TUGASTEKNIK PENGERINGAN DAN PENDINGINAN

(Tugas Pemodelan dalam Pengeringan)Oleh:

Prisilia Ratna Setyaningrum

240110120059

DEPARTEMEN TEKNIK DAN MANAJEMEN INDUSTRI PERTANIAN

FAKULTAS TEKNOLOGI INDUSTRI PERTANIAN

UNIVERSITAS PADJADJARAN

JATINANGOR

2015

SOAL TUGAS PEMODELAN DALAM PENGERINGAN1. Diketahui data pengeringan gabah sbb;

Drying at 80oCModel 1Model 2

Time (hr)MC(t)ToC (db)ToC (wb)%RHIMCEMCMRln MRLN (-ln MR)ln time00,40774,5544,1150,4070,051,000

0#NUM!#NUM!0,50,2774,5544,1150,4070,050,616-0,48411-0,72545-0,6931510,17274,5544,1150,4070,050,342-1,073710,07112401,50,12974,5544,1150,4070,050,221-1,508290,4109750,40546520,10674,5544,1150,4070,050,157-1,852380,6164740,6931472,50,08574,5544,1150,4070,050,098-2,322390,8425960,91629130,06974,5544,1150,4070,050,053-2,93331,0761271,0986123,50,05974,5544,1150,4070,050,025-3,680511,3030521,25276340,0574,5544,1150,4070,050,000#NUM!#NUM!1,386294

Drying at 90oCModel 1Model 2

Time (hr)MC(t)ToC (db)ToC (wb)%RHIMCEMCMRln MRLN (-ln MR)ln time

0,010,4070,40,4100,4070,031,0000#NUM!-4,60517

0,250,3490,350,4100,4070,030,846-0,16705-1,78944-1,38629

0,50,3030,30,4100,4070,030,724-0,32277-1,1308-0,69315

0,750,2640,260,4100,4070,030,621-0,47692-0,7404-0,28768

10,2350,230,4100,4070,030,544-0,60924-0,495550

1,250,2070,210,4100,4070,030,469-0,7561-0,279590,223144

1,50,1840,180,4100,4070,030,408-0,89529-0,11060,405465

1,750,1640,160,4100,4070,030,355-1,034410,0338270,559616

20,1440,140,4100,4070,030,302-1,196050,1790220,693147

2,50,1090,110,4100,4070,030,210-1,56280,4464770,916291

30,0910,090,4100,4070,030,162-1,821370,599591,098612

3,50,0750,070,4100,4070,030,119-2,125580,7540461,252763

40,0640,060,4100,4070,030,090-2,405880,8779181,386294

1. Hitunglah konstanta pengeringan pada suhu 80oC dan 90oC dengan pendekatan Simple log Model.

T (oC)1/T (1/oC)Slope (k)R2

800,0131,0020,991

900,0110,6030,999

2. Gambarkan kurva hubungan k vs. 1/T pada Simple Log Model

3. Berapa nilai konstanta pengeringan untuk suhu T= 87oC dengan Simple Log ModelK =(278,86 x 0,011)-2,4923

= 0,713Temperature (oC)1/Tk

k 87870,0110,713

4. Hitunglah konstanta pengeringan (k) dan n pada suhu suhu 80oC dan 90oC dengan pendekatan Modified log Model.

T (oC)1/T (1/oC)Slope (n)ln kkR2

800,0130,9941-0,00810,9920,993

900,0110,9651-0,47320,6230,9992

5. Gambarkan kurva hubungan ln k vs. 1/T dan n vs, 1/T pada Modified Log Model

6. Berapa nilai konstanta pengeringan (k) dan n untuk suhu T= 87oC dengan Modified Log ModelTemperature (oC)1/Tln kk

k 87870,011-0,3450,708

Temperature (oC)T1/Tn

k 8787870,0114940,973