The First Law of Thermodynamics

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SUMMARY

The discussion focuses on calculating the daily gas consumption required to maintain an indoor temperature of 23.5°C in a house with specific thermal properties. The average thermal conductivity is 0.480 W/m · °C, and the heat of combustion for natural gas is 9,300 kcal/m³. The participant correctly applied the rate of heat transfer formula, P = (kA ΔT)/Δx, to determine the total heat loss, but was advised to simplify the process by calculating the total area first. The final step involves converting the total energy loss into cubic meters of gas burned.

PREREQUISITES
  • Understanding of thermal conductivity and its units (W/m · °C)
  • Familiarity with the heat transfer equation P = (kA ΔT)/Δx
  • Knowledge of unit conversions between joules, calories, and kilocalories
  • Basic geometry for calculating surface areas of walls and roofs
NEXT STEPS
  • Research methods for calculating heat loss in buildings
  • Learn about thermal insulation materials and their properties
  • Explore energy efficiency strategies for residential heating
  • Study the principles of combustion and energy content of fuels
USEFUL FOR

Students studying thermodynamics, engineers involved in building design, and anyone interested in energy efficiency in residential heating systems.

rico22
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Homework Statement


The average thermal conductivity of the walls (including the windows) and roof of a house is 0.480 W/m · °C, and their average thickness is 24.7 cm. The house is heated with natural gas having a heat of combustion (that is, the energy provided per cubic meter of gas burned) of 9 300 kcal/m3. How many cubic meters of gas must be burned each day to maintain an inside temperature of 23.5°C if the outside temperature is 0.0°C? Disregard radiation and the energy lost by heat through the ground.

the Side walls are 5m x 8 m and the front and back walls are 5m x 10 m respectively. The roof has an incline of 37 degrees between the roof and the side of the house.



Homework Equations


rate of heat transfer P = (k(thermal conductivity)A ΔT)/Δx


The Attempt at a Solution


I tried finding the rate of heat transfer for the house. side wall, front wall and the roof and I added them up. then I found how many total joules were transferred for the day as this give me Joules/ sec. After I got this I converted it into cal by multiplying it by .239 and then kcal. finally I divided the total amount of kcal by the 9300 kcal/m3 given. I just want to know if I am on the right track or am I way off? Thanks in advance.
 
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Yes, that's the right procedure. For the first step, I would have just added the areas rather than calculating the rates separately then adding them up. Can't check the details without seeing your working.
 
Thanks for the reply
 

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