Water-cooled heat exchange
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Knowing the inlet and outlet temperatures of the cooling water, how can one calculate the heat removed by the cooling water?
Just calculate the flow rate; multiply the flow rate by the specific heat of water, and you’ve got it, buddy
The specific heat of water is equal to 1 cal·g-1℃-1 (which is 4.1868×103 J·kJ-1·K-1). Remember, buddy: just multiply it by the amount of circulation and you’re good to go
Heat = mass * specific heat * temperature difference
Generally, one first determines the required heat transfer amount, and then calculates the cooling water volume! ! ! That’s putting the cart before the horse, isn’t it?~~~~~~~~~~
I think the original poster might be using it for verification purposes; I’ll mention a less common scenario: if the outlet temperature reaches the boiling point, don’t forget to take the latent heat into account.
I agree with what was said above; what the original poster mentioned should be an analysis of whether it’s a verification issue or an actual system problem. For medium water, the change in specific heat capacity at constant pressure is small; it should be possible to calculate it easily by measuring the flow rate in the corresponding pipeline, provided that the instruments used for measuring the flow rate are accurate. For the setting of the temperature difference between supply and return water in traditional designs, it is rather complicated to analyze the characteristics of a given system as well as its energy-saving potential.
The conditions provided by the original poster are too limited to allow for any calculation; the cooling water flow rate is unknown, and the heat transfer amount calculated based on the flow rate and inlet/outlet temperatures will be higher than the actual value, as there is certainly a certain margin in the water flow rate. It is better to calculate the heat transfer amount based on the process parameters and requirements of the medium being cooled
This question is too simple; CpMdt will do.
The heat dissipated through the cooling tower is: flow rate of water × specific heat of water × temperature difference between inlet and outlet water. The water flow rate is essentially the sum of the flow rates of the pumps in operation (measured in KG/S). The specific heat capacity of water is 4.1868 KJ/(KG*degree Celsius), and the temperature difference between the inlet and outlet water refers to the difference in temperature of the water entering and leaving the cooling tower.
Heat flow rate = Flow rate (volume) X Density X Specific heat of the fluid X Temperature difference!
Water flow rate multiplied by specific heat multiplied by temperature difference
Product of the enthalpy difference of hot water at the inlet and outlet and the flow rate