How to Calculate KW Requirements for Typical Heater Applications
cheap generic viagra Tank Heating Calculations
When selecting a heater for tank heating application you must first determine whether the application requires that the temperature be maintained or if the temperature needs to be raised. Below are the calculations for each application. You can also visit our website and utilize our online calculator; look for the free calculator link near the top of the page.
https://stetsonpainting.com/whychooseus/ viagra without prescription Maintain Temperature
To calculate the KW required to maintain the temperature of a tank you will need to determine the tanks surface area, process temperature to be maintained, minimum ambient temperature and the R-value of the insulation.
Surface area:
Round tank –
A (ft²) = (2 x p x r x h) + (2 x p x r²)
p = 3.14
r = radius (ft)
h = height (ft)
Rectangular tank –
A (ft²) = 2 x [(l x w) + (l x h) + h x w)]
l = length (ft)
w = width (ft)
h = height (ft)
After determining the tanks surface area the maintain KW can be calculated as follows:
KW = (A x (1/R) x ΔT(°F) x SF)/3412
A = surface area
R = R-value of the insulation
- Use 0.5 as the R-value of an uninsulated steel tank
- See the chart below for typical examples
- R-value = thickness (in.)/k-factor
ΔT = difference between the process set point temperature and lowest ambient temperature
SF = safety factor, recommended 1.2
3412 = conversion of BTU to KW
http://sanfordbiggers.com/bio canadian pharmacy viagra Table 1
over the counter viagra Insulation Type | R-Value/inch of thickness |
Fiberglass | R-3 |
Mineral fiber | R-3.7 |
Calcium silicate | R-2 |
Open-cell polyurethane foam | R-3.6 |
Closed-cell polyurethane foam | R-6 |
Polyisocyanurate spray foam | R-6 |
Example:
A 42’ diameter x 40’ tall crude oil tank with R-6 insulation needs to be maintained at 75°F given a minimum ambient temperature of 10°F.
A = (2 x 3.14 x 21 x 40) + (2 x 3.14 x 21²)
A = 8044.68 ft²
KW = (8044.68 x 1/6 x 65 x 1.2)/3412
KW = 30.65
Raise Temperature
The KW calculation to raise the temperature of a material in a tank (heat-up) starts with the same information required in the maintain application. Additionally we’ll need the weight of the material to be heated, the specific heat of the material and the time required to heat the material from its start temperature to its end temperature. The KW calculation to raise the temperature is as follows:
KWtotal = KWheat-up + KWmaintain
KWheat-up = [(M x Cp x ΔT x SF)/3412]/t
M = weight of the material in pounds
Cp = Specific Heat, see examples in the chart
ΔT = difference between the process set point (end) temperature and the start temperature
SF = safety factor, recommended 1.2
3412 = conversion of BTU to KW
t = time in hours
KWmaintain = (A x (1/R) x ΔT(°F) x SF)/3412
A = surface area
R = R-value of the insulation
- Use 0.5 as the R-value of an uninsulated steel tank
ΔT = difference between the process set point temperature and lowest ambient temperature
SF = safety factor, recommended 1.2
3412 = conversion of BTU to KW
Example:
A 4’ x 6’ x 12’ tank with 1800 gallons of water needs to be heated from 60°F to 95°F in 3 hours. The tank has R-4 insulation and the minimum ambient temperature is 0°F.
To begin we need to convert the gallons of water to pounds:
lbs = G x D1
G = gallons
D1 = lbs per gallon from the chart below
lbs = 1800 x 8.34
lbs = 15,012
If the volume of the tank is stated in cubic feet (ft³) the formula looks like this:
lbs = C x D2
C = Cubic feet of material
D2 = lbs per ft³ from the chart below
Table 2
Material | D1
lbs/gallon |
D2
lbs/ft³ |
Specific Heat |
water | 8.34 | 62.4 | 1 |
#1 fuel oil | 6.8 | 50.5 | 0.47 |
#2 fuel oil | 7.2 | 53.9 | 0.44 |
#3,4 fuel oil | 7.5 | 55.7 | 0.425 |
#5,6 fuel oil | 7.9 | 58.9 | 0.41 |
Bunker C | 8.15 | 61 | 0.5 |
SAE 10-50 weight oil | 7.4 | 55.4 | 0.43 |
ethylene glycol | 9.4 | 70 | 0.55 |
50% ethylene glycol/water | 8.8 | 65.8 | 0.76 |
air | – | 0.073 | 0.24 |
nitrogen | – | 0.073 | 0.25 |
KWheat-up = [(15,012 x 1 x 35 x 1.2)/3412]/3
KWheat-up = 61.6
plus
KWmaintain = (288 x 1/4 x 95 x 1.2)/3412
KWmaintain = 2.4
KWtotal = 64
Calculations for heating air in a duct
Once the volume of air in standard cubic feet per minute (SCFM) and the required temperature rise in °F (ΔT) are known, the required kilowatt rating (KW) of the heater can be determined from the following formula:
KW = (SCFM x ΔT)/3193
Note that CFM is given at standard conditions (SCFM): 80° F and normal atmospheric pressure of 15 psi. The CFM at a higher pressure (P) and inlet air temperature (T) may be calculated as follows:
SCFM = ACFM x (P/15) x [540/(T+460)]
Example:
A drying oven, operating at 25 psia (10 psi gauge pressure), recirculates 3000 CFM of air per minute through a heater which raises its temperature from 350 to 400° F.
To select an appropriate heater:
Step 1: Convert 3000 CFM at 25 psia and 350° F to CFM at standard conditions using the above formula:
3000 x (25/15) x [540/(350°F+460)] = 3333 SCFM
Step 2: Calculate the required KW:
[3333 SCFM x (400°F-350°F)]/3193 = 52 KW
Calculations for circulation heater applications
When calculating the power required to heat a material flowing through a circulation heater, the KW equation shown below can be applied. This equation is based on the criteria that there is no vaporization occurring in the heater. The KW equation incorporates a 20% safety factor, allowing for heat losses of the jacket and piping, variation in voltage and wattage tolerance of the elements.
KW = (M x ΔT x x Cp x S.F.)/3412
Where:
KW = power in kilowatts
M = flow rate in lbs/hr
ΔT = temperature rise in °F (The difference between the minimum inlet temperature and maximum outlet temperature.)
Cp = specific heat in BTU/lb °F
S.F. = safety factor, 1.2
3412 = conversion of BTU to KWH
Water heating example:
We have 8GPM of water with an inlet temperature of 65°F and outlet temperature of 95°F. First, convert the flow rate to lbs/hr.
8 gal | x | 1 ft³ | x | 60 min | = | 64.17 ft³/hr |
min | 7.48 Gal | 1 hr |
Convert to lbs/hr, obtain the density and specific heat from table 2 above.
64.17 ft³/hr x 62.4 lbs/ft³ = 4004 lbs/hr
Now calculate the KW:
KW | = | 4004 lbs/hr x (95-65)°F x 1 BTU/lb °F x 1.2 |
3412 | ||
KW | = | 42 |
Gas heating example:
Air is flowing at 187 CFM and 5 PSIG pressure. It needs to be heated from an inlet temperature of 90°F to an outlet temperature of 250°F. First, convert the flow rate to SCFM using the formula given earlier.
187 x (20/15) x [540/(90°F+460)] = 243.7 SCFM
Convert to lbs/hr, again referring to table 2 for the density and specific heat.
243.7 SFCM | x | 60 min | x | 0.073 lbs | = | 1067.4 lbs/hr |
1 hr | ft³ |
Now calculate the KW:
KW | = | 1067.4 lbs/hr x (250-90)°F x 0.24 BTU/lb °F x 1.2 |
3412 | ||
KW | = | 14.4 |
Hi,I check your blogs named “How to Calculate KW Requirements for Typical Heater Applications – Accutherm” like every week.Your humoristic style is awesome, keep doing what you’re doing! And you can look our website about proxy list.
good explanation on How to Calculate KW Requirements for Typical Heater Applications
you have given nice formula
can I have the same Examples in Kg, Degree Celsius, KW, Meters,…. Dimensions system
my mail is
nshahbo@yahoo.com