Pap.pptx

Diketahui P N blanket V tank sg v P z1 Q elbow D

= = = = = = = =

g = 5 psig 25000 gal 0,81 1,125 cp 30 psig 10 ft 20 gpm 6 x 90◦ 2,067 Appendix D-16 2 in sch 40

32,2

Jawab 𝑄 𝑥 8,33 𝑣=

𝑙𝑏 𝑥 𝑠𝑔 𝑔𝑎𝑙

62,3 𝑥 0,81 3,355 𝑖𝑛2

60

𝑠 /144 𝑚𝑖𝑛

= 1,912959025 lb /in² s atau 1,912959 ft/s

Sehingga 𝑣2 2𝑔

= 0,056823

Reynolds number 𝑅𝑒 =

50,6 𝑄 𝜌 𝑑𝜇

= 21961,43 (aliran turbulen)

fc assumsi commercial steel ε 0,0018 Fd D 2,067 ε/D 0,000870827

0,006799327 0,027197308

equation : 4-35 𝜀/𝐷 6,7 𝐴= + 3,7 𝑅𝑒

0,9

= 0,000920804

1

𝜀 5,02 = −4 log − log 𝐴 = 12,12738143 3,7𝐷 𝑅𝑒 √𝑓𝑐 𝑓𝐷 = 4𝑓𝐶 = 0,027197308

equation 4-37 head loss (in pipe only) ℎ𝑓 = Fd

15 (𝑣)2 D/12 (2)(𝑔)

= 0,134580895

K 0,5Figure 4.14 Square-edged inlet β 1 𝐾 = 8 𝐹𝑡 Ft K ΣK

0,019Table 4-6 0,152 0,652

Frictional head loss (Fitting and Pipe entrance) ℎ𝑓 = 𝐾 𝑣 2 /2g = 0,037048708 ft fluid Total friction loss

Σ hf

0,171629603ft kerosene

Pressure drop per 100 ft from Eq 4-65

△𝑃

𝑝𝑠𝑖 𝑓𝑡 100

= 0,0216 𝑓 ρ 𝑄2 / 𝑑 5

Pressure drop per 100 ft K 0,675

𝐾 𝑝𝑖𝑝𝑒 = 4 𝑓𝐹 K pipe Sum K

𝐿 𝐷

= 𝑓𝐷

0,314275367psi/100 ft 𝐿 𝐷

2,368415777 3,043415777

Friction head loss (Fitting and pipe entrance)

ℎ𝑓 = 𝐾 𝑣 2 /2g hf

39,76866241

0,039769ft of fluid

Using Hooper`s 2-k method 𝐾𝑓 = Pipe ID k1 K∞ Kf Kf total presentase

𝐾1 𝑅𝑒

+ 𝐾∞ (1 +

1 𝐼𝐷𝑖𝑛𝑐ℎ

)

2,067 inch 300 0,1 0,162039607 Gate Valve 0,507 Square-edged 0,669039607 0,890887845

Energy balance gc conversi factor g Accelerasi

32,174 lbm/lbf 32,174 ft/s2

ft/s2

The general energy balance equation 𝑃2−𝑃1 ρ

+ g (z2-z1) + 1/2 (v22 - 𝑣12 ) + 𝑒𝑓 + w = 0

Pressure (P2) 𝑃2 = 𝑃1 +

ρ 𝑔 (z1-z2) 144 𝑔𝑐

+

ρ 1 ( ) 2 𝑔𝑐 144

(0-

𝑣22 )

-

ρ 𝑒𝑡 144 𝑔𝑐

𝑃2 = 5 𝑝𝑠𝑖𝑔 + 𝑠𝑡𝑎𝑡𝑖𝑐 ℎ𝑒𝑎𝑑 + 𝑘𝑖𝑛𝑒𝑡𝑖𝑐 𝑒𝑛𝑒𝑟𝑔𝑦 ℎ𝑒𝑎𝑑 + 𝑓𝑟𝑖𝑐𝑡𝑖𝑜𝑛 𝑙𝑜𝑠𝑠 𝑖𝑛 𝑠𝑢𝑐𝑡𝑖𝑜𝑛 𝑙𝑖𝑛𝑒

𝑒𝑡 = ( ρ 𝑒𝑓

144 𝑔𝑐

𝐿 𝑓𝐷 𝐷

+ Σ 𝐾𝑓 )

𝑣2 2

= 5,526473218

= 0,060194053

𝑙𝑏𝑚 𝑓𝑡 1 𝑓𝑡2 𝑓𝑡3 𝑠2 𝑙𝑏𝑚 𝑓𝑡 𝑖𝑛2 𝑙𝑏𝑓 𝑠2

atau lbf/in