Heat Rejection Calculator:
Enter the values of of specific heat capacity of the fluid, c(J/kg. °C), density of the fluid, d(kg/m3), flow rate of the fluid, F(m3/s) and temperature difference, T(°C) to determine the value of Heat Rejected, Q(J).
Heat Rejection Formula:
Heat rejection refers to the process of transferring excess heat from a system to a surrounding environment. This process is critical in a variety of engineering applications such as HVAC systems, automotive cooling, power plants, and refrigeration. Heat rejection is necessary to maintain the proper temperature for system operation and ensure safety, efficiency, and longevity.
In practical terms, heat rejection often occurs through components such as heat exchangers, radiators, or cooling towers, where a fluid (typically water or air) absorbs heat from a system and expels it to the environment. The efficiency of heat rejection depends on factors such as the heat capacity of the fluid, the flow rate, and the temperature difference between the system and the surrounding environment.
This equation calculates the amount of heat that a given mass of fluid can absorb or reject based on its specific heat, flow rate, and the temperature change it experiences.
Heat Rejected, Q(J) in Joules is calculated by the product of specific heat capacity of the fluid, c(J/kg. °C) in Joules per kilogram degree Celsius, density of the fluid, d(kg/m3) in kilograms per cubic metres, flow rate of the fluid, F(m3/s) in cubic metres per second and temperature difference, T(°C) in degree Celsius.
Heat Rejected, Q(J) = c(J/kg. °C) * d(kg/m3) * F(m3/s) * T(°C)
Q(J) = heat rejected in Joules, J.
c(J/kg. °C) = specific heat capacity in Joules per kilogram degree Celsius, J/kg.°C.
d(kg/m3) = density of the fluid in kilograms per cubic metres, kg/m3.
F(m3/s) = flow rate of the fluid in cubic metres per second, m3/s.
T(°C) = temperature difference in degree Celsius, °C.
Heat Rejection Calculation:
1.A cooling system uses water with a flow rate of 0.5 m3/s, a specific heat capacity of 4.18 kJ/kg·°C, a density of 1000 kg/m3 and a temperature difference of 10°C. Calculate the heat rejected.
Given: c(J/kg. °C) = 4.18 kJ/kg·°C, d(kg/m3) = 1000 kg/m3, F(m3/s) = 0.5 m3/s, T(°C) = 10°C.
Heat Rejected, Q(J) = c(J/kg. °C) * d(kg/m3) * F(m3/s) * T(°C)
Q(J) = 4.18 * 1000 * 0.5 * 10
Q(J) = 20900J.
2.A cooling system rejects 25,000 W of heat using water with a flow rate of 0.6 m3/s, a specific heat capacity 4.18 kJ/kg·°C, and a density =1000 kg/m3. Find the temperature difference.
Given: c(J/kg. °C) = 4.18 kJ/kg·°C, d(kg/m3) = 1000 kg/m3, F(m3/s) = 0.6 m3/s, Q(J) = 25000J.
Heat Rejected, Q(J) = c(J/kg. °C) * d(kg/m3) * F(m3/s) * T(°C)
T(°C) = Q(J) / c(J/kg. °C) * d(kg/m3) * F(m3/s)
T(°C) = 25000 / 4.18 * 1000 * 0.6
T(°C) = 25000 / 2508000
T(°C) = 9.97°C.