Engineering Excel

 

Given

: 

Barrel circular Area = 4.60 ft2, flow rate of water into the barrel = 36.8 lb / sec, flow rate of water from the leak in the bottom of the barrel = 9.2*h  lb/sec (9.2 is the ‘leak rate coefficient C’), water density = 62.4 lb / ft3.

Exit flow rate is a function of water level.

 

Find :  ( 15 pts) Define the five (5) constant variables of Area, Flow in, Coefficient for flow out, density of water,

   
 
                   

and appropriate time step size in seconds as shown in class and as done in Excel Assignment 02.

                       
The time step size should be at least 2.0 seconds, but not more than 10.0 seconds. NOT I DONT NEED WORDS DEFINITION FOR THIS QUESTION EVERTHING SHOULD BE BY NUMBERS AND EQUETION. 

            ( 5 pts)  Calculate the coefficients C1 and C2 to be used in the solution to the differential equation.     

            ( 5 pts)  Calculate the “Time Constant”  tau =  τ

            ( 5 pts)  Column tabulation of times beginning with time = 0.0 seconds   

            (20 pts) Column tabulation of water level in the barrel using the time-step simulation or approximations.  

            (15 pts) Column tabulation of water level in the barrel using the theoretical solution to the governing equation.

            (10 pts) Column tabulation of percent error between simulation and theoretical. 

            (  5 pts)  The maximum water level in the Barrel. Explicitly articulate this as you did with the Optimum Angle.     

            (20 pts) Titled and Labeled Graph showing Simulation (as a Dotted Line), Theoretical (as a Solid Line),

                        and Percent Error  (as any line).  

 

Solution: 

The Theoretical Water Level or height is :   h(t) = [ C2 / C1 ] * [ 1.0 – exp (- C1*t  ) ]   

: 

    OR    h(t) = [ C2 / C1 ] * [ 1.0 – exp (-t / τ  ) ]   

 

Where

  C1 = Leak Coeff. / density / Area             and           C2 = Flow rate in / density / Area

 

Include your name on the actual spreadsheet also.

 

Everyone does his/her own assignment; this is not like a lab. Ask questions during class or send an e-mail titled “Question about Excel 03 Assignment”.  Let me know if the points do not sum to 100 pts. 

 

Submit results to me as a spreadsheet file; this worked very well for previous Excel assignments. Send an e-mail with the Excel file attached and Excel 03 in the subject of the e-mail. All files are due by Monday, November 25, 2013  11:00pm.  But do not send the e-mail before 6:00pm; this will keep all the e-mails together in my file. They will arrive in one nice batch.  That means send the e-mail to me after 6;00pm and before 11;00pm.  

Thomas J. Crawford, PhD, PE
November 08, 2013
A. Economics:

Pay Back Period or Pay Back Time. ( MAY HAVE ERRORS, TYPOS !!)
If we make this Investment to improve Efficiency or Decrease Costs, how much Time is required to recover the Investment Costs and Expenditures?
Example: Coal–Fired Fossil Power Plant.
Given: Power = 2000 MWth in the Boiler, current efficiency = eff = 38.0 % Cost of Power Output = $0.082/kwhr, (Neglect Fuel and O&M costs)
Find: Pay Back Time if upgrades are made to increase eff to 40.0%, and outage time is 3 months. (boiler tubes replaces, condenser tubes cleaned, turbine upgrade, precipitators, pulverizers, etc. )
Solution:
Analogous to the Tortoise and the Hare running a race.
Considering only the costs of power, neglecting replacement power, Fuel Savings and Capital Investment Costs, etc.
Amount = Rate * time
We set Amount (1) = Amount (2)  R1 * ( t + Δt ) = R2 * t; Δt = outage time
t = R1 * Δt we need R1 and R2, the rates that money is generated.
R2 – R1
R1 = (0.38 * 2000) * 1000 * 0.082 = $62, 320 / hr. or 62, 320 $ / hr.
R2 = (0.40 * 2000) * 1000 * 0.082 = $65, 600 / hr. 65, 600 $ / hr.
t = 62,320 * ( 3 * 30 * 24 )  t = 41,040 hrs = 4.68 years
( 65,600 – 62,320 )
Check:
A1 = R1 * ( t + Δt ) = 62,320 * ( 41,040 + 3*30*24) = $ 2.692E09 = $ 2.692 billion
A2 = R2 * ( t ) = 65.600 * ( 41,040 ) = $ 2.692E09 = $ 2.692 billion
Lecture Data Analysis 04: Page 1
Data Analysis: Economics, Thermodynamics
ENGR1181_Lect_Data_Anal04.ppt

Thomas J. Crawford, PhD, PE
November 02, 2013
A. Economics:, continued

Pay Back Period or Pay Back Time, continued.
Solution:
Analogous to the Tortoise and the Hare running a race.
Both are trying to cover some distance in the same time; the tortoise runs steadily, while the Hare or Rabbit runs rapidly, but is also shutdown for periods.
Note: the cost per day of Outage is below.
Lecture Data Analysis 04: Page 2
Data Analysis: Economics, Thermodynamics
ENGR1181_Lect_Data_Anal04.ppt
Find: Extra Cost for One Additional Day of Outage.
Solution:
t = R1 * ( Δt + 1*24 ) = 62,320 * ( 2160 + 24 ) = 41,496 hrs
R2 – R1 ( 65,600 – 62,320 )
Extra time to catch up = 41,496 – 41,040 hrs = 456 hrs
Extra cost to catch up = 456 * ( 65,600 – 62,320 ) = $1.496E06 = $ 1.5 million
This is 456 hrs (19 days) at the old rate R1 instead of the New Rate R2.
R1 ( Tortoise )

t + Δt
R2 (Hare)

Outage
t
Δt

Thomas J. Crawford, PhD, PE
November 02, 2013
A. Economics:, continued

Pay Back Period or Pay Back Time, continued.
Example: Coal–Fired Fossil Power Plant. continued
Notice:
Total time = t + t = 41,040 + 3*30*24 = 41,040 + 2160 = 43200 hrs and
( 41,040 / 43,200 ) = ( 0.38 / 0.40 ) or
R1 / R2 = eff1 / eff2 because R1 / eff1 = R2 / eff2

Δt = outage time = 3 months = 3*30*24 = 2160 hrs
Lecture Data Analysis 04: Page 3
Data Analysis: Economics, Thermodynamics
ENGR1181_Lect_Data_Anal04.ppt

Thomas J. Crawford, PhD, PE
November 11, 2013
A. Economics:, continued

Pay Back Period or Pay Back Time.
If we make this Investment to improve Efficiency or Decrease Costs, how much Time is required to recover the Investment Costs and Expenditures?
Example: Coal–Fired Fossil Power Plant.
Given: Power = 2000 MWth in the Boiler, current efficiency = eff = 38.0 % Cost of Power Output = $0.082 /kwhr, Cost of Replacement Power = $0.07 /kwhr (Neglect Fuel, O&M, and Capitol costs)
Find: Pay Back Time if upgrades are made to increase eff to 40.0%, and outage time is again 3 months. .
Solution:
Similar to previous Example. Considering only the costs of power and replacement power; neglecting Capital Investment Costs, etc.
Amount = Rate * time
We set Amount (1) = Amount (2)  R1 * (t + Δt ) = R2 * t + Rp * Δt
t = R1 * Δt – Rp * Δt we have R1 and R2, need Rp; Rates of Money.
R2 – R1
R1 = $ 62, 320 $ / hr. R2 = $65, 600 / hr.
Rp = 0.38 * 2000 * 1000 * ( 0.082 – 0.070 )  Rp = $ 9120 / hr.
t = ( 62,320 – 9120 ) * ( 3 * 30 * 24 )  t = 35,034 hrs = 4.00 years
( 65,600 – 62,320 )
Check: ( 3*30*24) = 2160 hrs
A1 = R1 * ( t + Δt ) = 62,320 * ( 35,034 + 2,160 ) = $ 3.0665 E09 = $ 2.318 billion
A2 = R2 * t + Rp * Δt = 65,600 * 35,034 + 9120 * 2160 = $ 2.318 E09 OK
Lecture Data Analysis 04: Page 4
Data Analysis: Economics, Thermodynamics
ENGR1181_Lect_Data_Anal04.ppt

Thomas J. Crawford, PhD, PE
November 11, 2013
A. Economics:, continued

Pay Back Period or Pay Back Time.
Example: Coal–Fired Fossil Power Plant.
Given: Power = 2000 MWth in the Boiler, current efficiency = eff = 38.0 % Cost of Power Output = $0.082/kwhr, Cost of Replacement Power = $0.07 / kwhr Capital Costs = $18.4 E06 (Neglect Fuel and O&M costs)
Find: Pay Back Time if upgrades are made to increase eff to 40.0%, and outage time is again 3 months. .
Solution:
Similar to previous Examples. Considering only the costs of power, replacement power, and Capital Investment Costs.
Amount = Rate * time
We set Amount (1) = Amount (2)  R1 * (t + Δt ) = R2 * t + Rp * Δt – Cp
t = R1 * Δt – Rp * Δt + Cp we have R1, R2, and Rp; Rates of Money.
R2 – R1
R1 = $ 62, 320 $ / hr. R2 = $65, 600 / hr. Rp = $ 9120 / hr. Cp = $ 18.4E06
t = ( 62,320 – 9120 ) * ( 3 * 30 * 24 ) + 18.4E06  t = 40,644 hrs = 4.64 years
( 65,600 – 62,320 )
Check: ( 3*30*24) = 2160 hrs
A1 = R1 * ( t + Δt ) = 62,320 * ( 40,644 + 2,160 ) = $ 2.668 E09
A2 = R2 * t + Rp * Δt – Cp = 65,600 * 40,644 + 9120 * 2160 – 18.4E06 =
$ 2.668 E09 OK
Lecture Data Analysis 04: Page 5
Data Analysis: Economics, Thermodynamics
ENGR1181_Lect_Data_Anal04.ppt

Thomas J. Crawford, PhD, PE
November 11, 2013
A. Economics:, continued

Pay Back Period or Pay Back Time.
Example: Coal–Fired Fossil Power Plant. ( MAY HAVE ERRORS, TYPOS )
Given: Power = 2000 MWth in the Boiler, current efficiency = eff = 38.0 % Cost of Power Output = $0.082/kwhr, Cost of Replacement Power = $0.07 / kwhr Capital Costs = $18.4 E06, Fuel Costs = $1.92 / million Btu (Neglect O&M costs)
Find: Pay Back Time if upgrades are made to increase eff to 40.0%, and outage time is again 3 months. .
Solution:
Similar to previous Examples. Considering only the costs of power, replacement power, Capital Investment Costs, and Fuel Costs (Savings).
Amount = Rate * time
We set Amount (1) = Amount (2)  R1 * (t + Δt ) = R2*t + Rp * Δt +Rf * Δt – Cp
t = R1 * Δt + Rp * Δt + Cp – Rf * Δt we have R1, R2, Rp; Rates of Money.
R2 – R1
R1 = $ 62, 320 $ / hr. R2 = $65, 600 / hr. Rp = $ 9120 / hr.
Rf = $13,104 / hr ( explained page 7) Cp = $ 18.4E06
t = ( 62,320 – 9120 – 13,104 ) * ( 3 * 30 * 24 ) + 18.4E06  t = 32,014 hrs =
( 65,600 – 62,320 ) t = 3.65 years
Check: ( 3*30*24) = 2160 hrs
A1 = R1 * ( t + Δt ) = 62,320 * ( 32,014 + 2,160 ) = $ 2.130 E09
A2 = R2 * t + Rp * Δt + Rf * Δt – Cp =
= 65,600 *32,014 + ( 9120 + 13,104 ) * 2160 – 18.4E06 =
$ 2.130 E09 OK
Lecture Data Analysis 04: Page 6
Data Analysis: Economics, Thermodynamics
ENGR1181_Lect_Data_Anal04.ppt

Thomas J. Crawford, PhD, PE
November 11, 2013
A. Economics:, continued

Pay Back Period or Pay Back Time.
Example: Coal–Fired Fossil Power Plant.
Given: Same Example
Find: Fuel Rate for equation on page 6 if coal costs $1.92 / million Btu .
Solution:
38% efficiency  the Heat Rate. 9700 Btu / kwhr is a reasonable rate.
The conversion factor is 3412 Btu / kwhr and the definition of efficiency.
38% eff.  38 kwe / 100 kwth
3412 Btu * 100 kwth = 8980 Btu / kwhr (electric) (very low)
1.0 kwhr 38 kwe
Rf = 2000 * 0.38 * 1000 * 8980 * $1.92 / 1.00E06 = $ 13,104 / hr
units: —- kwe ——– ( Btu / kwhr ) ( $ / Btu ) = $ / hr
Lecture Data Analysis 04: Page 7
Data Analysis: Economics, Thermodynamics
ENGR1181_Lect_Data_Anal04.ppt

Thomas J. Crawford, PhD, PE
November 02, 2013
A. Economics: continued

Comparing Fuel Costs, the Costs of different types of Fuels
Example: Electric Power.
Given: Coal Cost = $1.94 / million Btu, Natural gas = $4 / million Btu
Find: ??? Costs per tone; still working on this problem.
Solution:
Example: Electric Light Bulbs.
Given: Cost of Electricity = 0.12 / kwhr Incandescent Bulb = $1.02, life = 800 hrs,
Florescent = $12.50, life = 8000 hrs
Find: Payback period for changing over the light bulbs
STILL WORKING ON THIS PROBLEM
Solution:

WORKING HERE NOV 05, 2013
Lecture Data Analysis 04: Page 6
Data Analysis: Economics, Thermodynamics
ENGR1181_Lect_Data_Anal04.ppt

Thomas J. Crawford, PhD, PE
November 09, 2013
A. Economics: continued

Costs of Stand-by Equipment:
Compare the Inventory cost of various parts or pieces of equipment vs. the cost of replacement power and lost revenue due to equipment failure.
Example: Electric Power. WORKING TO DEVELOP SOMETHING HERE
Given:
Find: ??? Costs of Each
Solution:
Example: House and Furnishings or replacement Parts.
Given: Replacement Lights, Garbage Disposal, various sizes of Batteries, Candles, Bottled Water, canned food needed in Emergency situations.
Find: Cost of Inventory; compare to consequences of not having Replacements and backups.
STILL WORKING ON THIS PROBLEM
Solution:
Example: Automobile and Replacement Parts.
Given: Replacement Lights, Battery, Air Filter, Extra Gasoline, Oil, Water, Tires, Fuel Pump, Water Pump, Alternator, Spark Plugs, Seats, Windows, etc.
Find: Cost of Inventory; compare to consequences of not having Replacements and backups.
STILL WORKING ON THIS PROBLEM
Solution:
Lecture Data Analysis 04: Page 6
Data Analysis: Economics, Thermodynamics
ENGR1181_Lect_Data_Anal04.ppt

Thomas J. Crawford, PhD, PE
November 02, 2013
B. Thermodynamics:, continued

Thermodynamic Efficiencies,
Moran 4th ed., Chapter 5, pp 215 – 218.
1. Heat Engine
eff = η = desired output = W = QH – QC
required input = QH QH
QH = W + QC  W = QH – QC
Carnot efficiency for a Reversible Cycle
eff = η = TH – TC η = 1.0 – TH / TC
TH
2. Refrigerator.
C.O.P = β = output = QC = QC
input W QH – QC
QH = W + QC  W = QH – QC
For a Reversible Cycle
C.O.P = β = __ TC _
TH – TC
3. Heat Pump.
C.O.P = γ = output = QH = QH
input W QH – QC
QH = W + QC  W = QH – QC
For a Reversible Cycle
C.O.P = γ = __ TH _
TH – TC
Lecture Data Analysis 04: Page 7
Data Analysis: Economics, Thermodynamics
ENGR1181_Lect_Data_Anal04.ppt
Eng.
Eng.
Eng.
TH
TC
QH
W
QC

TH
TC
QH
W
QC

TH
TC
QH
W
QC

Thomas J. Crawford, PhD, PE
November 02, 2013
B. Thermodynamics:, continued

WORKING HERE NOV. 02, 2013

Example: Coal–Fire Fossil Power Plant. continued
Notice:
Lecture Data Analysis 04: Page 8
Data Analysis: Economics, Thermodynamics
ENGR1181_Lect_Data_Anal04.ppt