Series HRSH Cooling Capacity Calculation Required Cooling Capacity Calculation Example 1: When the heat generation amount in the user’s 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 user’s equipment.* q Derive the heat generation amount from the power consumption. Power consumption P: 20 [kW] Q = P = 20 [kW] Cooling capacity = Considering a safety factor of 20%, 20 [kW] x 1.2 = w Derive the heat generation amount from the power supply output. Power supply output VI: 20 [kVA] Q = P = V x I x Power factor In this example, using a power factor of 0.85: = 20 [kVA] x 0.85 = 17 [kW] Cooling capacity = Considering a safety factor of 20%, 17 [kW] x 1.2 = * The above examples calculate the heat generation amount based on the power consumption. The actual heat generation amount may differ due to the structure of the user’s equipment. Be sure to check it carefully. Example 2: When the heat generation amount in the user’s equipment is not known. Obtain the temperature difference between inlet and outlet by circulating the circulating fluid inside the user’s equipment. Heat generation amount by user’s equipment Q : Unknown [W] ([J/s]) Circulating fluid : Tap water* 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 : 70 [L/min] Circulating fluid specific heat C : 4.186 x 103 [J/(kg・K)] Circulating fluid outlet temperature T1 : 293 [K] (20 [°C]) Circulating fluid return temperature T2 : 297 [K] (24 [°C]) Circulating fluid temperature difference iT : 4 [K] (= T2 . T1) Conversion factor: minutes to seconds (SI units) : 60 [s/min] * Refer to page 29 for the typical physical property value of tap water or other circulating fluids. Q = qm x C x (T2 . T1) = = = 19535 [J/s] . 19535 [W] = 19.5 [kW] Cooling capacity = Considering a safety factor of 20%, 19.5 [kW] x 1.2 = 24 [kW] 20.4 [kW] ρ x qv x C x iT 60 1 x 70 x 4.186 x 103 x 4.0 60 23.4 [kW] e Derive the heat generation amount from the output. Output (shaft power etc.) W: 13 [kW] Q = P = In this example, using an efficiency of 0.7: = = 18.6 [kW] Cooling capacity = Considering a safety factor of 20%, 18.6 [kW] x 1.2 = W Efficiency 13 0.7 22.3 [kW] Example of conventional measurement units (Reference) Heat generation amount by user’s equipment Q : Unknown [cal/h] → [W] Circulating fluid : Tap water* 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 : 70 [L/min] Circulating fluid specific heat C : 1.0 x 103 [cal/(kgf・°C)] Circulating fluid outlet temperature T1 : 20 [°C] Circulating fluid return temperature T2 : 24 [°C] Circulating fluid temperature difference iT : 4 [°C] (= T2 . T1) Conversion factor: hours to minutes : 60 [min/h] Conversion factor: kcal/h to kW : 860 [(cal/h)/W] Q = = = = . 19534 [W] = 19.5 [kW] Cooling capacity = Considering a safety factor of 20%, 19.5 [kW] x 1.2 = qm x C x (T2 . T1) 860 �､ x qv x 60 x C x iT 860 1 x 70 x 60 x 1.0 x 103 x 4.0 860 16800000 [cal/h] 860 23.4 [kW] HRSH250-A 28 Q: Heat generation amount User’s equipment I: Current Power consumption V: Power supply voltage P Thermo-chiller User’s equipment T1: Outlet temperature T2: Return temperature qv: Circulating fluid flow rate ΔT = T2 . T1 Q: Heat generation amount