Homework One (Solution) - KSU Facultyfac.ksu.edu.sa/sites/default/files/pge_362_hw1_solution.pdf ·...

5
King Saud University Petroleum and Natural Gas Engineering PGE 362: Properies of Reservior Fluids Posted: Sunday, September 21, 2014 Due: Sunday, October 12, 2014 ________________________________________________________________________ Homework One (Solution) Solve the following problems from (The Properties of Petroleum Fluids, McCain) book. 1. 1-2 2. 1-5 3. 1-12 4. 1-13 5. 3-1 6. 3-2

Transcript of Homework One (Solution) - KSU Facultyfac.ksu.edu.sa/sites/default/files/pge_362_hw1_solution.pdf ·...

Page 1: Homework One (Solution) - KSU Facultyfac.ksu.edu.sa/sites/default/files/pge_362_hw1_solution.pdf · Homework One (Solution) ... like an ideal. Solution Given: ρ = 0.103 lb.ft3 P

King Saud University

Petroleum and Natural Gas Engineering

PGE 362: Properies of Reservior Fluids

Posted: Sunday, September 21, 2014 Due: Sunday, October 12, 2014

________________________________________________________________________

Homework One (Solution)

Solve the following problems from (The Properties of Petroleum Fluids,

McCain) book.

1. 1-2

2. 1-5

3. 1-12

4. 1-13

5. 3-1

6. 3-2

Page 2: Homework One (Solution) - KSU Facultyfac.ksu.edu.sa/sites/default/files/pge_362_hw1_solution.pdf · Homework One (Solution) ... like an ideal. Solution Given: ρ = 0.103 lb.ft3 P

Solution

HC Type Molecular Structure

Cyclopentane naphthene

2,2-Dimethylbutane paraffin

2,3-Dimethylbutane paraffin

2-Methylpentane paraffin

3-Methylpentane paraffin

n-Hexane paraffin

Methylcyclopentane naphthene

Iso-propylbenzene aromatic

n-propylbenzene aromatic

Page 3: Homework One (Solution) - KSU Facultyfac.ksu.edu.sa/sites/default/files/pge_362_hw1_solution.pdf · Homework One (Solution) ... like an ideal. Solution Given: ρ = 0.103 lb.ft3 P

Solution

Video

http://www.youtube.com/watch?v=AzD3lGl05fM

Solution

Boiling Point:

1. Increase with increasing number of carbons?

2. Decrease with chain branching?

Page 4: Homework One (Solution) - KSU Facultyfac.ksu.edu.sa/sites/default/files/pge_362_hw1_solution.pdf · Homework One (Solution) ... like an ideal. Solution Given: ρ = 0.103 lb.ft3 P

Solution

Molecular Structure Name

CH3CH2SH Ethanol

CH3CH2NH2 Ethylamine

CH3CH2CH2CH2COOH Pentanoic acid

Cyclopentanol

Cyclopentyl alchohol

Hydrocyclopentane

Acetone

Propanone

Solution

𝑃𝑉 = 𝑛 𝑅𝑇

(100)𝑉 = (1)(10.73)(90 + 460)

𝑉 = (1)(10.73)(90 + 460)

(100)= 59.015 𝑓𝑡3

Page 5: Homework One (Solution) - KSU Facultyfac.ksu.edu.sa/sites/default/files/pge_362_hw1_solution.pdf · Homework One (Solution) ... like an ideal. Solution Given: ρ = 0.103 lb.ft3 P

like an ideal.

Solution

Given:

ρ = 0.103 lb.ft3

P = 14.7 psia

T = 100 °F = 100 + 460 = 560 °R

CmH2m

𝑃𝑉 = 𝑛𝑅𝑇

14.7 𝑉 =𝑤𝑡

𝑀𝑊× 10.73 × 560

𝑀𝑊 =𝑤𝑡

𝑉× 10.73 × 560/14.7

𝑀𝑊 = 𝜌 × 10.73 × 560/14.7

𝑀𝑊 = 0.103 × 10.73 ×560

14.7

𝑀𝑊 = 42

𝑀𝑊 = 12𝑚 + 2𝑚 = 42

14𝑚 = 42

𝑚 =42

14= 3

C3H6