HVAC Design of a Shipping Container HomeJesse BruceME 432

Background-Shipping containers can be transformed cheaply into a habitable space-Easily connected and reinforced using a variety of materials -Highly Customizable

Background-Shipping Container home converted into a habitable space-Orlando, Florida-One Floor: 2 main rooms

-Common room(Kitchen/Living room)- Bathroom

-40ft x 8ft x 8.5ft tall-Façade facing South

-Window and door or W. side -3 Identical Windows in S Side

-Flat roof-Walls:

-Wood Exterior, Colten Steel, Spray Foam with Studs, Drywall

Human Factors/ Environmental Concern

-2 people living in home-MET less than 1.2 as per sedentary activity-Thermal Sensation at “0” for neutral (Fig 4-2, 4-3)-Follow EPA’s ambient air quality standards

-CO2 levels below 900 PPM as per owners Specs

-CO levels below 9PPM-Particulate below 220 /

-Fresh air per person to be 25 cfm/person

Cooling Load: Transmission through

Walls(Calculation #2.1)-Composite Wall

-5’’ wooden exterior-2’’ thick Colten steel-3’’ layer spray foam-Studs-1/2’’ Drywall

-Resistance Diagram to find heat transfer shown-All in Series except spray foam and studs-Cooling load= Heating Load-transmitted heat through walls= 1,384.62 BTU/hr

Cooling Load: Solar Radiation(Calculation #2.2) Orlando, FL

Windows Aug 21, 12:00pm= 11:00am EST EOT= -2.4, A,B,C using chart Find LST, hour angle, Delta Find Incident angle for vertical walls using eq. 7-8 Find Normal Direct Radiation using Eq. 7-15 Use eq. 7-26 to find total radiation, find diffused and direct radiation Then, find the transmitted and absorbed radiation Total absorbed radiation heat= 3,424.8 BTU/hr for windows Total Transmitted radiation heat= 851,231 BTU/jhr

Roof Incident Angle Complement to vertical incident angle to window Compute Total Radiation

Cooling Load: Infiltration(Calculation #2.3) Conditions

Avg wind speed: 7.2 mph Avg Temp: 91.4 degrees F Avg Humidity: 35%

Find for wind using wind speed, wall pressure coeff. and density Find Q for Windward wall by considering cracks of the door and assume window (K=2) for

average fitting Use Fig 6-7 and multiply answer by total length of crack Use same method for side walls with different pressure coefficient (Cp-.6) Find cooling load using total Q, specific volume, and enthalpy Windward Infiltration: 5.6cfm Windward heat infiltration: 8.2 BTU/hr Side Infiltration: .98cfm Side heat infiltration: 1.44 BTU/hr

Minimum Fresh Air Requirement(Calculation #3.1

Filter Details Efficiency 70% Location B (In Supply Duct) Ventilation efficiency: 0.65

Environmental factors/ Concerns Particulate to be below 220 / Avg human produces 125 / Outdoor contaminate concentration= 25 /

Use eq. 4-12 to find the rate of recirculation per person assuming needed outdoor air rate (25cfm/person)

Find new contaminate concentration using 4-12 accounting the cooling load Find total supply of fresh air (cfm) for 2 people Select smallest element (12x24x12) with a pressure loss of .01 in Water in table 4-3. and

calculate CFM using eq 4-10. Lastly, Calculate the number of elements

Air Conditioning Requirement(Calculation #4.1)

Plot known SA, RA, and OA on Psychometric chart using assumed humidity and temperatures.

Using minimum fresh air required () , find the return air ratio to find MA point (using total mass of air)

Find enthalpy of MA point and SA point using Chart to find Use answer and multiply by the total air mass to find the cooling load

and cooling coil capacity Take into account the heat of the fan to find the A/C cooling capacity

using the mass flow rate and specific volume of the return air Find SHR as the latent heat divided by the total cooling load Find Compressor Capacity using P-h diagram from stage 1-2 using

R134a

Psychometric Chart

Conclusion

Cooling Load for AC system based off of Radiation for walls, windows, and roof Infiltration through window and door cracks Human and appliance heat generation(Including AC fan)