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GEAR PUMP

P = Rated power = 70bar

Q=Discharge=80 lpm = 80 x 10 -360 m3/sec

To find,1) seln of power for motor power 2) Design of gear, shaft, bearing3) Casing & nos of bolts.

1)Seln of motor capacity

Motor Capacity= P x Q OverallWhere overall = mech x volmech = 0.85 to 0.9 vol = 0.92 to 0.96Motor Capacity =___watt?Std. motor seln PSG 5.124Speed lpm720 rpm < 70 lpm960 rpm < 100 lpm1440 rpm > 100 lpm

2) Design of Gear

A] Caln of module

1) Q= 4 (Do2-Di2) x b x N x vol x 10-6

Do = m(z+2) & Di = m(z-2) Assuming z=14Do=__m? Di=__m?

2)= bm = 4 to 8

3) bm = 6 or 7

4) m=__? in eq (i)Std module seln [PSG 8.2]

5) Exact value bexact=

m=G*, b=G*, Do=G*, Di=G* & D=Do+Di2 N=G*

B] Gear for strength & wear1) Pd= Normal power x S.F.

Where Normal power = motor power/2 & SF = 1.2Pd=__KW?

2) Vm = DN60 => Vm=__m/s

As3) Cv=formula based on Vm on PSG 8.51

Cv = __?

4)Ft = PdVm => Ft=__N?

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5) Fd=Ft x Cv [PSG 8.50]Fd = __N?

6) Fs = b.b.m.y eqn-(i)

7) y = (0.154-0.912Z) where {z=14 assumed for 20o full depth}

8) seln of matl 15Ni 2Cr Mo15ut= 950 N/mm2 & FOS = 5b=utFOS = 190 N/mm2

9) Fs = __N/m2 Fd ..safe

C] Caln of wear load1) Fw = Dp.b.k.Q [where D=Dp]

2) k = c21.4sin[ 1E1+ 1E2 ] = 20o , E1 =E2 = 2.15 x 105 N/mm2

3) c = 2.8 BHN-70= 91.0 N/mm2

4) k = __?

5) Q = 2ii+1 = 1

6) Fw = __? Fw Fd

3) Design of ShaftA] Driver shaft1) Ft = Given above (B4)

2) Fr= Fttan = 20o full depthFr= __N?

3) Ra = Do2 [Do = G* in 2A5]

4) Pmax = 1.1 x (P) => Pmax = __N/m2?

5)Fh = 1.635 Ra.b.Pmax => Fh = __N?

6) W = (Fh-Ft)2+ (Fr)2 => W=__N?7) BM = WL2 where L=2.5b => BM=__N-mm?

8) Pd x 103 = 2N[Mt]60

Mt = __N-mm?

9) Teq = (BM)2+ (Mt1)2Seln of matl C-45 & = 45N/mm2

Teq = __?

10) Teq = 16 x d3 x

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d= __mm?

B] Driven shaft

1) W = (Fh-Ft)2+ (Fr)2 => W=__N?

2) BM = WL2 => BM=__N-mm?

3) [Mt2] = G* above

4) Te = G* above

5) Te = 16 x d3 x d= __mm?

4) Design of Bearing

1) Fr= W2 W=G* in driven shaftFr = __N? & Fa = 0

2) Lmr = Lh X 60 X N106 where Lhr=5000-8000 hrs

Lmr= __mr?

3) Pe = (X x V x Fr+ y x Fa)x SX=1, V=1, S=1.2 => Pe= __kgf?

4) Lmr= (CPe)10/3 => C=__kgf?From PSG 4.34 for C= __kgf select ds=__?

5) Design of Casing

1) tr1 = [ 1+Pi1-Pi ]-1where r1 = Do2 & Pi = P x 1.2 => Pi=__N/mm2

selecting GCI-20 matl ut = 200 N/mm2 FOS=8

2) t = utFOS => t = __N/mm2

3) t = __mm?

6) Design of bolts

1) Bolt matl C-30 [PSG 1.9] & yt = 300N/mm2

2) t = ytFOS => FOS = 5 to 6 => t =__ N/mm2?

3) kb = 2 kc formula only.

4) Pmax = P x 1.2 => Pmax = __N/mm2

?

5) Po = 1.5 x P => Po = __N/mm2?

6) A = 4 Do2 + 8 Do2 + a Do => A=__mm2?

7) Fext = [Pmax] x A => Fext = __KN?

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8) Fo = Po x A => Fo = __KN?

9) Fi = 3 Fo => Fi = __KN?

10) Fb = 23 Fext => Fb=__KN?

11) Fb = Fi + Fb => Fb=__KN?

12) Pr Size10-30 M-1230-70 M-16Above 70 M-20

Select M16 & As = 157 mm2

13) t = FbAs X n => n= __bolts? Adopt higher n=__?

7) Design of Suction & Delivery Pipe DiameterSuction Pipe

1) Q=As x Vswhere Vs = (1to 3) m/s = 1.5m/sAs = __m

2?

2) As = 4 d2 => d=__m?

Delivery Pipe

1) Q= Ad x VdVd = (2 to 5)m/s = 3m/sAd = __m

2?

2) Ad = 4 d2 => d=__m?