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

## 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:

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

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

Three Modes of Heat Transfer [Bradshaw 1993]

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 4.186 J / g °C ySpecific heat of superheated water vapor c pw = 0.444 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]

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 = 0.076 lbm, 1 hr = 60 min

Fluid Phase Change - Latent Heat zenthalpy of superheated water vapor (at reference temperature) yh g,ref = 1061.2 Btu / lbm or 2501.3 J / g z q = m * h y[Btu] = [lbm] * [ Btu / lbm ] y[J] = [g] * [ J / g °C ]

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 = 0.076 lbm, and 1 hr = 60 min

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

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 )

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

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 105-140F 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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