FE Exam Preparation Design

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FE Exam Preparation

Design

Terry Ring

Exam Overview

• Multiple Choice Exam

• Data Available

– See fe_exam_specs.pdf

– See fe_chemical_engineering.pdf

Morning

– General Information

Design Topics in

VIII. Strength of Materials 7%

• A. Shear and moment diagrams

• B. Stress types (e.g., normal, shear, bending, torsion)

• C. Stress strain caused by:

– 1. axial loads

– 2. bending loads

– 3. torsion

– 4. shear

• D. Deformations (e.g., axial, bending, torsion)

• E. Combined stresses

• F. Columns

• G. Indeterminant analysis

• H. Plastic versus elastic deformation

VI. Engineering Economics 8%

• A. Discounted cash flow (e.g., equivalence, PW, equivalent annual FW, rate of return)

• B. Cost (e.g., incremental, average, sunk, estimating)

• C. Analyses (e.g., breakeven, benefit-cost)

• D. Uncertainty (e.g., expected value and risk)

Classic Problem - I

• What is the cost of a 400L copper tank to be pressurized to 40 MPa?

– What dimensions minimize cost?

• Tank Vol= πR

2

L

• Two Stresses

– Hoop Stress

– Longitudinal Stress

HOOP STRESS

• The hoop stress can be expressed as:

σh = P d / 2 t

(1)

where

σh = hoop stress (MPa, psi)

P = internal pressure in the tube or cylinder (MPa, psi) d = [2R] internal diameter of tube or cylinder (mm, in) t = tube or cylinder wall thickness (mm, in)

Longitudinal Stress

• The longitudinal stress can be expressed as:

σ l

= P d / 4 t

where

σ l

= longitudinal stress (MPa, psi)

(2)

Copper Strength

• Elastic Limit

• Elastic

Modulus

• Yield point

• Ultimate

Strength

• Failure

Design Criterion

• Hoop Stress < Yield Limit of Metal

Cost

• Volume of Metal= (πR

2 t+

πR

2 t+2

πRLt)

»

Top Bottom Wall

• Cost= Cost/kg ρ

Cu

Vol

• Minimize cost by minimizing metal volume subject to the constraint that the hoop stress is less than the yield limit for copper (60 MPa) and that the tank volume is 400 L.

• 60 MPa= P

2R / 2 t (1)

Solve for t= f(R) substitute into Vol of metal Equation minimize volume by taking deriviative w/r to R set equal to zero and solve for R. Use R to calculate L for the tank volume of 400L=

πR

2

L

Engineering Economics

• Discounted Cash Flow

• Cost (e.g., incremental, average, sunk, estimating)

• Analyses (e.g., breakeven, benefit-cost)

Classic Problem II

• What is the present value of $1,000,000 of profit for a process after the plant is built and is up and running. The construction and commissioning time is 3 years.

Assume an interest rate of 8% per year.

• F=P(1+i) n

• Solve for P in F=$1,000,000= P(1+0.08)

3

Afternoon

VIII. Process Design and Economic

Optimization 10%

A. Process flow diagrams (PFD)

B. Piping and instrumentation diagrams (P&ID)

C. Scale-up

D. Comparison of economic alternatives (e.g., net present value, discounted cash flow, rate of return)

E. Cost estimation

XI. Safety, Health, and

Environmental 5%

A. Hazardous properties of materials

(e.g., corrosive, flammable, toxic), including MSDS

B. Industrial hygiene (e.g., noise, PPE, ergonomics)

C. Process hazard analysis (e.g., using fault-tree analysis or event tree)

D. Overpressure and underpressure protection (e.g., relief, redundant control, intrinsically safe)

E. Storage and handling (e.g., inerting, spill containment)

F. Waste minimization

G. Waste treatment (e.g., air, water, solids)

Classic Problem I

• Determine the total capital investment in 2000 associated with a reactor with a production rate of 1,000,000 lb/yr that had a purchase price of

$10,230 in 1990 and was designed for production rate of 1,000 lb/yr. Use Marshal &

Swift installed equipment index.

• C

TC

=(1,000,000lb/yr/1,000lb/yr)

0.6

*

$

10,230(1108/924)*(455/100)

Production Rate Factor M&S Factor Total Capital

Classic Problem II

• Scale-up calculations for a commercial evaporator

Containing 38 mm OD tubes with 2 mm wall indicates that the boiling coefficient is 4,500 W/m

2

/K and the condensing steam coefficient is 3500 W/m

2

/k. The

Steam condenses at 150C. The vapor rate will be 1.2 kg/s. All other conditions are the same.

• 23. Base on outside area the overall heat transfer coefficient is most nearly:

• A) 1,870, B)1,970, C)2,060, D)2,130 W/m

2

/K

• 24. The outside area of the heat transfer surface is most nearly:

• A) 18, B) 33, C) 41, D) 44, m

2

.

Heat Transfer

• [1/U] = [1/h o

] + [D o ln(D o

/D i

)/(2k)] + [D o

/(h i

D i

)]

• U={

[1/h o

] + [D o ln(D o

/D i

)/(2k)] + [D o

/(h i

D i

)]

}

-1

• U≈{

[1/4500] + [38/(3500*36)]

}

-1

• Q=UA*ΔT

LM

• Evap. Rate=Q*ΔH vap

ρ

H

2

O

=1.2 kg/s

• ΔT=50ºC, Calculate A. What is ΔT

LM definition?

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