01-07-2017, 10:38 AM
Gas Liquefaction Cycles. [ LINDE, CLAUDE, HEYLND ]
From the author’s Engineering Collection, included in the ETSII3 module
This program calculates the enthalpy at all the significant stages of the most common liquefaction cycles: Linde, Heylandt or Claude. Program prompts for some input data such as the enthalpy of the saturated liquid and vapor and the entry conditions of the gas. In addition, the turbine isentropic efficiency is also required for the Heylandt and Claude cases. The program also calculates the liquefied fraction per circulating mol of gas.
The enthalpies at the points of known conditions can be obtained using the tables corresponding to the gas used in the cycle. The compressors used in the cycle are assumed to be isothermal - or at least that the final temperature is the same as the initial if an association of several compressors in series is used.
The equations used are listed below:
1. Linde Cycle.
y = (h2 – h1) / (h5 –h1)
h4 = y.h5 + (1-y). h6
2. Heylandt Cycle.
liquid fraction extracted per mol: y.(h6 – h1) = x.h2 + (1-x).h8' - h1
x.h5 = y.h6 +(x-y).h7
x.h3 = (x-y).h8'+ y.h6
h8'= h2 - eff.(h2-h8)
3. Claude Cycle.
liquid fraction extracted per mol: y .(h1 – h7) = (h1-h2) + (1-x).h3 – (1-x).h9”
x. h4 = y.h7 + (x-y). h9”
x. h6 = y. h7 + (x-y). h8
(1-y).h10 = h3 + h1 (1-y) – h2
Example 1:
Calculate the liquid fraction extracted per mol in a Linde cycle with the following input conditions {h1, h2, h5, h6}. The results are also given in the table.
Example 2.-
Calculate the liquid fraction extracted per mol in a Claude cycle with the following input conditions {h1, h2, h2, h7, h8, h9}. The fraction thru the turbine is (1-a) = 0.55; and the isentropic efficiency of the turbine is r = 0.7. The results are also given in the table.
From the author’s Engineering Collection, included in the ETSII3 module
This program calculates the enthalpy at all the significant stages of the most common liquefaction cycles: Linde, Heylandt or Claude. Program prompts for some input data such as the enthalpy of the saturated liquid and vapor and the entry conditions of the gas. In addition, the turbine isentropic efficiency is also required for the Heylandt and Claude cases. The program also calculates the liquefied fraction per circulating mol of gas.
The enthalpies at the points of known conditions can be obtained using the tables corresponding to the gas used in the cycle. The compressors used in the cycle are assumed to be isothermal - or at least that the final temperature is the same as the initial if an association of several compressors in series is used.
The equations used are listed below:
1. Linde Cycle.
y = (h2 – h1) / (h5 –h1)
h4 = y.h5 + (1-y). h6
2. Heylandt Cycle.
liquid fraction extracted per mol: y.(h6 – h1) = x.h2 + (1-x).h8' - h1
x.h5 = y.h6 +(x-y).h7
x.h3 = (x-y).h8'+ y.h6
h8'= h2 - eff.(h2-h8)
3. Claude Cycle.
liquid fraction extracted per mol: y .(h1 – h7) = (h1-h2) + (1-x).h3 – (1-x).h9”
x. h4 = y.h7 + (x-y). h9”
x. h6 = y. h7 + (x-y). h8
(1-y).h10 = h3 + h1 (1-y) – h2
Example 1:
Calculate the liquid fraction extracted per mol in a Linde cycle with the following input conditions {h1, h2, h5, h6}. The results are also given in the table.
Code:
Inputs: | Results:
----- -------- -----------------
H1 419.600 | y 0.0929
H2 380.600 | H3 183.5878
H5 0.000 | H4 183.5878
H6 202.400 |Example 2.-
Calculate the liquid fraction extracted per mol in a Claude cycle with the following input conditions {h1, h2, h2, h7, h8, h9}. The fraction thru the turbine is (1-a) = 0.55; and the isentropic efficiency of the turbine is r = 0.7. The results are also given in the table.
Code:
Inputs: | Results:
----------------|----------------
h1 423.900 | y 0.2971
h2 384.900 | h4 11.0813
h3 300.000 | h5 10.4270
h7 0.000 | h6 10.4270
h8 200.000 | h9' 212.8504
h9 183.400 | h10 283.8504