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Energy Transport in HVAC Systems

Energy Transport in HVAC Systems. HVAC Fluids Air Water Steam Refrigerant Transport Flow rate Initial conditions of fluid Final conditions of fluid. Heat Load Estimation. Loads have various sources Loads are dynamic Accuracy (close to real value) vs. precision

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Energy Transport in HVAC Systems

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  1. Energy Transport in HVAC Systems • HVAC Fluids • Air • Water • Steam • Refrigerant • Transport • Flow rate • Initial conditions of fluid • Final conditions of fluid

  2. Heat Load Estimation • Loads have various sources • Loads are dynamic • Accuracy (close to real value) vs. precision • Hand vs. computer calculations

  3. Three Modes of Heat Transfer [Bradshaw 1993]

  4. Temperature Change - Sensible Heat • Specific heat: amount of energy (sensible heat) needed to raise 1 unit of substance by 1 unit of temperature • Water cf = 1.0 Btu/lbm °F or 4.186 J/g °C • Specific heat of superheated water vapor cpw = 0.444 Btu/lbm °F or 1.86 J/g °C • Dry air cpa = 0.24 Btu/lbm °F or 1.0 J/g °C • q = m * c *  T • [Btu] = [lbm] * [Btu/lbm °F ] * [°T] • [J] = [g] * [J/g °C] * [°C]

  5. q/t = m/t * c *  T •  Q = M * c *  T where • Q = q/t heat flow • M = mass flow after substitution of constants in English units: • for water: Q = 500 * GPM *  T • using cf = 1.0 Btu/lbm °F, 1 gallon = 8.35 lbm, and 1 hr = 60 min • for air: Q = 1.1 * CFM *  T • using cpa = 0.24 Btu/lbm °F, 1 ft3 of dry air = 0.076 lbm, 1 hr = 60 min

  6. Fluid Phase Change - Latent Heat • enthalpy of superheated water vapor (at reference temperature) • hg,ref = 1061.2 Btu/lbmor 2501.3 J/g • q = m * h • [Btu] = [lbm] * [Btu/lbm] • [J] = [g] * [J/g °C]

  7. q/t = m/t * h •  Q = M * h • [Btu/hr] = [lbm/hr] * [Btu/lbm] after substitution of constants in English units: • Q = 1,000 * SFR • where SFR= steam flow rate [lbm/hr] • Q = 0.68 * CFM * (Wfinal-Winitial) • using 7,000 grain = 1 lbm, 1 ft3 of dry air = 0.076 lbm, and 1 hr = 60 min

  8. Temperature criteria • Anticipated extremes • Percent concept: statistically, weather will be colder (hotter) than the listed condition for only a specified percentage of the hours listed • Heating: 2160 hours/year from December – February • Cooling: 2928 hours/year from June – September

  9. Calculation of Heating Loads • Conduction Q = U * A *  T = 1/R * A *  T • Infiltration and Ventilation Q = 1.1 * CFM *  T • Humidification Q = 0.68 * CFM * (Wfinal-Winitial)

  10. Calculation of Cooling Loads • Conduction • Solar Effects Q = U * A * TETD where TETD total equivalent temp. difference, a function of orientation, time of day, absorption of surface, thermal mass of assembly • Outside Air Loads - Infiltration and Ventilation • Internal heat gains Q = 3.41 * P

  11. Hot water for heating Supply at 160F or 70C Drop by 20F or 10C Return at 140F or 60C Chilled water for cooling Supply at 40-50F or 5-10C Temperature rise of 10-15F or 5-8C Warm air for heating supply at 105-140F or 40-60C to maintain space at 75F or 24C Chilled air for cooling supply at 50-60F or 10-15C These take care of the sensible portion of heating and cooling. In addition, you need to control the humidity. Typical Operating Ranges

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