tugas heat transfer - menentukan k dan t
Post on 06-Jul-2015
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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
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