matlab中编写一个M文件,计算液体氯化氢在277.4K、200psia(1379kPa)时的逸度
时间: 2024-02-18 21:04:45 浏览: 22
可以使用Van der Waals方程来计算液体氯化氢的逸度,具体步骤如下:
1. 定义所需参数
```
T = 277.4; % 温度,单位:K
P = 200; % 压力,单位:psia
R = 8.314; % 气体常数,单位:J/(mol*K)
a = 4.672; % Van der Waals方程参数a,单位:(L^2*bar)/(mol^2)
b = 0.02661; % Van der Waals方程参数b,单位:L/mol
```
2. 计算逸度
```
term1 = P * b / (R * T);
term2 = a / (R^2 * T^2);
term3 = P * b^2 / (R^2 * T^2);
term4 = 2 * a * b / (R * T^2);
Z = roots([1, -(1+term1-term2), term1*term2-term3-term4, -term1*term2*b]);
V = Z(3) * R * T / P;
fugacity_coefficient = exp((P * (V - b) / (R * T)) - log(Z(3) - b) - ((2 * a) / (sqrt(8) * b * R * T)) * log((Z(3) + (1 + sqrt(2)) * b) / (Z(3) + (1 - sqrt(2)) * b)));
```
3. 显示结果
```
disp(['液体氯化氢在', num2str(T), 'K、', num2str(P), 'psia(1379kPa)时的逸度为:', num2str(fugacity_coefficient)]);
```
完整的M文件代码如下:
```
T = 277.4; % 温度,单位:K
P = 200; % 压力,单位:psia
R = 8.314; % 气体常数,单位:J/(mol*K)
a = 4.672; % Van der Waals方程参数a,单位:(L^2*bar)/(mol^2)
b = 0.02661; % Van der Waals方程参数b,单位:L/mol
term1 = P * b / (R * T);
term2 = a / (R^2 * T^2);
term3 = P * b^2 / (R^2 * T^2);
term4 = 2 * a * b / (R * T^2);
Z = roots([1, -(1+term1-term2), term1*term2-term3-term4, -term1*term2*b]);
V = Z(3) * R * T / P;
fugacity_coefficient = exp((P * (V - b) / (R * T)) - log(Z(3) - b) - ((2 * a) / (sqrt(8) * b * R * T)) * log((Z(3) + (1 + sqrt(2)) * b) / (Z(3) + (1 - sqrt(2)) * b)));
disp(['液体氯化氢在', num2str(T), 'K、', num2str(P), 'psia(1379kPa)时的逸度为:', num2str(fugacity_coefficient)]);
```
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