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ENGR5003 / Week 7
ENGR203 2017 Q B2Past paperOld spec ENGR203First25 marks30 min

ENGR203 Power and Heat 2017 Section B Q B2 (same question as ENGR263 2017 B2)[VALID] official answers

Answer ALL parts (a) - (f).

A Constantan fuse wire has a diameter of 0.6 mm. Assume Constantan has the following properties, resistivity =49×10−8 Ω m= 49 \times 10^{-8}\ \Omega\,\mathrm{m}, emissivity =0.12= 0.12, thermal conductivity =20 W m−1 K−1= 20\ \mathrm{W\,m^{-1}\,K^{-1}}, specific heat =390 J kg−1 K−1= 390\ \mathrm{J\,kg^{-1}\,K^{-1}}, density 8900 kg m−38900\ \mathrm{kg\,m^{-3}}.

For parts (a) and (b) assume that the wire cools by radiation losses alone to an effective black body surrounding of temperature 30∘C30^\circ\mathrm{C}.

Formulas you may need
  • Radiation to large surroundings Q˙net=εAσ(Tobj4−Tsur4)\dot Q_{net} = \varepsilon A\sigma(T_{obj}^4 - T_{sur}^4), σ=5.6704×10−8 W m−2 K−4\sigma = 5.6704 \times 10^{-8}\ \mathrm{W\,m^{-2}\,K^{-4}} (on the formula sheet)
  • General conduction equation ρcp∂T∂t=k∇2T+qv\rho c_p\dfrac{\partial T}{\partial t} = k\nabla^2 T + q_v (on the formula sheet)
  • Heat stored ΔQ=mcpΔT\Delta Q = mc_p\Delta T (on the formula sheet)
  • Wire resistance per unit length R′=ρe/AcR' = \rho_e/A_c; Joule heating Q˙=I2R\dot Q = I^2R; volumetric generation qv=I2ρe/Ac2q_v = I^2\rho_e/A_c^2 (learn this)
  • 1-D conduction with uniform generation, both ends at T0T_0: T(x)=T0+qv2kx(L−x)T(x) = T_0 + \dfrac{q_v}{2k}x(L - x), Tmax−T0=qvL28kT_{max} - T_0 = \dfrac{q_vL^2}{8k} (learn this)
  1. (a)
    Determine the wire temperature when it carries an r.m.s. current of 3 amps.
    [5]2:2
  2. (b)
    Determine the current required to heat the wire to its melting temperature 1210∘C1210^\circ\mathrm{C}.
    [4]2:2
  3. (c)
    Neglecting all heat losses and assuming heat capacity, electrical resistance and emissivity do not change with temperature then starting at 30∘C30^\circ\mathrm{C} and applying the current computed in (b) determine the time until the fuse melts.
    [4]2:2
  4. (d)
    If the wire could not lose heat by radiation or convection but only by conduction to its ends derive a differential equation that gives temperature along the wire.
    [5]2:1
  5. (e)
    Using this differential equation determine the shortest length of the wire which would just melt at its centre when it carries a current of 3 amps and the ends are held at 30∘C30^\circ\mathrm{C}.
    [4]2:1
  6. (f)
    Write down a partial differential equation for the temperature of the wire as a function of time when it carries current II and cools by both thermal radiation and conduction to its ends.
    [3]