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Equations Remarks Limitations

HHPt = (Pt x Q)/1714 Hydraulic horsepower Steady state, fieldunits

Re = 928VD/ Reynolds number Field units

= y + p Shear Rate Non newtonianfluids - Bingham

Plastic

= n Shear RatePseudoplastic fluid(n 1)

Non newtonian

fluids - Power law

Minimum pressure

required for flow

Non newtonian

(pipe), field units,

Minimum pressure

required for flow

Non newtonian

(annulus), field

units

P = (32vL)/d2 Hagen-Poiseiulle Isothermal,laminar newtonian

flow in pipes

dP/dL= v/1500d2 Pressure loss in flow Newtonian flow ina pipe, field units

Pressure loss in flow Newtonian flow in

an annulus , field

units

dP/dL = g300(d2-d1)

dP/dL = g300d

dPdL

v

= 1500

d22 d

21

+

d22 d

21

_

_

d2d1

ln

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Equations Remarks Limitations

Pressure loss in

laminar flowing Non

newtonian fluids

Pipes, field units -

Bingham Plastic

Pressure loss in

laminar flowing Non

newtonian fluids

Annulus, field units

- Bingham Plastic

Pressure loss in

laminar flowing Non

newtonian fluids

Pipes, field units -

Power Law

Pressure loss in

laminar flowing Nonnewtonian fluids

Annulus, field units

- Power Law

Newtonian flow in a

pipe, field units

Turbulent flow

Newtonian flow in a

pipe, field units

Turbulent flow -

smooth pipes,

moderate Re

Apparent viscosity,field units - Bingham

Plastic

Used in Recalculations

Apparent viscosity-

annulus, field units -

Bingham Plastic

Used in Re

calculations

Apparent viscosity,

field units - Power law

Used in Re

calculations

No friction, field units Velocity of

discharge from bit

nozzle

dP=

pv

dL 1,500d2 225d+

y

dP=

pv

dL 1000(d2-d1)2 200(d2-d1)+

y

dP =kv

dL 144,000d(1+n) 0.0416

3+1/nn

dP

=

kv

dL 144,000d(d2-d1)(1+n) 0.0208

2+1/nn

dP

= 4fv

2

dL 2d

dP=

0.75q

1.75

0.25

dL d4.75

e= + 6.66d

v

e= +

5(d2 _ d1)

v

v

a=

Kd 3 + 1/n(1 - n)

(1 - n)96 0.0416

n

vn=

Pb

8.074x10-4

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Equations Remarks Limitations

Compensated for

friction, field units

Velocity of

discharge from bit

nozzle

Multiple nozzle rock

bit

Nozzles are

assumed to have

the same size

Pressure loss across

bit

Valid for both

Newtonian and

Non Newtonian

liquids

Predicting optimumbit pressure

n is approximatedat 1.85

Maximum allowable

bit surface pressure

for optimization

(graphical)

Optimum circulating

pressure (graphical)

Optimum nozzle area

(graphical)

vn= C

d

Pb

8.074x10-4

vn=

q

3.117At

P

b=

8.311x10-q

Cd

2A

t

2

xQ2

564 An

2P

bit=

Pb= n

(n+1)

Pt

Pcirc.opt = W x Pmax

Nozzle area =Qopt23.75

Pmax

-Pcirc.opt