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las tion (kins = 0.04 W/(mK)) in between. ue depth of the beams is actually 5.625 , but we will take the thickness of the walls eiling to be Lwall = 18 cm to allow for the r and exterior covering. Assume that the is a cube of length L = 9.0 m on a side. he that the roof has very high conductivity, so e air in the attic is at the same temperature outside air. Ignore heat loss through the d. The effective thermal conductivity of the or ceiling) keff , is the area-weighted average thermal conductivities of the wooden beams wood glass o fiborglacs inculation that mako up nch of

las tion (kins = 0.04 W/(mK)) in between. ue depth of the beams is actually 5.625 , but we will take the thickness of the walls eiling to be Lwall = 18 cm to allow for the r and exterior covering. Assume that the is a cube of length L = 9.0 m on a side. he that the roof has very high conductivity, so e air in the attic is at the same temperature outside air. Ignore heat loss through the d. The effective thermal conductivity of the or ceiling) keff , is the area-weighted average thermal conductivities of the wooden beams wood glass o fiborglacs inculation that mako up nch of

College Physics
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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In this problem you will estimate the heat lost by a typical house, assuming that the temperature inside is Tin = 20°C and the temperature outside is Tout = 0°C. The walls and uppermost ceiling of a typical house are supported by 2 x 6-inch wooden beams (kwood = 0.12 W/(mK)) with fiberglass insulation (kins = 0.04 W/(mK)) in between. The true depth of the beams is actually 5.625 %3D %3D %3D inches, but we will take the thickness of the walls and ceiling to be Lwall = 18 cm to allow for the interior and exterior covering. Assume that the house is a cube of length L = 9.0 m on a side. Assume that the roof has very high conductivity, so that the air in the attic is at the same temperature as the outside air. Ignore heat loss through the ground. The effective thermal conductivity of the wall (or ceiling) keff , is the area-weighted average of the thermal conductivities of the wooden beams and the fiberglass insulation that make up each of them. Allowing for the fact that the 2 x 6 beams are actually only 1.625 inches wide and are spaced 16 inches center to center, a calculation of this conductivity for the walls yields keff = 0.048 W/(mK). For simplicity, assume that the ceiling also has the same value of keff -
Transcribed Image Text:In this problem you will estimate the heat lost by a typical house, assuming that the temperature inside is Tin = 20°C and the temperature outside is Tout = 0°C. The walls and uppermost ceiling of a typical house are supported by 2 x 6-inch wooden beams (kwood = 0.12 W/(mK)) with fiberglass insulation (kins = 0.04 W/(mK)) in between. The true depth of the beams is actually 5.625 %3D %3D %3D inches, but we will take the thickness of the walls and ceiling to be Lwall = 18 cm to allow for the interior and exterior covering. Assume that the house is a cube of length L = 9.0 m on a side. Assume that the roof has very high conductivity, so that the air in the attic is at the same temperature as the outside air. Ignore heat loss through the ground. The effective thermal conductivity of the wall (or ceiling) keff , is the area-weighted average of the thermal conductivities of the wooden beams and the fiberglass insulation that make up each of them. Allowing for the fact that the 2 x 6 beams are actually only 1.625 inches wide and are spaced 16 inches center to center, a calculation of this conductivity for the walls yields keff = 0.048 W/(mK). For simplicity, assume that the ceiling also has the same value of keff -
Let us assume that the winter consists of 150 days in which the outside temperature is 0° C. This will give the typical number of "heating degree days" observed in a winter along the northeastern US seaboard. (The cumulative number of heating degree days is given daily by the National Weather Service and is used by oil companies to determine when they should fill the tanks of their customers.) Given that a gallon (3.4 kg) of oil liberates Qg = 1.4 x 10° J when burned, how much oil will be needed to supply the heat lost by conduction from this house over a winter? Assume that the heating system is 75% efficient. Give your answer numerically in gallons to two significant figures. • View Available Hint(s) ? Gallons consumed = gallons per winter
Transcribed Image Text:Let us assume that the winter consists of 150 days in which the outside temperature is 0° C. This will give the typical number of "heating degree days" observed in a winter along the northeastern US seaboard. (The cumulative number of heating degree days is given daily by the National Weather Service and is used by oil companies to determine when they should fill the tanks of their customers.) Given that a gallon (3.4 kg) of oil liberates Qg = 1.4 x 10° J when burned, how much oil will be needed to supply the heat lost by conduction from this house over a winter? Assume that the heating system is 75% efficient. Give your answer numerically in gallons to two significant figures. • View Available Hint(s) ? Gallons consumed = gallons per winter
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