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Energy Transport in HVAC Systems zHVAC Fluids yAir yWater ySteam yRefrigerant zTransport yFlow rate yInitial conditions of fluid yFinal conditions of fluid.

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Presentation on theme: "Energy Transport in HVAC Systems zHVAC Fluids yAir yWater ySteam yRefrigerant zTransport yFlow rate yInitial conditions of fluid yFinal conditions of fluid."— Presentation transcript:

1 Energy Transport in HVAC Systems zHVAC Fluids yAir yWater ySteam yRefrigerant zTransport yFlow rate yInitial conditions of fluid yFinal conditions of fluid

2 Heat Load Estimation zLoads have various sources zLoads are dynamic zAccuracy (close to real value) vs. precision zHand vs. computer calculations

3 Three Modes of Heat Transfer [Bradshaw 1993]

4 Temperature Change - Sensible Heat zSpecific heat: amount of energy (sensible heat) needed to raise 1 unit of substance by 1 unit of temperature yWater c f = 1.0 Btu / lbm °F or J / g °C ySpecific heat of superheated water vapor c pw = Btu / lbm °F or 1.86 J / g °C yDry air c pa = 0.24 Btu / lbm °F or 1.0 J / g °C z q = m * c *  T y[Btu] = [lbm] * [ Btu / lbm °F ] * [°T] y[J] = [g] * [ J / g °C ] * [°C]

5 zq/t = m/t * c *  T z  Q = M * c *  T where yQ = q/t heat flow yM = mass flow after substitution of constants in English units: yfor water: Q = 500 * GPM *  T xusing c f = 1.0 Btu / lbm °F, 1 gallon = 8.35 lbm, and 1 hr = 60 min yfor air: Q = 1.1 * CFM *  T xusing c pa = 0.24 Btu / lbm °F, 1 ft 3 of dry air = lbm, 1 hr = 60 min

6 Fluid Phase Change - Latent Heat zenthalpy of superheated water vapor (at reference temperature) yh g,ref = Btu / lbm or J / g z q = m * h y[Btu] = [lbm] * [ Btu / lbm ] y[J] = [g] * [ J / g °C ]

7 zq/t = m/t * h z  Q = M * h y[Btu/hr] = [lbm/hr] * [Btu/lbm] after substitution of constants in English units: yQ = 1,000 * SFR xwhere SFR= steam flow rate [lbm/hr] yQ = 0.68 * CFM * (W final -W initial ) xusing 7,000 grain = 1 lbm, 1 ft 3 of dry air = lbm, and 1 hr = 60 min

8 Temperature criteria zAnticipated extremes zPercent concept: statistically, weather will be colder (hotter) than the listed condition for only a specified percentage of the hours listed yHeating: 2160 hours/year from December – February yCooling: 2928 hours/year from June – September

9 Calculation of Heating Loads zConduction Q = U * A *  T = 1/R * A *  T zInfiltration and Ventilation Q = 1.1 * CFM *  T zHumidification Q = 0.68 * CFM * (W final -W initial )

10 Calculation of Cooling Loads zConduction zSolar 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 zOutside Air Loads - Infiltration and Ventilation zInternal heat gains Q = 3.41 * P

11 Typical Operating Ranges zHot water for heating ySupply at 160F or 70C yDrop by 20F or 10C yReturn at 140F or 60C zChilled water for cooling ySupply at 40-50F or 5-10C yTemperature rise of 10-15F or 5-8C zWarm air for heating ysupply at F or 40-60C to maintain space at 75F or 24C zChilled air for cooling ysupply at 50-60F or 10-15C zThese take care of the sensible portion of heating and cooling. In addition, you need to control the humidity.


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