It is often convenient to heat buildings with air. Air heating systems may be cost effective if they can be made simple or if they can be combined with ventilation systems. But be aware that due to the low specific heat capacity of air, the use of air for heating purposes is very limited. Large heat loads requires large volumes of air with huge oversized ducts and fans as results. Transporting huge volumes of air also requires a lot of energy.
Required air volume in an air heating system can be calculated as
L = Q / (cp ρ (th - tr)) (1)where
L = air volume (m3/s)
Q = heat loss from the building (kW)
cp = specific heat capacity air - 1.005 (kJ/kgoC)
ρ = density of air - 1.2 (kg/m3)
th = heating air temperature (oC)
tr = room temperature (oC)
As a rule of thumb the heating supply temperature should be in the range 40-50oC. The air flow should be in the range 1-3 times the room volume.
(1) expressed in imperial units:
L = Q / (1.08 (th - tr)) (2)
where
Q = heat (btu/hr)
L = air volume (cfm)
th = heating air temperature (oF)
tr = room temperature (oF)
Air Heating - Temperature Rise Diagram
The diagrams below can be used to estimate heat required to rise temperature in air flows.
SI units - kW, m3/s and deg C
Imperial units - Btu/h, cfm and deg F
- 1 m3/s = 3,600 m3/h = 35.32 ft3/s = 2,118.9 ft3/min (cfm)
- 1 kW (kJ/s) = 859.9 kcal/h = 3,413 Btu/h
- T(oC) = 5/9[T(oF) - 32]
A building with a large room with heat loss 20 kW is heated with air with maximum temperature 50oC. The room temperature is 20oC. The required air volume can be calculated as
L = 20 / (1.005 1.2 (50 - 20))
= 0.55 m3/s
Refrigeration is the withdrawl of heat from a substance or space so that temperature lower than that of the natural surroundings is achieved.
Refrigeration may be produced by
- thermoelectric means
- vapor compression systems
- expansion of compressed gases
- throttling or unrestrained expansion of gases.
Vapor compression systems are employed in most refrigeration systems. Here, cooling is accomplished by evaporation of a liquid refrigerant under reduced pressure and temperature. The fluid enters the compressors at state 1 where the temperature is elevated by mechanical compression (state 2). The vapor condenses at this pressure, and the resultant heat is dissipated to the surrounding. The high pressure liquid (state 3) then passes through an expansion valve through which the fluid pressure is lowered. The low-pressure fluid enters the evaporator at state 4 where it evaporates by absorbing heat from the refrigerated space, and reenters the compressor. The whole cycle is repeated.
The Specific Heat Capacity is the amount of heat required to change a unit mass of a substance by one degree in temperature. The heat supplied to a unit mass can be expressed as
dQ = m c dt (1)
where
dQ = heat supplied (kJ, Btu)
m = mass (kg, lb)
c = Specific Heat Capacity (kJ/kgoC, Btu/lboF)
dt = temperature change (oC, oF)
Expressing Specific Heat Capacity using (1)
c = dQ / m dt (1b)
Converting between Common Units
* 1 Btu/lbmoF = 4186.8 J/kg K = 1 kcal/kgoC
Specific Heat Capacity Gases
There are two definitions of Specific Heat Capacity for vapors and gases:
cp = (δh/δT)p - Specific Heat Capacity at constant pressure (kJ/kgoC)
cv = ( δh/ δT)v - Specific Heat Capacity at constant volume (kJ/kgoC)
Gas Constant
The gas constant can be expressed as
R = cp - cv (2)
where
R = Gas Constant
Ratio of Specific Heat
The Ratio of Specific Heat Capacities is expressed
k = cp / cv (3)
The volumetric flow rate in a heating system can be expressed by the basic equation:
q = h / ( cp ρ dt ) (1)
where
q = volumetric flow rate
h = heat flow rate
cp = specific heat capacity
ρ = density
dt = temperature difference
The basic equation can be modified for the actual units - SI or imperial - and the liquids in use.
Volumetric Water Flow Rate in Imperial Units
For water with temperature 60oF flow rate can be expressed as:
q = h (7.48 gal/ft3) / ((1 Btu/lbm.oF) (62.34 lb/ft3) (60 min/h) dt) (2)
or
q = h / (500 dt) (2b)
where
q = water flow rate (gal/min)
h = heat flow rate (Btu/h)
dt = temperature difference (oF)
For more exact volumetric flow rates for hot water the properties of hot water should be used.
Water Mass Flow Rate in Imperial Units
Water mass flow can be expressed as:
m = h / ((1.2 Btu/lbm.oF) dt) (2c)
where
m = mass flow (lbm/h)
Volumetric Water Flow Rate in SI-Units
For a water heating system the volumetric flow can be expressed in SI-units as:
q = h / ((4.2 kg.oC) (1000 kg/m3) dt) (3)
where
q = water flow rate (m3/s)
h = heat flow rate (kW or kJ/s)
dt = temperature difference (oC)
For more exact volumetric flow rates for hot water the properties of hot water should be used.
Water Mass Flow Rate in SI-units
Mass flow of water can be expressed as:
m = h / ((4.2 kg.oC) dt) (3b)
where
m = mass flow rate (kg/s)
Example - Flow Rate in a Heating System
A water circulating heating systems delivers 230 kW with a temperature difference of 20oC.
The volumetric flow can be expressed as:
q = (230 kW) / ((4.2 kg.oC) (1000 kg/m3) (20oC))
= 2.7 10-3 m3/s
The mass flow can be expressed as:
m = (230 kW) / ((4.2 kJ/kg.oC) (20oC))
= 2.7 kg/s
These are defined by the cooling effect in Btu (not in tons) divided by the power use in watts (not in kW) for the peak day (EER), or the seasonal average day (SEER).
For room air conditioners, the commonly used efficiency ratio is the
EER - Energy Efficiency Ratio
For central air conditioners, it the ratio used is
SEER - Seasonal Energy Efficiency Ratio
These ratings are posted on the Energy Guide Label, which shall be attached to all new air conditioners.
Some of the air conditioner manufacturers participate in the voluntary EnergyStar labeling program where EnergyStar-labeled appliances mean that they have high EER and SEER ratings.
EER - Energy Efficiency Ratio
Room air conditioners in general range from 5,000 Btu per hour to 15,000 Btu per hour. Select room air conditioners with EER of at least 9.0 for mild climates. In a hot climates, select air conditioners with EER over 10.
SEER - Seasonal Energy Efficiency Ratio
For central air conditioners there is units with SEERs reaching nearly 17.
Example - EER
A cooling unit operating at 1 kW/ton would have an EER of 12,000 Btu divided by 1000 watts or 12. This is mathematically equivalent to multiplying the COP by 3.413. Therefore a small cooling unit operating at 1 kW (1000 watts) per ton is equivalent to a COP of 3.516, or an EER of 12.
The sensible heat in a heating or cooling process of air can be expressed as
hs = 1.08 q dt (1)
where
hs = sensible heat (btu/hr)
q = air volume flow (cfm, cubic feet per minute)
dt = temperature difference (oF)
Latent Heat
The latent heat due to moisture in the air can be expressed as:
hl = 0.68 q dwgr (2)
or
hl = 4,840 q dwlb (3)
where
hl= latent heat (btu/hr)
q = air volume flow (cfm, cubic feet per minute)
dwgr = humidity ratio difference (gram water/lb dry air)
dwlb = humidity ratio difference (lb water/lb dry air)
Total Heat - Latent and Sensible Heat
Total heat due to both temperature and moisture can be expressed as:
ht = 4.5 q dh (4)
where
ht= total heat (btu/hr)
q = air volume flow (cfm, cubic feet per minute)
dh = enthalpy difference (btu/lb dry air)
Total heat can also be expressed as:
ht = hs + hl
= 1.08 q dt + 0.68 q dwgr (4)
Example - Heating Air
An air flow of one cfm is heated from 32 to 52oF. Using (1) the sensible heat added to the air can be expressed as:
hs = 1.08 1(cfm) (52 - 32)(oF)
= 21.6 (btu/hr)
In a chilled-water system the air conditioner cools water to between 40 and 45oF (4 and 7oC). The chilled water is distributed throughout the building in a piping system and connected to air condition cooling units wherever needed.
Total Heat Removed
The total heat removed by air condition chilled-water installation can be expressed as
h = 500 q dt (1)
where
h = total heat removed (Btu/h)
q = water flow rate (gal/min)
dt = temperature difference (oF)
Evaporator Flow Rate
The evaporator water flow rate can be expressed as
qe = htons 24 / dt (2)
where
qe = evaporator water flow rate (gal/min)
htons = air condition cooling load (tons)
Condenser Flow Rate
The condenser water flow rate can be expressed as
qc = htons 30 / dt (2)
where
qc = condenser water flow rate (gal/min)
htons = air condition cooling load (tons)
Density
Density is defined as an objects mass per unit volume. Mass is a property.
- Mass and Weight - the Difference! - What is weight and what is mass? An explanation of the difference between weight and mass.
ρ = m / V = 1 / vg (1)
where
ρ = density (kg/m3)
m = mass (kg)
V = volume (m3)
vg = specific volume (m3/kg)
The SI units for density are kg/m3. The imperial (BG) units are lb/ft3 (slugs/ft3). While people often use pounds per cubic foot as a measure of density in the U.S., pounds are really a measure of force, not mass. Slugs are the correct measure of mass. You can multiply slugs by 32.2 for a rough value in pounds.
- Unit converter for other units
property constant at a given temperature and density can help to identify a substance.
- Densities and material properties for common materials
Relative density of a substance is the ratio of the substance to the density of water, i.e.
Example - Use the Density to Identify the Material:
An unknown liquid substance has a mass of 18.5 g and occupies a volume of 23.4 ml. (milliliter).
The density can be calculated as
ρ = [18.5 (g) / 1000 (g/kg)] / [23.4 (ml) / 1000 (ml/l) 1000 (l/m3) ]
= 18.5 10-3 (kg) / 23.4 10-6 (m3)
= 790 kg/m3
If we look up densities of some common substances, we can find that ethyl alcohol, or ethanol, has a density of 790 kg/m3. Our unknown liquid may likely be ethyl alcohol!
Example - Use Density to Calculate the Mass of a Volume
The density of titanium is 4507 kg/m3 . Calculate the mass of 0.17 m3 titanium!
m = 0.17 (m3) 4507 (kg/m3)
= 766.2 kg
Specific Weight
Specific Weight is defined as weight per unit volume. Weight is a force.
* Mass and Weight - the difference! - What is weight and what is mass? An explanation of the difference between weight and mass.
Specific Weight can be expressed as
γ = ρ g (2)
where
γ = specific weight (N/m3)
ρ = density (kg/m3)
g = acceleration of gravity (m/s2)
The SI-units of specific weight are N/m3. The imperial units are lb/ft3. The local acceleration g is under normal conditions 9.807 m/s2 in SI-units and 32.174 ft/s2 in imperial units.
Example - Specific Weight Water
Specific weight for water at 39 oF (4 oC) is 62.4 lb/ft3 (9.81 kN/m3) in imperial units. Specific weight in SI units can be calculates like
γ = 1000 kg/m3 9.81 m/s2
= 9810 N/m3
= 9.81 kN/m3