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Terry Ring

• Multiple Choice Exam

• Data Available

– See fe_exam_specs.pdf

– See fe_chemical_engineering.pdf

•

**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)

• 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

• 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)*

• The longitudinal stress can be expressed as:

•

*σ l*

*= P d / 4 t *

–

*where*

*σ l*

*= longitudinal stress (MPa, psi)*

*(2)*

• Elastic Limit

• Elastic

Modulus

• Yield point

• Ultimate

Strength

• Failure

• Hoop Stress < Yield Limit of Metal

• 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

• Discounted Cash Flow

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

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

• 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)

• 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

• 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

.

• [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?