Heating and Air Conditioning I Principles of Heating, Ventilating and Air Conditioning R.H. Howell, H.J. Sauer, and W.J. Coad ASHRAE, 2005 basic textbook/reference material For ME 421 John P. Renie Adjunct Professor – Spring 2009

Chapter 6 – Residential Cooling/Heating Heating Load Methodology. During winter months – sutained periods of cold with small variation. Heat loss constant – peak during early morning – worst case scenario – in absence of solar effect and the presence of people, lighting, appliance. Determine maximum heat loss of each room and space due to transmission and infiltration Design condition based on worse case – set-back thermostat may require some excess capacity Equations used given in Table 6-15

Chapter 6 – Residential Cooling/Heating Heating Load Methodology.

Chapter 6 – Residential Cooling/Heating Heating Load Methodology – General Procedure

Chapter 6 – Residential Cooling/Heating Heating Load Methodology – General Procedure

Chapter 6 – Residential Cooling/Heating Heating Load Methodology – General Procedure

Chapter 6 – Residential Cooling/Heating Heating Load Methodology – General Procedure

Chapter 6 – Residential Cooling/Heating Heating Load Methodology – General Procedure Conductive and convective heat transfer given by …

Chapter 6 – Residential Cooling/Heating Heating Load Methodology – Heating Design Conditions Ideal solution – maximum output capacity = most severe local weather conditions. Not economical – excess capacity for most of design life. Questions to ask concerning outside design temperature Type of structure? Heavy, medium, or light? Is structure insulated? Exposed to high winds? More glass area than usual? Nature of occupancy? Long period of low temperature – non occupancy? Daily temperature fluctuations? Any other heating devices in building? If design temperature difference is exceeded, the indoor temperature will fall – depends on thermal mass Effect of wind should be considered – effects poorly insulated walls and windows – lend to more infiltration

Chapter 6 – Residential Cooling/Heating Heating Load Methodology – Above Ground Exterior Surfaces Above ground surface exposed to outdoor temperatures

Chapter 6 – Residential Cooling/Heating Heating Load Methodology – Below-Grade Surfaces Basement walls and floors depends on temperature between inside and ground, materials, conductivity of surrounding earth Earth has time lag due to thermal inertia – not a steady state type of calculation – approximate method …

Chapter 6 – Residential Cooling/Heating Heating Load Methodology – Below-Grade Surfaces

Chapter 6 – Residential Cooling/Heating Heating Load Methodology – Below-Grade Surfaces

Chapter 6 – Residential Cooling/Heating Heating Load Methodology – Below-Grade Surfaces Thermal conductivity of ground varies widely dependent on the soil type and moisture content Typically, k = 0.8 Btu/hr-ft-F – calculate R = 1.47 for uninsulated concrete walls

Chapter 6 – Residential Cooling/Heating Heating Load Methodology – Below-Grade Surfaces

Chapter 6 – Residential Cooling/Heating Heating Load Methodology – On-Grade Surfaces Concrete surface heated from above or within. Most loss occurs through the perimeter – proportional to the perimeter length Simplified approach …

Chapter 6 – Residential Cooling/Heating Heating Load Methodology – On-Grade Surfaces

Chapter 6 – Residential Cooling/Heating Heating Load Methodology – On-Grade Surfaces

Chapter 6 – Residential Cooling/Heating Heating Load Methodology – Buffer Spaces Heat loss to adjacent unconditioned or semi-conditioned spaces based on partition temperature difference.

Chapter 6 – Residential Cooling/Heating Heating Load Methodology – Infiltration Heat Loss Divided into sensible and latent components Energy to raise temperature up to indoor temperature is the sensible component Energy quantity associated with the net loss of moisture from the space is the latent component. Example 6-3

Chapter 6 – Residential Cooling/Heating Example 6-4

Chapter 6 – Residential Cooling/Heating Example 6-5

Chapter 6 – Residential Cooling/Heating Example 6-5

Chapter 6 – Residential Cooling/Heating Example 6-5

Chapter 6 – Residential Cooling/Heating Example 6-5

Chapter 6 – Residential Cooling/Heating Example 6-6

Chapter 6 – Residential Cooling/Heating Example 6-6

Chapter 6 – Residential Cooling/Heating Example 6-7

Chapter 6 – Residential Cooling/Heating Heating Load Methodology – Heating Load Summary Sheet For a multi-space building – see Figure 6-2

Chapter 6 – Residential Cooling/Heating Heating Load Methodology – Heating Load Summary Sheet For a multi-space building – see Figure 6-2

Chapter 6 – Residential Cooling/Heating Heating Load Methodology – Heating Load Summary Sheet Nomenclature

Chapter 6 – Residential Cooling/Heating Heating Load Methodology – Heating Load Summary Sheet Nomenclature

Chapter 6 – Residential Cooling/Heating Heating Load Methodology – Heating Load Summary Sheet Nomenclature

Chapter 6 – Residential Cooling/Heating Heating Load Methodology – Heating Load Summary Sheet Nomenclature

Chapter 6 – Residential Cooling/Heating Heating Load Methodology – Heating Load Summary Sheet Nomenclature

Chapter 6 – Residential Cooling/Heating Example Load Calculation … Atlanta Georgia Home Floor plan

Chapter 6 – Residential Cooling/Heating Example Load Calculation … Atlanta Georgia Home House characteristics

Chapter 6 – Residential Cooling/Heating Example Load Calculation … Atlanta Georgia Home House characteristics

Chapter 6 – Residential Cooling/Heating Example Load Calculation … Atlanta Georgia Home Design Conditions

Chapter 6 – Residential Cooling/Heating Example Load Calculation … Atlanta Georgia Home Component Characteristics

Chapter 6 – Residential Cooling/Heating Example Load Calculation … Atlanta Georgia Home Opaque Surface Factors – Table 6-22

Chapter 6 – Residential Cooling/Heating Example Load Calculation … Atlanta Georgia Home Opaque Surface Factors – Table 6-22

Chapter 6 – Residential Cooling/Heating Example Load Calculation … Atlanta Georgia Home Window Factors – Table 6-23 Note that some references are refering to equation numbers in Chapter 29 of the 2005 Fundamentals – I am trying to acquire this ASAP.

Chapter 6 – Residential Cooling/Heating Example Load Calculation … Atlanta Georgia Home Window Factors – example of 3 ft – west facing window

Chapter 6 – Residential Cooling/Heating Example Load Calculation … Atlanta Georgia Home Window Factors – example of 3 ft – west facing window

Chapter 6 – Residential Cooling/Heating Example Load Calculation … Atlanta Georgia Home Envelope Loads

Chapter 6 – Residential Cooling/Heating Example Load Calculation … Atlanta Georgia Home Infiltration and Ventilation Procedure discussed in text Table 6-3 for unit leakage, determine A L = 77 in 2 Heating and cooling IDF estimated

Chapter 6 – Residential Cooling/Heating Example Load Calculation … Atlanta Georgia Home Air changes per hour determined Ventilation determined

Chapter 6 – Residential Cooling/Heating Example Load Calculation … Atlanta Georgia Home Determine sensible infiltration/ventilation loads

Chapter 6 – Residential Cooling/Heating Example Load Calculation … Atlanta Georgia Home Internal Gains Distribution Losses and Total Sensible Load

Chapter 6 – Residential Cooling/Heating Example Load Calculation … Atlanta Georgia Home Latent load determined to be 2,565 Btu/hr Equation 6-2 with C l = 4,840, Qvi,c = 87 cfm, Qbal,oth = 0, and  W = to give infiltration/ventilation latent load of 2187 Btu/hr. Equation 6-23 gives latent load from internal gains of 378 Btu/hr