Heat Transfer

Double-pipe heat exchanger energy balance

A double-pipe heat exchanger is analysed with two relations that must hold together: the energy balance and the rate equation. For a stream without phase change, the heat duty is Q=\dot m\,c_p\,\Delta T. In a perfectly insulated exchanger, the heat released by the hot stream equals the heat absorbed by the cold stream, so the duties calculated on the two sides must agree. The product C=\dot m\,c_p is the heat capacity rate in W K^{-1}. The stream with the smaller C undergoes the larger temperature change, and C_{min} sets the maximum possible duty, Q_{max}=C_{min}(T_{h,in}-T_{c,in}). The outlet temperatures must therefore be found from the energy balance before the log mean temperature difference is evaluated. A common mistake is to use the inlet temperature of the heated stream as its outlet temperature. Example: 1.5 kg s^{-1} of water with c_p=4.2 kJ kg^{-1} K^{-1} heated through 20 K needs Q=1.5\times4.2\times20=126 kW. When one stream condenses at a constant saturation temperature, its heat capacity rate is effectively infinite, its temperature stays fixed, and the duty is set by the latent heat, Q=\dot m_{steam}\lambda. This fixes the flow rate of the other stream. In a counterflow exchanger the terminal temperature differences are measured at opposite ends, and the number of transfer units is NTU=UA/C_{min}. The effectiveness is \varepsilon=Q/Q_{max}. When C_r=C_{min}/C_{max}=1, the relation reduces to NTU=\varepsilon/(1-\varepsilon), so an effectiveness of 0.75 corresponds to NTU = 3. The log mean temperature difference for counterflow is \text{LMTD}=(\Delta T_1-\Delta T_2)/\ln(\Delta T_1/\Delta T_2), and the area follows from A=Q/(U\,\text{LMTD}). In a cocurrent (parallel) arrangement, both fluids enter at the same end, so the terminal differences are measured there, and the LMTD is computed in the same way.

Heat exchanger overall coefficient and fouling

The overall heat transfer coefficient U lumps every resistance between the two fluids and is always referred to a stated area, usually the outside area A_o. For a tube wall it is written as \dfrac{1}{U_o}=\dfrac{1}{h_o}+R_{f,o}+\dfrac{r_o\ln(r_o/r_i)}{k}+\dfrac{r_o}{r_i}R_{f,i}+\dfrac{r_o}{r_i\,h_i}, so the resistances add in series. Fouling resistances R_f represent deposits on the surfaces, and they are added to the clean resistance. If the clean coefficient is 400 W m^{-2} K^{-1} and R_f=0.0005 m^2 K W^{-1}, then 1/U_{dirty}=0.0025+0.0005=0.003, giving U_{dirty}=333 W m^{-2} K^{-1}. The simplified relation 1/U=1/h_i+1/h_o is a shortcut that neglects the wall resistance and assumes the inner and outer areas are nearly equal. It is valid only for a thin wall of high thermal conductivity, so that the wall resistance \Delta x/k is negligible compared with the convective resistances. A thick wall or a low conductivity makes the shortcut invalid. The LMTD used with these coefficients is based on the terminal temperature differences. When the temperature profiles are linear along the length, the two end differences are evaluated directly and inserted into the log mean formula. A pair of end differences of 40 K and 10 K gives \text{LMTD}=30/\ln 4=21.6 K. If both fluid inlets are at the same end of the exchanger, the flow is parallel.

Nusselt film condensation theory

When a saturated vapour condenses on a cold surface and wets it, the condensate forms a continuous liquid film, and the film resistance controls the heat transfer. Nusselt's theory (1916) describes laminar film condensation on a vertical wall of constant temperature. Its assumptions are a laminar film, negligible inertia, conduction across the film with a linear temperature profile, and a film that moves under the balance of gravity and viscous shear. The average coefficient is \bar h = 0.943\left[\dfrac{\rho_l(\rho_l-\rho_v)\,g\,h_{fg}\,k^3}{\mu L\,\Delta T}\right]^{1/4}. This film-condensation relation is attributed to Nusselt theory, whereas Dittus-Boelter and Sieder-Tate are forced-convection correlations for single-phase flow. Because the film thickens in the downward direction, the local coefficient falls with distance x from the top edge. For laminar film condensation, h(x)\propto x^{-1/4}, which is equivalent to h\propto\Gamma^{-1/3}, since the local condensate flow rate per unit width satisfies \Gamma\propto x^{3/4}. Averaging over the height L gives \bar h=\frac{1}{L}\int_0^L Cx^{-1/4}dx=\frac{4}{3}CL^{-1/4}, while the bottom local value is CL^{-1/4}. The ratio of the average to the bottom value is therefore 4/3, and the average is always larger than the value at the bottom of the plate.

Fin efficiency of straight fins

A fin is a extended surface, and its efficiency \eta_f is the ratio of the actual heat flow from the fin to the heat flow that would occur if the entire fin were at base temperature. For a straight fin of uniform cross section with an adiabatic tip, \eta_f=\tanh(mL)/(mL), where m=\sqrt{hP/(kA_c)}. For a circular fin of radius R, m=\sqrt{2h/(kR)}, so with Bi=hR/k based on the radius, mL=\sqrt{2\,Bi}\,(L/R). A fin with mL=1 has \eta_f=\tanh 1=0.76. An aspect ratio that is large or a small Biot number makes the fin close to isothermal and raises the efficiency. Fin effectiveness \varepsilon_f is the ratio of the fin heat rate to the heat rate that would flow through the base area without the fin, \varepsilon_f=q_f/(hA_c\theta_b). It is related to efficiency by \varepsilon_f=\eta_f A_f/A_c, where A_f is the fin surface area and A_c is the base cross-sectional area. A fin with a convecting tip is handled by using the corrected length L_c=L+D/4 for a circular fin, and the surface area is then A_f=\pi D L_c. Effectiveness is therefore a useful way to size a fin when a target heat rate is given.

Critical insulation radius of cylinder

Adding insulation to a cylinder increases the conduction resistance through the layer but decreases the convective resistance, because the outer surface area grows. The critical radius of insulation is r_{cr}=k/h, where k is the insulation conductivity and h the outside convection coefficient. For outer radii below r_{cr}, adding insulation increases heat loss, while for outer radii above r_{cr}, it reduces heat loss. This matters most for small wires, whose radius can be below the critical value. For a wire of radius r_1 with insulation out to r_2, insulation reduces heat loss relative to the bare wire only when the total insulated resistance exceeds the bare convective resistance. Writing R_{ins}=\dfrac{\ln(r_2/r_1)}{2\pi k}+\dfrac{1}{2\pi h r_2} and R_{bare}=\dfrac{1}{2\pi h r_1}, the condition R_{ins}>R_{bare} gives k<\dfrac{h\,r_1\ln(r_2/r_1)}{1-r_1/r_2}. The upper limit on k is therefore the maximum conductivity that still reduces heat loss. A common example is an electrical wire where the heat generated equals VI and flows through the insulation to ambient air. The interface temperature is T_i=T_\infty+Q(R_{cond}+R_{conv}), where the two resistances are in series. For example, with Q=5 W, R_{cond}=0.2 K W^{-1}, R_{conv}=0.6 K W^{-1} and T_\infty=25 °C, the interface is at 25+5(0.8)=29 °C.

Radiation shields between parallel plates

A radiation shield is a thin surface placed between two large parallel plates to reduce the net radiative exchange. When the shield and both plates have the same emissivity \varepsilon and N shields are installed, the net flux is q=\dfrac{\sigma(T_1^4-T_2^4)}{(N+1)(2/\varepsilon-1)}, so each shield adds one more gap in series and divides the flux by N+1. With equal emissivity on all surfaces, a single shield settles at the fourth-power mean of the two plate temperatures, T_s^4=\dfrac{T_1^4+T_2^4}{2}, when its conduction resistance is negligible. For plates at 600 K and 400 K, T_s^4=\dfrac{600^4+400^4}{2} gives T_s\approx 528 K. The result is independent of the common emissivity value. When the number of shields is chosen for a material limit, the hottest shield is the one next to the hot wall. Its temperature is T_1^4=T_h^4-\dfrac{T_h^4-T_c^4}{N+1}, so each added shield lowers the hottest shield temperature, and the largest number of shields is the largest N that keeps this temperature below the limit.

Conduction thermal resistance of slab

Steady one-dimensional conduction is analysed with the thermal resistance concept, which is the analogue of electrical resistance. For a plane slab of thickness L, conductivity k and area A, Fourier's law gives Q=kA(T_1-T_2)/L. The thermal resistance is defined as R_{th}=\dfrac{T_1-T_2}{Q}=\dfrac{L}{kA}, with units of K W^{-1}. The resistance grows with thickness and falls with both conductivity and area. Its reciprocal kA/L is the conductance, which is a common source of error when the expression is inverted. Resistances in series add, so a composite wall has R_{total}=\sum L_i/(k_iA), and the heat rate is Q=\Delta T/R_{total}. For example, a 0.02 m slab with k=50 W m^{-1} K^{-1}, A=0.1 m^2 and \Delta T=100 K carries Q=kA\Delta T/L=50\times0.1\times100/0.02=25 kW. The same relation applies to any slab whose faces are held at fixed temperatures and whose heat flow is perpendicular to the faces.

Laminar flat plate boundary layer heat transfer

For laminar flow over an isothermal flat plate, the local Nusselt number is Nu_x=0.332\,Re_x^{1/2}Pr^{1/3}, so the local coefficient varies as h_x\propto x^{-1/2}. The average coefficient over a length l is \bar h_l=\frac{1}{l}\int_0^l h_x\,dx=\frac{2C}{\sqrt l}, where C is the constant in h_x=C x^{-1/2}. This means that the average over the length is twice the local value at the trailing edge. If the same plate is viewed with flow along the other dimension, \bar h_w=2C/\sqrt w, and the ratio of the two averages is \sqrt{l/w}. For a plate with l=4 m and w=1 m, this ratio is 2. The thermal boundary layer is related to the velocity boundary layer. The Blasius velocity thickness is \delta=4.91\,x/\sqrt{Re_x}=4.91\sqrt{\nu x/V}. For Pr\gtrsim 0.6, the thermal thickness follows \delta_t\approx\delta\,Pr^{-1/3}. In water, with Pr\approx 7, the thermal layer is about half the velocity layer. Given a target thermal thickness, the distance x is obtained by inverting \delta_t=4.91\,Pr^{-1/3}\sqrt{\nu x/V}.