Ammonia cooler calculation
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The inlet conditions of the ammonia cooler are 10°C, 0.6 MPa, and 7.5 t/h. The outlet temperature of the ammonia cooler is -40°C, with a pressure of 0.0718 MPa and a flow rate of 7.5 t/h. What is the calculated cooling capacity in kilowatts? Please list in detail the calculation methods and steps? Thank you
The enthalpy of liquid ammonia at 10℃ and 0.6 MPa is -957.57643 kcal/kg, while the enthalpy of gaseous ammonia at -40℃ and 0.0718 MPa is -676.09185 kcal/kg. Therefore, ((-676.09185 – (-957.57643))) × 7500 / 859.8452 = -2455 KW.
Reply to 2# sun0225: Should the heat of vaporization be taken into consideration? If so, how should it be calculated? Because liquid ammonia turns into gaseous ammonia when it vaporizes
When using enthalpy values for calculations, the heat of vaporization does not need to be taken into account, right? But it needs to be considered when using specific heat capacities
What method is used for cooling, so as to reach -40?
What is 859.8452? Please explain it
For specific details, refer to the Process Technology and Design Manual for Small Ammonia Plants, which contains detailed calculation steps
Reply to 6# zhengrd: For unit conversion, KCal/kg needs to be changed to KJ/h=KW; one of the values in this conversion is obtained using the constant 3600/(4.18).