Heat Exchanger Types and the Energy Balance
Heat Exchanger Types and the Energy Balance
Heat exchangers transfer thermal energy between two fluids. They appear in HVAC systems, power plants, chemical processes, engines, and refrigeration. The fundamental design equation links heat duty, area, and temperature difference.
Double-pipe (concentric tube): One fluid inside, one in the annulus. Simplest type. Parallel-flow or counter-flow.
Shell-and-tube: Tubes bundled inside a shell. Baffles direct shell-side flow. 1, 2, or 4 tube passes.
Cross-flow: Fluids flow perpendicular. Common in finned-tube radiators, HVAC coils.
Plate (compact): Stacked corrugated plates. High area-to-volume ratio.
At steady state, heat lost by the hot fluid equals heat gained by the cold fluid (no losses):
$$\dot{Q} = \dot{m}_h c_{p,h}(T_{h,\text{in}} - T_{h,\text{out}}) = \dot{m}_c c_{p,c}(T_{c,\text{out}} - T_{c,\text{in}})$$
We often define the capacity rate: $C = \dot{m}c_p$ (W/K), so $\dot{Q} = C_h \Delta T_h = C_c \Delta T_c$.
$$\frac{1}{UA} = \frac{1}{h_i A_i} + R''_{f,i}\frac{1}{A_i} + \frac{\ln(D_o/D_i)}{2\pi k L} + R''_{f,o}\frac{1}{A_o} + \frac{1}{h_o A_o}$$
For thin-walled tubes ($A_i \approx A_o$): $\frac{1}{U} = \frac{1}{h_i} + R''_f + \frac{1}{h_o}$
Typical $U$ values (W/(m²·K)): Water-to-water: 850–1700 • Steam-to-water: 1000–6000 • Gas-to-gas: 10–40
Hot water ($\dot{m} = 2$ kg/s, $c_p = 4180$) cools from 90°C to 60°C. Cold water ($\dot{m} = 3$ kg/s, $c_p = 4180$) enters at 20°C. Find $\dot{Q}$ and $T_{c,\text{out}}$.
$\dot{Q} = 2 \times 4180 \times (90 - 60) = 250{,}800$ W = 250.8 kW
$T_{c,\text{out}} = 20 + \frac{250{,}800}{3 \times 4180} = 20 + 20 = 40$°C
Inner convection $h_i = 3000$ W/(m²·K), outer $h_o = 500$. No fouling, thin wall. Find $U$.
$\frac{1}{U} = \frac{1}{3000} + \frac{1}{500} = 0.000333 + 0.002 = 0.002333$
$U = 429$ W/(m²·K)
The outer (lower) coefficient dominates — this is common when one fluid is gas and the other is liquid.
Oil ($\dot{m} = 1.5$ kg/s, $c_p = 2100$) is cooled by water ($\dot{m} = 2$ kg/s, $c_p = 4180$). Which is $C_{\min}$?
$C_h = 1.5 \times 2100 = 3150$ W/K (oil)
$C_c = 2 \times 4180 = 8360$ W/K (water)
$C_{\min} = 3150$ W/K (oil side). The fluid with $C_{\min}$ experiences the larger temperature change.
Practice Problems
Show Answer Key
1. $\dot{Q} = 4 \times 4180 \times 40 = 668{,}800$ W = 668.8 kW
2. $1/U = 1/8000 + 1/200 = 0.005125$. $U = 195$ W/(m²·K). Outer side controls.
3. $C_h = 525$ W/K, $C_c = 804$ W/K. $C_{\min} = 525$ (hot gas).
4. $\Delta T = 100{,}000/5000 = 20$°C
5. High $U$ means efficient heat transfer — typically liquid-to-liquid or phase change.
6. $U = 500$ W/(m²·K)