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ENGR5003 / Week 10
ENGR203 2017 Q B3(b)-(d)Past paperOld spec ENGR2032:120 marks24 min

ENGR203 Power and Heat 2017 Section B Q B3 (same question as ENGR263 2017 B3)[PARTIAL] official answers Left out: (a) (5 marks), describe how a thermocouple is constructed and used, is instrumentation and not on the current ENGR5003 syllabus.

Answer ALL parts (a) - (d). [Part (a), 5 marks, on thermocouple construction, is not on the current syllabus and is omitted here.]

Formulas you may need
  • 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 balances Q˙=m˙hcphΔTh=m˙ccpcΔTc\dot Q = \dot m_h c_{ph}\Delta T_h = \dot m_c c_{pc}\Delta T_c (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})}; valid for parallel and counter flow, and for cross flow if one stream's temperature change is small (on the formula sheet)
  • Effectiveness ε=Q˙Cmin(Thi−Tci)\varepsilon = \dfrac{\dot Q}{C_{min}(T_{hi} - T_{ci})}, NTU=UA/Cmin\mathrm{NTU} = UA/C_{min} (on the formula sheet)
  1. (b)
    For a counter flow, liquid to liquid heat exchanger describe in words the key factors which limit the rate of energy transfer from one liquid stream to the other.
    [4]Third
  2. (c)
    When would you expect the Logarithmic Mean Temperature Difference (LMTD) method to give a reasonably accurate estimate for heat transfer in a cross flow heat exchanger.
    [2]Third
  3. (d)
    Water is passed through an aluminium tube of length 8 m, bore 20 mm and wall thickness 1 mm. The tube is cooled by air at 15∘C15^\circ\mathrm{C} blown perpendicularly to its outside surface. 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=85∘CT_{w1} = 85^\circ\mathrm{C}
    Mass flow rate of waterm˙w=0.3 kg s−1\dot m_w = 0.3\ \mathrm{kg\,s^{-1}}
    Mass flow rate of airm˙a=0.8 kg s−1\dot m_a = 0.8\ \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=2400 W m−2 K−1h_w = 2400\ \mathrm{W\,m^{-2}\,K^{-1}}
    Air side heat transfer coefficientha=180 W m−2 K−1h_a = 180\ \mathrm{W\,m^{-2}\,K^{-1}}
    [14]