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ENGR5003 / Week 9
ENGR203 2018 Q B2(b)-(c)Past paperOld spec ENGR2032:119 marks23 min

ENGR203 Power and Heat 2018 Section B Q B2[PARTIAL] Left out: Part (a) (6 marks), describe an experiment to measure the specific heat capacity of a liquid and its accuracy limits, is laboratory measurement technique that is not in the current ENGR5003 syllabus, so it is left out.

Answer ALL parts (a) - (c). [Part (a), 6 marks, on measuring the specific heat capacity of a liquid, is not on the current syllabus and is omitted here.]

Formulas you may need
  • Gr=βgρ2ΔTL3μ2\mathrm{Gr} = \dfrac{\beta g \rho^2 \Delta T L^3}{\mu^2}, Pr=μcpk\mathrm{Pr} = \dfrac{\mu c_p}{k}, Nu=hLk\mathrm{Nu} = \dfrac{hL}{k} (on the formula sheet)
  • Horizontal cylinder in a liquid: Nu=0.53(Gr Pr)0.25\mathrm{Nu} = 0.53(\mathrm{Gr}\,\mathrm{Pr})^{0.25} with L=DL = D (given in the question)
  • Q˙=hA ΔT\dot Q = hA\,\Delta T (on the formula sheet)
  • Tube-wall resistance between two fluids: Rth=1L[1h1πD1+ln⁡(D2/D1)2πk+1h2πD2]R_{th} = \dfrac{1}{L}\left[\dfrac{1}{h_1\pi D_1} + \dfrac{\ln(D_2/D_1)}{2\pi k} + \dfrac{1}{h_2\pi D_2}\right], UA=1/RthUA = 1/R_{th} (on the formula sheet)
  • Stream energy balance Q˙=m˙cpΔT\dot Q = \dot m c_p \Delta T (on the formula sheet)
  • LMTD: Q˙=UA ΔTmean\dot Q = UA\,\Delta T_{mean}, ΔTmean=ΔTmin−ΔTmaxln⁡(ΔTmin/ΔTmax)\Delta T_{mean} = \dfrac{\Delta T_{min} - \Delta T_{max}}{\ln(\Delta T_{min}/\Delta T_{max})}; usable for cross flow when one stream's temperature change is small (on the formula sheet)
  1. (b)
    A 1.8 kW immersion heater for heating water in a large tank is to be made from a straight horizontal sheathed element of diameter 11 mm. (A sheathed element is an electrical resistance wire embedded in magnesium oxide powder and enclosed in a stainless steel tube, there will be one in your kettle bent to a compact shape.) The tank has a thermostat that controls the water temperature to 55∘C55^\circ\mathrm{C}. If the surface temperature of the sheathed element is not to exceed 99∘C99^\circ\mathrm{C}, calculate the length of tube required assuming that convective losses from the element can be determined by the non-dimensional equation Nu=0.53(Gr×Pr)0.25\mathrm{Nu} = 0.53(\mathrm{Gr} \times \mathrm{Pr})^{0.25}. (For water at 77∘C77^\circ\mathrm{C} take μ=0.370×10−3 kg m−1 s−1\mu = 0.370 \times 10^{-3}\ \mathrm{kg\,m^{-1}\,s^{-1}}, cp=4.196 kJ kg−1 K−1c_p = 4.196\ \mathrm{kJ\,kg^{-1}\,K^{-1}}, ρ=974 kg m−3\rho = 974\ \mathrm{kg\,m^{-3}}, k=0.666 W m−1 K−1k = 0.666\ \mathrm{W\,m^{-1}\,K^{-1}}, β=0.615×10−3 K−1\beta = 0.615 \times 10^{-3}\ \mathrm{K^{-1}}, g=9.81 m s−2g = 9.81\ \mathrm{m\,s^{-2}}.)
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  2. (c)
    Water is passed through an aluminium tube of length 12 m, bore 22 mm and wall thickness 1 mm. The tube is cooled by air at 15∘C15^\circ\mathrm{C} blown perpendicularly to its outside surface. (Assume that any folds in the tube do not affect air side heat transfer to adjacent segments.) Using the LMTD method and data below estimate the water outlet temperature.
    QuantityValue
    Water inlet temperatureTw1=90∘CT_{w1} = 90^\circ\mathrm{C}
    Mass flow rate of waterm˙w=0.05 kg s−1\dot m_w = 0.05\ \mathrm{kg\,s^{-1}}
    Mass flow rate of airm˙a=0.7 kg s−1\dot m_a = 0.7\ \mathrm{kg\,s^{-1}}
    Heat capacity of watercpw=4190 J kg−1 K−1c_{pw} = 4190\ \mathrm{J\,kg^{-1}\,K^{-1}}
    Heat capacity of aircpa=1006 J kg−1 K−1c_{pa} = 1006\ \mathrm{J\,kg^{-1}\,K^{-1}}
    Water side heat transfer coefficienthw=2200 W m−2 K−1h_w = 2200\ \mathrm{W\,m^{-2}\,K^{-1}}
    Air side heat transfer coefficientha=140 W m−2 K−1h_a = 140\ \mathrm{W\,m^{-2}\,K^{-1}}
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