35 | P a g eC h e s s48-In order to conserve energy, it is decided to increase the R-value of an outside wall from 0.36 (m^2*Celsius)/(J/s) by addingan inexpensive layer of sheathing on the inside of the wall whichhas an R-value of 0.62 (m^2*Celsius)/(J/s). When thetemperature difference between the inside and outside is 20°C,by what percentage is the rate of energy transfer through the wallreduced by this added insulation?a. 37%b. 58%c. 63%d. this cannot be found without knowing the area of the wallANS:C
Question
35 | P a g eC h e s s48-In order to conserve energy, it is decided to increase the R-value of an outside wall from 0.36 (m^2Celsius)/(J/s) by addingan inexpensive layer of sheathing on the inside of the wall whichhas an R-value of 0.62 (m^2Celsius)/(J/s). When thetemperature difference between the inside and outside is 20°C,by what percentage is the rate of energy transfer through the wallreduced by this added insulation?a. 37%b. 58%c. 63%d. this cannot be found without knowing the area of the wallANS:C
Solution
The R-value is a measure of thermal resistance used in the building and construction industry. The higher the R-value, the greater the insulation's effectiveness.
Here's how to solve the problem:
-
First, find the total R-value for the wall with the added insulation. This is done by adding the R-value of the wall and the R-value of the insulation.
R_total = R_wall + R_insulation R_total = 0.36 (m^2Celsius)/(J/s) + 0.62 (m^2Celsius)/(J/s) = 0.98 (m^2*Celsius)/(J/s)
-
The rate of energy transfer (Q) through a wall can be calculated using the formula:
Q = ΔT / R
where ΔT is the temperature difference across the wall and R is the total R-value of the wall.
-
Calculate the initial rate of energy transfer (Q_initial) using the original R-value of the wall:
Q_initial = ΔT / R_wall Q_initial = 20°C / 0.36 (m^2*Celsius)/(J/s)
-
Calculate the final rate of energy transfer (Q_final) using the total R-value of the wall with the added insulation:
Q_final = ΔT / R_total Q_final = 20°C / 0.98 (m^2*Celsius)/(J/s)
-
The percentage reduction in the rate of energy transfer due to the added insulation can be calculated using the formula:
Reduction% = ((Q_initial - Q_final) / Q_initial) * 100%
After calculating, you will find that the rate of energy transfer through the wall is reduced by approximately 63% with the added insulation. So, the answer is (c) 63%.
Similar Questions
4: A wall 0.12 m thick having thermal diffusivity of 15 10-6 m2/s is initially at temperature of 85oC.Suddenly one face temperature lowered to a temperature of 20 oC while other face is perfectly insulated.Using explicit finite difference method, write descretization equation for space and time steps of 30 mmand 300 s respectively.
A wall 0.12 m thick having thermal diffusivity of 15 10-6 m2/s is initially at temperature of 85oC.Suddenly one face temperature lowered to a temperature of 20 oC while other face is perfectly insulated.Using explicit finite difference method, write descretization equation for space and time steps of 30 mmand 300 s respectively.
An air-conditioning system operating on the reversed Carnot cycle is required to remove heat from the house at a rate of 32 kJ/s to maintain its temperature constant at 20 degrees C. If the temperature of the outdoors is 35 degrees C, the power required to operate this air-conditioning system isQuestion 5Select one:a.0.58 kWb.3.20 kWc.1.56 kWd.2.26 kWe.1.64 kW
What is ΔSsurr when 40kJ of heat is removed from the system at 200K?Group of answer choices200 J/K- 0.2 J/K- 200 J/K0.2 J/K
Which of the following will NOT decrease the rate at which thermal energy is lost from a house?Replacing single-glazed windows with double or triple-glazed onesInstalling loft insulationIncreasing the temperature inside the houseUsing draught excluders at the bottoms of external doors
Upgrade your grade with Knowee
Get personalized homework help. Review tough concepts in more detail, or go deeper into your topic by exploring other relevant questions.