UJIAN ULANG PERBAIKAN NILAI

AERODINAMIKA

Nama Kiki Nuralam

NIM 1106089

Mata Kuliah Aerodinamika

Dosen Drs. Yayat M.Pd

Tanggal 10-08-2013

1. Data :

µ = 80 m

det

P = 1 atm

T = 250C = 2980K

CD = 1,28

X1 = 30% x 6m = 1,8m

X2 = 60% x 6m = 3,6m

Masalah :

a. lapisan batas ()

b. daerah kritis lapisan batas lamina dan turbulen ?

c. besarnya drag friction ?

jawab :

µ = μ0(T 25T 0 )

3/2

x ¿+110,4

T 25+110,4

= 1,7894x10-5 x ( 298288 )

3 /2

x 288+110,4298+110,4

= 1,7894x10-5 x 1,05 x 0,97

= 1,8225x10-5 kg/m.dt

ρ = P

R .T =

101325287,05 x 298

= 1,184kg/m3

A). Rex1 = ρ . v . x1

µ =

1,184 x80 x 1,8

1,8225 x10−5 = 9,35x106 (Turbulen)

Rex2 = ρ . v . x2

µ =

1,184 x80 x 3,6

1,8225 x10−5 = 1,87x106 (Laminer)

= 0,385 . x2

ℜ x0,2 = 0,385 x1,8¿¿

= 0,027m

= 5,2. x2

ℜ x0,2 = 5,2 x3,6

√1,87 x 106 = 0,013 m

B). Jarak keritis Turbulen:

Rex1 -3x106 = 9,35x106 - 3x106 = 6,35x106 m

Jarak keritis Laminer:

3x106 - Rex2 = 3x106 – 1,87x106 = 1,13x106 m

C). Drag Friction aliran laminer

Cf = 1,328

√ℜ x2

= 1,328

√1,87 x 106 = 9,711x10−4

2. Data:LE = 1,2

LW = 3 m

W = 3500kg

C1 = 0,01

CD = 0,68

Cs = 0,03

V = 100 Kmjam

= 27,8 mdt

= 100.000 m

jam

P = 6 m

L = 3 m

T = 2,75 m

Masalah :

A. Beban pada tiap roda ?

B. L, D dan SF ?

C. Rm, Ym dan Pm ?

D. Total tekanan RLT ?

E. Hp ?

Jawab :

A. ∑ MA = 0 W . LE - RB . LW = 0

RB = W .Lw

Lw

= 3500 .1,2

3

= 1400 Kg

Jadi :

RB = ½ . RB

= ½ . 1400

= 700 Kg

∑ MB = 0 RA . LW – W ( LW - LE ) = 0

RA = W .¿¿¿

= 3500 .(3– 1,2)

3

= 2100 Kg

Jadi :

RA = ½ . RA

= ½ . 2100

= 1050 Kg

B. D = ½ . ρ . v-2 . A .CD

= ½ . 1,184 . 772,84 . ( T . L ) . 0,86

= ½ . 1,184 . 772,84 . ( 2,75 . 3 ) . 0,86

= ½ . 1,184 . 772,84 . 8,25 . 0,86

= 3246,11 N

L = ½ . ρ . v-2 . A .CL

= ½ . ρ . v-2 . ( P . L ) .CL

= ½ . 1,184 . 772,84 . ( 6 . 3 ) . 0,01

= ½ . 1,184 . 772,84 . 18 . 0,01

= 82,35 N

SF = ½ . ρ . v-2 . A .CSF

= ½ . ρ . v-2 . ( T . P ) .CSF

= ½ . 1,184 . 772,84 . ( 2,75 . 6 ) . 0,03

= ½ . 1,184 . 16,5 . 0,03

= 0,29304 N

C. RM = SF – L

= 0,319 – 71,373

= - 71,054 N

YM = SF + D

= 0, 29304 + 3246

= 951,20 N

PM = D – L

= 3246,11 – 82,35

= 3163,76 N

D. Fr = (0,0041 + 0,000041 x V) Ch

= 0,0041 + 0,000041 x (100.000mh

) x 1

= 4,1

RRL = (Fr . W) + (12

. ρ . V . CD . A) + (W . sin θ)

= (4,1 . 3500) + (12

. 1,184 . 100000 . 0,68 . 18) + (3500 . sin 30)

= 14364,35 + 1739059,2 + 1750

= 1755173,55 Kg