UJIAN NERACA MASA DAN ENERGI

MATA KULIAH : NERACA MASA DAN ENERGIDOSEN PENGAMPU : Ir. SUMARNO M.Si

OLEH:

ANDY MAILIAN

PROGRAM MAGISTER TEKNIK KIMIAPROGRAM PASCASARJANAUNIVERSITAS DIPONEGORO

2011

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DEPARTEMEN PENDIDIKAN NASIONALUNIVERSITAS DIPONEGOROPROGRAM PASCASARJANAPROGRAM MAGISTER TEKNIK KIMIA Jl. Prof. H. Soedarto, SH. Kampus Tembalang Semarang 50239Telp. (024) 7460058, Fax. (024) 7460055

HALAMAN PENGESAHAN

Tugas ini telah diselesaikan oleh:

Nama : ANDY MAILIAN

Program : KHUSUS

Jepara, 16 Agustus 2011

Mengetahui,Dosen Pengampu

Ir. SUMARNO M.Si.

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Soal :

A catalytic reactor is used to produce formaldehyde from methanol in the reaction: CH3 OH → HCHO + H2

A single-pass conversion of 60.0% is achieved in the reactor. The methanol in the reactor product is separated from the formaldehyde and hydrogen in a multiple-unit process. The production rate of formaldehyde is 900.0 kg/h:

a. Calculate the required feed rate of methanol to the process (kmol/h) if there is no recycle.

b. Suppose the recovered methanol is recycled to the reactor and the single-pass conversion remains 60%. Without doing any calculations, prove that you have enough information to determine the required fresh feed rate of methanol (kmol/h) and rate (kmol/h) at which methanol enters and leaves the reactor. Then perform the calculations.

c. The single-pass conversion in the reactor, Xsp, affects the costs of the reactor (Cr) and the separation process and recycle line (Cs). What effect would you expect an increased Xsp would have on each of these costs for a fixed formaldehyde production rate? (Hint: To get a 100% single-pass conversion you would need an infinitely large reactor, and lowering the single-pass conversion leads to a need to process greater amounts of fluid through both process units and the recycle line.) What would you expect a plot of (Cr + Cs) versus Xsp to look like? What does the design specification Xsp = 60% probably represent?

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JAWABAN SOAL :

(a) Diagram Proses :

Hidrogen : Formaldehide = 1:1

Mol 900 kg/h Formaldehid produk :

n3,HCOH = ( 900 kg/h / 30 kg/mol)

= 30 kmol/h

Jika konversi reaktor adalah 60 % maka Mol Metanol adalah :

n3,HCOH = ( 30 kmol/h / 0.6)

= 50 kmol CH3OH/h

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n2 CH3OHn2 HCHOn2, H2

Proses pemisahan

reaktor

n1,CH3OH

n3,HCOHn3, H2

n4,CH3OH

(b) Diagram Proses :

Jadi produk yang diinginkan

n4,HCHO = 30 kmol/h sama dengan n3,HCHO = 30 kmol/h

Hidrogen : Formaldehide = 1:1

Konversi 60 %.

Maka fresh feed :

n2,CH3OH = ( 30 kmol/h / 0.6)

= 50 kmol /h Denngan 60% konversi, maka yang tidak bereaksi = 40 %Maka n3 HCHO = 0.40 x 50.0

= 20 kg/mol

Material Balance untuk total siystem :mol of n1CH3OH masuk = mol n4HCHO diproduksi

= 30 kg/mol

Maka :

n2,CH3OH = n1,CH3OH + n5,CH3OH

n5,CH3OH = 50 – 30

= 20 kmol/h

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n5,CH3OH

n3 CH3OHn3 HCHOn3, H2

Proses pemisahan

reaktor

n1,CH3OH

n4,HCOHn4, H2

n2,CH3OH

(c) Peningkantan single-pass conversion (Xsp),mengakibatkan penambahan biaya dari reaktor, karena harus dibuat lebih besar untuk untuk menampung reaktan yang masuk. Tetapi bisa mengakomodir sebahagian reaktan yang tidak bereaksi untuk didaur ulang.

Orang akan berharap bahwa Plot (Cr + Cs) vs XSP harus berkurang dengan meningkatnya meningkat XSP dari 0,0, dan harus melewati sebuah jumlah minimum sebelum menjadi nilan Xsp yang lebih tinggi.

Spesifikasi desain yang Xsp = 60% mungkin mewakili dimana jumlah minimum terjadi.

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