Hasil Perhitungan New.docx

I.

HASIL PERHITUNGAN 1. Dengan Tangki -

Metode 1 Kp =

∆𝑂𝑢𝑡𝑝𝑢𝑡 ∆𝐼𝑛𝑝𝑢𝑡 𝑃𝑉 −𝑃𝑉

1 = %𝑃𝑂2 −%𝑃𝑂 2

=

1

56,2−75,5 55−40

= -1,2867 τ

𝐽𝑎𝑟𝑎𝑘

= 𝐾𝑒𝑐𝑒𝑝𝑎𝑡𝑎𝑛 1,6 𝑐𝑚

= 12 𝑐𝑚/𝑚𝑖𝑛 = 0,1333 menit G(s) = τ

𝐾𝑝

𝑠 +1

𝑦 𝑥

=τ

y

=τ

𝐾𝑝

𝑠 +1

𝐾𝑝

∙𝑥

𝑠 +1

y(s) = τ

𝐾𝑝

∙

𝑠 +1

y(t) = L-1 τ

∆𝑜𝑢𝑡

𝐾𝑝

𝑠 +1

𝑠

∙

∆𝑜𝑢𝑡 𝑠 𝑡

y(t) = ∆𝑜𝑢𝑡 ∙ 𝐾𝑝(1 − 𝑒 −τ ) 𝑡

= (𝑃𝑉2 − 𝑃𝑉1 ) ∙ 𝐾𝑝( 1 − 𝑒 −τ ) = (56,2 − 75,5)(−1,2867)( 1 − 𝑒 = (-19,3) (−1,2867) (0,0895) = -2,2226/ menit

-

Metode 2 C(t) = 0,632 ∙ ∆𝐶𝑠 = 0,632 ∙ 19 = 12,008

τ

𝐽𝑎𝑟𝑎𝑘

= 𝐾𝑒𝑐𝑒𝑝𝑎𝑡𝑎𝑛

−

0,0125 0,1333

)

1,5 𝑐𝑚

= 12 𝑐𝑚/𝑚𝑖𝑛 = 0,125 menit G(s) = τ

𝐾𝑝

𝑠 +1

𝑦 𝑥

=τ

y

=τ

𝐾𝑝

𝑠 +1

𝐾𝑝

∙𝑥

𝑠 +1

y(s) = τ

𝐾𝑝

∙

𝑠 +1

y(t) = L-1 τ

∆𝑜𝑢𝑡

𝐾𝑝

𝑠 +1

𝑠

∙

∆𝑜𝑢𝑡 𝑠 𝑡

y(t) = ∆𝑜𝑢𝑡 ∙ 𝐾𝑝 (1 − 𝑒 −τ ) 𝑡

= (𝑃𝑉2 − 𝑃𝑉1 ) ∙ 𝐾𝑝( 1 − 𝑒 −τ ) = (75,5 − 56,2)(−1,2867)( 1 − 𝑒 = (-19,3) (−1,2867) (0,0952) = 2,3641/ menit

-

Metode 3 t2

= 0,632 ∙ ∆𝐶𝑠 = 0,632 ∙ 19 = 12,008

t1

= 0,283 ∙ ∆𝐶𝑠 = 0,283 ∙ 19 = 5,377

τ2

𝐽𝑎𝑟𝑎𝑘

= 𝐾𝑒𝑐𝑒𝑝𝑎𝑡𝑎𝑛 =

1,5 12

= 0,125 τ1

𝐽𝑎𝑟𝑎𝑘

= 𝐾𝑒𝑐𝑒𝑝𝑎𝑡𝑎𝑛 =

0,5 12

= 0,0416 τ

3

= 2 (τ2 − τ1 )

−

0,0125 0,125

)

3

= 2 (0,125 − 0,0416) 3

= 2 (0,0834) = 0,1251 menit G(s) = τ

𝐾𝑝

𝑠 +1

𝑦 𝑥

=τ

y

=τ

𝐾𝑝

𝑠 +1

𝐾𝑝

∙𝑥

𝑠 +1

y(s) = τ

𝐾𝑝

∙

𝑠 +1

y(t) = L-1 τ

∆𝑜𝑢𝑡 𝑠

𝐾𝑝

𝑠 +1

∙

∆𝑜𝑢𝑡 𝑠 𝑡

y(t) = ∆𝑜𝑢𝑡 ∙ 𝐾𝑝 (1 − 𝑒 −τ ) 𝑡

= (𝑃𝑉2 − 𝑃𝑉1 ) ∙ 𝐾𝑝( 1 − 𝑒 −τ ) = (75,5 − 56,2)(−1,2867)( 1 − 𝑒 = (-19,3) (−1,2867) (0,0951) = 2,3616/ menit

2. Tanpa Tangki -

Metode 1 Kp =

∆𝑂𝑢𝑡𝑝𝑢𝑡 ∆𝐼𝑛𝑝𝑢𝑡 𝑃𝑉 −𝑃𝑉

1 = %𝑃𝑂2 −%𝑃𝑂 2

=

1

56,7−76 55−40

= -1,2867 τ

𝐽𝑎𝑟𝑎𝑘

= 𝐾𝑒𝑐𝑒𝑝𝑎𝑡𝑎𝑛 0,35 𝑐𝑚

= 60 𝑐𝑚/𝑚𝑖𝑛 = 0,00583 menit

−

0,0125 0,1251

)

G(s) = τ

𝐾𝑝

𝑠 +1

𝑦 𝑥

=τ

y

=τ

𝐾𝑝

𝑠 +1

𝐾𝑝

∙𝑥

𝑠 +1

y(s) = τ

𝐾𝑝

∙

𝑠 +1

y(t) = L-1 τ

∆𝑜𝑢𝑡 𝑠

𝐾𝑝

∙

𝑠 +1

∆𝑜𝑢𝑡 𝑠 𝑡

y(t) = ∆𝑜𝑢𝑡 ∙ 𝐾𝑝(1 − 𝑒 −τ ) 𝑡

= (𝑃𝑉2 − 𝑃𝑉1 ) ∙ 𝐾𝑝( 1 − 𝑒 −τ ) = (76 − 56,7)(−1,2867)( 1 − 𝑒 = (-19,3) (-1,2867) (0,3487) = 8,6594/ menit

-

Metode 2 C(t) = 0,632 ∙ ∆𝐶𝑠 = 0,632 ∙ 19,5 = 12,324 τ

𝐽𝑎𝑟𝑎𝑘

= 𝐾𝑒𝑐𝑒𝑝𝑎𝑡𝑎𝑛 0,15 𝑐𝑚

= 60 𝑐𝑚/𝑚𝑖𝑛 = 0,0025 menit G(s) = τ

𝐾𝑝

𝑠 +1

𝑦 𝑥

=τ

y

=τ

𝐾𝑝

𝑠 +1

𝐾𝑝

𝑠 +1

y(s) = τ

𝐾𝑝

𝑠 +1

∙𝑥 ∙

∆𝑜𝑢𝑡 𝑠

−

0,0025 0,00583

)

y(t) = L-1 τ

𝐾𝑝

𝑠 +1

∙

∆𝑜𝑢𝑡 𝑠 𝑡

y(t) = ∆𝑜𝑢𝑡 ∙ 𝐾𝑝(1 − 𝑒 −τ ) 𝑡

= (𝑃𝑉2 − 𝑃𝑉1 ) ∙ 𝐾𝑝( 1 − 𝑒 −τ ) = (76 − 56,7)(−1,2867)( 1 − 𝑒 = (-19,3) (-1,2867) (0,6321) = 15,6971/ menit

-

Metode 3 t2

= 0,632 ∙ ∆𝐶𝑠 = 0,632 ∙ 19,5 = 12,324

t1

= 0,283 ∙ ∆𝐶𝑠 = 0,283 ∙ 19,5 = 5,5185

τ2

𝐽𝑎𝑟𝑎𝑘

= 𝐾𝑒𝑐𝑒𝑝𝑎𝑡𝑎𝑛 =

0,15 60

= 0,0025 τ1

𝐽𝑎𝑟𝑎𝑘

= 𝐾𝑒𝑐𝑒𝑝𝑎𝑡𝑎𝑛 =

0,05 60

= 0,00083 τ

3

= (τ2 − τ1 ) 2 3

= 2 (0,0025 − 0,00083) 3

= 2 (0,00167) = 0,002502 menit

−

0,0025 0,0025

)

G(s) = τ

𝐾𝑝

𝑠 +1

𝑦 𝑥

=τ

y

=τ

𝐾𝑝

𝑠 +1

𝐾𝑝

∙𝑥

𝑠 +1

y(s) = τ

𝐾𝑝

∙

𝑠 +1

y(t) = L-1 τ

∆𝑜𝑢𝑡

𝐾𝑝

𝑠 +1

𝑠

∙

∆𝑜𝑢𝑡 𝑠 𝑡

y(t) = ∆𝑜𝑢𝑡 ∙ 𝐾𝑝 (1 − 𝑒 −τ ) 𝑡

= (𝑃𝑉2 − 𝑃𝑉1 ) ∙ 𝐾𝑝( 1 − 𝑒 −τ ) = (76 − 56,7)(−1,2867)( 1 − 𝑒 = (-19,3) (-1,2867) (0,6381) = 15,8461/ menit

−

0,0025 0,002502

)