perhitungan tekanan fluida 1

VII. PERHITUNGAN

A. Secara Digital

1. PIPA P1-P2a) Laju Alir Volume 500 (L/h) Kehilangan Tekanan

∆P = 2,67 mbarKonversi Satuan :

2,67 mbar x 1 ¿̄1000mbar

x 1,0 x105 pa

1 ¿̄ ¿¿= 267 Pa = 267

kgms2

Volume Aliran / DebitQ = 500 L/h

500L/h X1dm3

1 Lx 1m3

103dm3x 1h3600 s = 1,388 x 10-4 m3/s

Kecepatan Q = 1,388 x 10-4 m3/s ; A = 5,64 x 10-4 m2

Sehingga, V = QA =1,388 x 10−4m

3/s5,64 x10−4m2

= 0,2460 m/s

Bilangan Reynolds

Re =V dv =

0 ,2460m / s .0,0268m

1 x10−6 m2

s= 6592,8

Koefisien Kehilangan Tekanan ( λ )

λ = ∆ PΠ2d5

8 ρQ2L

λ=267 kg

ms2.(227 )

2

.1,3825 x10−8m5

8.999 kgm3.(1,388 x10−4 m3

s)2

.1,5m

λ= 0,157

Kehilangan Tekanan Teoritis

∆P = ½ λ ρV2

d L

= 12 . 0,157 . 999

kgm3

. 0,060516m

2

s2

0,0268m . 1,50 m= 265,62 kg / ms2

Kehilangan tekanan Langsung

J = ∆ PL=265,62 Pa1,50m = 177,08 Pa/m

% kesalahan= |∆ P teori−∆ P prak∆ P teori | x 100

= |265,62−266265,62 | x 100 = 0,14

Friksif = 6 π µ r v

f = 6 . (227 ) . 9.99 x 10−4 kg

ms.8.65 x 10−3 . 0.2460

ms

= 4,008 x 10−5 kg.m/ s2

b) Laju Alir Volume 1000 (L/h) Kehilangan Tekanan


4,67 mbar x 1 ¿̄1000mbar

x 1,0 x105 pa

1 ¿̄ ¿¿= 467 Pa = 467

kgms2


1000L/h X1dm3

1 Lx 1m3

103dm3x 1h3600 s = 2,78 x 10-4 m3/s

Kecepatan

V = QA =2,78 x10−4m

3/ s5,64 x 10−4m2

= 0,4929 m/s

Bilangan Reynolds

Re =V dv

= 0 ,4929m /s .0,0268m

1 x10−6m2

s= 13209,72


λ = ∆ PΠ2d5

8 ρQ2L

λ=467 kg

ms2.( 227 )

2

.1,3825 x10−8m5

8.999 kgm3.(2,78 x10−4 m3

s)2

.1,5m

λ= 0,0688


∆P = ½ λ ρV2

d L

= 12 . 0,0688 . 999

kgm3

. 0,24295041m

2

s2

0,0268m . 1,50 m

= 467,3 kg / ms2


J = ∆ PL =467 Pa1,50m = 311,33 Pa/m

% kesalahan =|∆ P teori−∆ P prak∆ P teori | x 100

= |467,3−467467,3 | x 100 = 0,064


f = 6 . (227 ) . 9.99 x 10−4 kg

ms.8.65 x 10−3 . 0.4929

ms

= 8,031 x 10−5 kg.m/ s2

c) Laju Alir Volume 1500 (L/h)

Kehilangan Tekanan ∆P = 7,33 mbar

Konversi Satuan :

7,33 mbar x 1 ¿̄1000mbar

x 1,0 x105 pa

1 ¿̄ ¿¿= 733 Pa = 733

kgms2


1500L/h X1dm3

1 Lx 1m3

103dm3x 1h3600 s = 4,167 x 10-4 m3/s

Kecepatan

V = QA =4,167 x10−4m

3 /s5,64 x10−4m2

= 0,7388 m/s

Bilangan Reynolds

Re =V dv

= 0 ,7388m / s .0,0268m

1x 10−6 m2

s= 19799,84


λ = ∆ PΠ2d5

8 ρQ2L

λ=733 kg

ms2.(227 )

2

.1,3825 x10−8m5

8.999 kgm3.(4,167 x10−4m3

s)2

.1,5m

λ= 0,0480


∆P = ½ λ ρV2

d L

= 12 . 0,0480 . 999

kgm3

. 0,54582544 m

2

s2

0,0268m . 1,50 m

= 732,46 kg / ms2


J = ∆ PL =732,46Pa1,50m = 488,30 Pa/m


= |732,46−733732,46 | x 100 = 0,07


f = 6 . (227 ) . 9.99 x 10−4 kg

ms.8.65 x 10−3 . 0.7388

ms

= 1,20 x 10−4 kg.m/ s2


∆P = 12 mbarKonversi Satuan :

12 mbar x 1 ¿̄1000mbar

x 1,0 x105 pa

1 ¿̄ ¿¿= 1200 Pa = 1200

kgms2


500L/h X1dm3

1 Lx 1m3

103dm3x 1h3600 s = 1,388 x 10-4 m3/s

Kecepatan

V = QA = 1,388 x 10−4m

3/s2,35 x 10−4m2

= 0,5906 m/s

Bilangan Reynolds

Re =V dv

= 0 ,5906m/ s .0,0173m

1 x10−6 m2

s= 10217,38


λ = ∆ PΠ2d5

8 ρQ2L

λ=1200 kg

ms2.(227 )

2

.1,5496 x10−9m5

8.999 kgm3.(1,388 x10−4 m3

s)2

.1,5m= 0,0795


∆P = ½ λ ρV2

d L

= 12 . 0,0795 . 999

kgm3

. 0,3488m

2

s2

0,0173m . 1,50 m

= 1200,94 kg / ms2


J = ∆ PL =1200,94 Pa1,50m = 800,62 Pa/m


= |1200,94−12001200,94 | x 100 = 0,078


f = 6 . (227 ) . 9.99 x 10−4 kg

ms.8.65 x 10−3 . 0,5906

ms

= 9,60 x 10−5 kg.m/ s2



26,67 mbar x 1 ¿̄1000mbar

x 1,0 x105 pa

1 ¿̄ ¿¿= 2667 Pa = 2667

kgms2

Volume Aliran / Debit

1000L/h X1dm3

1 Lx 1m3

103dm3x 1h3600 s = 2,78 x 10-4 m3/s

Kecepatan

V = QA =2,78 x10−4m

3/ s2,35 x10−4m2

= 1,18 m/s

Bilangan Reynolds

Re =V dv

= 1,18m/ s .0,0173m

1 x10−6 m2

s= 20414


λ = ∆ PΠ2d5

8 ρQ2L

λ=2667 kg

ms2.( 227 )

2

.1,5496 x 10−9m5

8.999 kgm3.(2,78 x10−4m3

s)2

.1,5m= 0,044


∆P = ½ λ ρV2

d L

= 12 . 0,044 . 999

kgm3

. 1,3924m

2

s2

0,0173m . 1,50 m

= 2653,36 kg / ms2


J = ∆ PL =2653,36Pa1,50m = 1768,90 Pa/m


= |2653,36−26672653,36 | x 100 = 0,51


f = 6 . (227 ) . 9.99 x 10−4 kg

ms.8.65 x 10−3 . 1,18

ms

= 1,92 x 10−4 kg.m/ s2

c) Laju Alir Volume 1500 (L/h) Kehilangan Tekanan


49,33 mbar x 1 ¿̄1000mbar

x 1,0 x105 pa

1 ¿̄ ¿¿= 4933 Pa = 4933

kgms2


1500L/h X1dm3

1 Lx 1m3

103dm3x 1h3600 s = 4,167 x 10-4 m3/s

Kecepatan

V = QA =4,167 x10−4m

3 /s2,35 x10−4m2

= 1,77 m/s

Bilangan Reynolds

Re = V dv

= 1,77m /s .0,0173m

1 x10−6m2

s= 30621


λ = ∆ PΠ2d5

8 ρQ2L

λ=4933 kg

ms2.( 227 )

2

.1,5496 x 10−9m5

8.999 kgm3.(4,167 x10−4m3

s)2

.1,5m= 0,036


∆P = ½ λ ρV2

d L

= 12 . 0,036 . 999

kgm3

. 3,1329m

2

s2

0,0173m . 1,50 m = 4884,60 kg / ms2


J = ∆ PL = 4884,60 Pa1,50m = 3256,4 Pa/m


= |4884,60−493334884,60 | x 100 = 0,99


f = 6 . (227 ) . 9.99 x 10−4 kg

ms.8.65 x 10−3 . 1,77

ms

= 2,88 x 10−4 kg.m/ s2



14,33 mbar x 1 ¿̄1000mbar

x 1,0 x105 pa

1 ¿̄ ¿¿= 1433 Pa = 1433

kgms2


500L/h X1dm3

1 Lx 1m3

103dm3x 1h3600 s = 1,388 x 10-4 m3/s

Kecepatan

V = QA =1,388x 10−4m

3/s2,35x 10−4m2

= 0,5906 m/s

Bilangan Reynolds

Re =V dv =

0 ,5906m/ s .0,0173m

1 x10−6 m2

s= 10217,38


λ = ∆ PΠ2d5

8 ρQ2L

λ=1433 kg

ms2.(227 )

2

.1,5496 x10−9m5

8.999 kgm3.(1,388 x10−4 m3

s)2

.1,5m

λ= 0,0949


∆P = ½ λ ρV2

d L

= 12 . 0,949 . 999

kgm3

. 0,3488m

2

s2

0,0173m . 1,50 m = 1433,58 kg / ms2


J = ∆ PL =1433,58Pa1,50m = 955,72 Pa/m

% kesalahan =|∆ P teori−∆ P prak∆ P teori | x 100

= |1433,58−14331433,58 | x 100 = 0,04


f = 6 . (227 ) . 9.99 x 10−4 kg

ms.8.65 x 10−3 . 0,5906

ms

= 9,60 x 10−5 kg.m/ s2




x 1,0 x105 pa

1 ¿̄ ¿¿= 3100 Pa = 3100

kgms2


1000L/h X1dm3

1 Lx 1m3

103dm3x 1h3600 s = 2,78 x 10-4 m3/s

Kecepatan

V = QA =2,78 x10−4m

3/ s2,35 x10−4m2

= 1,18 m/s

Bilangan Reynolds

Re =V dv =

1,18m/ s .0,0173m

1 x10−6 m2

s= 20414


λ = ∆ PΠ2d5

8 ρQ2L

λ=3100 kg

ms2.( 227 )

2

.1,5496 x 10−9m5

8.999 kgm3.(2,78 x 10−4m

3

s)2

.1,5m= 0,0512


∆P = ½ λ ρV2

d L

= 12 . 0,0512 . 999

kgm3

. 1,3924m

2

s2

0,0173m . 1,50 m = 3087,55 kg / ms2


J = ∆ PL

=3087,55Pa1,50m = 2058,37 Pa/m

% kesalahan= |∆ P te ori−∆ P prak∆P teori | x 100

= |3087,55−31003087,55 | x 100 = 0,40


f = 6 . (227 ) . 9.99 x 10−4 kg

ms.8.65 x 10−3 . 1,18

ms

= 1,92 x 10−4 kg.m/ s2

c) Laju Alir Volume 1500 (L/h) Kehilangan Tekanan



x 1,0 x105 pa

1 ¿̄ ¿¿= 5600 Pa = 5600

kgms2


Q = 1500L/h X 1dm3

1 Lx 1m3

103dm3x 1h3600 s = 4,167 x 10-4 m3/s

Kecepatan

V = QA =4,167 x10−4m

3 /s2,35 x10−4m2

= 1,77 m/s

Bilangan Reynolds

Re =V dv =

1,77m /s .0,0173m

1 x10−6m2

s = 30621


λ = ∆ PΠ2d5

8 ρQ2L

λ=5600 kg

ms2.( 227 )

2

.1,5496 x10−9m5

8.999 kgm3.(4,167 x10−4m3

s)2

.1,5m= 0,041


∆P = ½ λ ρV2

d L

= 12 . 0,041 . 999

kgm3

. 3,1329m

2

s2

0,0173m . 1,50 m = 5563,02 kg / ms2


J = ∆ PL =5563,02Pa1,50m = 3708,68 Pa/m

% kesalahan = |∆ P teori−∆ P prak∆ P teori | x 100

= |5563,02−56005563,02 | x 100 = 0,66


f = 6 . (227 ) . 9.99 x 10−4 kg

ms.8.65 x 10−3 . 1,77

ms

= 2,88 x 10−4 kg.m/ s2

B. Secara Manual


∆P = (26,7 – 14,5) cm H2O= 12,2 cm H2O Konversi Satuan

12,2 cm H2O x10mmH 2O1cmH 2O

x 1mmHg13,6 cmH 2O

x1 ¿̄760mmHg

x 1000mbar1 ¿̄ ¿

¿ = 11,8034

mbar

11,8034 mbar x 1 ¿̄1000mbar

x 1,0 x105 pa

1 ¿̄ ¿¿= 1180,34 Pa = 1180,34

kgms2


Q = 500L/h X 1dm3

1 Lx 1m3

103dm3x 1h3600 s = 1,388 x 10-4 m3/s

Kecepatan

V = QA =1,388x 10−4m

3/s2,35x 10−4m2

= 0,5906 m/s

Bilangan Reynolds

Re =V dv

= 0 ,5906m/ s .0,0173m

1 x10−6 m2

s= 10217,38


λ = ∆ PΠ2d5

8 ρQ2L

λ=1180,34 kg

ms2.( 227 )

2

.1,5496 x10−9m5

8.999 kgm3. (1,388 x 10−4 m

3

s)2

.1,5m= 0,0782


∆P = ½ λ ρV2

d L

= 12 . 0,0782 . 999

kgm3

. 0,3488m

2

s2

0,0173m . 1,50 m

= 1181,30 kg / ms2


J = ∆ PL =1181,30Pa1,50m = 787,53 Pa/m


= |1181,30−1180,341181,30 | x 100 = 0,08


f = 6 . (227 ) . 9.99 x 10−4 kg

ms.8.65 x 10−3 . 0,5906

ms

= 9,60 x 10−5 kg.m/ s2


∆P = (26,7 – 14,5) cm H2O= 25,8 cm H2O

Konversi Satuan

25,8 cm H2O x 10mmH 2O1cmH 2O

x 1mmHg13,6cmH 2O

x1 ¿̄760mmHg

x 1000mbar1 ¿̄ ¿

¿ = 24,9613

mbar

24,9613 mbar x 1 ¿̄1000mbar

x 1,0 x105 pa

1 ¿̄ ¿¿= 1180,34 Pa = 2496,13

kgms2


Q = 1000L/h X 1dm3

1 Lx 1m3

103dm3x 1h3600 s = 2,78 x 10-4 m3/s

Kecepatan

V = QA =2,78 x10−4m

3/ s2,35 x10−4m2

= 1,18 m/s

Bilangan Reynolds

Re =V dv

= 1,18m/ s .0,0173m

1 x10−6 m2

s= 20414


λ = ∆ PΠ2d5

8 ρQ2L

λ= 2496,13 kg

ms2.(227 )

2

.1,5496 x 10−9m5

8.999 kgm3.(2,78 x10− 4m

3

s)2

.1,5m= 0,0412


∆P = ½ λ ρV2

d L

= 12 . 0,0412 . 999

kgm3

. 1,3924m

2

s2

0,0173m . 1,50 m = 2484,516 kg / ms2


J = ∆ PL =2484,516Pa1,50m = 1656,344 Pa/m

% kesalahan=|2484,516−2496,132484,516 | x 100 = 0,47


f = 6 . (227 ) . 9.99 x 10−4 kg

ms.8.65 x 10−3 . 1,18

ms

= 1,92 x 10−4 kg.m/ s2

c) Laju Alir Volume 1500 (L/h) Kehilangan Tekanan ∆P = (57,7 – 9) cm H2O= 48,7 cm H2O

48,7 cm H2O x 10mmH 2O1cmH 2O

x 1mmHg13,6 cmH 2O

x1 ¿̄760mmHg

x 1000mbar1 ¿̄ ¿

¿ = 47,11

mbar=4711 Pa


1500L/h X1dm3

1 Lx 1m3

103dm3x 1h3600 s = 4,167 x 10-4 m3/s

Kecepatan

V = QA = 4,167 x10−4m

3 /s2,35 x10−4m2

= 1,77 m/s

Bilangan Reynolds

Re =V dv

= 1,77m /s .0,0173m

1 x10−6m2

s= 30621


λ = ∆ PΠ2d5

8 ρQ2L

λ=4711 kg

ms2.( 227 )

2

.1,5496 x10−9m5

8.999 kgm3.(4,167 x10−4m3

s)2

.1,5m= 0,0346


∆P = ½ λ ρV2

d L

= 12 . 0,0346 . 999

kgm3

. 3,1329m

2

s2

0,0173m . 1,50 m = 4694,65 kg / ms2


J = ∆ PL =4694,65 Pa1,50m = 3129,76 Pa/m


= |4694,65−47114694,65 | x 100 = 0,34


f = 6 . (227 ) . 9.99 x 10−4 kg

ms.8.65 x 10−3 . 1,77

ms

= 2,88 x 10−4 kg.m/ s2

2. PIPA P19-P20 a) Laju Alir Volume 500 (L/h)

Kehilangan Tekanan ∆P = 12cm H2O

12cm H2Ox10mmH 2O1cmH 2O

x 1mmHg13,6 cmH 2O

x1 ¿̄760mmHg

x 1000mbar1 ¿̄ ¿

¿=11,6099

mbar=1160,99 Pa Volume Aliran / Debit

Q = 500L/h X 1dm3

1 Lx 1m3

103dm3x 1h3600 s = 1,388 x 10-4 m3/s

Kecepatan

V = QA =1,388x 10−4m

3/s2,35x 10−4m2

= 0,5906 m/s

Bilangan Reynolds

Re =V dv

= 0 ,5906m/ s .0,0173m

1 x10−6 m2

s= 10217,38


λ = ∆ PΠ2d5

8 ρQ2L

λ=1160,99 kg

ms2.( 227 )

2

.1,5496 x 10−9m5

8.999 kgm3.(1,388 x10−4 m3

s)2

.1,5m

= 0,0769


∆P = ½ λ ρV2

d L

= 12 . 0,0769. 999

kgm3

. 0,3488m

2

s2

0,0173m . 1,50 m

= 1161,67 kg / ms2


J = ∆ PL =1161,67Pa1,50m = 774,446 Pa/m


= |1161,67 –1160,991161,67 | x 100 = 0,06


f = 6 . (227 ) . 9.99 x 10−4 kg

ms.8.65 x 10−3 . 0,5906

ms

= 9,60 x 10−5 kg.m/ s2


∆P = 30,2 cm H2O

Konversi Satuan

30,2 cm H2Ox 10mmH 2O1cmH 2O

x 1mmHg13,6 cmH 2O

x1 ¿̄760mmHg

x 1000mbar1 ¿̄ ¿

¿

= 29,2182 mbar = 2921,82 Pa


Q = 1000L/h X 1dm3

1 Lx 1m3

103dm3x 1h3600 s = 2,78 x 10-4 m3/s

Kecepatan

V = QA =2,78 x10−4m

3/ s2,35 x10−4m2

= 1,18 m/s

Bilangan Reynolds

Re =V dv

= 1,18m/ s .0,0173m

1 x10−6 m2

s= 20414


λ = ∆ PΠ2d5

8 ρQ2L

λ=2921,82 kg

ms2.( 227 )

2

.1,5496 x10−9m5

8.999 kgm3. (2,78x 10−4 m

3

s)2

.1,5m= 0,0482


∆P = ½ λ ρV2

d L

= 12 . 0,0482 . 999

kgm3

. 1,3924m

2

s2

0,0173m . 1,50 m = 2906,64 kg / ms2


J = ∆ PL

=2906,64 Pa1,50m = 1937, 76 Pa/m

% kesalahan=|2906,64−2921,822906,64 | x 100 = 0,52


f = 6 . (227 ) . 9.99 x 10−4 kg

ms.8.65 x 10−3 . 1,18

ms

= 1,92 x 10−4 kg.m/ s2

perhitungan tekanan fluida 1

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