Calculation of the average specific heat capacity in heat balance
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For methanol, with an inlet temperature of 30 degrees and an outlet temperature of 120 degrees where the substance is in gas phase, and a constant pressure of 0.53 MPa(g), should the average specific heat capacity of methanol at this condition be taken as the specific heat capacity at the average temperature, or as the average of the specific heat capacities at the two temperatures?
This post was last edited by wiseboy on 2011-4-6 15:09. (1) The input is a liquid and the output is a gas; the average heat capacity of the two phases is meaningless. (2) Even when calculating heat, it needs to be calculated separately: temperature rise of the liquid phase, evaporation of the saturated liquid, and temperature rise of the vapor. . . .
Is the constant-pressure heat capacity not suitable? I’ve forgotten some of the basic knowledge.
According to what’s said on the 2nd floor, it is calculated by score ranges. Because a phase transition occurred.
This post was last edited by wiseboy on 2011-4-6 at 15:58. The complete answer is as follows: According to my calculations, the temperature of 30°C at the inlet is that of a liquid, while the temperature of 120°C at the outlet corresponds to saturated steam. In the liquid phase, Cp = (3.619 + 4.123)/2 = 3.871 kJ/(kg·°C). The latent heat of vaporization is Hv = 970.7 kJ/kg. Using a segmented calculation, with each kilogram as the basis, and by employing the average specific heat of the liquid (which has an approximate nature), the heat absorbed from the inlet to the outlet is: 3.871 × (120 – 30) + 970.7 = 348.4 + 970.7 = 1319.1 kJ. With other more precise calculation methods, rather than using the average specific heat (you don’t need to ask about this method), the heat absorbed per kilogram from the inlet to the outlet is 1312 kJ. The difference between the two is very small, indicating that the average specific heat method is feasible. But the question you raised is about the average specific heat of import and export, which is the average specific heat of the two phases; this concept is meaningless or incorrect in this context. Only the \"average specific heat of liquids\" exists.
Learning*, thank you :victory:
This post was last edited by cnsttop on 2011-4-7 09:35. Calculated based on the specific enthalpy value of methanol: at 30°C, it is 100 KJ/KG; at 120°C, it is 319 KJ/KG. At 120°C in the gas phase: 1474 KJ/KG. Is the total heat of absorption therefore (1474–100) KJ/KG?
The method is correct. But it seems the specific figures aren’t very accurate.
The specific enthalpy value at 30 degrees could not be found; it is 100 JK/KG at 40 degrees
According to what’s said on the 2nd floor, it is calculated by score ranges.
Phase transitions definitely occur in segments; the original poster needs to clarify the basic concepts