Latent Heat Lecture Notes
.pdf
School
University of Massachusetts, Amherst**We aren't endorsed by this school
Course
BCT 211
Subject
Chemistry
Date
Dec 18, 2024
Pages
2
Uploaded by CorporalMink4812
BCT 211 Notes. 9/29/20 Latent Heat and Enthalpy 1 Latent Heat of phase change and total enthalpy • Sensible Heat: heat you can sense • Latent Heat: the energy needed to change a substance to a higher state of matter. This same energy is released from the substance when the change of state (or phase) is reversed. • Total Enthalpy: both sensible heat and latent heat combined • Applications to humidity in buildings, the refrigeration cycle used in heat pumps (e.g., refrigerators, air conditioners, dehumidifiers), and understanding how steam heat works, and why condensing boilers are efficient. Condensing Gas boilers and furnaces take advantage of large latent heat in steam to capture most of the energy in exhaust gases. Evaporative cooling: Add moisture to air while maintaining same total enthalpy (adiabatic). Sensible temperature drops as heat is transferred to the phase change of liquid water into water vapor. Latent Cooling load (amount of energy required to remove moisture from air) • Rate of energy use to remove a latent load depends on: • Humidity ratio (how much moisture/unit of air) ω• Air flow (how much air) • Enthalpy of condensation per unit of air (h)
BCT 211 Notes. 9/29/20 Latent Heat and Enthalpy 2 Simplified Equation: QL= Δω x CFM x 60 x 0.075 x 1076Consolidate constants: 60 x 0.075 x 1076 = 4842 QL= Δω x CFM x 4842Your psychrometric chart may have ω in “grains/lb” There are 7000grains/lb, so 4842/7000=.68 QL= Δω x CFM x 0.68 Example:• An air conditioned house with air leakage at a rate of 100 CFM with indoor set point at 75°F and 50% RH. Outside air is 88°F and 50% RH. How much latent heat must be removed each hour due to air leakage? • Solution: • Humidity ratio (ω) lbs water / lb dry air (or kg/kg) • Outside: 88°F, 50%Rh: ω = 0.0148 • Set Point: 75°F, 50% Rh: ω = 0.0094 • ∆ ω = 0.0148 –0.0094 = 0.0054 • QL= Δω x CFM x 4842• ω(inside)= 0.0094 • ω(outside)=0.0148 • Δω=0.0054 • QL= 0.0054 x 100 x 4842• = 2614.68Btu/h Alternative approach: QR=ma[(h1-h2)-(⍵1-⍵2)hf2) QR = refrigeration energy required ma= mass of air h1 = enthalpy at state point 1 h2 = enthalpy at state point 2 ⍵1 = humidity ratio at state point 1 ⍵2 = humidity ratio at state point 2 hf2 = enthalpy in condensate