Applications of Heat Transfer

Fins

A fin is an extended surface that increases the area for convection, so that more heat leaves a hot surface. Heat flow $Q = hA\Delta T$ can be increased by raising $h$, $\Delta T$ or $A$; fins increase $A$ when $h$ is low (gas side). Examples: engine cylinder fins, radiator tubes, heat sinks.

Types: straight (rectangular, triangular, parabolic profile), pin (cylindrical or tapered), annular (circumferential) fins on tubes.

Derivation (uniform fin, insulated tip)

Assumptions: steady one-dimensional conduction along the fin, constant $k$ and $h$, uniform fin temperature across thickness. Let $A_c$ be the cross-section, $P$ the perimeter, $T_b$ the base temperature and $\theta = T - T_\infty$.

Energy balance on a slice $dx$: heat conducted in = heat conducted out + heat convected:

$$-kA_c\frac{dT}{dx} = -kA_c\left(\frac{dT}{dx}+\frac{d^2T}{dx^2}dx\right)+hP\,dx\,(T-T_\infty)$$ $$\frac{d^2\theta}{dx^2}-m^2\theta = 0,\qquad m=\sqrt{\frac{hP}{kA_c}}$$

Solution: $\theta = C_1e^{mx}+C_2e^{-mx}$. Boundary conditions: $\theta = \theta_b$ at $x = 0$; $d\theta/dx = 0$ at $x = L$ (insulated tip). This gives

$$\frac{\theta}{\theta_b} = \frac{\cosh m(L-x)}{\cosh mL}$$

Heat dissipated equals conduction at the base:

$$Q_{fin} = -kA_c\frac{d\theta}{dx}\Big|_{x=0} = \sqrt{hPkA_c}\;\theta_b\tanh(mL)$$

Other tip conditions: infinitely long fin $Q = \sqrt{hPkA_c}\,\theta_b$; convecting tip, use corrected length $L_c = L + A_c/P$ in the above formula.

Efficiency and effectiveness

$$\eta_{fin}=\frac{Q_{fin}}{Q_{max}}=\frac{Q_{fin}}{hPL\,\theta_b}=\frac{\tanh mL}{mL} \qquad \varepsilon_{fin}=\frac{Q_{fin}}{hA_c\theta_b}$$

Efficiency is the actual heat loss over the heat that would be lost if the entire fin were at the base temperature. Effectiveness is the heat with the fin over the heat without the fin.

A fin is worthwhile if $\varepsilon_{fin} > 2$ (preferably much greater). This needs high $k$ (copper, aluminium), thin closely spaced fins, and low $h$ on the fin side (gas, free convection). Fins are useless when $h$ is high, e.g. boiling or condensation.

Data: $D = 0.01$ m, $L = 0.05$ m, $\theta_b = 120 - 30 = 90$ K.

$$P = \pi D = 0.031416\ \text{m},\qquad A_c = \frac{\pi D^2}{4}=7.854\times10^{-5}\ \text{m}^2$$ $$m = \sqrt{\frac{hP}{kA_c}} = \sqrt{\frac{4h}{kD}} = \sqrt{\frac{4\times25}{200\times0.01}} = 7.071\ \text{m}^{-1}$$ $$mL = 7.071\times0.05 = 0.3536,\qquad \tanh(mL)=0.3395,\qquad \cosh(mL)=1.0630$$

(a) Heat dissipated

$$\begin{aligned} \sqrt{hPkA_c} &= \sqrt{25\times0.031416\times200\times7.854\times10^{-5}} = 0.1111\ \text{W/K}\\ Q &= 0.1111\times90\times0.3395 = 3.39\ \text{W} \end{aligned}$$

(b) Tip temperature

$$T_{tip} = T_\infty + \frac{\theta_b}{\cosh mL} = 30 + \frac{90}{1.0630} = 114.7\ ^\circ\text{C}$$

(c) Efficiency

$$\eta = \frac{\tanh mL}{mL} = \frac{0.3395}{0.3536} = 0.960 = 96\%$$

(Check: $Q_{max} = hPL\theta_b = 25\times0.031416\times0.05\times90 = 3.534$ W, and $3.394/3.534 = 0.960$.)

(d) Effectiveness

$$\varepsilon = \frac{Q}{hA_c\theta_b} = \frac{3.394}{25\times7.854\times10^{-5}\times90} = 19.2$$

An effectiveness of 19 shows the fin is well worth using.

Answer: $Q \approx 3.39$ W; $T_{tip} \approx 114.7\ ^\circ$C; $\eta \approx 96\%$; $\varepsilon \approx 19$.

(a) Classification

By flow arrangement

• Parallel flow: both fluids enter at the same end and flow in the same direction. The temperature difference is largest at inlet; the cold outlet can never exceed the hot outlet.
• Counterflow: fluids flow in opposite directions. Gives the highest mean temperature difference and effectiveness (oil coolers, regenerative systems).
• Cross flow: fluids flow at right angles (car radiator, air coolers), mixed or unmixed.

By construction

• Double pipe (pipe-in-pipe): simple, for small duties.
• Shell and tube: a bundle of tubes in a shell with baffles; most common in power and process plants (condensers, boilers).
• Compact (plate-fin): large area per volume, for gases (aircraft, automobiles).
• Plate type: corrugated plates, easy to clean (food, dairy).
• Direct contact (cooling tower) and regenerators (stored heat) are other categories.

(b) LMTD for counterflow

Assumptions: steady state; constant $U$, $c_{ph}$ and $c_{pc}$; no heat loss to surroundings; no phase change; negligible axial conduction.

 Th1 -->--------------------->-- Th2
 Tc2 --<---------------------<-- Tc1
      dA element at distance x

For an element of area $dA$: with $\Delta T = T_h - T_c$:

$$dQ = -\dot m_hc_h\,dT_h = -\dot m_cc_c\,dT_c = U\,\Delta T\,dA$$

Taking $x$ along the hot-fluid direction, $T_h$ and $T_c$ both fall as $x$ increases. Writing $C_h = \dot m_hc_h$ and $C_c = \dot m_cc_c$:

$$d(\Delta T) = dT_h - dT_c = -dQ\left(\frac{1}{C_h}-\frac{1}{C_c}\right)$$

Substituting $dQ = U\Delta T\,dA$:

$$\frac{d(\Delta T)}{\Delta T} = -U\left(\frac{1}{C_h}-\frac{1}{C_c}\right)dA$$

Integrate over the full area, from end 1 ($\Delta T_1 = T_{h1}-T_{c2}$) to end 2 ($\Delta T_2 = T_{h2}-T_{c1}$):

$$\ln\frac{\Delta T_2}{\Delta T_1} = -UA\left(\frac{1}{C_h}-\frac{1}{C_c}\right)$$

The total heat is $Q = C_h(T_{h1}-T_{h2}) = C_c(T_{c2}-T_{c1})$, so $\dfrac{1}{C_h}-\dfrac{1}{C_c} = \dfrac{(T_{h1}-T_{h2})-(T_{c2}-T_{c1})}{Q} = \dfrac{\Delta T_1-\Delta T_2}{Q}$. Therefore

$$\ln\frac{\Delta T_2}{\Delta T_1} = -\frac{UA(\Delta T_1-\Delta T_2)}{Q} \;\Rightarrow\; Q = UA\,\frac{\Delta T_1-\Delta T_2}{\ln(\Delta T_1/\Delta T_2)} = UA\,\Delta T_{lm}$$ $$\boxed{\Delta T_{lm}=\frac{\Delta T_1-\Delta T_2}{\ln(\Delta T_1/\Delta T_2)}}$$

Same expression holds for parallel flow, with $\Delta T_1$, $\Delta T_2$ taken at the inlet and outlet ends. If $\Delta T_1 = \Delta T_2$ then $\Delta T_{lm} = \Delta T_1$. For cross-flow and multi-pass exchangers, $Q = UAF\Delta T_{lm,cf}$ with a correction factor $F<1$.

(a) Heat duty

$$Q = \dot m_hc_h(T_{h1}-T_{h2}) = 2\times2100\times(100-60) = 168\,000\ \text{W} = 168\ \text{kW}$$

(b) Water outlet

$$T_{c2} = T_{c1}+\frac{Q}{\dot m_cc_c} = 20 + \frac{168\,000}{1.2\times4180} = 20 + 33.49 = 53.5\ ^\circ\text{C}$$

(c) Counterflow LMTD and area

 Oil   100 ----------> 60
 Water  53.5 <---------- 20

$\Delta T_1 = T_{h1}-T_{c2} = 100 - 53.5 = 46.5$ K, $\Delta T_2 = T_{h2}-T_{c1} = 60 - 20 = 40$ K.

$$\Delta T_{lm} = \frac{46.5-40}{\ln(46.5/40)} = \frac{6.51}{0.1505} = 43.2\ \text{K}$$ $$A = \frac{Q}{U\,\Delta T_{lm}} = \frac{168\,000}{400\times43.17} = 9.73\ \text{m}^2$$

(d) Parallel flow

Both fluids enter at the same end: $\Delta T_1 = 100 - 20 = 80$ K, $\Delta T_2 = 60 - 53.5 = 6.5$ K.

$$\Delta T_{lm,p} = \frac{80-6.5}{\ln(80/6.5)} = \frac{73.5}{2.510} = 29.3\ \text{K}$$ $$A_p = \frac{168\,000}{400\times29.29} = 14.3\ \text{m}^2$$

Parallel flow needs about 47 percent more area for the same duty, showing why counterflow is preferred.

Answer: $Q = 168$ kW; $T_{c2} = 53.5$ °C; $\Delta T_{lm} = 43.2$ K and $A = 9.73$ m$^2$ (counterflow); $A = 14.3$ m$^2$ (parallel flow).

The effectiveness-NTU method

Effectiveness $\varepsilon$ is the ratio of the actual heat transfer to the maximum possible:

$$\varepsilon=\frac{Q}{Q_{max}},\qquad Q_{max}=C_{min}\,(T_{h,in}-T_{c,in})$$

where $C = \dot m c_p$ is the heat capacity rate. Number of transfer units $NTU = UA/C_{min}$ and capacity ratio $C_r = C_{min}/C_{max}$. For each flow arrangement, $\varepsilon = f(NTU, C_r)$.

It is preferred when the outlet temperatures are unknown (rating or performance problems): the LMTD method would need trial and error, while the NTU method is direct. LMTD is easier for sizing when all four temperatures are known.

Solution

Heat capacity rates:

$$C_h = 0.8\times2100 = 1680\ \text{W/K},\qquad C_c = 0.6\times4180 = 2508\ \text{W/K}$$

So $C_{min} = C_h = 1680$ W/K (oil), $C_{max} = 2508$ W/K.

$$C_r = \frac{1680}{2508} = 0.670,\qquad NTU = \frac{UA}{C_{min}} = \frac{350\times3}{1680} = 0.625$$

Effectiveness (counterflow):

$$\varepsilon = \frac{1-e^{-0.625(1-0.670)}}{1-0.670\,e^{-0.625(1-0.670)}} = \frac{1-0.8134}{1-0.5449} = \frac{0.1866}{0.4551}= 0.410$$

Heat transfer rate:

$$Q = \varepsilon C_{min}(T_{h,in}-T_{c,in}) = 0.4097\times1680\times(140-25) = 79\,160\ \text{W}\approx79.2\ \text{kW}$$

Outlet temperatures:

$$T_{h,out} = 140 - \frac{79\,160}{1680} = 92.9\ ^\circ\text{C},\qquad T_{c,out} = 25 + \frac{79\,160}{2508} = 56.6\ ^\circ\text{C}$$

(For comparison, a parallel-flow exchanger of the same size would have $\varepsilon = \frac{1-e^{-NTU(1+C_r)}}{1+C_r} = 0.388$.)

Answer: $\varepsilon \approx 0.41$; $Q \approx 79.2$ kW; $T_{h,out} \approx 92.9$ °C; $T_{c,out} \approx 56.6$ °C.