Transcript Slide 1

Exercise Problem No. 5
Figure below shows a diagram of fluid power system for a hydraulic press used to
extrude rubber patrs. The following data are known :
1. The fluid is oil (sg =0,93), Volume flow rate is 175 gal/min
2. Power input to the pump is 28.4 hp, Pump efficiency is 80 %
3. Energy loss from point 1 to 2 is 2.80 lb-ft/lb, from point 3 to 4 is 28.50 lb-ft/lb and
from point 5 to 6 is 3.50 lb-ft/lb
a). Compute the power removed from the fluid by the press
b). Compute the pressure at point 3 at the pump outlet
c). Compute the pressure at point 5 at the press outlet
Answer :
a). 21.16 hp
b). 219,1 psig
c). 1.01 psig
Jawab :
m
 0,3048m
3,281ft
m
h L  h L1 2  h L 3 4  h L 56  (2,80  28,50  3,50)  34,8 ft
 10,61m
3,281ft
N
  g  0,93(1000)(9,81)  9123,3 3
m
p6
V62
V62
p1
V12
 z1 
 hA  hR  hL 
 z6 
 h R  h A  h L  z6 

2g

2g
2g
p1  Po
p 6  Po
z1  0 V1  0 z 6  1 ft
m3
gal
m3
1
s
Q  175
 0,011 D 6  2 in  A 6  3,090x103 m 2
min 15850gal
s
2
745,7 W
P
16942,3
PA  0,8(28,4 )  22,72 hp
 16942,3W  h A Q  h A  A 
 168,82 m
hp
Q 9123,3(0,011)
V62
Q
0,011
m
5,562
V6  
 5,56


 1,576m
3
A 3,090x10
s
2g 2(9,81)
h R  168,82  10,61 0,3048 1,576  156,329m PR  h R Q  9123,3(156,329)(0,011)  15668,6 W
PA  15668,6 W
hp
 21,012hp
745,7 W
b).
h L1 2
m
m
 2,80 ft
 0,853m z 3  4 ft
 1,219m
3,281
3,281
V32 V62

 1,576m
2g 2 g
p3
V32
p1
V12
 z1 
 h A  h R  h L   z3 

2g

2g
p 3  p1
V32
 h A  h L1 2  z 3 
 168,82  1,219 0,853 1,576  165,172m

2g
psig
p 3  p1  165,172(9123,3)  1506913Pa
 218,6 psig
6895P a
c).
h L56  3,50 ft
m
m
 1,067m z 5  2 ft
 0,61m z 6  0,3048 V5  V6
3,281
3,281
p5
V52
p6
V62
 z5 
 h A  h R  h L   z6 

2g

2g
p5  p6
 z 6  h L56  z 5  0,3048 1,067  0,61  0,762m

psig
p 5  p 6  0,762(9123,3)  6950Pa
 1,01psig
6895P a
p6  po
Soal Kuis 2
Gambar di bawah ini adalah suatu bagian dari sistem pemadam kebakaran dimana
pompa menarik air pada 50oFdengan debit 1500 gal/min dari suatu reservoir dan
mengeluarkannya ke titik B. Kehilangan energi antara reservoir dan titik A adalah 0,65
lblb.ft/lb.
a). Tentukan kedalaman h untuk mempertahankan tekanan paling sedikit 5 psig pada
titik A. [12,8 ft]
b). Bila titik A tekanannya 5 psig tentukan daya pompa yang diperlukan agar tekanan di
B adalah 85 psig. Kehilangan energi antara pompa dan titik B adalah 28 lb.ft/lb
[90 hp]
N
3
lb
N
m
  62,4 3
 9803 3
lb
ft
m
3
ft
m3
gal
m3
s
Q  1500
0,0946
min 15850 gal
s
min
D  10in  A A  5,090x10 2
157,1
1
2
D  8 in  A B  3,226x10 2
VA 
Q
0,0946
m


1
,
859
A A 5,090x10 2
s
m
 0,198m
3,281
6895P a
p A  p o  5psig
 34475P a
psig
h L1  0,65ft
p1
V12
pA
VA2
 z1 
 h L1 
 zA 

2g

2g
p1  p o
V1  0 z A  0 z1  h
pA  po
VA2
h
 h L1 

2g
34475
1,8592
3,281ft

 0,198
 3,517  0,198 0,176  3,891m
 12,8 ft
9803
2(9,81)
m
N
  9803 3
m
m3
Q  0,0946
s
m
 8,534m
3,281
6895P a
p A  p o  5psig
 34,475x103 P a
psig
h L 2  28ft
p A  34,475x103  101,3x103  1,358x105 P a
m
 7,62m
3,281ft
6895P a
p B  p 0  85psig
 586,075x103 P a
psig
z B  25ft
p B  586,075x103  101,3x103  687,375x103 P a
p2
V22
pB
VB2
 z2 
 h L2 
 zB 

2g

2g
z 2  0 V2  VB
p 2  p B   (h L 2  z B )  9803(8,534 7,62)  1,583x105 P a
p 2  1,583x105  687,375x103  8,45x105 P a
p 2  p A 8,45x105  1,358x105
 hA 

 72,345m

9803
hp
PA  h A Q  9803(72,345)(0,0946)  6,709x104 W
 90hp
745,7 W
pA
p
 hA  2


Exercise Problem No. 6
Professor Crocker is building a cabin on a hillside and has proposed the water system
shown in figure below. The distribution tank in the cabin maintain a pressure of 30 psig
above the water. There is an energy loss of 15.5 lb-ft/lb in the piping. The pump is
delivering 40 gal/min of water.
If the pump has an efficiency of 72%, what size motor is required to drive the pump ?
Answer : 42,8 hp