36、传热相关计算与分析

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分类专栏： MATLAB赋能化工计算文章标签：传热计算管壳式换热器套管式换热器

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传热相关计算与分析

1. 传热基础数据与计算示例

1.1 基础数据设定

在传热计算中，需要设定一系列基础数据，以下是一些常见的参数及代码示例：

Trt = [323 283]; Trs = [375 289]; % reference temperatures (tube ans shell)(K)
rhoreft = [988.1 999.7]; rhorefs = [798 885]; % densities (kg/m^3)
mureft = [0.6 1.26]; murefs = [0.258 0.679]; % viscosities (mNs/m^2)
xkreft = [0.64 0.603]; xkrefs = [0.126 0.163]; % thermal conductivities
cpreft = [4183 4195]; cprefs = [1980 1675]; % heat capacities (J/kg/K)
% Heat exchanger geometry
Do = 25.4; Di = 19.86; % tube outside diameter (Do,mm) and inside diameter (Di,mm)
Xkw = 45; % heat conductivity of tube wall (W/m/K)
L = 2; % tube length (m)
Ls = 27; % tube sheet thickness (mm)
rf = 0.025; % tube roughness(mm)
Nt = 86; % total number of tubes in tube bundle
Layout = 1; % tube layout (1:triangular, 2:in-line square, 3:rotated square)
Pt = 1.25*Do; % tube pitch (mm)
Nss = 0; % number of pairs of sealing strips
Npass = 4; % number of passes
Rdt = 0.00036; Rds = 0.00018; % fouling resistance (m^2*K/W)
Ds = 305; % shell inside diameter(mm)
Dotl = 294; % shell outside tube limit(mm)
Dsb = 4.45; % shell-baffle clearance(mm)
Lbin = 165; Lbout = 165; % inlet and outlet baffle spacing(mm)
Lbc = 450; % central baffle spacing(mm)
Lc = 0.25*Ds; % baffle cut(mm)
% Specification of key variables
Ti1 = 298; Ti2 = 303; % inlet and outlet temperatures for tube-side(K)
Ts1 = 353; % Shell-side(hot stream) inlet temperature(K)
Ws = 4.8; % Shell-side(hot stream) mass flow rate(kg/s)
fsT = 1; fsS = 1; % state of tube-side and shell-side fluids(1:liquid, 2:vapor)
ptype = 8; % problem type
Ts2 = 338; % guess outlet temperature of shell-side fluid
Wi = 5; % guess tube-side flow rate

1.2 计算结果

运行主脚本 hxnst 可以得到以下输出结果：

>> hxnst
Overall heat transfer coefficient: U = 125.291(W/m^2/K)
Heat transfer coefficient:
tube-side = 529.371(W/m^2/K), shell-side = 201.457(W/m^2/K)
Heat duty: Q = 80243(W)
Pressure drop: tube-side = 232564(Pa), shell-side = 26610.9(Pa)
Tube-side: Ti1 = 298(K), Ti2 = 303(K), flow rate = 3.77462(kg/s)
Shell-side: Ts1 = 353(K), Ts2 = 344.138(K), flow rate = 4.8(kg/s)

2. 套管式换热器

2.1 适用场景

套管式翅片管换热器适用于以下几种情况：
- 热负荷适中（即 UA < 100,000）。
- 其中一股物流为粘性液体。
- 流量较小时。

2.2 计算方法

2.2.1 管侧计算

液体情况 ：
- 当雷诺数 (Re) 满足一定条件时，传热系数计算公式为：
  [h_i = 0.023\left(\frac{Re}{1}\right)^{0.8}\left(\frac{Pr}{1}\right)^{0.333}\left(\frac{\mu}{\mu_w}\right)^{0.14}\frac{K}{D_{eq}}]
  其中，(Re = \frac{GD_{ti}}{\mu})，(D_{eq}) 为当量直径。
- 若 (Re < 10000)，则 (h_i = J Pr^{0.333}\frac{K}{D_{eq}})。
蒸汽情况 ：
[h_i = 0.0144\left(\frac{C_pG}{D_{eq}}\right)^{0.8}\left(\frac{h_{if}}{h_{i}}\right)^{0.2}]
其中，(h_{if}) 为考虑污垢修正后的传热膜系数。

2.2.2 壳侧计算（光管）

在管侧计算中使用 (D_{eq} = D_{si} - D_{to})。

2.2.3 总传热系数

总传热系数计算公式为：
[\frac{1}{U} = \frac{D_{to}}{D_{ti}}\frac{1}{K}\left(\frac{A_o}{A_i}\right) + \frac{1}{h_{if,i}}\left(\frac{A_o}{A_{tube}}\right) + \frac{1}{h_{if,shell}}\left(\frac{A_i}{A_o}\right)]

2.2.4 壳侧计算（翅片管）

在管侧计算中，需要重新设定一些参数：
[D_{eq} = \frac{4N_f}{\pi(D_{si} + D_{to}) - N\omega + A_f}]
[A_f = 2H_fN]
[C_s = \frac{\pi(D_{si}^2 - D_{to}^2)}{4}]
[N_f = C_s - \frac{\omega A_f}{2}]
[A_o = \pi D_{to} + A_f]
[X = H_f\sqrt{\frac{h_{ifs}}{6K_t\omega}}]
[\omega = \frac{\tanh(X)}{X}]
[\omega’ = \omega\frac{A_f}{A_o} + 1 - \frac{A_f}{A_o}]
[h_{ifd} = h_{ifs}\omega’]
[U_o = \frac{1}{\frac{D_{to} - D_{ti}}{K_t} + h_{ift}\frac{A_r}{A_o} + h_{ifd}}]

2.3 示例计算

以下是一个计算翅片管换热器传热系数的示例代码：

% htcoef.m : heat transfer coefficients (double-pipe finned-tube heat 
exchanger)
clear all;
% Data
Fft = 0.002; Kt = 0.366; mut = 0.72; Cpt = 1; Jt = 9.75; % Tube
Ffs = 0.002; Ks = 0.074; mus = 2.45; Cps = 0.518; % Shell
Dis = 3.068/12; Dos = 3.5/12; % shell size
Dit = 1.61/12; Dot = 1.9/12; % tube size
N = 24; Hf = 0.5/12; w = 0.035/12; % fin size
Gt = 26740; Gs = 18000; % flow rates
% Tube-side :
Deq = Dit;
Pr = Cpt*(2.4191*mut)/Kt; % 1 cP = 2.4191 lb/(ft-hr)
Re = Gt*Deq/(2.4191*mut);
if (Re < 10000)
     hit = Jt* Pr^0.333 * Kt/Deq;
else
     hit = 0.023 * Re^0.8 * Pr^0.333 * Kt / Deq;
end
hift = 1 / (1/hit + Fft);
% Shell-side (bare tube):
Deq = Dis - Dot; Pr = Cps*(2.4191*mus)/Ks; % 1 cP = 2.4191 lb/(ft-hr)
Re = Gs*Deq/(2.4191*mus);
if (Re < 10000)
     his = Jt* Pr^0.333 * Ks/Deq;
else
     his = 0.023 * Re^0.8 * Pr^0.333 * Ks / Deq;
end
hifs = 1 / (1/his + Ffs);
% Shell-side (finned-tube):
Af = 2*Hf*N; Cs = pi*(Dis^2 - Dot^2)/4; Nf = Cs - w*Af/2;
Deq = 4*Nf/(pi*(Dis+Dot) - N*w + Af);
Ao = pi*Dot + Af; X = Hf*sqrt(hifs/(6*Kt*w));
e = (exp(X) - exp(-X))/(exp(X) + exp(-X))/X;
ep = e*Af/Ao + 1 - Af/Ao; hifd = hifs*ep;
Ar = pi*Dit/(pi*Dot + Af); Uo = 1/((Dot-Dit)/Kt + hift*Ar + hifd);
% Output
fprintf('\nShell-side heat transfer coefficient: %10.7f\n', hifd);
fprintf('Tube-side heat transfer coefficient: %10.7f\n', hift);
fprintf('Fin efficiency: %10.7f\n', ep);
fprintf('Overall heat transfer coefficient: %10.7f\n', Uo);

运行该脚本得到的结果如下：

>> htcoef
Shell-side heat transfer coefficient: 12.3624311
Tube-side heat transfer coefficient: 41.0441459
Fin efficiency: 0.5070751
Overall heat transfer coefficient: 0.0516646

3. 管道伴热

3.1 适用场景

输送粘性流体的管道需要通过伴热来维持高温。
含有蒸汽的管道也可能需要伴热，以防止组分冷凝。
当管道散热导致的冷却无法接受时，伴热就变得必要。

3.2 计算方法

通过绝缘管道的传热涉及四个热阻：管道内壁的膜阻、管壁的热阻、绝缘层的热阻以及绝缘层外的空气膜阻。计算这些量的方程如下：
- 平均温度计算：
[T_{m,avg} = \frac{T_{mi} + T_{mo}}{2}]
[T_{a} = \frac{T_{m,avg} + T_{p}}{2}]
- 热传递计算：
[Q = \frac{2\pi k(T_{i} - T_{a})}{\ln(\frac{d_{o}}{d_{i}})}]
[Q = h_{air}\pi d_{o}(T_{s} - T_{a})]
[x = \frac{h_{a}d_{o}}{24K_{i}}\ln(\frac{d_{o}}{d_{i}})]
[T_{s} = \frac{T_{a} + xT_{air}}{x + 1}]
[w_{f} = 2.814 - 0.0003885714(T_{s} - T_{air}) - 0.0000012857(T_{s} - T_{air})^2]
[h_{a} = \frac{564w_{f}}{d_{o}^{0.19}(273 - T_{s} + T_{air})}]
[Q_{t} = QL]
[W = \frac{Q_{t}}{C_{p}(T_{mi} - T_{mo})}]
[n_{t} = \frac{Q}{a(T_{m,avg} - T_{p})}]
[n_{tc} = \frac{Q}{b(T_{m,avg} - T_{p})}]

其中，各参数含义如下表所示：
|参数|含义|
|----|----|
|(T_{m,avg})|加热介质的平均温度 (°F)|
|(T_{mi})|加热介质的入口温度 (°F)|
|(T_{mo})|加热介质的出口温度 (°F)|
|(d_{i})|绝缘层的内径 (in.)|
|(d_{o})|绝缘层的外径 (in.)|
|(r_{o})|管道的外径 (in.)|
|(t_{A})|伴热直径的余量 (in.)|
|(t_{k})|绝缘层厚度 (in.)|
|(T_{a})|管道和伴热的平均温度 (°F)|
|(T_{air})|空气温度 (°F)|
|(T_{s})|绝缘层外表面温度 (°F)|
|(k_{i})|绝缘材料的热导率 (Btu/(ft °F hr))|
|(h_{a})|空气的膜传热系数 (Btu/(ft² °F hr))|
|(h_{rc})|对流和辐射的综合传热系数 (Btu/(ft² °F hr))|
|(w_{f})|风因子|
|(Q_{t})|管道的总热损失 (Btu/hr)|
|(L)|管道总长度 (ft)|
|(W)|热介质的流量 (lb/lr)|
|(C_{p})|热介质的比热容 (Btu/(lb °F))|
|(n_{t})|不使用传热水泥时所需的伴热数量|
|(n_{tc})|使用传热水泥时所需的伴热数量|

3.3 计算流程

graph TD;
    A[假设 \(h_{a}\) 并计算 \(T_{s}\)] --> B[计算新的 \(h_{a}\) 并更新风因子];
    B --> C[计算新的 \(T_{s}\)];
    C --> D{ \(h_{a}\) 是否收敛? };
    D -- 否 --> B;
    D -- 是 --> E[计算 \(Q\), \(W\), \(n_{t}\), \(n_{tc}\)];

3.4 示例代码

% pipeTracer.m: determine heat loss and tracing requirements
clear all
% Input data
Tmi = input('Inlet temperature of heating medium(deg.F): ');
Tmo = input('outlet temperature of heating medium(deg.F): ');
Cp = input('Specific heat of heating medium(Btu/lb/F): ');
Tp = input('Pipe temperature(deg.F): ');
Tair = input('Air temperature(deg.F): ');
ro = input('Outside diameter of pipe(in): ');
L = input('Pipe length(ft): ');
tt = input('Tracer size(in): ');
tk = input('Insulation thickness(in): ');
Ki = input('Thermal conductivity of insulation(Btu/(hr-ft-F)): ');
% Parameters:
epsh = 1e-4; tA = 1.25;
Tmavg = (Tmi +Tmo)/2; Ta = (Tmavg + Tp)/2;
Di = ro + tA; Do = Di + 2*tk;
if (tt <= 3/8)
       a = 0.295; b = 3.44;
elseif (tt <= 0.5)
       a = 0.393; b = 4.58;
else
       a = 0.49; b = 5.73;
end
% Determine ha (guess initial value)
ha = 4; hai = 0;
while (abs(ha-hai) >= epsh)
hai = ha; x = (ha*Do/24/Ki)*log(Do/Di);
Ts = (Ta + x*Tair)/(x+1);
wf = 2.814-0.0003885714*(Ts-Tair)-0.0000012857*(Ts-Tair)^2;
ha = 564*wf/(Do^0.19 * (273-Ts+Tair));
end
% Heat loss and tracers
Q = ha*pi*Do*(Ts-Tair)/12; Qt = Q*L;
W = Qt / (Cp*(Tmi-Tmo)); nt = Q / (a*(Tmavg-Tp));
ntc = Q / (b*(Tmavg-Tp));
% Output
fprintf('\nFilm heat transfer coefficient to air = %10.7f Btu/(hr-F-ft^2)\n', ha);
fprintf('Total heat loss = %12.5f Btu/hr\n', Qt);
fprintf('Outside surface temperature = %12.5f deg.F\n', Ts);
fprintf('Flow rate of hot medium = %12.5f Btu/hr\n', W);
fprintf('Tracers(without heat-transfer cement) = %3.1f\n', nt);
fprintf('Tracers(with heat-transfer cement) = %3.1f\n', ntc);

3.5 示例结果

>> pipeTracer
Inlet temperature of heating medium(deg.F): 625
outlet temperature of heating medium(deg.F): 550
Specific heat of heating medium(Btu/lb/F): 0.53
Pipe temperature(deg.F): 500
Air temperature(deg.F): 0
Outside diameter of pipe(in): 6.065
Pipe length(ft): 100
Tracer size(in): 0.5
Insulation thickness(in): 2.5
Thermal conductivity of insulation(Btu/(hr-ft-F)): 0.037
Film heat transfer coefficient to air = 3.8641643 Btu/(hr-F-ft^2)
Total heat loss = 23428.95335 Btu/hr
Outside surface temperature = 18.80591 deg.F
Flow rate of hot medium = 589.40763 Btu/hr
Tracers(without heat-transfer cement) = 6.8
Tracers(with heat-transfer cement) = 0.6

4. 空气冷却器

4.1 适用场景

空气冷却器可用于冷却高压物流，此时流体在管内流动。在空气冷却器的设计中，需要确定每排管的数量、管排数、空气出口温度、管的总表面积、冷却器迎风面积、空气流量、风扇功率和管束宽度等参数。

4.2 计算方法

设计计算中使用了各种关联方程，具体如下：
- 管排数计算：
[R = 3.167876 + 3.794767\ln\left(\frac{T_{PI} - T_{AI}}{U}\right)]
- 空气出口温度估算：
[T_{AO}’ = 0.005U\left(\frac{T_{PO} + T_{PI}}{2} - T_{AI}\right) + T_{AI}]
- 总传热面积计算：
[A = \frac{Q}{U\times LMTD}]
[LMTD = \frac{(T_{PO} - T_{AI}) - (T_{PI} - T_{AO})}{\ln\left(\frac{T_{PO} - T_{AI}}{T_{PI} - T_{AO}}\right)}]
- 其他参数计算：
[F_V = 720.85418\times0.953^R]
[F_A = \frac{A}{1.2557445\times R^{1.0030502}}]
[F = F_A\times F_V]
[BHP = \frac{F_A}{7.4212329 + 12.5342466R}]
[W_t = F_A\times(36.4 + 9.35R)]
[W = \frac{F_A}{L}]

其中，各参数含义如下表所示：
|参数|含义|
|----|----|
| (R) | 管排数 |
| (T_{PI}) | 流体入口温度 (°F) |
| (T_{AI}) | 空气入口温度 (°F) |
| (A) | 裸管表面积 (ft²) |
| (F_A) | 管束迎风面积 (ft²) |
| (F_V) | 空气迎面风速 (ft/min) |
| (T_{AO}’) | 估算的空气出口温度 (°F) |
| (U) | 总传热系数 (Btu/(hr·ft²·°F)) |
| (T_{PO}) | 流体出口温度 (°F) |
| (LMTD) | 对数平均温差 (°F) |
| (Q) | 热负荷 (Btu/hr) |
| (T_{AO}) | 计算的空气出口温度 (°F) |
| (F) | 管外空气流量 (std ft³/min) |
| (BHP) | 风扇功率 |
| (W_t) | 空气冷却器重量 (lb) |
| (W) | 管束宽度 (ft) |
| (L) | 管长 (ft) |

4.3 计算流程

graph TD;
    A[输入数据 \(T_{PI}\), \(T_{PO}\), \(T_{AI}\), \(U\), \(Q\), \(L\)] --> B[计算 \(R\)];
    B --> C[计算 \(F_V\), \(T_{AO}'\) 并设 \(T_{AO}=T_{AO}'\)];
    C --> D[计算 \(LMTD\), \(A\), \(F_A\), \(T_{AO}\)];
    D --> E{ \(|T_{AO} - T_{AO}'|\) 是否小于 \(\epsilon\)? };
    E -- 否 --> C;
    E -- 是 --> F[计算 \(F\), \(BHP\), \(W_t\), \(W\)];

4.4 示例代码

% airCooler.m: preliminary design of air cooler
clear all;
% Data input
Tpi = input('Inlet fluid temperature(deg.F) : ');
Tpo = input('Outlet fluid temperature(deg.F) : ');
Tai = input('Inlet air temperature(deg.F) : ');
U = input('Overall heat transfer coefficient(Btu/hr/ft^2/deg.F) : ');
Q = input('Heat load(Btu/hr) : ');
L = input('Tube length(ft) : ');
% Parameters:
epst = 1e-4; R = 3.167876 + 3.794767*log((Tpi-Tai)/U);
Fv = 720.85418 * 0.953^R; Tao = 0.005*U*((Tpo + Tpi)/2 - Tai) + Tai;
Taop = Tao + 10;
% Check convergence
while(abs(Tao-Taop) >= epst)
    Lmtd = ((Tpo-Tai)-(Tpi-Tao)) / (log((Tpo-Tai)/(Tpi-Tao)));
    A = Q/(U*Lmtd); Fa = A/(1.2557445*(R^1.0030502));
    Taop = Tao; Tao = Q/(1.08*Fa*Fv) + Tai;
end
% Output variables
F = Fa*Fv; Hp = Fa/(7.4212329+12.5342466*R);
Wt = Fa*(36.4+9.35*R); W = Fa/L;
% Print results
fprintf('\nOutlet air temperature = %10.5f deg.F\n', Tao);
fprintf('Air flow over tube = %12.5f ft^3/min\n', F);
fprintf('Tube bundle width = %9.5f ft\n', W);
fprintf('Log-mean temperature difference = %10.5f deg.F\n', Lmtd);
fprintf('Air cooler weight = %12.5f lb\n', Wt);
fprintf('Bare-tube surface area = %12.5f ft^2\n', A);
fprintf('Fan horsepower = %9.5f\n', Hp);

4.5 示例结果

>> airCooler
Inlet fluid temperature(deg.F) : 160
Outlet fluid temperature(deg.F) : 125
Inlet air temperature(deg.F) : 95
Overall heat transfer coefficient(Btu/hr/ft^2/deg.F) : 55
Heat load(Btu/hr) : 4550000
Tube length(ft) : 40
Outlet air temperature = 110.78583 deg.F
Air flow over tube = 266882.56061 ft^3/min
Tube bundle width = 11.11469 ft
Log-mean temperature difference = 38.81778 deg.F
Air cooler weight = 31986.71312 lb
Bare-tube surface area = 2131.16960 ft^2
Fan horsepower = 8.07255

5. 问题求解示例

5.1 问题列表

以下是一些传热相关的问题及简要描述：
1. 问题 1 ：低压饱和蒸汽在保温的 2 英寸 40 号钢管中流动，已知蒸汽侧和空气侧传热系数、空气温度等参数，求不同保温厚度下的热通量以及保持特定热通量所需的保温厚度。
2. 问题 2 ：水平钢管输送蒸汽，已知保温材料热导率与温度的函数关系以及自然对流下的外部传热系数公式，求无保温和特定保温厚度下的每米管长热损失。
3. 问题 3 ：二维金属板的温度分布满足二维椭圆型偏微分方程，已知边界条件，求板内温度分布。
4. 问题 4 ：同样是二维金属板，在不同边界条件下（部分边绝缘，部分边恒温），求板内温度分布。
5. 问题 5 ：用四程管壳式换热器冷却液态苯，已知冷却水流量、进出口温度等参数，求换热器内的传热情况。

这些问题涵盖了不同场景下的传热计算，通过运用前面介绍的传热知识和计算方法，可以逐步求解这些问题，进一步加深对传热过程的理解和掌握。