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This Old House By: Steve White Ryan Anderson

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Thermal Resistance of the House Denoted by R Value It is the ability of an object or material to resist heat transfer through it. Heat into the house (in the form of natural gas) Heat out of the house

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Governing Equation q= T*A s /R eqv. To find the equivalent thermal resistance of the house, solve for R eqv. R eqv. depends on the materials used to construct the house, the dimensions of the walls and roof, the temperature difference from inside to outside the house, as well as many other factors including the size, shape and quantity of windows.

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Data & Calculations Decatherms Used Watts *Adjusted Watts Temperature Difference R Value$ Spent October3.51424.6511111.2288.3731.89004126.25 November12.75169.4484032.1714.8730.92523795.25 December14.86024.2394698.90619.8731.060865111 January13.95657.94413.16220.8731.186392104.25 February11.74762.4053714.67617.3731.17313387.75 March8.53459.8672698.69612.3731.15004463.75 April6.32564.3722000.218.8731.11272647.25 May2.2895.495698.48613.8731.39085916.5 Total : $552 *Adjusted Watts Watts Decatherms Used $ Spent% Savings October0.418650.5367310.3307572322.4890.55 November0.743650.9533970.5875256554.4195.37 December0.993651.273910.7850398275.8994.70 January1.043651.3380130.8245426626.1894.07 February0.868651.1136540.6862827415.1594.13 March0.618650.7931410.488768573.6794.25 April0.443650.5687820.3505086492.6394.44 May0.193650.2482690.1529944771.1593.05 Total : $31.5494.25 Collected Data from monthly bills and Temperatures from Almanac Calculations based on an R Value of 20 * Adjustment based on 78% efficient furnace

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Actual House Walls constructed of bricks and plaster in the 1950s Low quality single pane windows Cost $552 to heat from October to May

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Standard R Value Home Constructed of newer materials with an overall R value of 20 (which is typical for newly built homes) Would cost $31.54 to heat the home from October to May

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Application By raising the thermal resistance of a home, the cost of heating can be reduced significantly In this instance, the heating cost could be lowered by 94.25% by simply raising the thermal resistance to standard levels

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