tugas heat transfer - menentukan k dan t

Nama: Rolina Z Tambunan NIM:

No. 1 Komposisi Susu:

Rumus-rumus:

Air : 87.50%

Protein : 3.70% Lemak : 3.70% Laktosa : 4.60% Abu : 0.50% T=283 K, Diperoleh: Pers. 3-9

Kw : 0.588042 Kp : 0.190486

Kf : 0.153078 Kc : 0.214841 Ka : 0.343319 Pers. 12-18

w : 997.8357

p : 1324.719

f : 921.4143

c : 1595.995

a : 2420.994

Pers. 10 Xvw : 0.008769

Xvp : 0.000279 Xvf : 0.000402 Xvc : 0.000288 Xva : 0.000021 Choi dan okos (Pers.2):

k=

0.005340 W/moC

No. 2

T 2 = 43oC, T 1 = 45oC, x2 = 4 mm, x1 = 2mm

Gradien suhu T / x=(T 2-T1)/(x2-x1)

=(43-45)/(4-2)*1000

= -1000.00

Pers (23) : T = - (-1000)(x1-x)+T1

kw = 0.57109 + 0.0017625T-6.7306x10

-6T

2 .......................3)

kp = 0.1788 + 0.0011958 T – 2.7178 x 10

-6T

2 ...................5)

kf = 0.1807 – 0.0027604 T – 1.7749 x 10

-7 T

2 .................. 6)

kc = 0.2014 + 0.0013874 T – 4.3312 x 10

-6 T

2 ...................7)

ka = 0.3296 + 0.001401 T – 2.9069 x 10

-6T

2 ......................9)

Satuan untuk nilai k = W/moK. w = air murni, ic = es, p = protein, f =

fat/lemak, c = karbohidrat, fi = fiber, a = abu

Densitas komponen dihitung dengna rumus :

w

= 998.18 + 0.0031439T – 0.0037574 T2

........12)

p

= 1329.9 – 0.51814 T ...................................14)

f

= 925.59 – 0.41757 T ..................................15)

c

= 1599.1 – 0.31046 T ..................................16)

a

= 2423.8 – 0.28063 T ..................................18)

Xvi ditentukan dari fraksi massa (X

i), densitas

komponen (i) dan densitas komposit ( )

Choi dan okos (1987) : k = (k

iX

vi) ........................2)

ki = konduktivitas panas komponen murni

Xvi = fraksi volume dari tiap komponen murni

............23)

Di permukaan x = 0, dan pada x1=0.002, T 1 = 45, sehingga :

T = 1000(0.002)+45

T =

47oC