case study 1 - hydraulic jack analysis

The hydraulic jack has an input area of 10 mm2 and an output area of 40 mm2, and is used to move the output piston against a constant 2kN load with different output speed values. The input cylinder is connected to the output cylinder with a 4 mm2 pipe of length L. The fluid used in the jack is water at 20 °C (Density = 998 kg/m3 and viscosity is 1 x 10-3 Pa.s)

Plot the force F needed on the input piston as a function of the steady output piston speed. Generate three different curves on the same plot corresponding to (i) L = 5 m, (ii) L = 10 m and (iii) L = 20 m. Identify all the critical inflection points on the plot.

Under steady conditions, the acceleration of the weight is zero and using Newton’s 2nd law we have

∑ F=0 [1]

Po Ao−W=0 ,Po=W / Ao [2]

Using Bernoulli principle on the pipe,

Pi=Po+P loss [3]

with

Ploss=f ( LD )( ρ v p2

2 ) [4]

But F i=A i p i, and from continuity v p Ap=vo Ao .Using [2], [3] and [4], we have

F i=A i(WAo +f ( LD )( A pAo )2

( ρ vo2

2 )) [5]

Where

f={ 64ℜ ,ℜD<4000

−1.8 log( 6.9ℜ +( ε /D3.7 )1.11)

−2

,ℜD>2300}Points of discontinuity, inflection occur at ℜD=0 (motion starts) ℜD=2300 ( turbulent solution

starts) and ℜD=4000 (laminar solution ends). Using ℜD=ρ v pD

μ, and vo=

A pAov p, we have

vo=A pAo

μℜD

ρD

withD=√4 A/ π=√16 /π=2.26mm, the results in the table below are obtained:

L (m) ReD Vo (m/s) Fi (Laminar) (kN) Fi (Turbulent) (kN)5 2300 0.1021 500.3208 500.5261

4000 0.1776 500.5579 501.385610 2300 0.1021 500.6416 501.0522

4000 0.1776 501.1159 502.7713

20 2300 0.1021 501.2833 502.10444000 0.1776 502.2318 505.5425

rho = 998; % water density at 20 deg Cmu = 1e-3; % water viscosity at 20 deg CAp = 4e-6; %pipe areaDp = sqrt(4*Ap/pi); %pipe diameterAo = 40e-6; %output areaAi = 10e-6; %input areaWt = 2000; %Weight Lp = input('\n pipe length:'); % pipe length ReD = input('\n Reynolds number:'); % Reynolds Number Vp = ReD .* mu ./ (rho * Dp); % pipe flow speedVo = Vp .* (Ap/Ao); % output speed f_f_lam = 64 ./ ReD; % laminar friction factorf_f_trb = 0.316 ./ (ReD .^ 0.25); % turbulent friction factor Pl_lam = f_f_lam .* (Lp/Dp) .* 0.5 .* rho .* Vp .^ 2;Pl_trb = f_f_trb .* (Lp/Dp) .* 0.5 .* rho .* Vp .^ 2; Fi_lam = Ai .* ( (Wt/Ao) + Pl_lam); %laminar forceFi_trb = Ai .* ( (Wt/Ao) + Pl_trb); %turbulent force fprintf('Lp = %-2.2f \n', Lp);fprintf('ReD = %-5.1f \n',ReD);fprintf('Vo = %-2.4f \n',Vo);fprintf('Fi_lam = %-5.4f \n',Fi_lam);fprintf('Fi_trb = %-5.4f \n',Fi_trb);

rho = 998; % water density at 20 deg Cmu = 1e-3; % water viscosity at 20 deg CAp = 4e-6; %pipe areaDp = sqrt(4*Ap/pi); %pipe diameterAo = 40e-6; %output areaAi = 10e-6; %input areaWt = 2000; %Weight ReD_lam = 0:2300; % Range of Reynolds for laminar flowReD_crt = 2300:4000 ; %Range of Reynolds for critical flowReD_trb = 4000:6000; % Range of Reynolds for turbulent flow Vp_lam = ReD_lam .* mu ./ (rho * Dp); % laminar pipe speedVp_crt = ReD_crt .* mu ./ (rho * Dp); % critical pipe speedVp_trb = ReD_trb .* mu ./ (rho * Dp); % turbulent pipe speed Vo_lam = Vp_lam .* (Ap/Ao); % laminar output speedVo_crt = Vp_crt .* (Ap/Ao); % laminar output speedVo_trb = Vp_trb .* (Ap/Ao); % turbulent output speed f_f_lam = 64 ./ ReD_lam; %laminar friction factorf_f_trb = 0.316 ./ (ReD_trb .^ 0.25); %turbulent friction factor f_f_lam_crt = 64 ./ ReD_crt; %laminar friction factor in critical regionf_f_trb_crt = 0.316 ./ (ReD_crt .^ 0.25); %turbulent friction factor in critical region hold on

for Lp = [5, 10, 20]; %pipe lengths Pl_lam = f_f_lam .* (Lp/Dp) .* 0.5 .* rho .* Vp_lam .^ 2; Pl_lam_crt = f_f_lam_crt .* (Lp/Dp) .* 0.5 .* rho .* Vp_crt .^ 2; Pl_trb = f_f_trb .* (Lp/Dp) .* 0.5 .* rho .* Vp_trb .^ 2; Pl_trb_crt = f_f_trb_crt .* (Lp/Dp) .* 0.5 .* rho .* Vp_crt .^ 2; Fi_lam = Ai .* ( (Wt/Ao) + Pl_lam); %laminar force Fi_lam_crt = Ai .* ( (Wt/Ao) + Pl_lam_crt); %laminar force in critical region Fi_trb = Ai .* ( (Wt/Ao) + Pl_trb); %turbulent force Fi_trb_crt = Ai .* ( (Wt/Ao) + Pl_trb_crt); %turbulent force in critical region plot(Vo_lam, Fi_lam, Vo_crt, Fi_lam_crt, '--', Vo_crt, Fi_trb_crt, '--', Vo_trb, Fi_trb);end xlabel ('output speed (m/s)');ylabel ('Input Force (N)');

0 0.05 0.1 0.15 0.2 0.25 0.3 0.35500

502

504

506

508

510

512

output speed (m/s)

Inpu

t F

orc

e (N

)

For F = 1 kN and L = 20 m, derive an expression for the transient output speed, v, as a function of time, t, and find the time needed for the piston to reach 95% of the steady speed. Plot v(t) using appropriate scale.

Under transient conditions, we have

∑ F=m dvdt

[5]

Po Ao−W=m dvdt

[6]

Using Bernoulli principle on the pipe,

Po=P i−Ploss [7]

with pi=F i /A i and using the results of the plot in part (a) when Fi = 1 kN, the flow becomes fully

turbulent short time after motion starts. It can then be assumed that Ploss=f ( LD )( ρ v p2

2 ), where f

is independent of v.

mdvd t

=Ao( F iA i−f ( LD )( ρ v p2

2 ))−W [8]

From continuity v p=v ( Ao/A p ) .and the equation becomes

mdvdt

=Ao( F iA i−f ( LD )( AoA p )2

( ρ v2

2 ))−W [5]

dvdt

+Aom ( AoAp )

2

f ( LD )( ρ v2

2 )= Fim ( AoAi )−g [5]

Let c=Aom ( AoAp )

2

f ( LD )( ρ2 ) and =k=F im ( AoA i )−g, the equation may be written as:

dvdt

+c v2=k [5]

dv

k−c v2=dt [5]

∫vo

vdv

v2−k /c=−c∫

0

t

dt [5]

12√k /c

ln( v−√k /cv+√k /c )=−ct

v−√k /cv+√k /c

=−e−2√kct [5]

v (t )=√ kc (1−e−2√kct1+e−2√kc t ) [5]

0 0.5 1 1.5 20

0.5

1

1.5

2

2.5

time

out

pu

t sp

eed

To find the time needed for the output piston to reach a ratio, r v, of the steady velocity, vs=√k /c

v (t )√k /c

=rv

=( 1−e−2√kc t1+e−2√kc t )Solving for t

t= 12√kc

ln ( 1+r v1−r v )Given, k=

F im ( AoA i )−g=g

c=Aom ( AoAp )

2

f ( LD )( ρ2 )Using the values specified for the parameters above

c= g∗40×10−6

2000102 f ( 20

√16×10−6/π )(500)c=8.87g f

The steady speed is therefore, vs=√k /c=√1 /8.87 f

Assume c=2 ( f=0.024 ), vs=√k /c=√g /2=2.21 m/s, and with r v=95%

t= 12√2g

ln ( 1+0.951−0.95 )t=0.41s.