Required Cooling Capacity Calculation
Example 1: When the heat generation amount in the customer equipment is known.
The heat generation amount can be determined based on the power consumption or output of
the heat generating area — i.e. the area requiring cooling — within the customer equipment.
*
1
*
1 The examples above calculate the heat generation amount based on the power consumption.
The actual heat generation amount may differ due to the structure of the customer equipment.
Be sure to check it carefully.
Example 2: When the heat generation amount in the customer equipment is not known.
Obtain the temperature difference between inlet and outlet by circulating the circulating fluid inside the customer equipment.
Heat generation amount by customer equipment
Q : Unknown [W] ([J/s])
Circulating fluid
: Tap water
*
1
Circulating fluid mass flow rate
qm
: (=
ρ
x
qv
÷
60) [kg/s]
Circulating fluid density
ρ
: 1 [kg/L]
Circulating fluid (volume) flow rate
qv
: 35 [L/min]
Circulating fluid specific heat
C
: 4.186 x 10
3
[J/(kg·K)]
Circulating fluid outlet temperature
T
1
: 293 [K] (20 [
°
C])
Circulating fluid return temperature
T
2
: 296 [K] (23 [
°
C])
Circulating fluid temperature difference
i
T
: 3 [K] (=
T
2 –
T
1
)
Conversion factor: minutes to seconds (SI units) : 60 [s/min]
*
1 Refer to page 25 for the typical physical property value of tap water or other circulating fluids.
Q = qm
x C x (T
2
–
T
1
)
= =
= 7325 [J/s]
≈
7325 [W] = 7.3 [kW]
Cooling capacity = Considering a safety factor of 20%,
7.3 [kW] x 1.2 =
ρ
x qv x C x
i
T
60
1 x 35 x 4.186 x 10
3
x 3.0
60
8.8 [kW]
Thermo-chiller
Customer
equipment
∆
T = T
2
– T
1
Q: Heat generation
amount
T
1
: Outlet
temperature
T
2
: Return
temperature
qv: Circulating
fluid flow
rate
HRLE
Series
Cooling Capacity Calculation
e
Derive the heat generation amount from the output.
Output (shaft power, etc.) W: 5.1 [kW]
Q = P =
In this example, using an efficiency of 0.7:
= = 7.3 [kW]
Cooling capacity = Considering a safety factor
of 20%, 7.3 [kW] x 1.2 =
W
Efficiency
5.1
0.7
8.8 [kW]
r
Calculate based on the laser output.
Laser output power 3 [kW], conversion efficiency 30%
The oscillator’s power consumption is, 3 [kW] ÷ 0.3 = 10 [kW]
The cooling capacity required for the oscillator is,
10 [kW] − 3 [kW] = 7 [kW]
Considering a safety factor of 20%, 7 [kW] x 1.2 = 8.4 [kW]
w
Derive the heat generation amount from the power supply output.
Power supply output VI: 8.8 [kVA]
Q = P = V x I x Power factor
In this example, using a power factor of 0.85:
= 8.8 [kVA] x 0.85 = 7.5 [kW]
Cooling capacity = Considering a safety factor of
20%, 7.5 [kW] x 1.2 = 9.0 [kW]
q
Derive the heat generation amount from the
power consumption.
Power consumption P: 7 [kW]
Q = P = 7 [kW]
Cooling capacity = Considering a safety factor of
20%, 7 [kW] x 1.2 = 8.4 [kW]
HRLE090-A
Q: Heat generation amount
Customer
equipment
I: Current
Power consumption
V: Power supply
voltage
P
Example of conventional units (Reference)
Heat generation amount by customer equipment
Q : Unknown [cal/h]
→
[W]
Circulating fluid
: Tap water
*
1
Circulating fluid weight flow rate
qm : (=
ρ
x
qv x 60) [kgf/h]
Circulating fluid weight volume ratio
: 1 [kgf/L]
Circulating fluid (volume) flow rate
qv : 35 [L/min]
Circulating fluid specific heat
C
: 1.0 x 10
3
[cal/(kgf·
°
C)]
Circulating fluid outlet temperature
T
1
: 20 [
°
C]
Circulating fluid return temperature
T
2
: 23 [
°
C]
Circulating fluid temperature difference
i
T : 3 [
°
C] (=
T
2
–
T
1
)
Conversion factor: hours to minutes : 60 [min/h]
Conversion factor: kcal/h to kW
: 860 [(cal/h)/W]
Q =
=
=
≈
7325 [W] = 7.3 [kW]
Cooling capacity = Considering a safety factor of 20%,
7.3 [kW] x 1.2 =
q
m
x C x (T
2
– T
1
)
860
x q
v
x 60 x C x
i
T
860
1 x 35 x 60 x 1.0 x 10
3
x 3.0
860
8.8 [kW]
24
