me 414 : project 1

ME 414 : Project 1 Heating System for NASA North Pole Project

Team MembersAlan Benedict Jeffrey JonesLaura O’Hair

Aaron Randall

May 5, 2006

Problem Statement

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10

BLO

WER

UN

IT

OXY

GEN

TA

NK

Your job as a Thermal Fluid Systems engineer is to deliver the housing heating system in the North Pole.

4 occupants Oxygen supply tank or circulating fresh air from outside The outside temperature in North Pole is -40C and the desired temperature inside the housing is 25C.You have a space of 12” in the outside walls and 8” in the interior walls.

Deliverables

• Lowest blower cost measured by the least system pressure drop

• Least material cost measured by the number of sheets used

• Least labor cost per the labor rates given• Least operating cost measured by the

cost of maintenance items and monthly natural gas, oxygen, electricity, etc usage.

• Most comfort to occupants measured by the least flow rate variation between registers

Supply Air System

Return Air System

Heat Loss Calculation Assumptions

• All heat loss occurs through exterior walls and roof.

• The structure is perfectly sealed. No transfer of air.

• There is no heat transfer between rooms.• There is not heat transfer to or from the

basement.

Heat Loss Calculations

• Interior Temperature25˚C

• Exterior Temperature-40˚C

• Thermal Conductivity of Wall0.8W/m˚C

• Convection Coefficients– Interior surfaces

• Walls 4.2W/m2˚C• Roof 5.17W/m2˚C

– Exterior• All surfaces

34W/m2˚C

Heat Loss Calculations

Used Resistance Network

ResultsRoof

Walls

Heat LossRoom 1 7791.5WRoom 2 10118.9WRoom 3 6269.2WRoom 4 7380.7WRoom 5 8457.8WTotal 40018.1W

Heat Loss CalculationsWith Insulation Added

Resistance Network

Insulation Conductivityk=0.043W/m2˚C

Results

Roof

Walls

Heat LossRoom 1 489.9WRoom 2 634.3WRoom 3 394.4WRoom 4 466.7WRoom 5 534.8WTotal2520.1W

=8598.9Btu/h

Heat Loss Rates

• Heat loss rate through walls and roof:– 2520W

• Heat loss rate through heating of outside air:– 72W

Insulation Cost Benefit Analysis

• Cost to add insulation:– 12 inches in walls and roof– Total of 3501.1 ft3 insulation required– Cellulose insulation cost $0.387 per ft3

– Total cost to add insulation: $1354.07


• Heat loss rate without insulation:– 40,018.1W

• Heat loss rate with insulation:– 2,520.1W

• Heat loss rate reduction:– 37,498W or 93.7%


• 4 month cost to heat house without insulation– $20,903.11

• 4 month cost to heat house with insulation– $1,316.35

• 4 month savings:– $19,586.75

• Time to recover cost of insulating:– 8.4 days

Fresh Air or Oxygen Tank?

• 4 month analysis of using bottled O2

– 5.3592 x 10-4 m3/s O2 consumption rate– 3000L volume of O2 in tank at 1atm– $1,050 per bottle material– $75 per bottle labor

• COST–$2,112,750

Fresh Air or Oxygen Tank?

• 1.6 ft3/min addition of outside air to interior– 5.3592 x 10-4 m3/s occupants– 7.7794 x 10-4 m3/s burning gas– -40°C air temperature– $0.045/ft3 cost for natural gas

• COST–$32.28

Furnace and Blower

• Gibson KG6RA Series Specifications– 45000 Btu/h– 80% Efficiency– Cost of $543

Furnace and Blower

Blower Electrical Consumption and Cost for 4 months

• Electricity Consumption– 1/5 hp = 149.14W– 149.14W*2880hrs = 429.5kWhrs

• Operational Cost– 429.5kWhrs*$0.4/kWhr = $171.80

Materials

• Duct Diameter– 7.43 inches– 3 ducts per each 90” X 70” sheet

Materials

• Total sheets– 9

• 90 degree bends– 6

• Branches– 9

• Registers– 9

Material and Labor Costs

CIRCULAR DUCTS• Material:

– $2,250.00• Labor:

– $2,400.00• Total

– $4,650.00

SQUARE DUCTS• Material:

– $3,250.00• Labor:

– $2,600.00• Total

– $5,850.00

Problems not Overcome

• Flowmaster– Flow rates in pump do not coincide with

branch flow rate– Flow rates don’t produce results as

expected

Flow Output of Pump Lower than First Branch

Register size vs. output discrepancy

Diameter in inches Output in Ft^3/min.30 3.331.301 3.262.302 3.192.303 3.122.304 3.053.305 12.93.306 14.89.307 17.11.308 19.74.309 22.72.310 14.02

Conclusion

• Least Pressure Drop not achievable through Flowmaster

• Least material cost calculated at $4147

• Least labor cost calculated at $2400• Least operating cost calculated at

$1488• Flow rate variation between registers

not achievable through Flowmaster

Questions?

ME 414 : Project 2Heat Exchanger Optimization

Team MembersAlan Benedict Jeffrey JonesLaura O’Hair

Aaron Randall

May 5, 2006

Problem Statement

Design a heat exchanger to meet the customer requirements for heat transfer and maximum dimensions, while optimizing the weight and pressure losses in both the tube and shell sides.

Project Definition

• Chemical Specifications:– Temperature must be reduced from 35°C to

25°C– Mass flow rate is 80,000 kg/hr– Material properties closely approximate that of

water• Cooling Water Specifications:

– Treated city water at 20°C– Mass flow rate is not fixed– Exit temperature is function of design

Customer Requirements

• Must cool the chemical from 35 C to 25 C• Heat exchanger length can not exceed 7m• Heat exchanger shell diameter can not

exceed 2m• Minimize heat exchanger shell and tube

weight • Minimize heat exchanger pressure drop

Initial Design Specifications

Shell Fluid water (given) Baffle spacing n/a

Tube Fluid water (given) Baffle cut n/a

Mass flow rate shell 38.9 kg/s Shell ID 1.5m

Mass flow rate tube 22.2 kg/s (given) Shell thickness 0.002m

Temp. shell in 20ºC Shell material stainless

Temp. shell out calculated Tube material stainless

Temp. tube in 35ºC (given) Nusselt shell Dittus

Temp. tube out 25ºC (given) Nusselt tube Petukkov-Kirillov

Friction factor tube 0.0001 Tube OD 0.0254m

Friction factor shell 0.0001 Tube thickness 0.00491m

Reverse tube/shell no Tube length 4.6m

Counter flow yes Tube pitch calculated

# tube passes 1 Tube config. square

# shell passes 1 Tube layout 90º

Baffle no

Initial Results

• Desired heat transfer rate of 928,502W

• Calculated heat transfer rate of 924,068W

• Difference of 4,434W• Desired-to-calculated ratio 0.995

First DOE ResultsM

ean

of Q

Cal

c 0.029210.02159

1400000

1200000

1000000

1.7251.275 5.293.91

213

1400000

1200000

1000000

213 10

0.40.1

1400000

1200000

1000000

Tube OD Shell ID Tube Length

Tube Mat Shell Mat Baffle

Baffle Space

Main Effects Plot (data means) for Q Calc

Mea

n of

Wei

ght 0.029210.02159

25000

20000

15000

1.7251.275 5.293.91

213

25000

20000

15000

213 10

0.40.1

25000

20000

15000



Baffle Space

Main Effects Plot (data means) for Weight

Mea

n of

DP

Shel

l 0.029210.02159

300000

150000

01.7251.275 5.293.91

213

300000

150000

0213 10

0.40.1

300000

150000

0



Baffle Space

Main Effects Plot (data means) for DP Shell

Mea

n of

DP

Tube 0.029210.02159

80

60

40

1.7251.275 5.293.91

213

80

60

40

213 10

0.40.1

80

60

40



Baffle Space

Main Effects Plot (data means) for DP Tube

Initial Design Specifications

Shell Fluid water (given) Baffle spacing n/a

Tube Fluid water (given) Baffle cut n/a

Mass flow rate shell 38.9 kg/s Shell ID 1.5m

Mass flow rate tube 22.2 kg/s (given) Shell thickness 0.002m

Temp. shell in 20ºC Shell material stainless

Temp. shell out calculated Tube material stainless

Temp. tube in 35ºC (given) Nusselt shell Dittus

Temp. tube out 25ºC (given) Nusselt tube Petukkov-Kirillov

Friction factor tube 0.0001 Tube OD 0.0254m

Friction factor shell 0.0001 Tube thickness 0.00491m

Reverse tube/shell no Tube length 4.6m

Counter flow yes Tube pitch calculated

# tube passes 1 Tube config. square

# shell passes 1 Tube layout 90º

Baffle no

Final DOE Pareto ChartsTe

rm

Standardized Effect

ABADBC

CDB

ACDA

C

706050403020100

2.45Factor

Baffle

NameA Tube ODB Shell IDC Tube LengthD

Pareto Chart of the Standardized Effects(response is Q Calc, Alpha = .05)

Term

Standardized Effect

DBDAD

CDACBCAB

C

9080706050403020100

2.45Factor

Baffle

NameA Tube ODB Shell IDC Tube LengthD

Pareto Chart of the Standardized Effects(response is Weight, Alpha = .05)

Term

Standardized Effect

AD

A

C

CD

D

20151050

2.23Factor NameA Tube ODC Tube LengthD Baffle

Pareto Chart of the Standardized Effects(response is DP Shell, Alpha = .05)

Term

Standardized Effect

BC

AB

AC

B

C

A

35302520151050

2.26Factor NameA Tube ODB Shell IDC Tube Length

Pareto Chart of the Standardized Effects(response is DP Tube, Alpha = .05)

Final DOE Optimization

Hi

Lo0.00000D

New

Cur

d = 0.99357

Targ: 928502.0Q Calc

d = 0.30544

MinimumDP Tube

d = 0.99933

MinimumDP Shell

d = 0.00000

MinimumWeight

y = 9.286E+05

y = 22.3640

y = 16.7174

y = 2.615E+04

0

1

1.50

3.90

1.70

2.0

0.0216

0.0292Shell ID Tube Len BaffleTube OD

[0.0222] [1.8959] [3.0639] 0Hi

Lo0.00000D

New

Cur

d = 0.77655

Targ: 928502.0Q Calc

d = 0.53962

MinimumDP Tube

d = 0.00000

MinimumDP Shell

d = 0.00000

MinimumWeight

y = 9.239E+05

y = 16.5096

y = 1.404E+04

y = 2.245E+04

0

1

1.50

3.90

1.70

2.0

0.0216

0.0292Shell ID Tube Len BaffleTube OD

[0.0254] [1.8576] [2.9038] 1

Without Baffles With Baffles

Specifications for Optimized Heat Exchanger

• Counter flow design• Stainless steel material for shell and tube• Single pass shell• Single pass tube• Tube OD of 2.22cm (standard size)• Tube length of 3.06m• Tube thickness of 2.40mm• Tube pitch of 3.18cm• Square tube configuration with 90° layout angle• Shell ID of 1.90m• No baffles

Final Results

Initial 1st DOE Final DOE

Heat Transfer (kW) 924.1 1020.8 929.9

Tube-Side Pressure Loss (Pa) 37.54 8.15 22.36

Shell-Side Pressure Loss (Pa) 23.7 14,500 16.72

Weight (kg) 22,902 32,035 26,150

Conclusion

• Met requirement to cool the chemical from 35 C to 25 C

• Tube length of 3.06m 3.06m<7m• Shell diameter of 1.9m 1.9m<2m• Minimized heat exchanger shell and

tube weight 26,150 kg• Minimized pressure drop

– Shell side 16.72 Pa– Tube side 22.36 Pa

Questions?

me 414 : project 1

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