Conduct this experiment to determine the heat loss coefficient of your house or apartment in \(\mathrm{W} /{ }^{\circ} \mathrm{C}\) or \(\mathrm{Btu} / \mathrm{h} \cdot{ }^{\circ} \mathrm{F}\). First make sure that the conditions in the house are steady and the house is at the set temperature of the thermostat. Use an outdoor thermometer to monitor outdoor temperature. One evening, using a watch or timer, determine how long the heater was on during a 3 -h period and the average outdoor temperature during that period. Then using the heat output rating of your heater, determine the amount of heat supplied. Also, estimate the amount of heat generation in the house during that period by noting the number of people, the total wattage of lights that were on, and the heat generated by the appliances and equipment. Using that information, calculate the average rate of heat loss from the house and the heat loss coefficient.

Short Answer

Expert verified
Answer: To determine the heat loss coefficient of a house or apartment, follow these steps: 1. Survey the house: Ensure the house is at the set temperature of the thermostat and note any heat-generating factors like people, lights, and appliances. 2. Collect data: Monitor outdoor temperature and the time the heater was on for a 3-hour period. 3. Calculate heat supplied: Multiply the heat output rating by the time the heater was on (in hours). 4. Estimate heat generation: Assess the heat generated by people, lights, and appliances during the 3-hour period. 5. Calculate the average rate of heat loss: Subtract the total heat generation from the heat supplied by the heater. 6. Determine the heat loss coefficient: Calculate the temperature difference between the inside and outside of the house, then divide the average rate of heat loss by the temperature difference. The heat loss coefficient will be in units of W/°C or Btu/(h·°F).

Step by step solution

01

Surveying the house

First, make sure that the conditions in the house are steady, and that the house is at the set temperature of the thermostat. Take note of how many people are in the house, the wattage of all lights that are on, and any appliances and equipment that generate heat.
02

Collecting data

Monitor the outdoor temperature using an outdoor thermometer. Over a 3-hour period, use a watch or timer to determine how long the heater was on. Also, keep track of the average outdoor temperature during that period.
03

Calculating heat supplied

Using the heat output rating of your heater, which can usually be found in the manufacturer's documentation or on a label on the heater itself, calculate the amount of heat supplied during the time the heater was on. To do this, simply multiply the heat output rating by the time the heater was on (in hours). This will give you the heat supplied in either W or Btu/h.
04

Estimating heat generation

Estimate the amount of heat generation in the house during the 3-hour period by considering the following sources: - Number of people: Multiply the number of people by 100 W (average heat generation per person) or 340 Btu/h. - Lights: Multiply the total wattage of all lights that were on by the number of hours they were on, and convert to Btu/h if necessary. - Appliances and equipment: Estimate the heat generated by appliances and equipment, consulting manufacturer documentation or online resources for typical heat generation values. Add up the heat generation from all these sources to obtain the total heat generation within the house during the 3-hour period.
05

Calculating the average rate of heat loss

To find the average rate of heat loss, subtract the total heat generation from the heat supplied by the heater.
06

Determining the heat loss coefficient

The heat loss coefficient is the ratio between the average rate of heat loss and the temperature difference between the inside and outside of the house. Calculate the temperature difference by subtracting the average outdoor temperature from the thermostat set temperature. Then, divide the average rate of heat loss by the temperature difference to determine the heat loss coefficient in W/°C or Btu/(h·°F).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Consider a person who is trying to keep cool on a hot summer day by turning a fan on and exposing his body to air flow. The air temperature is \(32^{\circ} \mathrm{C}\), and the fan is blowing air at a velocity of \(5 \mathrm{~m} / \mathrm{s}\). The surrounding surfaces are at \(40^{\circ} \mathrm{C}\), and the emissivity of the person can be taken to be \(0.9\). If the person is doing light work and generating sensible heat at a rate of \(90 \mathrm{~W}\), determine the average temperature of the outer surface (skin or clothing) of the person. The average human body can be treated as a 30 -cm-diameter cylinder with an exposed surface area of \(1.7 \mathrm{~m}^{2}\). Evaluate the air properties at film temperature of \(35^{\circ} \mathrm{C}\) and \(1 \mathrm{~atm}\).

In an experiment, the local heat transfer over a flat plate were correlated in the form of local Nusselt number as expressed by the following correlation $$ \mathrm{Nu}_{x}=0.035 \mathrm{Re}_{x}^{0.8} \operatorname{Pr}^{1 / 3} $$ Determine the ratio of the average convection heat transfer coefficient \((h)\) over the entire plate length to the local convection heat transfer coefficient \(\left(h_{x}\right)\) at \(x=L\).

A \(1.8\)-m-diameter spherical tank of negligible thickness contains iced water at \(0^{\circ} \mathrm{C}\). Air at \(25^{\circ} \mathrm{C}\) flows over the tank with a velocity of \(7 \mathrm{~m} / \mathrm{s}\). Determine the rate of heat transfer to the tank and the rate at which ice melts. The heat of fusion of water at \(0^{\circ} \mathrm{C}\) is \(333.7 \mathrm{~kJ} / \mathrm{kg}\).

Air \((k=0.028 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, \operatorname{Pr}=0.7)\) at \(50^{\circ} \mathrm{C}\) flows along a 1 -m-long flat plate whose temperature is maintained at \(20^{\circ} \mathrm{C}\) with a velocity such that the Reynolds number at the end of the plate is 10,000 . The heat transfer per unit width between the plate and air is (a) \(20 \mathrm{~W} / \mathrm{m}\) (b) \(30 \mathrm{~W} / \mathrm{m}\) (c) \(40 \mathrm{~W} / \mathrm{m}\) (d) \(50 \mathrm{~W} / \mathrm{m}\) (e) \(60 \mathrm{~W} / \mathrm{m}\)

Air at 1 atm is flowing in parallel over a \(3-\mathrm{m}-\) long flat plate with a velocity of \(7 \mathrm{~m} / \mathrm{s}\). The air has a free stream temperature of \(120^{\circ} \mathrm{C}\) and the surface temperature of the plate is maintained at \(20^{\circ} \mathrm{C}\). Determine the distance \(x\) from the leading edge of the plate where the critical Reynolds number \(\left(\operatorname{Re}_{c r}=5 \times 10^{5}\right)\) is reached. Then, using the EES (or other) software, evaluate the local convection heat transfer coefficient along the plate. By varying the location along the plate for \(0.2 \leq x \leq 3 \mathrm{~m}\), plot the local convection heat transfer coefficient as a function of \(x\), and discuss the results.

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free