hasil laprak 3 tungku
DESCRIPTION
xTRANSCRIPT
BAB IV
HASIL PERCOBAAN DAN PEMBAHASAN
4.1. Hasil Percobaan
4.1.1 Waktu Pengukuran
To = nyala awal
T1 = nyala api stabil, dilakukan pengukuran pada t1,1 hingga t1,n hingga
kenaikan suhu (∆T) 50°C.
T2 = selesai (bahan bakar habis)
4.1.2 Perhitungan
1. Selimut Tungku
D = 25 cm
t = 30 cm
d = 8,9 cm
A1 = (Dt) – (14
π d2¿
= (π .0,25 .0,3) – (14. π .0,0892)
= 0,22939 m2
T1,0 = 25,9 oC
T1,1 = 50,87 oC
QL1,1 = h.A1.∆T
= 50 Wm2 K
x 0,22939m2x (323,87−298,9)K
= 286,393415 Watt
QL1 = QL 1,1+...+QL1 , n
n=¿286,393415 Watt
2. Alas Tungku
D = 25 cm
A2 = 14
π D2
= 14. π .0,252
= 0,04908 m2
T2,0 = 25,83 oC
T2,1 = 61 oC
QL2,1 = h.A2.∆T
= 50 Wm2 K
x 0,04908m2 x(334−298,83)K
= 86,30718 Watt
QL2 = QL 2,1+ ...+QL2 , n
n=¿86,30718 Watt
3. Atas Tungku
s = -
t = 3 cm
T3,0 = T ∞ = 25,93 oC
T3,1 = Tg = 470 oC
m udara = V udara . ρ udara
= 1.19282 x 10−3 m3 x 1.2 kg/m3 = 1.43139 x 10−3 kg
V udara = luas alas panci x tinggi celah
= (14
π (0.225m)2 ¿ x0.03m=1.19282 x10−3 m3
QL3,1 = mudara .Cpudara .∆T
t
= 1.43139 x10−3 kg x 1003.5 Jkg
K x¿¿
QL3 = 3,54367 Watt
QL3 = QL 3,1+...+QL3 , n
n=¿3,54367 Watt
4. Selimut Panci
D = 22,5 cm
t = 13,5 cm
A4 = Dt
= x (0,225) m x (0,135)m
= 0,09542 m2
T4,0 = 25,9 oC
T4,1 = 65,33 oC
QL4,1 = h . A .∆T
= 50 Wm2 K
x 0.09542m2x (338,33−298,9)K
= 188,12053 Watt
QL4 = QL 4,1+...+QL 4 , n
n=¿188,12053 Watt
5. Tutup Panci
D = 25 cm
t = 1,5 cm
A = (14
π D2 ¿– (14
π d2¿
= (14
π (0.25m)2 ¿−( 14
π (0.15m)2)
= 0.03142 m2
T5,0 = 25,93 oC
T5,1 = 70,67 oC
QL5,1 = h.A5.∆T
= 50 Wm2 K
x 0.03142m2 x (343,67−298,93)K
= 70,28654 Watt
6. Efisiensi
Efisiensi Pembakaran
QG = nilaikalor x mbb
t
= 15000KJ /kg x1kg
1452 s = 10,33057 kW
Qe1 = QG – (QL1 + QL2 + QL3)
= 10,33057 W – (286,393 W + 86,307 W + 3,543 W)
= - 365,91242 W
η1 = Qe1
Q6=365,91242W
10,33057Wx100 % = 35,42035 %
Efisiensi Pemasakan
Qe2 = Qe1 – (QL4 + QL5)
= 365,91242 W – (188,12053 W + 70,28654 W)
= 107,50535 W
Efisiensi Total
ηtot = η1x η2
= Qe2
QG =
152,181417W270.018622W x 100% = 56,3596 %
4.2. Pembahasan