ε 2 Using e 6m 2F l μl K A w ρ l � Using equation 67 4 Nmw m

# Ε 2 using e 6m 2f l μl k a w ρ l ? using equation

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ε ) 2 ------------------ Using equation 6.8 =6.46e-6m 2 F l = μ l K A w ρ l λ ---------------------------------- Using equation 6.7 =4 N/m/w-m Vapor frictional coefficient F v For wrapped screen for circular vapor flow passage the coefficient of drag; f v R ev =16 Vapor flow area, A v =.25 π D v 2 =1.767e-4m 2 F v = ( f v R ev ) μ v 2 r h,v A v λ Where r h ,v 2 = ( D v 2 ) 2 = 81 e 6 m 2 , -----------------------------Using equation 6.10 =2.634e-6 N/m/w-m Capillary heat transfer factor, ( QL ) c ,max ( QL ) c ,max = p c p g F l + F v -------------------- Using equation 6.11 = 1707.89w/m Capillary heat transfer limit, ( Qc ) c ,max Q c ,max = ( QL ) c ,max [ .5 L e + L a + .5 l c ] ----- Using equation 6.14 =1148.2w 7.3 .3 Sonic Limitation , Qs , Max Q S, MAX = A V ρ O λ [ γ V R V T V 2 ( γ V + 1 ) ] 0.5 ------------------ Using equation 6.12 Vapor specific heat ratio γ V =1.33 Vapor constant =467 ( J Kg . K ) Q S, MAX =11.36 KW 7.3. 4 .Entrainment Limit Q e, Max Q E, MAX = A V λ [ σρ V 2 r h, S ] 0 .5 ------------ Using equation 6.13 r h,s =wick surface hydraulic radius = 1 2 N d 2 N=1.968*10 3 =1.27*10 -4 Q E, MAX =2.426 KW 7.3.5 Boiling Limitation , Q b, Max Q B,MAX = 2 πL e K e T V λρ V ln ( r i r v ) [ 2 σ r n ρ C ] ----------------- Using equation 6.15 Where K e = K l [ K l + K w ( 1 ε ) ( K l K w ) ( K l + K w ) + ( 1 ε ) ( K l K w ) ] K l =0.649 ε = 0.59 K w =385 W m 2 . K K e =1.55 r n =2.54*10 -7 m Substituting all these values; Q B, MAX =9.274 Kw From the above calculation, all the limits are above the amount of the energy needed at the evaporator stage. This means the energy at the evaporator stage can be transferred without any failure in the heat pipe. 7.4 Design of Heat Pipe Container The following are taken into consideration for the design of the heat pipe: Material of heat used for the heat pipe is copper Working fluid temperature =333.15K Ultimate tensile strength of copper at 333.15K =186.135*10 6 N m 2 = 186.165 MP The ASME code specifies that the maximum allowable stress at any temperature (UTS) at that temperature. I.e. UTS f max 4 For rounded tubes in which the wall thickness is loss than 10% of the diameter, the maximum pressure stress is closely approximated by simple expression f max = pd o 2 t ---------------- Using equation 6.5 Where f max is the maximum hoop stress in the wall P is pressure differential across the wall d o is tube outside thickness Substituting the following values to the above equation d o =20 mm P amb =101.235 KPa Psat @60 o c =19.94 KPa t=1mm P=101.325-19.94=81.385 KPa t d o = 1 20 = 0.05 = 5% < 10% Hence the above formula can be used f max = 81.31 20 2 = 813..1 10 3 N m 2 To check UTS f max 4 186.165*10 6 813 10 3 = 229 229 ¿ 4 The above result indicates that the assumed heat pipe container dimensions are acceptable. 8. FIN (COLLECTOR) DESIGN The following parameters are taken into consideration for design of this part: -Available thickness of the fin material =0.8 mm -Fin material selected for this purpose is Aluminum because of the following two reasons; Low cost Easily corrugated to hold heat pipes -Thermal conductivity of aluminum K fe =221 w mm 2 k -Collector clip diameter D c =21.6 mm -Energy at the evaporator stage = 400W=Q e -Emmitance of glass ε g = .88 -Ambient temperature Ta=23 o c -Emitance of plate =0.27 -Working fluid temperature =333.15 K= Tv -Average wind speed, V w =5 m s -Collector tilt angle =15 o =B -Incident solar radiation I = 900 w m 2 -Effective transmittance absorbance ( ατ ) ε = 0.9 ( ατ ) -Number of glass covers is taken to be 2 to prevent high convection losses Assumptions taken for this design are : -Wind of fin =0.45m  #### You've reached the end of your free preview.

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