Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

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
Question: Calculate the heat loss coefficient of a house using the heater's runtime, heat output rating, thermostat setting, average outdoor temperature, number of people, wattage of lights, and heat generated by appliances and equipment during a 3-hour period.

Step by step solution

01

Determine the heater's runtime

Measure the time during the 3-hour period that the heater is on using a watch or timer.
02

Calculate total heat supplied by the heater

Multiply the heater's runtime (in hours) with its heat output rating to obtain the total amount of heat supplied by the heater during that period. \(\mathrm{Heat \thinspace supplied} = \mathrm{Runtime} \times \mathrm{Heat \thinspace output \thinspace rating}\)
03

Estimate the heat generated inside the house

Estimate the heat generation during the given period by taking into account: - The number of people: Assign a value of about \(100\,\mathrm{W}\) per person as the average heat generated by a seated person - The total wattage of lights that were on - The heat generated by the appliances and equipment used: Check the wattage of these items on their label or user's manual Add these values to get the total heat generation in the house.
04

Calculate the average rate of heat loss

Subtract the total heat generated inside the house from the total heat supplied by the heater. Divide this value by 3 hours to find the average rate of heat loss during this period. \(\mathrm{Average \thinspace rate \thinspace of \thinspace heat \thinspace loss} = \frac{\mathrm{Total \thinspace heat \thinspace supplied \thinspace by \thinspace heater} - \mathrm{Total \thinspace heat \thinspace generated \thinspace inside \thinspace house}}{3\,\mathrm{hours}}\)
05

Calculate the heat loss coefficient

Divide the average rate of heat loss by the difference between the indoor temperature (thermostat setting) and the average outdoor temperature during the 3-hour period to find the heat loss coefficient. \(\mathrm{Heat \thinspace loss \thinspace coefficient} = \frac{\mathrm{Average \thinspace rate \thinspace of \thinspace heat \thinspace loss}}{\mathrm{Indoor \thinspace temperature} - \mathrm{Average \thinspace outdoor \thinspace temperature}}\)Express the heat loss coefficient in \(\mathrm{W}/{ }^{\circ}\mathrm{C}\) or \(\mathrm{Btu}/\mathrm{h}\cdot{ }^{\circ}\mathrm{F}\).

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

A \(0.55\)-m-internal-diameter spherical tank made of 1 -cm-thick stainless steel \((k=15 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) is used to store iced water at \(0^{\circ} \mathrm{C}\). The tank is located outdoors at $30^{\circ} \mathrm{C}\( and is subjected to winds at \)8 \mathrm{~km} / \mathrm{h}$. Assuming the entire steel tank to be at \(0^{\circ} \mathrm{C}\) and thus its thermal resistance to be negligible, determine \((a)\) the rate of heat transfer to the iced water in the tank and (b) the amount of ice at $0^{\circ} \mathrm{C}\( that melts during a \)24-\mathrm{h}$ period. The heat of fusion of water at atmospheric pressure is \(h_{i f}=333.7 \mathrm{~kJ} / \mathrm{kg}\). Disregard any heat transfer by radiation.

A thin, square, flat plate has \(1.2 \mathrm{~m}\) on each side. Air at \(10^{\circ} \mathrm{C}\) flows over the top and bottom surfaces of a very rough plate in a direction parallel to one edge, with a velocity of $48 \mathrm{~m} / \mathrm{s}$. The surface of the plate is maintained at a constant temperature of \(54^{\circ} \mathrm{C}\). The plate is mounted on a scale that measures a drag force of \(1.5 \mathrm{~N}\). Determine the total heat transfer rate from the plate to the air.

A heated long cylindrical rod is placed in a crossflow of air at $20^{\circ} \mathrm{C}(1 \mathrm{~atm})\( with velocity of \)10 \mathrm{~m} / \mathrm{s}$. The rod has a diameter of \(5 \mathrm{~mm}\), and its surface has an emissivity of \(0.95\). If the surrounding temperature is \(20^{\circ} \mathrm{C}\) and the heat flux dissipated from the rod is \(16,000 \mathrm{~W} / \mathrm{m}^{2}\), determine the surface temperature of the rod. Evaluate the air properties at \(70^{\circ} \mathrm{C}\).

In flow over blunt bodies such as a cylinder, how does the pressure drag differ from the friction drag?

Engine oil at \(105^{\circ} \mathrm{C}\) flows over the surface of a flat plate whose temperature is \(15^{\circ} \mathrm{C}\) with a velocity of $1.5 \mathrm{~m} / \mathrm{s}\(. The local drag force per unit surface area \)0.8 \mathrm{~m}$ from the leading edge of the plate is (a) \(21.8 \mathrm{~N} / \mathrm{m}^{2}\) (b) \(14.3 \mathrm{~N} / \mathrm{m}^{2}\) (c) \(10.9 \mathrm{~N} / \mathrm{m}^{2}\) (d) \(8.5 \mathrm{~N} / \mathrm{m}^{2}\) (e) \(5.5 \mathrm{~N} / \mathrm{m}^{2}\) (For oil, use $\nu=8.565 \times 10^{-5} \mathrm{~m}^{2} / \mathrm{s}, \rho=864 \mathrm{~kg} / \mathrm{m}^{3}$ )

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