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FE Exam Preparation DesignPowerPoint Presentation

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

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### FE Exam PreparationDesign

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)

- VIII. Strength of Materials 7%

Classic Problem - I

- What is the cost of a 400L copper tank to be pressurized to 40 MPa?
- What dimensions minimize cost?
- Tank Vol= πR2L
- Two Stresses
- Hoop Stress
- Longitudinal Stress

- What dimensions minimize cost?

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:
- σl = P d / 4 t (2)

- whereσl = longitudinal stress (MPa, psi)

Copper Strength

- Elastic Limit
- Elastic Modulus
- Yield point
- Ultimate Strength
- Failure

Design Criterion

- Hoop Stress < Yield Limit of Metal

Cost Cost= Cost/kg ρCuVol 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= πR2L

- Volume of Metal= (πR2t+ πR2t+2πRLt)
- Top Bottom Wall

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

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)

AfternoonClassic 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.
- CTC=(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/m2/K and the condensing steam coefficient is 3500 W/m2/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/m2/K
- 24. The outside area of the heat transfer surface is most nearly:
- A) 18, B) 33, C) 41, D) 44, m2.

Heat Transfer

- [1/U] = [1/ho] + [Doln(Do/Di)/(2k)] + [Do/(hiDi)]
- U={[1/ho] + [Doln(Do/Di)/(2k)] + [Do/(hiDi)]}-1
- U≈{[1/4500] + [38/(3500*36)]}-1
- Q=UA*ΔTLM
- Evap. Rate=Q*ΔHvapρH2O =1.2 kg/s
- ΔT=50ºC, Calculate A. What is ΔTLM definition?

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