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Answer:Consider the fluid flow in laminar Boundary layer.Flow is steady and two dimensional.Taking a differential volume of length dx, height dy and unit depth in the z-direction.Applying Consenvation of energyEnergy in - Enengy out = Change in Energyy of the SystemEin - Eout = DsystemAt steady flow{:[Delta" Esystem "=0],[" Ein "=" Eout "]:}No work is involued here. Energy transfen will takes placl by heat and mass.The rate of energy transfer to the control volume by mass in x-directionLet m^(m)= mass flow rate of fluid{:[=" mass flow rate of fluid "],[=" P.uty ")" (velocity)(areal "],[=(dy.1)]:}e= Energy of the fluid per unit mass CSScanned with CamScanner =h=C_(p)TWhere,{:[h=" Specific enthalpy "],[c_(p)=" Specific neat capacity at constant "],[],[T=" Pressure "],[" Temperature "]:}Now,{:[(" Ein "-" Eout ")_("mass ",x){:=me_(x)-[(m^(˙))e_(x)+(del(me)/(del x))_(x)*dx]],[=-(del((me^(˙)))_(x))/(del x)*dx],[=-rho(del)/(del x)[rho u(dy*1)C_(p)T]dx],[=-rhoC_(p)(u(del T)/(del x)+T(del u)/(del x))dxdy]:}Rate of energy transfer to the control volume by mass in y-direction(E_("un ")-" Eout ")_("mass "y)=(me^(˙))_(y)-[((me^(˙)))_(y)+(del((m^(˙))e)_(y))/(del y)dy]Here,m^(˙)=p theta(dx*1){:[=>=-(del((m^(˙))e)_(y))/(del y)*dy],[=-(del(Pv(dx*1)C_(P))/(del y)dy],[=-PC_(P)(v(del T)/(del y)+T(del v)/(del y))dxdy]:}Adding eqn (1) and eqn (II)En ... See the full answer