Tugas Peristiwa Perpindahan

Latihan Soal Bab 6 Perpindahan antarfasa system isotermal

Kelompok 7 :

Didik Sudarsono(1206242555)

Dita Amelia Putri(1206201965)

Indah Kemala(1206202122)

Rizky Ramadhan(1206201920)

Wildan Nurasad(1206202160)

DEPARTEMEN TEKNIK KIMIA FAKULTAS TEKNIK UNIVERSITAS INDONESIA DEPOK 2013No. 6B1

Water at 68oF is to be pumped through 95 ft of standard 3-in. Pipe (Internal diameter3.068 in ) into an over head reservoir, shown in the figure below.

(a) What pressure is needed at the outlet of the pump to supply water to the overhead reservoir at a rate 18 gal/min? (At 68oF the viscosity of water is 1.002 cp and the density is0.9982 g m-1)(b) What percentage of the pressure drop is needed for overcoming the pipe friction?

Diketahui : = 95 = 2896.34 cm

3 = 3.068 = 7.793

= 18 = 1135.624 = 0.9983

3 = 22.4 3 = 1.002 = 0.01002 . Ditanya :

a) Po-PL (P)b) Persentase penurunan tekanan yang diperlukan untuk mengatasi gesekan pipa

Jawab :

a) Mencari P Menghitung nilai < >< v > = Menghitung nilai Re

=

113 5 . 624 3 14 3. 14 (7.793 )2

= 23.818

=

< >

3 =7.793 23.818 2 0.9983 0.01002 . = 18492.827= 1.84 104

Menentukan nilai f dari Re melalui grafikRe = 1.84 104 f = 0.0064

Menentukan nilai P 1 2 = ( ) + 4 2 = 0.9983

980

(50 30 ,48 sin 45 ) + 2 [(

) + (2 457.2 )] 3

2

1

2896.34 cm7.793

30.9983

(23.818 )2 0.0064 = 1491377.431 .

= 15.2

b. . What percentage of the pressure drop is needed for overcoming the pipe friction

1) Mencari persentase pressure drop menggunakan persamaan :

2) Sehingga diperoleh nilai persentase pressure drop untuk menangani gaya friksi sebesar 16.1 %

No. 6C1

How many gal/hour of water at 68oF can be delivered trough a 1320 ft length of smooth 6.0 in iD pipe under a pressure differe 0.25 Psi.? (a) solve by method A example6.2-2; (b) solve by methid B of example 6.2-2; (c) compare the valve of obtained herewith that obtained in problem 5.B, part e . assume that the pipe is hydrautically smooth.

Diketahui :

L= 1320 ft iD = 6.0 inch= 0.5 ft P = 0.25 Psi = 1.16 103 . 2 = 62.4 3 = 6.73 104 . Ditanya :Tentukan Q ( g al ) dengan menggunakan metode A dan B untuk menentukan Re danhourmembandingkannya ?

Jawab :

Menggunakan Metode A

.

(po pL)Re =

2L

(0.5 )(62.4 3 )=6.73 104 .

(1.16 103 . 2)(0.5 )2 1320 62.4 3

= 2.74 103

3.6 104Dengan metode A , nilai Re yang didapatkan adalah 3.6 104

Menggunakan Metode BData sebelumnya , Re = 2.74 103Jika f =1 maka Re = 2.74 103 dan pada f = 0.01, Re = 2.74 104 .Dengan mengamil intersection dari f, didapat nilai Re = 3.6 104

3.6 104

Nilai Re yang didapat dari kedua metode tersebut adalah sama.

Menghitung Kecepatan aliran :

3. 6 104 . 6. 72 10 4 . = (. ) =

0.5 62.4

3

= 0.775 /

Debit air yang mengalir :2 = 4= (3.14)(0.5)24

0.775

3= 0.153 =68 gal /menit x 60 menit / 1 hour = 4080 gal /hour

6D1.

A hollow steel sphere, 5.00 mm in diameter, with a mass of 0.5 g, is released in a column of liquid and attains a terminal velocity of 0.5 cm sec-1 . The liquid density is 0.9 gm cm-3 . The local acceleration of gravity is 980.7 cm sec-2 . The sphere is far enough from the containing walls so that their effect may be neglected.a. Compute the drag force in dyneb. Compute the coefficient (friction factor)c. Determine the viscosity of the liquid in centipoises

DIketahui : = 0.5 = 0.05 = 0.5 1 = 0.9 3 = 980.7 cm 2Ditanya :a) = ?b) = ?c) = ?Jawab

a. Gaya gravitasi mendorong sebuah bola kebawah Ketika memasuki fluida. Pada saat beradadalam fluida, bola mendapat gaya apung atau F buoyant. Dari kedua gaya ini, akan menghasilkan gaya kinetic Fk yang besarnya sama dengan drag force, tetapi berlainan arah.

=

4 0.5 (3 2

3)

= = . . 2 0.9 3 980.7cm 2 0.05 980.7 cm 2= 57.8 . cm 2 49 . cm 2= 8.8 . cm 2= 8.8 b. Faktor friksi (f)2 1 = ( ) (4 2

2) 2 =( 2 ) ( 2)4 8 8.8 . cm 2= (3.14)(0.5 )2)(0.9 3)(0.5 1)2)= 398.58c. Viscositas FluidaFaktor friksi 398.58 Re < 2100 (laminar), sehingga

= 24

= 24 = 24 = 0.06021

398.58

maka,

1 3 = = 0. 5 0. 5 0.9 = 3.7369 11 = 373.69

0.06021

6E1

a) How does one make drag calculations, using fig. 6.3-1, if the sphere diameter is unknown? Show how a trial-and-error solution can be avoided.b) Rework problem 2.C, using Fig. 6.3-1.c) Rework (b) when the gas velocity is 10 ft sec-1

Jawab ;

a. Mencari nilai diameter partikel . . .? Dengan metode Trial and eror = 4 (

) . . . . (1)3 2 = . . . .(2)Dengan tabel

Trial Dp (m)f(Dp)Re (Dp)Dp

1.12 x 10-40.263930.951.12 x 10-4

Langkah langkaha) Masukkan nillai trial Dpb) Mensubtitusi nilai Dp ke fungsi f(Dp) (pers. 1)c) Setelah mendapatkan nilai f(Dp), menetukan nilai Re dengan membaca grafik dibawah ini !d) Menguji nilai Re ke pers. (2)e) Jika nilai hasil Dp pada persamaan (2) sama pada nilai Trial Dp , maka perhitungan trial Dp tersebut benar atau mendapatkan angka yang teapt.

Keterangan gambar :

b. Diketahui

3 = 30.5 = 7.2 104

3 = 1.2 = 2.6 104 . Ditanya : Dp (diameter partikel)

4= 3

3

( )

4

4 980 2 ( 2 .6 10 ) . 1 .2 3 = 2

3 3 (30.5 /)3 (7.2 104 )= 27.7 = 27.7

Dengan sebuah garis yang mempunyai kemiringan 1, = 1 = 27.7 = 0.1 = 2.77

sehingga didapat Re = 0.95 (Re sebenarnya).

. = =

0 . 95 (2 . 6 10 4 . )

= 0.01125 = 112.5 .

30.5 (7.2 104 )c. Ketika 10 ft/s berapa nilai D nya ?

= 10 ft xs

30 . 48 = 304.81

4= 3

3

( )

34

4 980 2 ( 2 .6 10 ) . 1 .2 3 =

3 3 (304.8 )

(7.2 104

23 )

= 27.8 103 = 27.8 103

3Dengan cara yang sama pada bagian b, didapat Re = 75. .

75 2 .6 104 .

6F1.

= =

.

304.8 7.2 104

= 890

A column of 146 in.2 cross section and 73 in. high is packed with spherical particles of diameter 2mm. When a pressure difference of 158 psi is maintained across the bed, a 60 pecent aqueous sucrose solution at 20oC flows through the bed at a rate of 244 lb min-1. Atthis temperature, the viscosity of the solution is = 56.5 cp and its density is = 1.2865 g cm-3. What is the void fraction of the bed? Discuss the usefulness of this method of obtaining the void fraction.

Diketahui :

P = 158 psi = 158 x 68947 dyne cm-2 Dp = 0.2 cm = 56.5 cp = 0.562 . = 1.2865 g cm-3 Kolom ; Luas (A) = 146 in2 ; Tinggi (h) = 73 in Bola , diameter = 2 mm

Ditanya :a) void fraction of the bed ()Jawab :

Menghitung nilai vo3244 / 1 453.59 1 0 =

60

1.2865 2= 1.522

146 2 ( 2 . 54 )Berdasarkan persamaan Blake-Konezy :

2 3 0 =

150

(1)2

3

150 0 (1 )2 =

23(1 )2 =3

150 0.565 . 73 2.54 2. 54 1.522 = 0.0549(0.2 )2 158 68947 dyne cm 2(1 )2 = 0.05493 = 0.0549(1 )2

= 0.30