SOAL- SOAL SISTEM PENGATURAN 2
BAB 1 PRINSIP-PRINSIP ALAT KONTROL
1.1
Sebuah sistem kontrol kecepatan mempunyai mempunyai daerah pengukuran pengukuran 220 ± 460 mm/s. Jika setpoint adalah 327 mm/s dan harga terukur adalah 294 mm/s, hitungkah eror persentase.
1.2
Parameter kontrol adalah kecepatan sebuah motor yang berubah-ubah berubah-ubah dari 800 rpm sampai 1750 rpm. Jika kecepatan dikontrol oleh sinyal dc 25 ± 50 v, hitung : (a). Kecepatan yang dihasilkan oleh masukkan masukkan sebesar sebesar 38 v. v. (b). Kecepatan dihitung sebagai persentase dari daerah pengukuran penuh.
1.3
Sebuah tangki silindris berdiameter 5 m dikosongkan dikosongkan oleh aliran keluar 3 yang konstan dengan laju 1,0 m /menit. Sebuah alat kontrol dua posisi dipakai untuk untuk membuka dan menutup menutup sebuah katup Pengisi dengan dengan laju 3 aliran buka sebesar 2,0 m /menit. Untuk kontrol ketinggian daerah netral adalah 1m dan setpoint adalah 12m, (a). Hitung periode sikling (b). Grafik ketinggian cairan versus waktu.
1.4.
Untuk contoh 1.4, buktikan bahwa daerah netral ± 2% 2% menghasilkan hasilhasil dari limit isolasi isolas i dan periode yang diberikan untuk contoh tersebut.
1.5.
Tangki di soal 1.3 diatas memakai memakai alat kontrol kontrol 3 posisi dengan laju sebesar 2,0 m³/menit untuk untuk ketinggian cairan cairan yang lebih rendah dari 11 m, m, 0,5 m³/menit untuk ketinggian antara 11 m dan 12 m, serta 0, m³/menit untuk ketinggian yang lebih besar dari 12 m. Carilah periode sikling. Jika kelambatan adalah 5 menit, plot ketinggian ketinggian cairan dimulai pada 13 m.
1.6.
Untuk alat kontrol kontrol proporsional, proporsional, variabel yang dikontrol adalah suhu suhu proses dengan daerah pengukuran 50 - 130º C dan setpoin sebesar 73,5º C untuk mana keluaran alat kontrol adalah sebesar 50 %. Carilah offset proporsional yang dihasilkan dari suatu perubahan bagi keluaran alat kontrol sebesar 55 % guna menjaga setpoin jika penguatan proporsional adalah : (a). 0,1 ; (b). ( b). 0,7 ; (c). 2,0 ; (d). 5,0.
Febry Mantika (342 08 039)
1
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- SOAL SIST E E SOAL -
1.7.
Unt
PE G ATU R AN ATU R AN 2
ilah persentase kel aran alat kontro l ap li asi dar i soa l 1.6, car il
dengan set po poin 73,5º
C dan penguatan proporsional sebesar 2 yang
dihasilkan oleh : (a). 61º C ; (b). 122º C ; (c). (82 + 5 t)ºC. 1.8.
Sebuah alat kontrol integral mempunya i tindakan reset sebesar 2,2 men it.
Car ililah konstanta alat kontrol integral dalam sع . Car ililah keluaran dar i alat kontrol ini untuk eror konstan sebesar 2,2 %.
Gambar 1.25
Gambar Untuk Soa l 1.10 1.9.
Baga imanakah sebuah a lat kontrol der i ati e dengan Kd = 44 s akan member ikan tanggapan bagi sebuah eror yang berubah sebaga i Ep = 2,2 sin (0,04 t) ?
1.10.
Sebuah a lat kontrol proporsional mempunya i penguatan Kp = 2. P lot tanggapan eror pada gambar 1.25 jika P (0) = 50 %.
1.11.
Sebuah a lat kontrol PI mempunya i kons tanta Kp = 2,2 sع, P (0) = 40 %. Plot tanggapan terhadap eror pada gambar 1.25.
1.12.
Sebuah a lat kontrol PD mempunya i konstanta Kp = 2, Kp = 2 s, Po = 40 %. Plot tanggapan bagi eror untuk gambar 1.25.
1.13.
Sebuah a lat kontrol PID mempunya i konstanta Kp = 2, Kd = 2 s, Kr = 2,2 ti yang di ber sع , dan P (0) = 40 %. J ika erornya ada lah seper ti ber ikan digambar 1.25, p lot keluaran alat kontrolnya.
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SOAL- SOAL SISTEM PENGATURAN 2
Carilah : (a). penguatan proporsional (b). penguatan integral (c). waktu dimana keluaran alat kontrol akan mencapai 0 % setelah muncul eror sebesar + 1,5 % dan tetap bernilai demikian. Keluaran alat kontrol pada waktu eror muncul adalah 72 %.
Gambar 1.26 untuk Soal 1.16
1.15. Andaikan tindakan laju ditambahkan ditambahkan kea lat kontrol kontrol di soal 1.14 dengan penguatan laju sebesar 0,1, menit. Spesifikasikan penguatan derivative dan tentukan waktu pada mana keluaran keluaran alat kontrol kontrol mencapai 0 % dengan penambahan cara ini. 1.16. Sebuah alat control control PI dipergunakan untuk mengontrol aliran dengan daerah pengukuran 20 gal/menit sampai 100 gal/menit. Setpoint adalah 60 gal/menit. Keluaran alat kontrol mengendalikan sebuah katup dengan tekanan sebesar 3 ± 15 psi untuk 0 sampai 100 % keluaran. Pengaturan alat kontrol adalah langsung, langsung, Kp = 0,9 %/% , Kr = 0,4 % ( % - menit menit ) . Plot tekanan pneumatic pada katup untuk aliran pada gambar 8.26. Tekanan adalah sebesar 10,8 psi pada saat t = 0 .
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SOAL- SOAL SISTEM PENGATURAN 2
Penyelesaian : 1.1. Dik :
Cm = 294 mm / s
Cmaks = 460 mm / s Cmin = 220 mm / s Csp = 327 mm / s it : EP ......?
D
Peny :
Cm - Cap EP = ( -------------------------------- ------- ) . 100 % Cmaks - Cmin 294 - 327 = ( -------------------------------- ------- ) . 100 100 % 460 - 220 EP = - 7,53 %
1.2.
ik : S maks = 1750 rpm
D
Smin = 800 rpm Vmaks = 50 V Vmin = 25 V it : (a). S ......?
D
(b). P ......? Peny : (a). Mencari kemiringan kurva m, dan potongan dengan So dari hubungan linier antar tegangan V dan kecepatan S, di mana :
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SOAL- SOAL SISTEM PENGATURAN 2
Asumsi :
Sp = m V + So engan mengetahui Sp dan V dua posisi yang diketahui, kita tuliskan dua persamaan :
D
800 = 25 m + So 1750 = 50 m So Penyelesaian kedua persamaan secara serentak, akan memperoleh m = 38 rpm/V dan So = -150 rpm. Jadi, pada 310 rpm kita peroleh : S = 38 . (38) + (-150) Sehingga S = 1294 rpm (b). Jika keluaran alat kontrol ini dinyatakan sebagai persentase dari daerah 800 ± 1750 rpm, maka : Sp - Smin P = ----------------------------------------------------- . 100 Smaks ± Smin 1294 - 800 = ---------------------------------- ------------------ . 100 % = 52 % 1750 - 800 1.3.
ik: K sp sp = 12 m
D
K 1 = 1,0 m3/menit K 2 = 2,0 m3/menit H = 1 m it : a). Hitung Periode sikling .....?
D
b). Grafik Tinggi Cairan Versus Waktu ......?
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SOAL- SOAL SISTEM PENGATURAN 2
Peny : a).
Pada H = 1 m, dimana H = daerah netral. Maka, dapat kita menggunakan persamaan di bawah untuk menentukan
besar periode sikling sikli ng dari masing ± mas masing ing alat kontrol : Asumsi :
K 1 (t) = K sp sp (ta) ± 2 (t ± t a), dan K 2 (t) = K sp sp (ta) + 4 (t ± t a) Sehingga pada persamaan ; K 1 (t) = K (t a) ± 2 (t ± t a), maka dapat dihitung ; 1 = 12 (ta ) ± 2 (t ± t a ) 12 (ta) = 1 ta = 0,083 menit
y
-2t + 2ta = 1 = (-2t) + 2(4) = 1 -2t = -7 t = 3,5 menit
dan pada persamaan ; K 2 (t) = K (t a) + 4 (t ± t a) 2
= 12 (ta ) + 4 (t ± t a)
12 ta = 2 ta = 0,17 menit
4t - 4ta = 2,0 = (4t) ± 4 (0,17) = 2,0 4t = 2,68
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SOAL- SOAL SISTEM PENGATURAN 2
b).
Grafik Tinggi Cairan Versus Waktu dapat dilihat di bawah ;
ik : T = 323 K
1.4.
D
Netral =
2%
Penyelesaian : T1 = 323 + 6,46 = 329,46 ( setpoint atas ) T2 = 323 ± 6,46 = 316,54 ( setpoint bawah ) Overshoot Set poin atas
329 328 327 326 325 ) K ( u h u S
324 323 322 321 320 319 318
5
10
15
20
25
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SOAL- SOAL SISTEM PENGATURAN 2
it : a). Periode sikling .....?
D
b). Jika kelambatan 5 menit, Plot ketinggian cairan mulai dari 13 m ......? Peny : a). Pada H = HH ± HL
= 12 ± 11 m = 1 m, dimana H = daerah netral.
Maka, dapat kita menggunakan persamaan di bawah untuk menentukan besar periode sikling sikli ng dari masing ± mas masing ing alat kontrol : K 1 (t) = K (t a) ± 2 (t ± t a), dan K 2 (t) = K (t a) + 4 (t ± t a) Sehingga pada persamaan ; K 1 (t) = K (t a) ± 2 (t ± t a), maka dapat dihitung ; 2,0 t = 0,5 (t(t a) ± 2 (t ± t a ) y
0,5 (ta) = 2,0 ta = 4 menit -2t + 2ta = 2,0 = (-2t) + 2(4) = 2,0 -2t = -6 t = 3 menit
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SOAL- SOAL SISTEM PENGATURAN 2
t = 0 menit. b).
1.6.
Plot ketinggian cairan pada 13 m jika kelambatan 5 menit;
T = 50 ± 130 oC
ik :
D
Setpoint = 73,5 oC PO = 50 % P= 55 % it : a). EP pada Kp = 0,1 .....?
D
b). EP pada Kp = 0,7 .....? c). EP pada Kp = 2,0 .....? d). EP pada Kp = 5,0 .....? Peny : P = K P + EP + PO Sehingga dapat dihitung ; a). EP =
x 100 %
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SOAL- SOAL SISTEM PENGATURAN 2
=
%
= 0,025 %
x 100 % = %
d). EP =
= 0,01 % 1.7.
ik : Setpoint/P O = 73,5 oC
D
Kp = 2 it : a). % Ep pada T/P = 61 oC ......?
D
b). % Ep pada T/P = 122 oC ......? c). % Ep pada T/P = 82 + 5t oC ......? Peny :
x 100 % = %
a). % EP =
= - 6,25 % b). % EP =
x 100 %
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SOAL- SOAL SISTEM PENGATURAN 2
1.8.
ik :
t = 2,2 menit
D
EP = 2,2 % it : a). Kr / KI ......?
D
b). P (t) pada EP = 2,2 % ......? Peny :
a). Kr /KI =
b). P = KI
= 7,58 x 10 -3 s-1
= KI
Sehingga,
P = KI + p (o)
= 7,58 x 10 -3 + p (o) = 7,58 x 10 -3 .(0,022) t + p (o) = 1,67 x 10 -4t + p (o)
1.9.
ik : K = 44s
D
D
Ep = 2,2 sin (0,04t)
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SOAL- SOAL SISTEM PENGATURAN 2
1.10.
ik : K P = 2
D
P(o) = 50 % = 0,5. it : E p .....?
D
Peny : engan melihat grafik error pada gambar 1.25, maka untuk ;
D
E p1 berada pada titik (0,0) dan (2,2) E p2 berada pada titik (2,2) dan (3,0) E p3 berada pada titik (3,0) dan (4,-3) E p4 berada pada titik (4,-3) dan (6,0)
Untuk P1 = Kp.EP1 + P(o) = 2.EP1 + 0,5
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SOAL- SOAL SISTEM PENGATURAN 2
Persamaan garis :
= =
y = 2x + 6 sehingga ; P = 2 (-2t + 6) + 0,5 = -4t + 12 + 0,5 = -4t + 12,5 Untuk t = 2 , maka P = -4 (2) + 12,5 = 4,5 Untuk t = 3 , maka P = -4 (3) + 12,5 = 0,5
Untuk P3 = Kp.EP3 + P(o)
y = -2t + 6
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SOAL- SOAL SISTEM PENGATURAN 2
Untuk P4 = Kp.EP4 + P(o) = 2.EP4 + 0,5 Persamaan garis :
= = y= x-3
sehingga ;
P = 2 ( t - 3) + 0,5 = 3t - 6 + 0,5 = 3t - 6,5 Untuk t = 4 , maka P = 3 (4) - 6,5= 5,5
y= t-3
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SOAL- SOAL SISTEM PENGATURAN 2
K i = 2,2 P(o) = 40 % = 0,4. it : E p .....?
D
Peny : engan melihat grafik error pada gambar 1.25, maka untuk ;
D
E p1 berada pada titik (0,0) dan (2,2) E p2 berada pada titik (2,2) dan (3,0) E p3 berada pada titik (3,0) dan (4,-3) E p4 berada pada titik (4,-3) dan (6,0)
Untuk P1 = Kp.EP1 + K pK i E p1 dt + P(o) = 2.EP1 + 2.2,2 E p1 dt + 0,4
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SOAL- SOAL SISTEM PENGATURAN 2
Untuk P2 = Kp.EP2 + K pK i E p2 dt + P(o) = 2.EP2 +2. 2,2 E p2 dt + P(o) Persamaan garis :
= =
y = 2x + 6 sehingga ; P = 2.(-2t + 6) + 4,4 (-2t + 6) dt + P(o)
= -4t + 12 + 4,4 . - t2 + 6t + 0,4 = -4,4t2 +2t + 12,4
y = -2t + 6
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SOAL- SOAL SISTEM PENGATURAN 2
= -6t + 18 + 4,4. - t2 + 9t + 0,4 = -6,6t2 + 3t + 18,4 Untuk t = 3 , maka P = -6,6.(3) 2 + 3.(3) + 18,4 = -32 Untuk t = 4 , maka P = -6,6 (4) 2 + 3.(4) + 18,4 = -75,2
Untuk P4 = Kp.EP4 + K pK i E p4 dt + P(o) = 2.EP4 + 2. 2,2 E p4 dt + P(o)
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SOAL- SOAL SISTEM PENGATURAN 2
t=2
E p1 = 13,2
t=2
E p2 = 7,6
P2
E p3 = -75,2
t=4
E p4 = 270,4
t=6
E p4 = 566,8
P4 t=3
1.12.
t=4
ik : K P = 2
D
E p2 = 5,2
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SOAL- SOAL SISTEM PENGATURAN 2
y=x
y=t
sehingga ; P = 2.EP1 + 4. = 2 (t) + 4.
+ 0,4
+ 0,4 = 2t + 4,4
Untuk t = 0 , maka P = 2 (0) + 4,4 = 4,4 Untuk t = 2 , maka P = 2 (2) + 4,4 = 12,4
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SOAL- SOAL SISTEM PENGATURAN 2
Persamaan garis :
= =
y = -3x + 9 sehingga ; P = 2.(-3t + 9) + 4.
+ 0,4
y = -3t + 9
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SOAL- SOAL SISTEM PENGATURAN 2
= 3t - 6 + 6 + 0,4 = 3t + 0,4 Untuk t = 4 , maka P = 3.(4) + 0,4 = 12,4 Untuk t = 6 , maka P = 3.(6) + 0,4 = 18,4 Jadi Plot tanggapan errornya adalah sebagai sebaga i berikut : t=0
E p1 = 4,4
P1
t=3
E p3 = -11,6
t=4
E p3 = -17,6
P3 t=2
E p1 = 12,4
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SOAL- SOAL SISTEM PENGATURAN 2
= 2. (t) + 4,4 (t). dt + 4
+ 40
= 2. t + 4,4 ( + 4 (1) + 40
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SOAL- SOAL SISTEM PENGATURAN 2
Jadi = 4 P2 = -5t ± 5,5t2 + 74,8 = -5 (4) ± 5,5 (4) 2 + 74,8
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SOAL- SOAL SISTEM PENGATURAN 2
1.14.
ik : PB = 20,12
D
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SOAL- SOAL SISTEM PENGATURAN 2
b).
P = K p .E p . t + P(o)
0,72 = 6,7 . (0,015) t
P(o) = 0
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SOAL- SOAL SISTEM PENGATURAN 2
Penyelesaian Penyelesaian kedua persamaan persa maan secara seren s erentak, tak, akan memperoleh m = 6,7 6,7 gal/menit / psi dan Qo = -0,5 gal/menit. Jadi, pada Q m = 80 gal /menit dapat diperoleh :
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Trusted by over 1 million members
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Trusted by over 1 million members
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Trusted by over 1 million members
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Trusted by over 1 million members
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Trusted by over 1 million members
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Trusted by over 1 million members
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Trusted by over 1 million members
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Trusted by over 1 million members
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Trusted by over 1 million members
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Trusted by over 1 million members
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Trusted by over 1 million members
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Trusted by over 1 million members
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Trusted by over 1 million members
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Trusted by over 1 million members
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Trusted by over 1 million members
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Trusted by over 1 million members
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