Heat Transfer Practice Exam — Free

Steady one-dimensional conduction through layers in series, with convection at each face, is a series thermal-resistance network. Per unit area:

$$R''_{conv,in} = \frac{1}{h_{in}} = \frac{1}{10} = 0.100,\qquad R''_{brick} = \frac{L}{k} = \frac{0.200}{0.70} = 0.286$$ $$R''_{ins} = \frac{0.050}{0.040} = 1.250,\qquad R''_{conv,out} = \frac{1}{25} = 0.040\qquad(\text{all in m}^2\cdot\text{K/W})$$

Total: $R''_{tot} = 0.100 + 0.286 + 1.250 + 0.040 = 1.676\,\text{m}^2\cdot\text{K/W}$.

The overall temperature difference drives the flux through the whole network:

$$q'' = \frac{T_{in} - T_{out}}{R''_{tot}} = \frac{20 - (-10)}{1.676} = 17.9\,\text{W/m}^2$$

Because the same flux crosses every resistance, the intermediate temperatures follow at once: the inner wall surface is at $T_{s,in} = 20 - 17.9(0.100) = 18.2$ °C and the brick–insulation interface is at $T_{int} = 18.2 - 17.9(0.286) = 13.1$ °C. The insulation, which holds three-quarters of the total resistance, takes most of the temperature drop (about 22 °C), as expected.

The tempting answer 19.5 W/m² omits the two convection resistances; they are small here but not negligible (8% of the total). 24.0 W/m² uses the insulation alone and 105 W/m² the brick alone.

$q'' = 17.9\,\text{W/m}^2$ (from inside to outside)

Validity: the lumped method applies when the Biot number based on the characteristic length $L_c = V/A_s$ is below 0.1. For a sphere $L_c = r_o/3$:

$$L_c = \frac{0.005}{3} = 1.667 \times 10^{-3}\,\text{m},\qquad Bi = \frac{hL_c}{k} = \frac{200(1.667 \times 10^{-3})}{400} = 8.3 \times 10^{-4}$$

$Bi \ll 0.1$, so internal conduction is so fast compared with surface convection that the sphere stays essentially uniform in temperature, and the lumped energy balance $\rho c V\,dT/dt = -hA_s(T - T_\infty)$ applies.

Its solution is an exponential decay with time constant $\tau = \rho c V/(hA_s) = \rho c L_c/h$:

$$\tau = \frac{8933(385)(1.667 \times 10^{-3})}{200} = 28.66\,\text{s}$$ $$\frac{T - T_\infty}{T_i - T_\infty} = e^{-t/\tau} \;\Rightarrow\; \frac{50 - 30}{150 - 30} = \frac{1}{6} = e^{-t/28.66}$$ $$t = 28.66\,\ln 6 = 28.66(1.792) = 51.4\,\text{s}$$

The sphere loses five-sixths of its initial temperature excess in about 1.8 time constants, which is the expected behaviour for exponential cooling (one time constant removes 63% of the excess, two remove 86%).

$Bi \approx 8 \times 10^{-4}$, so lumped analysis is valid; $t \approx 51\,\text{s}$

(a) The heat given up by the oil follows from an energy balance on the hot stream:

$$\dot Q = \dot m_h c_{p,h}(T_{h,in} - T_{h,out}) = 0.50(2100)(120 - 60) = 63\,000\,\text{W} = 63.0\,\text{kW}$$

(b) With no losses the water absorbs all of it:

$$T_{c,out} = T_{c,in} + \frac{\dot Q}{\dot m_c c_{p,c}} = 20 + \frac{63\,000}{0.40(4180)} = 20 + 37.7 = 57.7\,^\circ\text{C}$$

The water warms less than the oil cools because its capacity rate ($\dot m c_p = 1672$ W/K) exceeds the oil's (1050 W/K).

(c) In counterflow the hot inlet faces the cold outlet and the hot outlet faces the cold inlet, so the end temperature differences are

$$\Delta T_1 = T_{h,in} - T_{c,out} = 120 - 57.7 = 62.3\,^\circ\text{C},\qquad \Delta T_2 = T_{h,out} - T_{c,in} = 60 - 20 = 40.0\,^\circ\text{C}$$ $$\Delta T_{lm} = \frac{\Delta T_1 - \Delta T_2}{\ln(\Delta T_1/\Delta T_2)} = \frac{62.3 - 40.0}{\ln(62.3/40.0)} = \frac{22.3}{0.443} = 50.3\,^\circ\text{C}$$

No correction factor is needed for a true counterflow double-pipe exchanger ($F = 1$).

(d) From the rate equation $\dot Q = UA\,\Delta T_{lm}$:

$$A = \frac{\dot Q}{U\,\Delta T_{lm}} = \frac{63\,000}{300(50.3)} = 4.17\,\text{m}^2$$

The log mean (50.3 °C) lies between the two end differences and slightly below their arithmetic mean (51.2 °C), as it always does. A parallel-flow arrangement with the same four temperatures would give a log mean of only 26.0 °C and need almost twice the area.

(a) 63.0 kW; (b) $T_{c,out} = 57.7$ °C; (c) $\Delta T_{lm} = 50.3$ °C; (d) $A = 4.17\,\text{m}^2$