Sensitivity to backpressure

Resistance to freeze damage

Dirt sensitivity

Installation versatility

Air venting

Responsiveness to changing load Very responsive Very responsive

Predominant failure mode

Discharge mode

Tend to decline in efficiency as backpressure exceed 50% of the inlet pressure

Mechanical traps do not easily lend themselves to flexibility of use

Some models can be installed successfully in horizontal or vertical line

Some models can be installed successfully in horizontal or vertical line

Pass air very quicly

Thermodynamic disc trap and bucket trap release air much more slowlyy

Must first cool slightly before they can open wider to pass a greater amount of condensate

Resistance to shock vibration and water hammer

The closed-float are fragile and damage-prone

Bimetallic thermostatic generally are very rugged

Generally are very rugged

Bellows thermostatic trap will fail either open or closed (depending on the design of bellows)

Thermodynamic and bimetal trap fail open when they are worn out

Cyclic thermodynamic disc and bucket trap are easier to check for proper maintenance and better at passing dirt particles

Condensate discharge temperature relative to saturation curve

Most thermodynamic trap follows saturation curve closely

Page 12: Kalkulasi Steam Trap MFO Rev 00

Ease of maintenance

Third level criteria

Product availability

Post-sales service

Warranty

Price

Magniture of condensate sub-cooling

Supplementing accessories or features

Page 13: Kalkulasi Steam Trap MFO Rev 00

Atmospheric tank system-gravity drainage(see Equipment Type 1, Class A)

Pemanasan dalam tangki diasumsikan hanya disebabkan oleh heat loss dari tangki karena pengaruh lingkungan

Page 14: Kalkulasi Steam Trap MFO Rev 00

(Hot condensate near steam temperature)

Note

When properly sized and installed

When properly sized and installed

When properly sized and installed

350

Page 15: Kalkulasi Steam Trap MFO Rev 00

The faster the air is vented, the more quickly equipment is brought up to temperature

A trap that fails open is more desirable than one that fails closed in order to preserve the process

Continuous draining float trap is especially responsive to rapidly changing condensate loads and does not contribute to pressure surges in return system

Page 16: Kalkulasi Steam Trap MFO Rev 00

A trap with little subcooling will discharge condensate within 2-3 degrees of steam temperature, while a trap with large subcooling will discharge condensate with 30 or more degree F of steam temperature. It is desirable to discharge condensate as soon as it forms to achieve steady temperature control

Page 17: Kalkulasi Steam Trap MFO Rev 00
Page 18: Kalkulasi Steam Trap MFO Rev 00