An uninsulated hot water pipe is suspended horizontally in a room. (a) (b) Using the data...

   

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An uninsulated hot water pipe is suspended horizontally in a room. (a) (b) Using the data provided and properties of air from the table provided, determine the convective heat transfer rate from the pipe to the room per unit length of pipe. If the surface temperature of the pipe was increased to 36°C, would you expect the heat transfer rate by convection to double? Explain your reasoning. Data for Question A1. Pipe surface temperature Pipe external diameter Room air temperature (c) Explain why expressions for free convection utilise the Grashoff Number whilst those for forced convection utilise the Reynolds Number. free convection over horizontal tube forced convection over horizontal tube where 80 °C 80 mm 24 °C 104 Gr Pr <107 107 Gr Pr <1012 Nu = Re<40 4,000<Re<40,000 40,000<Re<400,000 hD k 40<Re<4,000 Ra= Gr Pr Nup=0.48 Raº Gr= Nup = 0.125 Raº 0.25 0.385 Nu, =0.193 Re" 0.333 Bgp²D³ At 4² Nup=0.911Re Pr0333 Nup = 0.683 Re0.466 Pr0333 0.618. [70%] Nup = 0.027 Re Prº Re= [10%] [20%] Pr0.333 0.805 0.333 puD H An uninsulated hot water pipe is suspended horizontally in a room. (a) (b) Using the data provided and properties of air from the table provided, determine the convective heat transfer rate from the pipe to the room per unit length of pipe. If the surface temperature of the pipe was increased to 36°C, would you expect the heat transfer rate by convection to double? Explain your reasoning. Data for Question A1. Pipe surface temperature Pipe external diameter Room air temperature (c) Explain why expressions for free convection utilise the Grashoff Number whilst those for forced convection utilise the Reynolds Number. free convection over horizontal tube forced convection over horizontal tube where 80 °C 80 mm 24 °C 104 Gr Pr <107 107 Gr Pr <1012 Nu = Re<40 4,000<Re<40,000 40,000<Re<400,000 hD k 40<Re<4,000 Ra= Gr Pr Nup=0.48 Raº Gr= Nup = 0.125 Raº 0.25 0.385 Nu, =0.193 Re" 0.333 Bgp²D³ At 4² Nup=0.911Re Pr0333 Nup = 0.683 Re0.466 Pr0333 0.618. [70%] Nup = 0.027 Re Prº Re= [10%] [20%] Pr0.333 0.805 0.333 puD H