Hydraulic Circuit Analysis

Problem 5.37M:

For the sistem of figure 5.33, if P1=7 bar, solve for P2. The pipe is 15 m long, has a 38 mm ID

Solution:

Rumus: V = Q/A

UNITS METRIC

Input Output

7 1.76

15 670

0.04 f 0.096

0 19.6

throughout, and lies in a horizontal plane.Q = 0.002 m/s of oil (sg = 0.9 and v = 0.0001m/s).

NR = VD / V

f = 64 / NR

Le = 2 (KD / f)elbow+(KD / f)valve+Lpipe

HL = f x (Le/D) x (V/2g)

= 1000 x sg x g

P (N/m) = x HL

P2 (bar) = P - P1

P1 (bar) V (m/s)

Lpipe (m) NR

Dpipe (m)

Q (m/s) Le

0.9 7.80

0 8829

3.14 68841

10 0.69

0.75 6.31

sg HL

v (cs)

P (N/m)

K valve P (bar)

K elbow P2 (bar)

Hydraulic Circuit Analysis

Problem 5.38E:

while moving in the extending direction.Take frictional pressure losses into account. The pump

produces a pressure increase of 1000 psi from the inlet port to the discharge port and a flow rate of

40 gpm.

Pipe No. Length ( ft ) Pipe No. Length ( ft )

1 2 1.5 8 5 1.0

2 6 1.5 9 5 0.75

3 2 1.5 10 5 0.75

4 50 1.0 11 60 0.75

5 10 1.0 12 10 0.75

6 5 1.0 13 20 0.75

7 5 1.0

Fig. 5.34Solution:

Rumus:

V =0.408 Q D

x h F = P x A

Input Output

0 7.25 906.67

50 16.3 1360

8 12.24 1020

4 29.01 1813

1000 21.8 1360

40 0.0706

0.75 0.0471

Elbow 90 0.0627

3.14 0.0353

0.0471

5.84 2.03

195 67.8

19.26 6.69

83.6 29.0

509 177

45303 30

6919 38383

For the fluid power system in Fig. 5.34, determine the external load F that the hydraulic can sustain

Dia (in) Dia (in)

HL = (f .L / D) v/2g

Q return line(gpm) = Q( dpiston- drod ) / drpiston

NR = V x D f = 64 / NR

P =

(ft/s) V (ft/s) 1,2,3 NR 1,2,3 (lb/ft) V (ft/s) 4,5,6 NR 1,2,3

dpiston(in) V (ft/s) 7,8 NR 7,8

drod(in) V (ft/s) 9,10 NR 9,10

P (psi) V (ft/s) 11,12,13 NR 11,12,13

Q (gpm) f 1,2,3

K factor f 4,5,6

f 7,8

f 9,10

f 11,12,13

HL(ft)1, 2,3 P (psi) 1,2,3

HL(ft) 4,5,6 P (psi) 4,5,6

HL (ft) 7,8 P (psi) 7,8

HL (ft) 9,10 P (psi) 9,10

HL(ft) 11,12,13 P (psi) 11,12,13

F(lb)ext. Q return line(gpm)

F(lb)retr. F (lb)

Hydraulic Circuit Analysis

Problem 5.39E:For the sistem of problem 5.38, determine the heat-generation rate due to frictional pressure losses.

Solution :

Rumus:

HPloss = P (psi) x Q (gpm)1714

HPloss = P(psi)1, 2,3 + P(psi) 4,5,6 + P(psi) 9,10 x Q (gpm) +1714 1714

Sinse 1 HP = 42.4 BTU/min

Heat generation rate =42.4 x HPloss BTU/min or BTU/hr

UNITS METRIC

Input Output

P (psi) 1,2,3 2.03 HPloss 5.57

68 Q gener.(btu/min) 236

6.69 Q gener.(btu/hr) ###

29.0

P (psi) 11,12,13 177

Q (gpm) 40

Qreturn (gpm) 30

P(psi) 7,8 + P(psi) 11,12,13, x Q (gpm)

P (psi) 4,5,6

P (psi) 7,8

P (psi) 9,10

5.37M5.38E5.39E