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Ans: Datagiven: quad D=1cm{:[L=10cm],[T_(omega)=50^(@)C],[T_(0i)=10^(@)C],[V=1.3m//s]:}Properties of oilReynolds number of flow:{:[R_(eD)=(rho_(0)VD)/(40)],[=(890 xx1.3 xx(1xx10^(-2)))/(0.1)],[=115.7quad" (which is less than 2300 thus "],[quad" flow inside pipe is Laminar.) "]:}Assumptions: rarr Effect of entry length is neglected and flow is throught to be fully developed throughout the length.for circuer tabe steady state conditions for circular tube, laminer flow empirical relationship for Nu number is given by: (constent wall temperature){:[Nu=3.66],[=>(hL_(c))/(k_(0))=3.66quad(L_(c)=D)],[=>h=(3.66 xx0.15)/((1xx10^(-2)))=>h=54.9w//m^(2)K]:}Let temperature of oil at a distance x from entry be T_(x). (as shown)for the dlement shown in above schematic, by writting energy balance at steady state: E^(˙)_("in ")=E^(˙)_("out "){:[=>quadQ^(˙)conv+m^(˙)c_(p_(0))T_(x)=m^(˙)c_(p_(0))(T_(x)+dT_(x))],[=>quad h(dA)xx(T_(omega)-T_(x))=m^(˙)c_(p_(0))(T_(x)+dT_(x))-m^(˙)c_(p_(0))T_(x)],[=>quad hx(pi Ddx)(T_(w)-T_(x))=m^(˙)c_(p_(0))dT_(x)],[=>quad(pi hD ... See the full answer