Lecture 5- Heating Loads

8/12/2019 Lecture 5- Heating Loads

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Khairul Nizar Ismail

EAT 257

7 March 2013


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Objectives Find R and U values for building compenent

Select appropriate indoor and outdoor design

condition Calculate room and building heat transfer losses

Determine room and building heating loads


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The Energy Equation (First Law of

Themodynamics) The first Law of Themodynamics is a principle that

may be stated in various ways for instance energy canneither be created nor destroyed or there isconservation of energy in nature

This principle is used extensively in the HVACindustry especially when stated as an energy balance :

The change in total energy in a system equals theenergy added to the system minus the energy removedfrom the system


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Energy Equation The energy balance can be expressed as an equation,

called the Energy Equation:

Ech

= Ein

- Eout

Ech= Change in stored energy

Ein = Energy added (entering) the system

Eout= Energy removed from (leaving) the system


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Example 1A hot water heating convector in a office is supplying

4000 BTU/hr of heat. Heat is being transferred fromair to the outdoors at rate of 6500 BTU/hr. What willhappen in the room? What size electric heater shouldthe office temporarily use to solve the emergency?


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Solution 1Ech= EinEout

Ech= 4000 BTU/hr 6500 BTU/hr

Ech= - 2500 BTU/hr

The negative sign means the room air energy isdecreasing

The room temperature will be dropped

A solution is to install an electric heater that will makeup the heat loss of 2500 BTU/hr


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Cont Solution 1 The capacity of an electric heater is nomarlly

expressed in watts (W) or kilowatts (KW) rather thenBTU/hr

The heater should therefore have the followingcapacity

3410 BTU/hr = 1000 watts

2500 BTU/hr x 1000 watts = 733 watts3410 BTU/hr


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Example 2A classroom has 1000 watts of lighting and some small

motors with a total output of 10 HP. All of the energyin the lighting and from the motors is converted intoheat. What is the increase in enthalpy of the classroomair from the sources?


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Solution 2 The energy added to the classroom air will increse its

enthalpy

Applying the Energy Equation & converting all units toBTU/hr

Ech= EinEout

Ech= 1000 watts x 3.41 BTU/hr + 10 HP x 2545 BTU/hr - 0

1 watts 1 HPEch= 28, 860 BTU/hr


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Defination Conduction- is the form of heat transfer through a

body that occurs without any movement of the body, itis result of molecular or electron action

Convection- is the form of heat transfer that resultsfrom gross movement of liquids or gases

Thermal Radiation- is the form of heat transfer that

occurs between two separated bodies as a result of ameans called electromagnetic radiation, sometimescalled wave radiation


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Heat Conduction

Key Question:

How does heat passthrough differentmaterials?


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Heat Transfer The science of how heat flows is called heat transfer.

There are three ways heat transfer works: conduction,convection, and radiation.

Heat flow depends on the temperature difference.


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Thermal Equilibrium Two bodies are in thermal

equilibriumwith each other

when they have the sametemperature.

In nature, heat alwaysflowsfrom hot to cold until

thermal equilibrium isreached.


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Heat Conduction Conductionis the transfer of heat through materials

by the direct contact of matter.

Dense metals like copper and aluminum are very goodthermal conductors.


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Heat ConductionA thermal insulatoris a material that conducts heat

poorly.

Heat flows very slowly through the plastic so that thetemperature of your hand does not rise very much.


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Heat Conduction Styrofoam gets its

insulating ability bytrapping spaces of airin bubbles.

Solids usually are betterheat conductors than

liquids, and liquids arebetter conductors thangases.


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Heat Conduction The ability to conduct heat

often depends more on thestructure of a material than

on the material itself. Solid glass is a thermal

conductor when it isformed into a beaker or

cup.

When glass is spun intofine fibers, the trapped airmakes a thermal insulator.


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Thermal Conductivity The thermal conductivityof a material describes how

well the material conducts heat.


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Thermal

Conductivity

Heat conduction insolids and liquidsworks by transferringenergy through bonds

between atoms ormolecules.


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There are three ways heat transferworks: conduction, convection,

and radiation
http://localhost/var/www/apps/conversion/tmp/scratch_5/Real%20Player/What%20is%20Heat%20Transfer,%20Understanding%20Leakage,%20Conduction,%20Convection%20and%20Radiation%20-%20YouTube.flvhttp://localhost/var/www/apps/conversion/tmp/scratch_5/Real%20Player/What%20is%20Heat%20Transfer,%20Understanding%20Leakage,%20Conduction,%20Convection%20and%20Radiation%20-%20YouTube.flvhttp://localhost/var/www/apps/conversion/tmp/scratch_5/Real%20Player/What%20is%20Heat%20Transfer,%20Understanding%20Leakage,%20Conduction,%20Convection%20and%20Radiation%20-%20YouTube.flvhttp://localhost/var/www/apps/conversion/tmp/scratch_5/Real%20Player/What%20is%20Heat%20Transfer,%20Understanding%20Leakage,%20Conduction,%20Convection%20and%20Radiation%20-%20YouTube.flvhttp://localhost/var/www/apps/conversion/tmp/scratch_5/Real%20Player/What%20is%20Heat%20Transfer,%20Understanding%20Leakage,%20Conduction,%20Convection%20and%20Radiation%20-%20YouTube.flvhttp://localhost/var/www/apps/conversion/tmp/scratch_5/Real%20Player/What%20is%20Heat%20Transfer,%20Understanding%20Leakage,%20Conduction,%20Convection%20and%20Radiation%20-%20YouTube.flv


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Rate of Heat Transfer The rate at which heat is conducted through any

material depends on three factors:

1) The temperature difference across which the heatflows

2) The area of the surface through which heat is flowing

3)The thermal resistance (R) of the material to heat

transfer


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Heat transfer equation Q = 1 x A x TD

R

Q = heat tranfer rate (BTU/hr)

R = thermal resistance of material

A = surface area through which heat flow

TD = temperature difference across heat flow


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Example 3 Determine the heat transmission coefficient for

building wall section:

Face brick : 0.44

Concrete : 1.11

Insulation : 11.0

Drywall : 0.32


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Solution 3 U-Factor = 1/R

U-factor = 1/ 0.44+1.11+11.0+0.32

U-factor = 1/12.87

U-factor = 0.077 BTU/h/sq ft/F


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Example 4 Determine the heat loss through the wall described in

Example 3. The wall is 30 feet wide by 12 feet high. Theheating design indoor air is 70 F and heating design

outdoor air is 5 F


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Solution 4 Q = U x A x T

Q = 0.077 BTU/hr/sq ft/F x 360 sq ft x 65 F

Q = 1801.8 BTU/hr

The heat loss through the wall 1801.8 BTU/hr


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Load calculator
http://localhost/var/www/apps/conversion/tmp/scratch_5/Real%20Player/Calculating%20Air%20Conditioning%20load%20with%20MyVirtualHome%20-%20YouTube.flvhttp://localhost/var/www/apps/conversion/tmp/scratch_5/Real%20Player/Calculating%20Air%20Conditioning%20load%20with%20MyVirtualHome%20-%20YouTube.flv

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Lecture 5- Heating Loads