Thermo-chiller
Water bath
V
After 15 minutes,
cool 32
°
C down to 20
°
C.
20
°
C
Q x
t: Heat capacity [kJ]
Required Cooling Capacity Calculation
Example 3: When there is no heat generation, and when cooling the object below a certain temperature and period of time.
Heat quantity by cooled substance (per unit time)
Q : Unknown [W] ([J/s])
Cooled substance
: Water
Cooled substance mass
m
: (=
ρ
x
V) [kg]
Cooled substance density
ρ
: 1 [kg/L]
Cooled substance total volume
V
: 150 [L]
Cooled substance specific heat
C
: 4.186 x 10
3
[J/(kg·K)]
Cooled substance temperature when cooling begins
T
0
: 303 [K] (30 [
°
C])
Cooled substance temperature after t hour
T
t
: 293 [K] (20 [
°
C])
Cooling temperature difference
i
T
: 10 [K] (=
T
0 –
T
t
)
Cooling time
i
t
: 900 [s] (= 15 [min])
*
Refer to the following for the typical physical property values by circulating fluid.
Q = =
= = 6977 [J/s]
≈
7.0 [kW]
Cooling capacity = Considering a safety factor of 20%,
7.0 [kW] x 1.2 =
m x C x (T
0
– T
t
)
i
t
ρ
x V x C x
i
T
i
t
1 x 150 x 4.186 x 10
3
x 10
900
8.4 [kW]
Precautions on Cooling Capacity Calculation
1. Heating capacity
When the circulating fluid temperature is set above room temperature, it needs to be heated by the thermo-chiller. The heating capacity depends on the circulating
fluid temperature. Consider the radiation rate and heat capacity of the customer equipment and check beforehand if the required heating capacity is provided.
2. Pump capacity
<Circulating fluid flow rate>
Circulating fluid flow rate varies depending on the circulating fluid discharge pressure. Consider the installation height difference
between the thermo-chiller and the customer equipment, and the piping resistance such as circulating fluid pipings, or piping size, or
piping curves in the machine. Check beforehand if the required flow is achieved, using the pump capacity curves.
<Circulating fluid discharge pressure>
Circulating fluid discharge pressure has the possibility to increase up to the max. pressure in the pump capacity curves.
Check beforehand if the circulating fluid pipings or circulating fluid circuit of the customer equipment are fully durable against this pressure.
Circulating Fluid Typical Physical Property Values
1. This catalog uses the following values for density and specific heat in calculating the required cooling capacity.
Density
ρ
: 1 [kg/L] (or, using conventional units, weight volume ratio
= 1 [kgf/L] )
Specific heat
C: 4.19 x 10
3
[J/(kg·K)] (or, using conventional units, 1 x 10
3
[cal/(kgf·
°
C)])
2. Values for density and specific heat change slightly according to temperature shown below. Use this as a reference.
Water
Physical property
value
Temperature
Density
ρ
[kg/L]
Specific heat C
[J/(kg·K)]
Conventional units
Weight volume ratio
[kgf/L] Specific heat C [cal/(kgf·
°
C)]
5
°
C
1.00
4.2 x 10
3
1.00
1 x 10
3
10
°
C
1.00
4.19 x 10
3
1.00
1 x 10
3
15
°
C
1.00
4.19 x 10
3
1.00
1 x 10
3
20
°
C
1.00
4.18 x 10
3
1.00
1 x 10
3
25
°
C
1.00
4.18 x 10
3
1.00
1 x 10
3
30
°
C
1.00
4.18 x 10
3
1.00
1 x 10
3
35
°
C
0.99
4.18 x 10
3
0.99
1 x 10
3
40
°
C
0.99
4.18 x 10
3
0.99
1 x 10
3
15% Ethylene Glycol Aqueous Solution
Physical property
value
Temperature
Density
ρ
[kg/L]
Specific heat C
[J/(kg·K)]
Conventional units
Weight volume ratio
[kgf/L] Specific heat C [cal/(kgf·
°
C)]
5
°
C
1.02
3.91 x 10
3
1.02
0.93 x 10
3
10
°
C
1.02
3.91 x 10
3
1.02
0.93 x 10
3
15
°
C
1.02
3.91 x 10
3
1.02
0.93 x 10
3
20
°
C
1.01
3.91 x 10
3
1.01
0.93 x 10
3
25
°
C
1.01
3.91 x 10
3
1.01
0.93 x 10
3
30
°
C
1.01
3.91 x 10
3
1.01
0.94 x 10
3
35
°
C
1.01
3.91 x 10
3
1.01
0.94 x 10
3
40
°
C
1.01
3.92 x 10
3
1.01
0.94 x 10
3
*
Shown above are reference values. Contact circulating fluid supplier for details.
HRLE090-A
Example of conventional units (Reference)
Heat quantity by cooled substance (per unit time)
Q : Unknown [cal/h]
→
[W]
Cooled substance
: Water
Cooled substance weight
m
: (=
ρ
x
V) [kgf]
Cooled substance weight volume ratio
: 1 [kgf/L]
Cooled substance total volume
V
: 150 [L]
Cooled substance specific heat
C
: 1.0 x 10
3
[cal/(kgf·
°
C)]
Cooled substance temperature when cooling begins
T
0
: 30 [
°
C]
Cooled substance temperature after t hour
T
t
: 20 [
°
C]
Cooling temperature difference
i
T
: 10 [
°
C] (=
T
0
–
T
t
)
Cooling time
i
t
: 15 [min]
Conversion factor: hours to minutes
: 60 [min/h]
Conversion factor: kcal/h to kW
: 860 [(cal/h)/W]
Q = =
=
≈
6977 [W] = 7.0 [kW]
Cooling capacity = Considering a safety factor of 20%,
7.0 [kW] x 1.2 =
m x C x (T
0
– T
t
)
i
t x 860
x V x 60 x C x
i
T
i
t x 860
1 x 150 x 60 x 1.0 x 10
3
x 10
15 x 860
8.4 [kW]
*
This is the calculated value by changing the fluid temperature only.
Thus, it varies substantially depending on the water bath or piping shape.
25
HRLE
Series
