《环境工程学一》作业

  • 2019-03-17 12:00
  • 2020-01-03 23:02

第 1 次作业

1.

$\ce{SO2}$:$w_V = \cfrac{w_mV_m}{M} = \cfrac{0.15 \times 22.4}{64} = \pu{52.5 ppb}$;

$\ce{NO2}$:$w_V = \cfrac{w_mV_m}{M} = \cfrac{0.12 \times 22.4}{46} = \pu{58.4 ppb}$;

$\ce{CO}$:$w_V = \cfrac{w_mV_m}{M} = \cfrac{4.00 \times 22.4}{28} = \pu{3.2 ppm}$。

2.

(1)

$w_m = \cfrac{w_VM}{V_m} = \cfrac {1.50 \times 10^{-4} \times 154 \times 1000}{22.4}\pu{g/m3_N} = \pu{1.03 g/m3_N}$;
$c = \cfrac{w_V}{V_m} = \cfrac {1.50 \times 10^{-4} \times 1000}{22.4} \pu{g/m3_N} = \pu{67.0 mol/m3_N}$。

(2)

$m = qw_mt = \cfrac {10 \times 1.03 \times 86400}{1000} \pu{kg} = \pu{890 kg}$。

3.

取 $\pu{100 g}$ 重油,则各元素的含量为:$\ce{C}$:$\pu{85.5 g} \rightarrow \pu{7.12 mol}$,$\ce{H}$:$\pu{11.3 g} \rightarrow \pu{11.2 mol}$,$\ce{O}$:$\pu{2.0 g} \rightarrow \pu{0.125 mol}$,$\ce{N}$:$\pu{0.2 g} \rightarrow \pu{0.0143 mol}$,$\ce{S}$:$\pu{1.0 g} \rightarrow \pu{0.03125 mol}$。

(1)

①理论耗氧量:$\cfrac{7.12 + \cfrac {1}{4} \times 11.2 + 0.03125 - \cfrac {1}{2} \times 0.125}{0.1}\pu{mol/kg} = \pu{98.9 mol/kg}$,理论空气量:$4.78 \times \pu{98.9 mol/kg} = \pu{473 mol/kg}$,即 $\cfrac{473 \times 22.4}{1000}\pu{m3/kg} = \pu{10.6 m3/kg}$;

理论烟气量:$\left(\cfrac{7.12 + \cfrac{1}{2} \times 11.2 + \cfrac {1}{2} \times 0.0143 + 0.03125}{0.1} + 3.78 \times 98.9\right)\pu{mol/kg} = \pu{501 mol/kg}$,即 $\cfrac{501 \times 22.4}{1000}\pu{m3/kg} = \pu{11.2 m3/kg}$。

(2)

理论干烟气量:$\left(501 - \cfrac{\cfrac{1}{2}\times 11.2}{0.1}\right)\pu{mol/kg} = \pu{445 mol/kg}$,即 $\cfrac{445 \times 22.4}{1000}\pu{m3/kg} = \pu{9.97 m3/kg}$。

$\ce{SO2}$浓度:$\cfrac {0.03125}{0.1\times445} = 7.02\times10^{-4}$,$\ce{CO2}$最大浓度:$\cfrac {7.12}{0.1\times331} = 0.16$。

(3)

实际空气量:$1.1 \times \pu{10.6 m3/kg} = \pu{11.7 m3/kg}$,实际烟气量:$\pu{11.2 + 0.1 \times 10.6 m3/kg} = \pu{12.3 m3/kg}$。

4.

取 $\pu{100 g}$ 燃煤,则各元素的含量为:$\ce{C}$:$\pu{65.7 g} \rightarrow \pu{5.475 mol}$,$\ce{H}$:$\pu{3.2 g} \rightarrow \pu{3.17 mol}$,$\ce{O}$:$\pu{2.3 g} \rightarrow \pu{0.144 mol}$,$\ce{S}$:$\pu{1.7 g} \rightarrow \pu{0.0531 mol}$,水分:$\pu{9.0 g} \rightarrow \pu{0.500 mol}$。

(1)

理论耗氧量:$\cfrac{5.475 + \cfrac{1}{4}\times 3.17 + 0.0531 - \cfrac{1}{2}\times 0.144}{0.1}\pu{mol/kg} = \pu{62.5 mol/kg}$,即 $\cfrac{62.5 \times 22.4}{1000}\pu{m3/kg} = \pu{1.4 m3/kg}$;

理论空气量:$4.78 \times \pu{1.4 m3/kg} = \pu{6.7 m3/kg}$;

理论烟气量:$\left(\cfrac{5.475 + \cfrac {1}{2}\times 3.17 + 0.0531 + 0.500 }{0.1} + 3.78 \times 62.5\right)\pu{mol/kg} = \pu{312.1 mol/kg}$,即 $\cfrac{312.1 \times 22.4}{1000}\pu{m3/kg} = \pu{7.0 m3/kg}$;

$\ce{SO2}$ 的浓度:$\cfrac{0.0531 \times 22.4}{0.1 \times 1000 \times 7.0} = 1.70 \times 10^{-3}$。

(2)

灰分含量:$\cfrac{18.1 \times 10 \times 80~\%}{7.0}\times \pu{10^6 mg/m3} = \pu{2.07E4 mg/m3}$。

(3)

$\pu{1 t}$ 燃煤含有 $\ce{S}$ 含量为 $\pu{17 kg}$,物质的量 $\cfrac {17}{32}\pu{mol} = \pu{0.531 mol}$; 则需要 $\ce{Ca}$ 质量为 $0.531 \times 40 \times \pu{1.7 kg} = \pu{36.1 kg}$,石灰石的质量为 $\cfrac {28.9}{35 ~\%}\pu{kg} = \pu{103.2 kg}$。

5.

(1)

烟道气中 $\ce{N2}$ 的体积分数为 $1 - 0.11 - 0.08 - 0.02 - 120 \times 10^{-6} = 0.79$,则空气过剩 $\alpha = \cfrac {0.08 - 0.5 \times 0.02}{0.264 \times 0.79 - 0.08 + 0.5 \times 0.02} = 0.505$。

(2)

$\pu{1m3}$ 该状态下气体含有气体物质的量为:$\cfrac {700 \times 133.322}{8.31 \times 443}\pu{mol} = \pu{25.35 mol}$,则 $\ce{SO2}$ 排放浓度为 $120 \times 10^{-6} \times 25.35 \times \pu{64E6 \mu g/m3} = \pu{1.94E5 \mu g/m3}$。

(3)

干烟道气排放流量为 $5663.37 \times (1 - 0.08) \pu{m3/min} = \pu{5210.30 m3/min}$,校准至标况:$\cfrac {5210.30 \times 700 \times 133.322 \times 273.15}{443 \times 101325} = \pu{2958. 99 m3/min}$。

(4)

标况下颗粒物浓度:$\cfrac {30.0 \times 5210.30}{2958.99}\pu{g/m3} = \pu{52.83 g/m3}$。

第 2 次作业

1.

(1)

由已知,列表,如表 2-1 所示:

表2-1 $G - x - \ln d_p$分布表
$G$$x$
标准正态分布对应随机变量
$d_p$
($\pu{\mu m}$)
$\ln d_p$
0.004-2.65206980820.693147181
0.065-1.51410188841.386294361
0.19-0.87789629561.791759469
0.47-0.075269862102.302585093
0.851.036433389202.995732274
0.9852.170090378403.688879454

作图:

y = 0.6236 x + 2.3419 r ² = 0.9999 0 0.5 1 1.5 2 2.5 3 3.5 4 -3 -2 -1 0 1 2 3 ln d p x G - x - ln d p 分布表

由图得:$r^2 = 0.9999$,说明该粉尘的粒径分布符合对数正态分布。

(2)

由图得:\(\ln \sigma = 0.6236\),几何标准差\(\sigma = \mathrm e^{0.6236} = 1.866\)。

质量中位直径\(\mathrm{MMD} = \mathrm e^{2.3419} = \pu{10.40 \mu m}\),

个数中位直径\(\mathrm{NMD} = \mathrm{MMD} \cdot \mathrm e^{-3 \ln^2 \sigma} = \pu{3.24 \mu m}\),

算术平均直径\(\bar{d_{\mathrm L}} = \mathrm{MMD} \cdot \mathrm e^{-5/2 \ln^2 \sigma} = \pu{3.93 \mu m}\),

表面积-体积平均直径\(\bar{d_{\mathrm{SV}}} = \mathrm{MMD} \cdot \mathrm e^{-1/2 \ln^2 \sigma} = \pu{8.56 \mu m}\)。

2.

(1)

处理气体流量:\(Q_{1, N} = \pu{10000 m3_N/h}\),\(Q_{2, N} = \pu{12000 m3_N/h}\),则\(Q_N = \cfrac 12 (Q_{1, N} + Q_{2, N}) = \pu{11000 m3_N/h}\)。

(2)

漏风率:\(\delta = \cfrac {Q_{1, N} - Q_{2, N}}{Q_{1, N}} \times 100 ~\% = -20 ~\%\)。

(3)

除尘效率:考虑漏风,\(\eta = 1 - \cfrac {\rho_{1, N} Q_{1, N}}{\rho_{2, N} Q_{2 N}} = 90.3~\%\)。不考虑漏风,\(\eta = 1 - \cfrac {\rho_{1, N}}{\rho_{2, N}} = 91.9~\%\)。

3.

进气口气流流速\(v = \cfrac {Q_{1, N}\cfrac{T_1}{T_0}}{S} = \pu{17.94 m/s}\),气体密度\(\rho = \cfrac {(p_0 - p)M}{RT} = \pu{0.831 kg/m3}\)。压力损失\(\Delta p = \cfrac {\varepsilon \rho v_1^2}{2} = \pu{1310 Pa}\)。

4.

(1)

总除尘效率:\(\eta = 1 - (1 - 80~\%)(1 - 95~\%) = 99~\%\)。

(2)

排放浓度:\(c = \cfrac{\rho_1(1-\eta)}{Q} = \cfrac{22.2 \times (1 - 99~\%)}{2.22}\pu{g/m3} = \pu{0.1 g/m3}\)。

(3)

排放量:\(\rho_2 = \rho_1(1-\eta) = 22.2 \times (1 - 99~\%)\pu{g/s} = \pu{0.222 g/s}\)。

5.

由公式:\(\eta_i = 1 - P\cfrac{g_{2i}}{g_{1i}}\),列表,如表 2-2 所示:

表2-2 分级除尘效率
粒径间隔/$\pu{μm}$<0.60.6~0.70.7~0.80.8~1.01~22~33~44~55~66~88~1010~1220~30
质量频率/$\%$进口/$g_{1i}$20.40.40.73.562413223118
出口/$g_{2i}$71231416296222.58.57
分级除尘效率/$\eta_i$0.930 0.950 0.900 0.914 0.920 0.947 0.976 0.991 0.980 0.980 0.983 0.985 0.983

作图:

0.89 0.9 0.91 0.92 0.93 0.94 0.95 0.96 0.97 0.98 0.99 1 0 5 10 15 20 25 30 η i d p i /μm 分级除尘效率曲线

6.

由公式:\(\eta = \sum\limits_{i} n_i\eta_i\),列表,如表 2-3 所示:

表2-3 分级除尘效率
平均粒径/$\pu{\mu m}$0.2512345678101420>23.5
质量频率/$\%$0.10.49.5202015118.55.55.540.80.2
分级效率/$\%$83047.56068.575818689.5959899100
$n_i\eta_i$0.000080.00120.0451250.120.1370.11250.08910.07310.0492250.052250.03920.007920.002

求得\(\eta = 72.87~\%\)。

7.

查表得\(\pu{387.5 K}\)、\(\pu{101325 Pa}\)下空气的黏度为\(\mu = \pu{2.25E-5 Pa*s}\),密度为\(\rho = \cfrac{pM}{RT} = \pu{0.912 kg/m3}\)。

(1)

直径\(d_1 = \pu{0.4 \mu m}\),近似计算坎宁汉修正系数\(C = 1 + \cfrac {0.165}{d_1} = 1.4125\),\(u_1 = \cfrac {d_1^2\rho_p}{18\mu}gC = \pu{1.26E-5 m/s}\)。\(h_1 = u_1t = \pu{3.78E-4 m}\)。

(2)

直径\(d_2 = \pu{40 \mu m}\),忽略坎宁汉修正,假设其位于斯托克斯区,\(u_2 = \cfrac {d_2^2\rho_p}{18\mu}g = \pu{8.93E-2 m/s}\),\(Re = \cfrac {d_2\rho u_2}{\mu} = 0.0145 < 1\),符合要求。\(h_2 = u_2t = \pu{2.68 m}\)。

(3)

直径\(d_3 = \pu{4000 \mu m}\),直接考虑位于湍流过渡区,\(u_3 = 1.74\sqrt{\cfrac{d_2(\rho_p - \rho)g}{\rho}} = \pu{17.34 m/s}\),\(Re = \cfrac {d_2\rho u_2}{\mu} = 281\),满足\(1 < Re < 500\),符合要求。\(h_3 = u_3t = \pu{520 m}\)。

8.

查表得\(\pu{293 K}\)、\(\pu{101325 Pa}\)下空气的黏度为\(\mu = \pu{1.81E-5 Pa*s}\),密度为\(\rho = \cfrac{pM}{RT} = \pu{1.206 kg/m3}\)。

考虑最小粒径的水泥颗粒,\(d_p = \pu{25 \mu m}\),忽略坎宁汉修正,假设其位于斯托克斯区,\(u_s = \cfrac {d_p^2\rho_p}{18\mu}g = \pu{3.69E-3 m/s}\),计算\(Re = \cfrac {d_2\rho u_2}{\mu} = 6.26\times 10^{-2} < 1\),符合要求。

则沉降时间\(t = \cfrac{h}{u_s} = \pu{122 s}\),最远距离\(s = v_0t = \pu{171 m}\)。

第 3 次作业

1.

恰好完全分离时,最大石英颗粒和最小角闪石颗粒应具有相同的终端沉降速率。代入牛顿区终端沉降速率公式 \(u_s = 1.74\sqrt{\cfrac {d_{\mathrm p}(\rho_{\mathrm p} - \rho)g}{\rho}}\) 得:\(\cfrac {d_{\mathrm p1}}{d_{\mathrm p2}} = \cfrac {\rho_{\mathrm p2} - \rho}{\rho_{\mathrm p1} - \rho} = \cfrac {3.5}{2.6} = 1.37\),即最大石英粒径与最小角闪石粒径的最大比值为 \(1.37\)。

2.

给定温度 \(\pu{293 K}\)、气压 \(\pu{101325 Pa}\) 条件下,求得空气密度 \(\rho = \cfrac {pM}{RT} = \pu{1.206 kg/m3}\),黏度 \(\mu = \pu{1.809E-5 Pa*s}\)。

先按斯托克斯区计算终端沉降速度,\(u_{\mathrm s} = \cfrac {d_{\mathrm p}^2(\rho_{\mathrm p} - \rho)g}{18 \mu} = \pu{2.23 m/s}\),雷诺数 \(Re_{\mathrm p} = \cfrac {d_{\mathrm p}\rho u}{\mu} = 29.7\),超出斯托克斯区范围,不符合假设。

按湍流过渡区计算终端沉降速度,\(u_{\mathrm s} = \cfrac {0.153d_{\mathrm p}^{1.14}(\rho_{\mathrm p} - \rho)^{0.714}g^{0.714}}{\mu^{0.428}\rho^{0.286}} = \pu{1.04 m/s}\),雷诺数 \(Re_{\mathrm p} = \cfrac {d_{\mathrm p}\rho u}{\mu} = 13.8\),符合湍流过渡区范围。

阻力系数 \(C_{\mathrm D} = \cfrac {18.5}{Re_{\mathrm p}^{0.6}} = 3.83\),阻力 \(F_{\mathrm D} = \cfrac 12C_{\mathrm D}A_{\mathrm p}\rho u^2 = \pu{7.76E-8 N}\)。

3.

给定温度 \(\pu{293 K}\)、气压 \(\pu{101325 Pa}\) 条件下,求得空气密度 \(\rho = \cfrac {pM}{RT} = \pu{1.206 kg/m3}\),黏度 \(\mu = \pu{1.809E-5 Pa*s}\)。

考虑最小粒径的水泥颗粒,\(d_{\mathrm p} = \pu{25 \mu m}\),忽略坎宁汉修正,假设其位于斯托克斯区,\(u_{\mathrm s} = \cfrac {d_{\mathrm p}^2(\rho_{\mathrm p} - \rho)g}{18\mu} = \pu{3.69E-3 m/s}\),计算 \(Re_{\mathrm p} = \cfrac {d_{\mathrm p}\rho u}{\mu} = 6.26\times 10^{-2} < 1\),符合要求。

则沉降时间 \(t = \cfrac{h}{u_{\mathrm s}} = \pu{122 s}\),最远距离 \(s = v_0t = \pu{171 m}\)。

4.

给定温度 \(\pu{433 K}\)、气压 \(\pu{101325 Pa}\) 条件下,求得空气密度 \(\rho = \cfrac {pM}{RT} = \pu{0.816 kg/m3}\),黏度 \(\mu = \pu{2.5E-5 Pa*s}\)。

\(d_{\mathrm p} = \pu{10 \mu m}\) 时,假设其位于斯托克斯区,\(u_{\mathrm s} = \cfrac {d_{\mathrm p}^2(\rho_{\mathrm p} - \rho)}{18\mu}\cfrac{u_{\mathrm t}^2}{R} = \pu{0.768 m/s}\),计算 \(Re_{\mathrm p} = 0.25\) 符合要求。

\(d_{\mathrm p} = \pu{500 \mu m}\) 时,假设其位于牛顿区,有 \(0.055\pi\rho d_{\mathrm p}^2u_{\mathrm s}^2 = \cfrac {\pi d_{\mathrm p}^3\rho_{\mathrm p}}{6}\cfrac{u_{\mathrm t}^2}{R}\),即 \(u_{\mathrm s} = 1.74\sqrt{\cfrac{d_{\mathrm p}(\rho_{\mathrm p} - \rho)}{\rho}\cfrac{u_{\mathrm t}^2}{R}} = \pu{80.1 m/s}\),计算 \(Re_{\mathrm p} = 1307\) 符合要求。

5.

在条件相同的情况下,多层重力沉降室除尘效率正比于层数,即 \(\cfrac {\eta_1}{\eta_2} = \cfrac {n_1 + 1}{n_2 + 1}\),则 \(n_2 + 1 = \cfrac {\eta_2(n_1 + 1)}{\eta_1} = \cfrac {0.8 \times 18}{0.649} = 22.2\),即设置 23 层可得到 \(80~\%\) 的操作效率。

6.

求得 \(\mu = \pu{0.067 kg/(m*h)} = \pu{1.86E-5 kg/(m*s)} = \pu{1.86E-5 Pa*s}\)。

假设位于斯托克斯区,\(d_{\mathrm min} = \sqrt{\cfrac {18\mu v_0H}{\rho_{\mathrm p}gL}} = \pu{8.39E-5 m} = \pu{83.9 \mu m} < \pu{100 \mu m}\),满足要求。

7.

\(q_V = \pu{3.61 L/min} = \pu{6.02E-5 m3/s}\)。

对粒径为 \(\pu{0.63 \mu m}\) 的粒子,估算其坎宁汉修正系数 \(C = 1 + \cfrac {0.165}{d_{\mathrm p}} = 1.26\),\(u_{\mathrm s} = \cfrac {d_{\mathrm p}^2\rho_{\mathrm p}gC}{18\mu} = \pu{1.57E-5 m/s}\),沉降效率 \(\eta = \cfrac {u_{\mathrm s}LW(n+1)}{q_V} = 52.3~\%\)。

对粒径为 \(\pu{0.83 \mu m}\) 的粒子,估算其坎宁汉修正系数 \(C = 1 + \cfrac {0.165}{d_{\mathrm p}} = 1.20\),\(u_{\mathrm s} = \cfrac {d_{\mathrm p}^2\rho_{\mathrm p}gC}{18\mu} = \pu{2.60E-5 m/s}\),沉降效率 \(\eta = \cfrac {u_{\mathrm s}LW(n+1)}{q_V} = 86.3~\%\)。

8.

按公式 \(\eta_i = \cfrac {\eta}{\eta + Pg_{2i}/g_{3i}}\),计算分级效率,如表 3-1 所示:

表3-1 分级效率
$d_{\mathrm p}$$d_{\mathrm pi}$$g_{3i}$$g_{2i}$$\eta_i \\ (\%)$
0 ~ 52.50.5765.59
5 ~ 107.51.412.949.41
10 ~ 1512.51.94.579.17
15 ~ 2017.52.12.190.00
20 ~ 2522.52.11.592.65
25 ~ 3027.520.796.26
30 ~ 3532.520.597.30
35 ~ 4037.520.497.83
40 ~ 4542.520.398.36
>45N/A841.199.9

按照上表计算结果作出分级效率曲线,如图 1 所示:

0.00% 10.00% 20.00% 30.00% 40.00% 50.00% 60.00% 70.00% 80.00% 90.00% 100.00% 0 5 10 15 20 25 30 35 40 45 分级效率曲线

由图得,分割粒径为 \(\pu{7.5 \mu m}\)。

9.

\(\eta_i = \cfrac {(d_{\mathrm pi}/d_{\mathrm c})^2}{1 + (d_{\mathrm pi}/d_{\mathrm c})^2} = \cfrac {(d_{\mathrm pi}/5)^2}{1 + (d_{\mathrm pi}/5)^2} = \cfrac {(d_{\mathrm pi})^2}{25 + (d_{\mathrm pi})^2}\),

\(\eta = \int_0^{+\infty} \eta_iq\mathrm dd_{\mathrm pi} = \int_0^{+\infty} \cfrac {(d_{\mathrm pi})^2}{25 + (d_{\mathrm pi})^2} q\mathrm dd_{\mathrm pi}\),

由于颗粒粒径分布符合对数正态分布,\(D_{\mathrm m} = \pu{20 \mu m}\),\(\sigma = 1.25\),则 \(q = \cfrac {1}{\sqrt{2\pi}d_{\mathrm pi}\ln\sigma_{\mathrm g}}\mathrm {exp}\left[-\left(\cfrac {\ln d_{\mathrm pi}/{d_{\mathrm g}}}{\sqrt 2\ln\sigma_{\mathrm g}}\right)^2\right] = \cfrac {1.79}{d_{\mathrm pi}}\mathrm e^{-10.04(\ln d_{\mathrm pi}/20)^2}\)。

积分可得 \(\eta = 96.3~\%\)。

10.

(1)

四块板子将电除尘器分成 \(3\) 个通道,则 \(q_V = 2/\pu{3 m3/s} = \pu{0.667 m3/s}\),板面积 \(A = 2\times\pu{3.66^2 m2} = \pu{26.8 m2}\)。则 \(\eta_i = 1 - \mathrm e^{-26.8\times0.122/0.667} = 99.3~\%\)。

(2)

流量为 \(50~\%\) 的通道达到最大速度,\(v_{\mathrm {max}} = 0.5\);平均速度为 \(\bar v = 1/3\),则二者比值为 \(\cfrac {0.5}{1/3} = 1.5\)。

查图 6-27 得校正系数 \(F_V = 1.75\),则通过率 \(P = (1 - \eta_i)F_V = 1.22~\%\),分级效率 \(\eta_i = 1 - P = 98.8~\%\)。

11.

(1)

代入 \(d_{\mathrm p} = \pu{0.9 \mu m}\) 时 \(\eta = 0.5\),求得 \(k = -\cfrac {\ln (1 - \eta)}{d_{\mathrm p}} = \pu{0.77 \mu m-1}\)。

按公式计算,如表 3-2 所示:

表3-2 分级除尘效率
质量分数/$\%$0 ~ 2020 ~ 4040 ~ 6060 ~ 8080 ~ 100
平均粒径/$\pu{\mu m}$3.5 8.0 13.0 19.0 45.0
分级效率/$\%$93.25%99.79%100.00%100.00%100.00%

则总分级效率 \(\eta = \sum \eta_ig_{1i} = 98.61~\% > 98~\%\)。

(2)

排放浓度为 \(\rho = 30 \times (1 - 98.61~\%)~\pu{g/m3} = \pu{0.417 g/m3} < \pu{0.5 g/m3}\),符合环境保护的相关规定。

(3)

满足使用者需要。

12.

给定温度 \(\pu{297 K}\),求得空气黏度 \(\mu = \pu{1.829E-5 Pa*s}\)。

当粒径为 \(d_{\mathrm p} = \pu{10 \mu m}\)、液滴直径为 \(d_{\mathrm D} = \pu{50 \mu m}\) 时,碰撞数 \(M = \sqrt{St} = \sqrt{\cfrac {d_{\mathrm p}^2\rho_{\mathrm p}\Delta u}{18\mu d_{\mathrm D}}} = 19.09\),\(R = d_{\mathrm p} / d_{\mathrm D} = 0.2\),\(\eta = \mathrm e^{-(0.018M^{0.5 + R}/R - 0.6R^2} = 50.39~\%\)。

同理分别求出粒径为 \(\pu{10 \mu m}\)、\(\pu{50 \mu m}\) 和液滴直径在 \(\pu{50 \mu m}\)、\(\pu{100 \mu m}\)、\(\pu{500 \mu m}\) 下的捕集效率,如表 3-3 所示:

表3-3 捕集效率
液滴直径/$\pu{\mu m}$粒径/$\pu{\mu m}$1050
5050.39%0.00%
10042.66%10.23%
50010.11%25.05%

13.

按公式 \(P = \mathrm {exp}\left(-\cfrac{6.1\times10^{-9}\rho_{\mathrm L}\rho_{\mathrm p}d_{\mathrm p}^2f^2\Delta p}{\mu_{\mathrm G}^2}\right)\),其中 \(\Delta p = -1.03\times10^{-3}v_{\mathrm T}^2\left(\cfrac {q_{V, \mathrm L}}{q_{V, \mathrm G}}\right) = \pu{96.5 cm}\ce{H_2O}\),则 \(P = \mathrm e^{-0.331d_{\mathrm p}^2}\)。求出各粒径分级效率,如表 3-4 所示:

表3-4 分级效率
$d_{\mathrm p}/\pu{\mu m}$$d_{\mathrm pi}/\pu{\mu m}$$g_i/\%$$\eta_i/\%$
< 0.10.050.010.1
0.1 ~ 0.50.30.212.9
0.5 ~ 1.00.750.7817.0
1.0 ~ 5.031394.9
5.0 ~ 10.07.516100.0
10.0 ~ 15.012.512100.0
15.0 ~ 20.017.58100.0
> 20.0N/A50100.00%

则总除尘效率 \(\eta = \sum\limits_i g_i\eta_i = 98.5~\%\)。

14.

给定温度 \(\pu{293 K}\)、气压 \(\pu{101325 Pa}\) 条件下,求得空气密度 \(\rho = \cfrac {pM}{RT} = \pu{1.206 kg/m3}\),黏度 \(\mu = \pu{1.809E-5 Pa*s}\)。

雨滴直径为 \(\pu{2 mm}\),位于牛顿区,终端沉降速率 \(u_{\mathrm D} = 1.74\sqrt{\cfrac {d_{\mathrm D}\rho_{\mathrm D}g}{\rho}} = \pu{7.02 m/s}\)。

颗粒物密度取 \(\rho_{\mathrm p} = \pu{2.0E3 kg/m3}\),则碰撞数 \(M = \sqrt{St} = \sqrt{\cfrac {d_{\mathrm p}^2\rho_{\mathrm p}\Delta u}{18\mu d_{\mathrm D}}} = 0.194\),查教材图 5-16 (3A) 得 \(\eta_{\mathrm t} = 15~\%\)。

由公式,每个雨滴下降过程中补给的颗粒物质量 \(M = \cfrac {\pi}{4}d_{\mathrm D}^2\Delta zc\eta_{\mathrm t} = \pu{1.13E-2 \mu g}\)。

雨滴自身质量 \(M_{\mathrm D} = \cfrac {\pi d_{\mathrm D}^3\rho_{\mathrm D}}{6} = \pu{4.19E3 \mu g}\),则比例为 \(\cfrac {1.13\times10^{-2}}{4.19\times10^{3}} = 2.7\times10^{-4}~\%\)。

15.

给定温度 \(\pu{300 K}\) 条件下,求得空气黏度 \(\mu = \pu{1.845E-5 Pa*s}\)。

由公式 \(\Delta p = \Delta p_{\mathrm 0} + \Delta p_{\mathrm p} = \Delta p_{\mathrm 0} + \cfrac {x_{\mathrm p}\mu_{\mathrm g}v}{K_{\mathrm p}}\),其中 \(x_{\mathrm p} = \cfrac {m}{\rho_{\mathrm p}S}\),则 \(\Delta p\) 和 \(m\) 成线性关系,斜率为 \(\cfrac {\mu_{\mathrm g}v}{K_{\mathrm p}\rho_{\mathrm p}S}\)。

对 \(\Delta p\) 和 \(m\) 作散点图如图 3-2 所示:

y = 13146 x + 616.51 r ² = 0.9905 0 200 400 600 800 1000 1200 1400 0 0.005 0.01 0.015 0.02 0.025 0.03 0.035 0.04 0.045 Δ p /Pa m /kg 回归曲线

求得回归曲线斜率为 \(k = \pu{13146 Pa/kg}\),则 \(K_{\mathrm p} = \cfrac {\mu_{\mathrm g}v}{k\rho_{\mathrm p}S} = \pu{3.51E-12 m2}\)。

16.

(1)

给定温度 \(\pu{293 K}\)、气压 \(\pu{101325 Pa}\) 条件下,求得空气密度 \(\rho = \cfrac {pM}{RT} = \pu{1.206 kg/m3}\),黏度 \(\mu = \pu{1.809E-5 Pa*s}\)。

代入公式 \(P = \mathrm{exp}\left(-\cfrac {7zv_{\mathrm s}D_{\mathrm {pa}}^2}{9D_{\mathrm c}^2\mu_{\mathrm G}\varepsilon}\right) = \mathrm e^{-4.30} = 0.0136\),则 \(\eta = 1 - P = 98.6~\%\)。

(2)

捕集效率为 \(99.9~\%\) 时,\(P = 0.001\),代入公式得 \(z = -\cfrac {9D_{\mathrm c}^2\mu_{\mathrm G}\varepsilon\ln P}{7v_{\mathrm s}D_{\mathrm {pa}}^2} = \pu{3.21 m}\)。

(3)

由《Air Pollution Control Engineering》公式,穿透率 \(P = \mathrm {exp}\left(-\cfrac {\pi Nv_{\mathrm c}D^2\rho_{\mathrm p}}{9W_i\mu}\right)\),取\(W_i = 0.25D_{\mathrm c}\),\(N = \cfrac {0.5z}{D_{\mathrm c}}\),\(v_{\mathrm c} = \cfrac {v_{\mathrm s}}{\varepsilon}\),\(D_{\mathrm {pa}} = D^2\rho_{\mathrm p}\),取 \(2\pi = 7\),即有 \(P = \mathrm{exp}\left(-\cfrac {7zv_{\mathrm s}D_{\mathrm {pa}}^2}{9D_{\mathrm c}^2\mu_{\mathrm G}\varepsilon}\right)\)。