如何在Matplotlib中仅绘制T_f函数340-373K范围的数值？

我尝试绘制函数T_f的特定范围数值，但不知如何操作。我使用np.arange生成了T_f方程所需的z数组，希望仅绘制x轴340-373K范围的数值。以下是我的代码：

import numpy as np
import matplotlib.pyplot as plt

case = int(input('Which Case [1 (PWR) or 2 (BWR)]? '))

if case == 1:  # PWR
    H = 3.80  # m
    Lc = 3.80  # m
    D_rod = 0.0095  # m
    pitch = 0.0125  # m

    G = 3460  # kg/m^2-s
    q_0 = 33  # kW/m - Linear heat rate
    P_0 = 15  # MPa - Initial pressure
    T_f_in = 551  # Kelvin - Inlet temperature
    T_sat = 373  # Kelvin
    cp = 4.22  # kJ/kg
    hfg = 2256.4  # kJ/kg
    g = 9.81  # m/s^2

    heated_parameter = np.pi * D_rod  # m
    area = pitch ** 2 - 0.25 * np.pi * D_rod ** 2  # m^2
    volume = 0.5 * np.pi * D_rod ** 2 * H  # m^3
    circumference = 2 * np.pi * (D_rod / 2)  # m
    rho_f = 958.4  # kg/m^3 - Water density
    mu_f = 0.00000000009888  # MPa/s - Water viscosity
    Re_L = (rho_f * volume * Lc) / mu_f  # Reynold's number
    f_1phaseL = 0.316 * Re_L ** -0.25  # Friction factor

    z = np.arange(0, 3.8, 0.001)

    Xe = -q_0 * heated_parameter / (G * circumference * area * hfg) * H / np.pi * np.cos(np.pi * z / H) + cp * (T_sat - T_f_in) / hfg
    #plt.plot(Xe, z, label="Equilibrium Quality")

    void = z*0
    #plt.plot(void, z, label="Void Fraction")

    X = z * 0
    #plt.plot(X, z, label="Quality")
    #plt.xlabel("Void Fraction, Equilibrium Quality, Quality")


    P = P_0 - ((heated_parameter / area * 0.5 * f_1phaseL * G ** 2 * z / rho_f) + rho_f * g * z) * 0.000001
    #plt.plot(P, z)
    #plt.xlabel("Pressure (MPa)")

    plt.vlines(x=373, ymin=0, ymax=3.8, label="Saturated Temperature", colors='Red')

    T_f = T_sat - (q_0 * heated_parameter * H / (cp * circumference * G * area * np.pi) * np.cos(np.pi * z / H))
    plt.plot(T_f, z, label="Fluid Temperature")


    plt.xlabel("Temperature (K)")
    plt.legend()
    plt.ylabel("z (meters)")
    plt.show()

解决方法 #

要仅绘制T_f在340-373K范围的数值，核心是通过布尔索引筛选出符合温度范围的T_f和对应的z值，再进行绘制：

1. 创建过滤掩码：生成布尔数组标记T_f在目标范围内的位置

mask = (T_f >= 340) & (T_f <= 373)

2. 用过滤后的数据绘图：替换原绘图语句，使用掩码筛选后的数组

plt.plot(T_f[mask], z[mask], label="Fluid Temperature (340-373K)")

3. 可选：锁定x轴范围：确保图表仅展示340-373K区间，避免无关数据干扰

plt.xlim(340, 373)

修改后的完整代码 #

import numpy as np
import matplotlib.pyplot as plt

case = int(input('Which Case [1 (PWR) or 2 (BWR)]? '))

if case == 1:  # PWR
    H = 3.80  # m
    Lc = 3.80  # m
    D_rod = 0.0095  # m
    pitch = 0.0125  # m

    G = 3460  # kg/m^2-s
    q_0 = 33  # kW/m - Linear heat rate
    P_0 = 15  # MPa - Initial pressure
    T_f_in = 551  # Kelvin - Inlet temperature
    T_sat = 373  # Kelvin
    cp = 4.22  # kJ/kg
    hfg = 2256.4  # kJ/kg
    g = 9.81  # m/s^2

    heated_parameter = np.pi * D_rod  # m
    area = pitch ** 2 - 0.25 * np.pi * D_rod ** 2  # m^2
    volume = 0.5 * np.pi * D_rod ** 2 * H  # m^3
    circumference = 2 * np.pi * (D_rod / 2)  # m
    rho_f = 958.4  # kg/m^3 - Water density
    mu_f = 0.00000000009888  # MPa/s - Water viscosity
    Re_L = (rho_f * volume * Lc) / mu_f  # Reynold's number
    f_1phaseL = 0.316 * Re_L ** -0.25  # Friction factor

    z = np.arange(0, 3.8, 0.001)

    Xe = -q_0 * heated_parameter / (G * circumference * area * hfg) * H / np.pi * np.cos(np.pi * z / H) + cp * (T_sat - T_f_in) / hfg
    #plt.plot(Xe, z, label="Equilibrium Quality")

    void = z*0
    #plt.plot(void, z, label="Void Fraction")

    X = z * 0
    #plt.plot(X, z, label="Quality")
    #plt.xlabel("Void Fraction, Equilibrium Quality, Quality")


    P = P_0 - ((heated_parameter / area * 0.5 * f_1phaseL * G ** 2 * z / rho_f) + rho_f * g * z) * 0.000001
    #plt.plot(P, z)
    #plt.xlabel("Pressure (MPa)")

    plt.vlines(x=373, ymin=0, ymax=3.8, label="Saturated Temperature", colors='Red')

    T_f = T_sat - (q_0 * heated_parameter * H / (cp * circumference * G * area * np.pi) * np.cos(np.pi * z / H))
    # 筛选340-373K范围的数据
    mask = (T_f >= 340) & (T_f <= 373)
    plt.plot(T_f[mask], z[mask], label="Fluid Temperature (340-373K)")
    # 锁定x轴显示范围
    plt.xlim(340, 373)


    plt.xlabel("Temperature (K)")
    plt.legend()
    plt.ylabel("z (meters)")
    plt.show()

内容的提问来源于stack exchange，提问作者Joseph Wunschel