22-Mec-A1 Applied Thermodynamics and Heat Transfer · May 2018
Question 5 of 8: Current-carrying wire with a ceramic sheath (radial conduction + generation)
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format: National Examination 16-Mec-A1, 3 hours, open book. Eight questions of equal value: Part A — Thermodynamics (Q1–Q4) and Part B — Heat Transfer (Q5–Q8). A complete paper is any five questions (three from one part and two from the other).
Reference texts: Çengel & Boles, Thermodynamics: An Engineering Approach (9th ed., McGraw-Hill) — first-law cycle analysis, wet-region steam properties, the ideal regenerative Rankine cycle with a closed feedwater heater, the air-standard Diesel cycle, and isentropic-efficiency compressor analysis; Çengel & Ghajar, Heat and Mass Transfer (6th ed.) and Incropera, DeWitt, Bergman & Lavine, Fundamentals of Heat and Mass Transfer (8th ed., Wiley) — steady conduction with internal generation in a composite cylinder, internal-flow convection at constant wall temperature, combined convection–radiation surface balances, and the LMTD method for counterflow heat exchangers. Water/steam properties are IAPWS-consistent; air is treated as an ideal gas with constant specific heats.
Question 5 — Current-carrying wire with a ceramic sheath (radial conduction + generation) (Part B, equal value)
Given. Long wire with uniform volumetric generation inside a ceramic sheath, cooled by convection.
Quantity
Value
Generation, $q'''$
50 W/cm³ = 50×10⁶ W/m³
Wire radius, $r_1$
0.2 cm = 0.002 m
Wire conductivity, $k_\text{wire}$
15 W/m·°C
Ceramic thickness / outer radius $r_2$
0.5 cm → $r_2=0.007$ m
Ceramic conductivity, $k_\text{ceramic}$
1.2 W/m·°C
Air / coefficient
$T_\infty=20$ °C, $h=50$ W/m²·°C
Find. the wire–ceramic interface temperature and the wire centreline temperature.
Figure 5 — All heat generated in the wire flows radially outward through the ceramic and then convects to the air. The interface and centreline temperatures are built up by adding the convective and conductive temperature drops to $T_\infty$, plus the parabolic rise inside the generating wire.
Approach. Compute the total heat generated per metre, add the convective drop to reach the ceramic surface, add the conductive drop across the ceramic to reach the interface, then add the internal parabolic rise of a solid generating cylinder to reach the centreline.
Heat generated per unit length. Only the wire generates:
$$q'_L=q'''\,\pi r_1^{2}=50\times10^{6}\,\pi(0.002)^2=628.3\ \text{W/m}$$
Ceramic surface temperature (convection). All of $q'_L$ leaves the outer surface by convection:
$$T_s=T_\infty+\frac{q'_L}{h\,(2\pi r_2)}=20+\frac{628.3}{50(2\pi\cdot0.007)}=20+285.7=305.7\ ^\circ\text{C}$$
Interface temperature (conduction across the ceramic). Cylindrical conduction with no generation in the ceramic:
$$T_\text{interface}=T_s+\frac{q'_L\,\ln(r_2/r_1)}{2\pi k_\text{ceramic}}=305.7+\frac{628.3\,\ln(0.007/0.002)}{2\pi(1.2)}=305.7+104.4$$
$T_\text{interface}\approx 410\ ^\circ\text{C}$
Centreline temperature (parabolic rise in the generating wire). For a solid cylinder with uniform generation the centre exceeds the surface (here the interface) by $q'''r_1^2/(4k_\text{wire})$:
$$T_\text{centerline}=T_\text{interface}+\frac{q'''\,r_1^{2}}{4k_\text{wire}}=410.1+\frac{50\times10^{6}(0.002)^2}{4(15)}=410.1+3.3$$
$T_\text{centerline}\approx 413\ ^\circ\text{C}$
