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PERPINDAHAN KALOR “ PROBLEM 7.11 ”



Oleh PARALEL C - KELOMPOK 4 Toifur Madani



(15031010079)



Firnanti Praditama



(19031010102)



Alvi Nur Diana



(19031010104)



Shiva Eric Z.N



(19031010106)



Kevin Christian Y



(19031010115)



Sheila Qothrunnada



(19031010116)



Muhammad Hakam



(19031010120)



Nabilah Pratama Putri



(19031010129)



Dosen Pengampu : Ir. Mu’tasim Billah, MS



PROGRAM STUDI TEKNIK KIMIA FAKULTAS TEKNIK UNIVERSITAS PEMBANGUNAN NASIONAL “VETERAN” JAWA TIMUR



Soal No. 7. 11 A 1-2 exchanger recovers heat from 10,000 lb/hr of boiler blowdown at 135 psig by heating raw water from 70 to 96°F. Raw water flows in the tubes. Available for the service is a 10.02 in. ID 1-2 exchanger having 52 tubes 3/4 in. OD, 16 BWG, 8'0" long. Baffles are spaced 2 in. apart, and the bundle is arranged for two tube passes. What are the pressure drops and fouling factors? Water t1 = 70oF



Steam T1



Heater



t2 = 96oF Diketahui:



Shell side



Tube side



ID = 10,02 in



Number n =52



Baffle space = 2 in



length = 8



Passes = 1



OD = 3/4 in BWG = 16 Passes = 2 Pitch = 1



Ditanya: a) Nilai pressure drop? b) Berapa nilai fouling factor (Rd)?



T2



Penyelesaian: 1.



Heat Balance Steam,



Q = 1000 × 321.8415 = 3218415 Btu/hr



Raw water,



W=



Q steam 3218415 Btu /hr = =¿123785.2 1×(t 2−t 1) 1×(96−70)℉



Q = W × 1 × (t2 - t1) = 123785.2 × 1 × (96 - 70)oF = 3218415 Btu/hr 2.



∆t Hot Fluid



Cold Fluid



Diff



350.17



Higher Temp



96



254.17



350.17



Lower Temp



70



280.17



0



Differences



26



-26



LMTD = R=



(∆ t 2−∆ t 1) (254.17−280.17)℉ = = 266.959 oF ln( ∆ t 2/∆ t 1) ln (254.17 /280.17)℉ 0 =0 26



∆t = LMTD = 266.959 oF