The MET 434 design project report should describe engineering objectives, underlying assumptions for the analysis, actual computations, and summary of results. Final submission length should be 10-12 pages, including any figures or graphs supporting students' designs. PROBLEM STATEMENT Design a home exterior to minimize heat gain during a summer day in Norfolk, VA. Requirements to be fulfilled: ? Home must be 196m2. One story is sufficient. ? All walls must be straight, but the shape may be any that the designer chooses. ? Walls for a one story home are 3m high. ? The interior must be maintained at 22oC. Make reasonable assumptions for the following: ? Temperature and sun received. ? Heat generation within the home, including any parasitic electrical loads. ? Maximum number of people occupying the home. Make sure you can account for any gatherings that may occur such as a birthday party. ? Wall cross-section and layout. Calculate heat loss through the walls and roof. Allowance must be made for six 1.5mx1m windows and two 1.5mx2m windows. Discuss losses associated with doors. Do not include these in your design calculations. Results should include: ? A sketch of the overall layout of the walls, including placement of windows. Discuss the choice of shape of the home and how this affects the heat transfer problem. ? A sketch of the wall cross-section. ? Calculation of thermal resistance of a wall with a window in it, as well as without a window in it. ? A discussion and supporting calculations of heat gained due to radiation. ? Estimate of the amount of heat required to be removed from the house in a day. ? Estimate of the cost to remove that heat based on reasonable thermodynamic analysis.

Use the pictures from part 1. These calculations are the thermal resistance for each type of wall section and the transfer rate as well.

Course Project Revised Wall 1: Insulation + window Ry = 0.78 . ( table 1 -1: Glass) KI= 6.043WK ( table!.1 : Fiber L6 = 50mm glass T2 = 32"C Cp= 30 mm window Insulation Area= 3 m x /m = 3 m? 150mm1 - 30 mm-1 LL 50mn ( 0.001)#7mm RI= KI . A- 0. 3576- 0.043-.3 me 2 = LP Rg . A = 30 mm (0.0012mm )= 0.0128 # 1 . 3 M 2 AT Q = (32 %- 21"C ) RTotal (0.3876 #5-0.0125 5) = 27. 4725 mok Ti = 32 ' C RI = KIA Re= Le 2/27/25 MET 434Course Project Part1 Wall 2: Joist + Window KW = 0.17 Will Table 1 -1 : " Wood and glass ) Ky= 0.78 Lic 71=2IT TL= 32'C Joist Cp = 30 mm Lf = 150 mm A = 3m. Im = 3m2 150 mm - 1-30mm R.= CP 30mm (0.00/m/mm ) Kg. A (0.78-15,( - 3 m 2) = 0-0125 - 150mm ( 0.0017mm ) R2 = (0.17-. 3 M2) = 0.19615 AT ( 32. T2= 32% Insulation (- Window 1 5oml - 150 mm - 1 30m wall 4: T.=210C T2= 32 Governing Equations KI = 0.043 wr(table 1-1 : Fiber glass) KW= 0. 17*K ( table 1-1: wood (oare)) R = K.A Kg = 0. 78-wic ( table 1-1: Glass ) Q= AT 4= Comm ( Length of Insulation Rtotal LF= 150mm ( Length of Structural Joist) LP= 30 mm ( window length ) T-32 Ti =zinc -Mm R= L6 Bys LP Aurea = HXV= 3m x /m = 3m2 KW .A 26 RI = KI . Aarea 50mm (0101 )him - 0.3876 W 0.043 Hik . 3 m 2 Lf R2 = KW ' Aarea 150mm ( 0.001 ) mom - . 196 1 w 0-17 ik . 3 m 2 Lp 30 mm ( 0.001 )2m R3 - Kg . Aurea 0.78 + 3 M 2 = 0.6128 AT (32 %-21'C ) Q = RTotal - 6. 387 ( + 0 . 1961 + 0 . 0128 ) = 18.4409 m.k