Ministry of Science and Technology

Department of Technical and Vocational Education

Petroleum Engineering Department

SAMPLE ANSWER FOR B.Tech: SECOND YEAR

PE – 04025 Reservoir Engineering

1.Calculate the gas deviation factor of B for gas from its composition, Molecular

weight, critical pressure and temperature.

Component Component of mole molecular weight Pc Tc

Mole fraction

Methane C1 0.8612 16.04 673 343

Ethane C2 0.0591 30.07 708 550

Propane C3 0.0358 44.09 617 666

Butane C4 0.0172 58.12 550 766

Pentane C5 0.0050 72.1 490 846

CO2 0.0010 44.01 1070 548

N2 0.0207 28.02 492 227

1 2 3 4 5 6 7 8

ComponentComp.Mol

FractMol.Wt Pc Tc (2) x(3) (2)x (4) (2)x (5)

MethaneC1 0.8612 16.04 673 343 13.81 579.59 295.35

Ethane C2 0.0591 30.07 708 550 1.78 41.84 32.51

Propane C3 0.0358 44.09 617 666 1.58 22.08 23.84

Butane C4 0.0172 58.12 550 766 1.00 9.46 13.18

Pentane C5 0.0050 72.15 490 846 0.36 2.45 4.23

CO2 0.0010 44.01 1070 548 0.04 1.07 0.55

N2 0.0207 28.02 492 227 0.58 10.18 4.7

1.0000 19.15 666.67 347.4

The sum of Col (6) is average molecular weight of the gas.

Mgas = 19.15

The sum of Col (6) is pseudocritical pressure.

Pc = 666.67psia

The sum of Col (7) is pseudocritical temperature.

Tc= 347.40° R

Specific Gravity (SG) =

Pseudoreduced pressure, Pr =

Pseudoreduced temperature, Tr =

From Fig. 1.3 , for Pr = 4.87 and Tr = 1.80 Z= 0.91

2. The following data are given for a gas condensate reservoir.

Initial reservoir pressure = 2740psia

Res tem = 215°F

Avg porosity = 25%

Average connate water = 30%

Daily tank oil = 242bbl

Oil gravity , at 60 F = 48.0° API

Daily separator gas = 3100MCF

Separator gas gravity = 0.650

Daily tank Gas =120MCF

Tank gas gravity = 1.20

Calculate (a) Initial oil in place per ac-ft .

(b) Initial gas in place per ac-ft

(c) Total daily gas condensate production.

(d) Total daily reservoir voidage.

Solution

Average gas gravity,

= 0.670

Specific gravity of Oil,

=

= 0.7883

Molecular wt. of tank oil, Mo =

Mo = = 144.5

R = =13,300SCF/bb

Specific gravity of well fluid,

=

= 0.893

From fig 1.2, for (condensate) Tc=425 ْR, and Pc=652psia

= =1.59, = 2740/652 = 4.20

From Fig 1.3, for Tr = 1.59 and Pr = 4.2 z = 0.82

Total initial gas in place per ac-ft , G = ,

G =

=1334MCF/ac-ft

The fraction of the total which is produced on the surface is

fg = =

=

=0.9483

Thus, Initial gas in place = 0.9483 x 1334

=1265MCF/ac-ft

Initial oil in place = = 95.1 bbl/ac-ft

Daily gas condensate production in SCF/day

= 3396MCF/day

Total daily reservoir voidage is

= =19,400 CF/day.

3(a) Calculate the stock tank barrels of oil initially in place in a combination drive

reservoir.

Given Data:

Volume of bulk oil zone = 112,000ac-ft

Volume of bulk gas zone = 19,600ac-ft

Initial reservoir pressure , = 2710psia

Initial FVF , = 1.3040bbl /STB

Initial gas volume factor, = 0.006266cuft/SCF

Initial dissolved GOR , = 526 SCF/ STB

Oil produced during during the interval, = 20MMSTB

Reservoir pressure at the end of interval ,= 2000psia

Average produced GOR, = 700SCF /STB

Two –phase FVF at 2000psia, = 1.4954bbl/STB

Volume of water encroached, = 11.58MMbbl

Volume of water produced , = 1.05MMSTB

FVF of the water , Bw = 1.028bbl / STB

Gas volume factor at 2000 psia , = 0.008479cuft/scf

solution

Bti = 1.3400 x 5.615= 7.5241cuft/cuft

Bt= 1.4954 x 5.516 = 8.3967 cuft / STB

We= 11.58 x 106 x 5.615 = 65.02 MMcuft

Wp = 1.05 x 106 x 1.028 x 5.615 = 6.0608 MMrescuft

Assuming the same prosity and connate water for the oil and gas zones,

Substituting in Eq (4.4)

N=

=

= 98.97MMSTB

If Bt is in barrel per stock tank barrel, then Bg must be in barrels per standard cubic

foot and We and Wp in barrel and the substitution is as follow;

N=

= 98.97MMSTB

3.(b) Define the following

a. Permeability

b. Porosity

c. Saturation

d. Resistivity

e.Density

Solution

(a) Permeability

Permeability is

1. a measure of the case with which fliud can pass through a porous rock

2. the fluid conductivity of a porous medium

3. the ability of fluid to flow within the interconnected pore network of a porous

medium.

Absolute permeability

The permeability of a rock is a rock is a property of a rock and not of the fluid

which flow through it, provided that the fluid 100 percent saturates the pore spaces of

the rock .This permeability at 100 percent saturation of asingle fluid is called the

absolute permeability of a rock.

Effective permeability

Effective permeability is the permeability of a rock to a particular fluid when

that fluid has pore saturation of less than 100 percent. The sum of effective

permeability is always less than absolute permeability.

Relative Permeability

Relative permeability is the ratio of the effective permeability to absolute

permeability.

(b) Porosity

Porosity is the condition of something that contains pore. The relative volume

of pore spaces between mineral grains as compared to total rock volume. Porosity

measures the capacity of rock to hold oil, gas and water.

(c) Saturation

Saturation is the percentage or fractional part of the total capacity that actually

holds any particular fluid. The extend to which porespaces in the formation contains

hydrocarbon or connate water. The extend to which gas is dissolved in the liquid

hydrocarbon in the formation.

(d) Resistivity

Resistivity is the electrical resistance offered to the passage of current; the

opposite of conductivity.

(e) Density

The density of a substance is defined as mass per unit volume.

4.Calculate the initial gas in place and the intital reserve of a gas reservoir from

pressure production data for a volumetric reservoir.

Given data;

Initial pressure = 3250psia

Reservoir temperature = 213 °F

Standard pressure = 15.025psia

Standard temperature = 60°F

Cumulative production = 1.00 x109 SCF

Average reservoir pressure = 2864psia

Gas deviation factor at 3250psia = 0.910

Gas deviation factor at 2864psia =0.888

Gas deviation factor at 500psia= 0.951

solution

Solve Eq (1.3) for the reservoir gas pore volume Vi

= -

= -

Vi = 56.17 MMcuft

The initial gas in place by Eq (1.5) is

G = x

=

=10.32 MMMSCF

The gas remaining at 500psia abandonment pressure is

Ga = x

=

= 1.52MMMSCF

The initial gas reserve based on a 500 psia abandonment pressure is the difference

between the initial gas in place and the gas remaining at 500psia or

= (10.32 – 1.52) x 109

= 8.8 x 109

= 8.8 MMMSCF

5. Calculate water influx and residual gas saturation in water drive gas reservoir .

Given data :

Bulk reservoir volume , initial = 415.3MMcuft

Average porosity = 0.172

Average connate water =0.25

Initial pressure =3200psia

Bgi = 0.005262cuft/SCF, 14.7 psia and 60° F

Final pressure = 2925psia

Bgf = 0.005700cuft /SCF , 14.7psi and 60°F

Cumulative water production = 15200bbl (surface)

Bw = 935.4MMSCF at 14.7psi and 60°F

Gp =1.03bbl / surface bbl

Bulk volume invaded by water at 2925psia = 13.04MMcuft

Solution

Initial Gas in place ,G=

=

= 10,180MMMSCF at 14.7 psia and 60 ْF

=

=(935.4 106 0.005700) –(10,180 109 (0.005700-0.005262) + (1.03

15200 5.615)

= 960,400 cuft

This much water has invaded 13.04 MMcuft of bulk rock, which initially contained

25percent connate water. Then the final water saturation of the flooded position of the

reservoir is

=

=0.68 or 68 percent

Then, the residual gas saturation , Sgr = 32 percent

6. Define the following:

(a) Molecular weights

(b) The mole or pound mole

(c) The perfect gas law

(d) Specific gas law

(e) Ideal gas

Solution

6. (a) Molecular weights

Molecular weights are the sum of atomic weight of atom forming the

molecules and since atomic weight of hydrogen is close to one , then it is inferred that

atomic weight of carbon is very close to 12.0. Further, it may be said that the

molecular weight of water is approximately 18, having two hydrogen atomic weight

of one each and one oxygen atom of weight 16.

(b) The mole or pound mole

From what has been said it should be clear that the same conditions of

temperature and pressure 2.016 pounds of hydrogen, 32.000lbs of oxygen, 16.04 of

methane, etc., will all (1) contain the same number of molecules and (2) occupy the

same volume. These quantities of gases (or liquids or solids for that matter), equal in

pounds to their molecular weights, are called moles, or more specifically pound

moles. A mole of any compound is the number of weight units equal to its molecular

weight. For example, a mole of methane is 16.04 weight units of methane. If it is

expressed in pound, 16.04 lbs of methane is a pound mole of methane.

(c) The Perfect Gas Law

We were able to calculate above, quite easily, that the 10,000cubic-foot tank

of methane contained 422.8lb because the gas happened to be at 14.7 psia and 60 F. if

the gas had been at other conditions, the problem would have been more difficult

because the volume would first have to be converted to the volume it would occupy

under the standard conditions. To make the problem more general it can be said that

there are three quantities which define the stage of a gas i.e: temperature, pressure and

volume. Boyles and Charles in experimenting with these variables found that for any

given quantity of gas the following law holds,

Where, P1, P2 and V1 are the temperature, pressure and volume of a gas in

stage 1 and P2, V2 and T2 in stage 2. In this equation, V1and V2 may be measured in

any units provided they are the same; P1 and P2 may be measured in any units

provided they are the same and provided they are absolute pressure ; T1 and T2 may be

measured in any units provided they are the same and provided are absolute

temperature.

(d) Specific gravity of gases

Since the density of a substance is defined as mass per unit volume, the

density of a gas ρg, at a given temperature and can be derived by substituting m/M for

and in the perfect gas law, where m is the pounds of gas and M is the molecular

weight

Because it is more convenient to measure the specific gravity of gas than the

gas density, specific gravity is more commonly used. Specific gravity of gases is

defined as the ratio of the density of gas at a given temperature and pressure to the

density of air at the same temperature and pressure, usually near 60 ْF and atmospheric

pressure.

(e) Ideal Gas

The ideal gas may be defined as one whose pressure will be exactly doubled if

its volume is reduced one half and whose pressure will be exactly doubled if, keeping

the volume constant, the absolute