When you get an Energy Audit on your home, inevitably, one of the greatest \"lea

Question

When you get an Energy Audit on your home, inevitably, one of the greatest "leak points" for energy (heat) is through windows and doors. We are interested in the heat losses associated with a house window. Each window measures 1.0 m^2, and the thickness of the glass window pane is 0.32 cm. The inside temperature is maintained at 20 degree C and the outside temperature is a chilly -10 degree C. As a starting point for analysis (BASE CASE), we will assume that the inside temperature is also the temperature on the surface of the glass pane facing the interior, and the outside temperature is also the temperature of the of the glass pane surface facing the outdoors (no convection). We will also assume a steady-state energy balance, and one-dimensional heat flux in the x-direction through the window. For all cases include the following: Sketch and label the problem. State all assumptions. Show all calculations, including UNITS. Determine thermal conductivities at the average system temperature (using tables in App. H and I). Calculate the resistances to heat transfer (R_i). Compare and comment on most significant. Calculate the rate of heat transfer (q_x) leaving the window. Calculate and sketch the temperature profile through the composite structure. Compare the new CASE to the BASE CASE (% increase/decrease in heat transfer rate). Comment Cost/Benefit Analysis based on Energy Loss Calculations BASE CASE: Single glass pane window described above CASE #1 Consider that there are boundary layers (convection) on the inside and outside of the window. Table 15.1 Inside: free convection h_t = 10 W/m^2-K; outside forced convection (wind) h_o = 50 W/m^2-k. CASE #2: Double-Pane window - air space between the two glass window panes is 2.5 cm thick. Assume stagnant air between window panes. CASE #3 Install DuckBrand Plastic Window Film Insulation A low cost alternative is to use standard 2 mil high density polyethylene shrink film (HDPE, k = 0.5 W/m-K). (1 mil = 0.001") DuckBrand Plastic Window Film Insulation Top window: 10 cm air space; Bottom window: 5cm air space (assume stagnant air) CASE #4 It has been shown that if the air space is > 2.5 cm, free convection in the air space will become important. Recalculate Case #3 assuming h_ar = 5 W/m^2-K in the air space.

Explanation / Answer

solution:

1)here for given window frame that each frma consist of two window pane in parallel to heat flow and givn data is

Tin=293 K

To=263 K

dT=To-Tin=30 K

2) where thermal conductivity from standard table we get that

Kair=.024 w/mk

Kglass=Kg=.8 w/mk

A1=A2=.5 m2

A=A1+A2=1 m2

3)here for base case we have that

structure diagram as heat flow is

Qin=(Q1pane+Q2pane)=Qout

if we neglect base convection at window we get resistance to flow as

1/R=(1/R1)+(1/R2)

R1=x1/Kg*A1

R2=x2/Kg*A2

for x1=x2=.32*10^-2 m

we get R=4*10^-3 K/m

where heat flow is given by

Q=(Tin-To)/R=7500 watt

where temperature profile is given by straight line from inside to outside

4)for case 1

we have that

here for base case we have that

structure diagram as heat flow is

Qin=Qconvectionin=(Q1pane+Q2pane)=Qconvectiono=Qout

if we consider base convection at window we get resistance to flow as

1/R0=(1/R1)+(1/R2)

R1=x1/Kg*A1

R2=x2/Kg*A2

for x1=x2=.32*10^-2 m

we get R0=4*10^-3 K/m

where total resistance as it is in series is given by

Rci=1/hiA=.1

Rco=.02

R=Rci+Rco+R0=.124 k/m

where heat flow is given by

Q=(Tin-To)/R=242.9354 watt

where temperature profile is given by straight line from inside to outside through all regions

5)for case 2

here for base case we have that

structure diagram as heat flow is

Qin=Qconvectioni=(Qpane1+Qpane2)=Qair=(Q1pane+Q2pane)=Qconvectiono=Qout

if we consider base convection at window we get resistance to flow as

1/R=(1/R1)+(1/R2)

R1=x1/Kg*A1

R2=x2/Kg*A2

for x1=x2=.32*10^-2 m

we get R0=4*10^-3 K/m

total resistance is given by

R1=x3/Kair*A=1.04166

Rci=.1

Rco=.02

R=2*R0+R1+Rci+Rco=1.1696 k/m

where heat flow is given by

Q=(Tin-To)/R=25.6497watt

where temperature profile is given by straight line from inside to outside

6)for case 3

by taking in consideration of for space of air more that 2.5 cm hence we have to add convection resistance for double pan window along with film of HDPe resistance

here for base case we have that

structure diagram as heat flow is

Qin=Qconvectioni=Qfilmi=(Q1pane+Q2pane)=[(Qca1+Qcca1+Qca1)+(Qca2+Qcca2+Qca2)]=(Qpane1+Qpnae2)=Qfilmo=Qconvectiono=Qout

window we get resistance for apne as to flow as

1/R0=(1/R1)+(1/R2)

R1=x1/Kg*A1

R2=x2/Kg*A2

for x1=x2=.32*10^-2 m

we get R0=4*10^-3 K/m

Rfilm=x4/Kf*A=.0508

Qair gap resiatance for pane 1

Rca1=1/ha*A1=.4

Rcca1=x5/ka*a1=8.33

for pane 1

R11=9.1333

for pane 2

Rcca2=x6/ka*A2=4.1666

R22=4.9666

pane resistances are

1/Rr=1/R11+1/R22

Rr=3.2171

total resistance is

R=Rr+2*Rfilm+2*Ro+Rci+Rco=3.4467 k/m

where heat flow is given by

Q=(Tin-To)/R=8.7039 watt

where temperature profile is given by straight line from inside to outside

7)in this way we get minimum resistance for base case and getting maximum resistance for case 3 and thus third case is more suitable to avoid energy wastage

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