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(1)Problem 2.1-1 (2-1 in text): Buried Tubes Figure P2.1-1 illustrates two tubes that are buried in the ground behind your house that transfer water to and from a wood burner. The left hand tube carries hot water from the burner back to your house at Tw,h = 135°F while the right hand tube carries cold water from your house back to the burner at Tw,c = 70°F. Both tubes have outer diameter Do = 0.75 inch and thickness th = 0.065 inch. The conductivity of the tubing material is kt = 0.22 W/m-K. The heat transfer coefficient between the water and the tube internal surface (in both tubes) is hw = 250 W/m2-K. The center to center distance between the tubes is w = 1.25 inch and the length of the tubes is L = 20 ft (into the page). The tubes are buried in soil that has conductivity ks = 0.30 W/m-K. ks = 0.30 W/m-K th = 0.065 inch. kt = 0.22 W/m-K. Tw,c = 70°F 2 hw = 250 W/m -K. Tw,h = 135°F 2 hw = 250 W/m -K w = 1.25 inch Do = 0.75 inch. Figure P2.1-1: Tubes buried in soil.. a.) Estimate the heat transfer from the hot water to the cold water due to their proximity to one another. The inputs are entered in EES: $UnitSystem SI MASS RAD PA K J $TABSTOPS 0.2 0.4 0.6 0.8 3.5 in D_o=0.75 [inch]*convert(inch,m) "outer diameter of tube" th=0.065 [inch]*convert(inch,m) "thickness of tube" T_hw=converttemp(F,K,135) "hot water temperature" T_cw=converttemp(F,K,70) "cold water temperature" L=20 [ft]*convert(ft,m) "length of tubes" w=1.25 [inch]*convert(inch,m) "center to center distance" k_t=0.22 [W/m-K] "conductivity of teflon" k_s=0.30 [W/m-K] "conductivity of sand" h_w=250 [W/m^2-K] "heat transfer coefficient between the water and tube inner surfaces". The heat transfer is resisted by convection within the tube, conduction through the tube, and conduction in the soil. The convection resistance is calculated according to: Rconv =. 1 hw L π ( Do − 2 th ). The conduction resistance associated with the tube is calculated according to:. (1).

(2) Rcond. ⎛ Do ⎞ ln ⎜ ⎟ Do − 2 th ⎠ ⎝ = 2 kt L π. R_conv=1/(h_w*L*pi*(D_o-2*th)) R_cond=ln(D_o/(D_o-2*th))/(2*pi*k_t*L). (2) "convection resistance" "conduction resistance". The shape factor for parallel, buried tubes (SF) is obtained using EES’ internal library and used to compute the resistance due to conduction through the soil:. Rtubetotube =. 1 SF ks. SF=SF_4(D_o,D_o,w,L) R_tubetotube=1/(SF*k_s). (3) "shape factor" "tube to tube resistance". The heat transfer rate is therefore: q =. Tw,h − Tw,c 2 Rconv + 2 Rcond + Rtubetotube. q_dot=(T_hw-T_cw)/(2*R_conv+2*R_cond+R_tubetotube). (4). "heat transfer rate". which leads to q = 137.3 W. b.) To do part (a) you should have needed to determine a shape factor; calculate an approximate value of the shape factor and compare it to the accepted value. The shape factor can be thought of as the ratio of the effective area for conduction, Aeff, to the effective length required for conduction, Leff. The effective area for conduction can be estimated according to:. Aeff ≈ w L. (5). while the length for conduction is approximately:. Leff ≈ w − Do. (6). and the approximate value of the shape factor should be:. SFapp ≈ A_eff=L*w. Aeff. (7). Leff "approximate area".

(3) L_eff=(w-D_o) SF_app=A_eff/L_eff. "approximate length" "approximate shape factor". The exact value of the shape factor returned by the function is SF = 17.4 m while the approximate value is SFapp = 15.2 m; these are sufficiently close for a sanity check. c.) Plot the rate of heat transfer from the hot water to the cold water as a function of the center to center distance between the tubes. The heat transfer rate is shown in Figure 2 as a function of the center to center distance between the tubes.. Figure 2: Heat transfer rate as a function of the center to center distance between the tubes.

(4) Problem 2.1-2 Currently, the low-pressure steam exhausted from a steam turbine at the power plant is condensed by heat transfer to cooling water. An alternative that has been proposed is to transport the steam via an underground pipe to a large building complex and use the steam for space heating. You have been asked to evaluate the feasibility of this proposal. The building complex is located 0.2 miles from the power plant. The pipe is made of uninsulated PVC (thermal conductivity of 0.19 W/m-K) with an inner diameter of 8.33 in and wall thickness 0.148 in. The pipe will be buried underground at a depth of 4 ft in soil that has an estimated thermal conductivity of 0.5 W/m-K. The steam leaves the power plant at 6.5 lbm/min, 8 psia with a 95% quality. The outdoor temperature is 5°F. Condensate is returned to the power plant in a separate pipe as, approximately, saturated liquid at 8 psia. You may neglect the resistance due to convection to the air and the steam. a.) Neglecting the inevitable pressure loss, estimate the state of the steam that is provided to the building complex. The inputs are entered in EES: $UnitSytem SI K J Pa $TabStops 0.2 0.4 3.5 in "known information" L=0.2 [mile]*convert(mile,m) D_i=8.33 [in]*convert(in,m) thk=0.148 [in]*convert(in,m) k_soil=0.5 [W/m-K] k_PVC=0.19 [W/m-K] T_air=convertTemp(F,K,5 [F]) P_steam=8 [psia]*convert(psia,Pa) T_steam=T_sat(Steam,P=P_steam) W=4 [ft]*convert(ft,m) m_dot=6.5 [lb_m/min]*convert(lb_m/min,kg/s). "pipe length" "pipe inner diameter" "pipe wall thickness" "soil thermal conductivity" "PVC thermal conductivity" "air temperature in K" "steam pressure" "steam temperature in K" "distance of pipe below ground level" "mass flow rate of the steam". The outer diameter of the pipe is calculated: Do = Di + 2 th. (1). where Di is the inner diameter and th is the pipe thickness. The resistance to conduction through the pipe is:. R pipe. ⎛D ⎞ ln ⎜ o ⎟ ⎝ Di ⎠ = 2 π k PVC L. (2). where kPVC is the conductivity of the PVC pipe and L is the length of the pipe. The resistance between the pipe surface and the surface of the soil is obtained using the shape factor (S) obtained using the function SF_2 in EES. Rsoil =. 1 S k soil. (3).

(5) The heat transfer from the pipe to the soil surface is: q =. (Tsteam − Tair ) R pipe + Rsoil. (4). where Tsteam is the temperature of the steam and Tair is the temperature of the air. D_o=D_i+2*thk R_pipe=ln(D_o/D_i)/(2*pi*k_pvc*L) R_soil=1/(k_soil*S) S=SF_2(D_o,W,L) q_dot_loss=(T_steam-T_air)/(R_pipe+R_soil). "pipe outer diameter" "resistance to conduction through pipe wall" "resistance of soil" "shape factor for buried pipe" "heat transfer rate through pipe wall". which leads to q = 107.3 kW. An energy balance on the steam leads to:. q = m ( is ,in − is ,out ). (5). where is,in and is,out are the inlet and outlet enthalpies of the steam, respectively. The outlet enthalpy is used to determine the outlet quality, xs,out: i_s_in=enthalpy(Steam,P=p_steam,x=0.95) q_dot_loss=m_dot*(i_s_in-i_s_out) x_out=quality(Steam,P=p_steam,h=i_s_out). "enthalpy of inlet steam" "energy balance on steam" "quality of leaving steam". which leads to xs,out = 0.673. b.) Are the thermal losses experienced in the underground pipe transport process significant in your opinion? Do you recommend insulating this pipe? Yes, the losses are significant. The quality of the steam is reduced from 95% to 67.3% and therefore you have lost approximately 30% of the heating content of the steam. The pipe should be insulated. c.) Provide a sanity check on the shape factor that you used to solve this problem. The shape factor represents, approximately, the ratio of the cross-sectional area for conduction to the length for conduction. A crude estimate of the shape factor for this problem is:. S app =. ⎛ ⎝. π ⎜ Do + W. W⎞ ⎟L 2⎠. (6). which leads to Sapp = 687.3 m; this is approximately in line with the value provided by the EES function, S = 651.8 m..

(6) Problem 2.1-3 (2-2 in text) A solar electric generation system (SEGS) employs molten salt as both the energy transport and storage fluid. The molten salt is heated to Tsalt = 500°C and stored in a buried semi-spherical tank. The top (flat) surface of the tank is at ground level. The diameter of the tank before insulation is applied Dt = 14 m. The outside surfaces of the tank are insulated with tins = 0.30 m thick fiberglass having a thermal conductivity of kins = 0.035 W/m-K. Sand having a thermal conductivity of ksand = 0.27 W/m-K surrounds the tank, except on its top surface. Estimate the rate of heat loss from this storage unit to the Tair = 25°C surroundings. You may neglect the resistance due to convection. The inputs are entered in EES: $UnitSytem SI K J Pa $TabStops 0.2 0.4 3.5 in "known information" t_ins=0.3 [m] D_t=14 [m] k_sand=0.27 [W/m-K] k_ins=0.035 [W/m-K] T_air=convertTemp(C,K,25 [C]) T_salt=converttemp(C,K,500 [C]). "insulation thickness" "tank diameter" "sand thermal conductivity" "insulation thermal conductivity" "air temperature in K" "salt temperature in K". The resistance between the bottom (buried, hemispherical) surface of the tank and the air is due to the conduction through the spherical shell of insulation. Rins ,bottom. ⎛ 1 ⎞ 1 + ⎜ ⎟ Dt / 2 ( Dt / 2 + tins ) ⎠ ⎝ = 2 π kins. (1). and the sand:. Rsand =. 1 k sand S. (2). where S is the shape factor, obtained using the EES function SF_22. The resistance from the top, unburied surface of the tank is only due to conduction through the insulation: Rins ,top =. tins D2 kins π t 4. The heat transfer from the bottom and top surfaces are calculated according to:. (3).

(7) qbottom =. (Tsalt − Tair ) Rsoil + Rins ,bottom. (4). and. qtop =. (Tsalt − Tair ) Rins ,top. (5). The total heat loss is:. qtotal = qtop + qbottom R_ins_bottom=(1/(D_t/2)-1/(D_t/2+t_ins))/(2*pi*k_ins) "resistance to conduction through insulation on bottom" S=SF_22(D_t+2*t_ins) "shape factor" R_sand=1/(k_sand*S) "resistance of sand" R_ins_top=t_ins/(k_ins*pi*D_t^2/4) "resistance to conduction through insulation on top" q_dot_bottom=(T_salt-T_air)/(R_sand+R_ins_bottom) "heat loss from bottom surface of tank" q_dot_top=(T_salt-T_air)/(R_ins_top) "heat loss from top surface of tank" q_dot_total=q_dot_bottom+q_dot_top "total heat loss". which leads to qtotal = 12.95 kW.. (6).

(8) Problem 2.1-4 A square extrusion is L = 1 m long and has outer dimension W = 3 cm. There is a D = 1 cm diameter hole aligned with the center of the extrusion. The material has conductivity k = 0.5 W/m-K. The external surface of the extrusion is exposed to air at Ta = 20ºC with heat transfer coefficient ha = 50 W/m2-K. The inner surface of the extrusion is exposed to water at Tw = 80 ºC with heat transfer coefficient hw = 150 W/m2-K. a.) Determine the rate of heat transfer between the water and the air. The inputs are entered in EES: $UnitSystem SI MASS RAD PA K J $Tabstops 0.2 0.4 0.6 3.5 in L=1 [m] D=1 [cm]*convert(cm,m) W=3 [cm]*convert(cm,m) k=0.5 [W/m-K] h_bar_a=50 [W/m^2-K] T_a=converttemp(C,K,20 [C]) h_bar_w=150 [W/m^2-K] T_w=converttemp(C,K,80 [C]). "diameter of hole" "width of extrusion" "conductivity" "heat transfer coefficient on air-side" "air temperature" "heat transfer coefficient on water-side" "water temperature". The resistance to convection to air is:. Rconv ,a =. 1 ha 4W L. (1). Rconv , w =. 1 hw π D L. (2). The resistance to convection to water is:. R_conv_a=1/(h_bar_a*4*W*L) R_conv_w=1/(h_bar_w*pi*D*L). "air side convection resistance" "water side convection resistance". The shape factor between the inner and outer surfaces of the extrusion (SF) is obtained from the SF_12 function in EES. The resistance to conduction is: Rcond =. 1 SF k. (3). The total rate of heat transfer is: q =. (Tw − Ta ) Rconv ,a + Rcond + Rconv , w. (4).

(9) SF=SF_12(D,W,L) R_cond=1/(SF*k) q_dot=(T_w-T_a)/(R_conv_a+R_cond+R_conv_w). "shape factor" "conduction resistance" "heat transfer rate". which leads to q = 79.7 W. b.) Carry out a sanity check on the value of the shape factor that you used in (a). The shape factor can be thought of as the ratio of the effective area to the effective length for conduction. The area for conduction is smallest at the inner surface and largest at the outer surface. The approximate area can be taken to be the average of these two:. Aapp =. π D L + 4W L. (5). 2. The effective length is largest from the surface of the hole to the corner and smallest from the surface of the hole to the middle of the edge. The average of these values is used: Lapp =. ⎡ 1 ⎢ (W − D ) + 2⎢ 2 ⎣. (. ). 2W − D ⎤ ⎥ ⎥ 2 ⎦. (6). The approximate shape factor is computed according to: SFapp = A_app=(pi*D*L+4*W*L)/2 L_app=((W-D)/2+(sqrt(2)*W-D)/2)/2 SF_app=A_app/L_app. Aapp. (7). Lapp "approximate area for conduction" "approximate length for conduction" "approximate shape factor". which leads to SFapp = 5.78 m which compares well to the shape factor obtained using the EES library, SF = 5.35 m..

(10) $UnitSystem SI MASS RAD PA K J $Tabstops 0.2 0.4 0.6 3.5 in "Inputs" th=1.5 [mm]*convert(mm,m) k=75 [W/m-K] e=1 [-] q``_s=900 [W/m^2] T_infinity=converttemp(C,K,10[C]) h_bar=5 [W/m^2-K] L=8 [cm]*convert(cm,m) T_w=converttemp(C,K,50 [C]) W=1 [m]. "thickness" "conductivity of collector plate" "emissivity of collector plate" "solar flux" "ambient temperature" "heat transfer coefficient" "half-width between tubes" "water temperature" "per unit length of collector". Nodes are placed along the length of the collector. Only the region from 0 < x < L is considered due to the symmetry of the system. Therefore, the position of each node is:. xi =. L ( i − 1) for i = 1..N ( N − 1). (8). L ( N − 1). (9). The distance between adjacent nodes is: Δx = N=21 [-] Dx=L/(N-1) duplicate i=1,N x[i]=L*(i-1)/(N-1) end. "number of nodes". "location of each node". Energy balances on each of the internal nodes leads to: q LHS ,i + q RHS ,i + qs ,i + qconv ,i + qrad ,i = 0 for i = 2.. ( N − 1). (10). where. q LHS ,i =. k W th (Ti −1 − Ti ) Δx. (11).

(11) q RHS ,i =. k W th (Ti +1 − Ti ) Δx. (12). qs ,i = W Δx q s′′. (13). qs ,i = W Δx h (T∞ − Ti ). (14). qrad ,i = W Δx ε σ (T∞4 − Ti 4 ). (15). "internal nodes" duplicate i=2,(N-1) q_dot_LHS[i]=k*W*th*(T[i-1]-T[i])/Dx "conduction from left hand side node" q_dot_RHS[i]=k*W*th*(T[i+1]-T[i])/Dx "conduction from right hand side node" q_dot_s[i]=W*Dx*q``_s "absorbed solar radiation" q_dot_conv[i]=W*Dx*h_bar*(T_infinity-T[i]) "convection" q_dot_rad[i]=W*Dx*e*sigma#*(T_infinity^4-T[i]^4) "radiation" q_dot_LHS[i]+q_dot_RHS[i]+q_dot_s[i]+q_dot_conv[i]+q_dot_rad[i]=0 "energy balance" end. The temperature of node 1 is assumed to be equal to the water temperature (neglecting any resistance to convection on the water-side):. T1 = Tw. (16). q LHS , N + qs , N + qconv , N + qrad , N = 0. (17). An energy balance on node N leads to:. where. q LHS , N =. k W th (TN −1 − TN ) Δx W Δx q s′′ 2. (19). W Δx h (T∞ − TN ) 2. (20). W Δx ε σ (T∞4 − TN4 ) 2. (21). qs , N = qs , N = qrad , N = "node 1" T[1]=T_w. (18).

(12) "node N" q_dot_LHS[N]=k*W*th*(T[N-1]-T[N])/Dx q_dot_s[N]=W*Dx*q``_s/2 q_dot_conv[N]=W*Dx*h_bar*(T_infinity-T[N])/2 q_dot_rad[N]=W*Dx*e*sigma#*(T_infinity^4-T[N]^4)/2 q_dot_LHS[N]+q_dot_s[N]+q_dot_conv[N]+q_dot_rad[N]=0. "conduction from left hand side node" "absorbed solar radiation" "convection" "radiation" "energy balance". The temperature distribution within the collector is shown in Figure 2: 333. Temperature (K). 331. 329. 327. 325. 323 0. 0.02. 0.04. 0.06. 0.08. 0.1. Position (m). Figure 2: Temperature as a function of position in the collector.. An energy balance on node 1 provides the rate of energy transfer to the water:. qwater = q RHS ,1 + qs ,1 + qconv ,1 + qrad ,1. (22). where. q RHS ,1 =. k W th (T2 − T1 ) Δx W Δx q s′′ 2. (24). W Δx h (T∞ − T1 ) 2. (25). qs ,1 = qs ,1 = qrad ,1 =. (23). W Δx ε σ (T∞4 − T14 ) 2. "node 1" q_dot_RHS[1]=k*W*th*(T[2]-T[1])/Dx q_dot_s[1]=W*Dx*q``_s/2 q_dot_conv[1]=W*Dx*h_bar*(T_infinity-T[1])/2 q_dot_rad[1]=W*Dx*e*sigma#*(T_infinity^4-T[1]^4)/2. "conduction from right hand side node" "absorbed solar radiation" "convection" "radiation". (26).

(13) q_dot_water=q_dot_RHS[1]+q_dot_s[1]+q_dot_conv[1]+q_dot_rad[1] "energy balance". which leads to qwater = 28.9 W. b.) Determine the efficiency of the collector; efficiency is defined as the ratio of the energy delivered to the water to the solar energy incident on the collector. The efficiency is calculated according to:. η= eta=q_dot_water/(L*W*q``_s). qwater LW. (27). "efficiency". which leads to η = 0.402. c.) Plot the efficiency as a function of the number of nodes used in the solution. Figure 3 illustrates the efficiency as a function of the number of nodes and shows that at least 20-30 nodes are required for numerical convergence. 0.409 0.408 0.407. Efficiency (-). 0.406 0.405 0.404 0.403 0.402 0.401 2. 10. 100. 225. Number of nodes. Figure 3: Efficiency as a function of the number of nodes.. d.) Plot the efficiency as a function of Tw - T∞. Explain your plot. Figure 4 illustrates the efficiency as a function of the water-to-ambient temperature difference. When the water temperature is low then the losses are low (but not zero, because the temperature of the copper plate is elevated by conduction). As the water temperature increases, the temperature of the plate increases and therefore the losses increase and efficiency drops. The drop in efficiency is dramatic for this type of unglazed collector and therefore the collector may be suitable for providing water heating for swimming pools (at low water temperature) but probably is not suitable for providing domestic hot water (at high water temperature)..

(14) 1 0.9 0.8. Efficiency. 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0 0. 10. 20. 30. 40. 50. 60. 70. 80. Water to ambient temperature difference (K). Figure 4: Solar collector efficiency as a function of the water-to-ambient temperature difference..

(15) Problem 2.1-5 A pipe carrying water for a ground source heat pump is buried horizontally in soil with conductivity k = 0.4 W/m-K. The center of the pipe is W = 6 ft below the surface of the ground. The pipe has inner diameter Di = 1.5 inch and outer diameter Do = 2 inch. The pipe is made of material with conductivity kp = 1.5 W/m-K. The water flowing through the pipe has temperature Tw = 35ºF with heat transfer coefficient hw = 200 W/m-K. The temperature of the surface of the soil is Ts = 0ºF. a.) Determine the rate of heat transfer between the water and the air per unit length of pipe. The inputs are entered in EES: $UnitSystem SI MASS RAD PA K J $Tabstops 0.2 0.4 0.6 3.5 in L=1 [m] D_i=1.5 [inch]*convert(inch,m) D_o=2.0 [inch]*convert(inch,m) k_p=1.5 [W/m-K] k=0.4 [W/m-K] W=6 [ft]*convert(ft,m) T_s=converttemp(F,K,0 [F]) T_w=converttemp(F,K,35 [F]) h_bar_w=200 [W/m^2-K]. "inner diameter of tube" "outer diameter of tube" "conductivity of plastic" "conductivity of soil" "depth of tube beneath soil" "temperature of soil surface" "water temperature" "convection coefficient with water". The resistance to convection to water is:. Rconv , w =. 1 hw π Di L. (1). The resistance to conduction through the pipe is:. Rcond , p. ⎛D ⎞ ln ⎜ o ⎟ D = ⎝ i⎠ 2π k p L. R_conv_w=1/(h_bar_w*pi*D_i*L) R_cond_p=ln(D_o/D_i)/(2*pi*k_p*L). (2). "resistance to convection with water" "resistance to conduction through pipe". The shape factor between the outer surface of the pipe and the surface of the soil (SF) is obtained from the SF_2 function in EES. The resistance to conduction through the soil is: Rcond , s = The total rate of heat transfer is:. 1 SF k. (3).

(16) q =. (Tw − Ta ). (4). Rconv , w + Rcond , p + Rcond , s. SF=SF_2(D_o,W,L) "shape factor between outer surface of tube and surface" R_cond_s=1/(SF*k) "resistance to conduction through soil" q_dot=(T_w-T_s)/(R_conv_w+R_cond_p+R_cond_s) "rate of heat transfer". which leads to q = 9.49 W/m. b.) Plot the heat transfer as a function of the depth of the pipe. Figure 1 illustrates the rate of heat transfer as a function of the depth of the pipe. 17. Rate of heat transfer (W). 16 15 14 13 12 11 10 9 8 7 0. 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. Depth of pipe (m) Figure 1: Rate of heat transfer as a function of the depth of the pipe (per unit length).. c.) Carry out a sanity check on the value of the shape factor that you used in (a). The shape factor can be thought of as the ratio of the effective area to the effective length for conduction. Assuming that the heat spreads at an angle of approximately 45º as it moves from the pipe to the surface, the approximate area for conduction is:. Aapp = 2 W L. (5). The effective length is the average of the length from the pipe to the surface perpendicularly and at 45º.. Lapp =. 2W +W 2. The approximate shape factor is computed according to:. (6).

(17) SFapp = A_app=sqrt(2)*W*L L_app=(sqrt(2)*W+W)/2 SF_app=A_app/L_app. Aapp Lapp. (7). "approximate area" "approximate length" "approximate shape factor". which leads to SFapp = 1.17 m which compares well to the shape factor obtained using the EES library, SF = 1.26 m..

(18) Problem 2.2-1: Model of Welding Process You are evaluating a technique for controlling the properties of welded joints by using aggressive liquid cooling. Figure P2.2-1 illustrates a cut-away view of two plates that are being welded together. Both edges of the plate are clamped and effectively held at temperatures Ts = 25°C. The top of the plate is exposed to a heat flux that varies with position x, measured from joint, according to: qm′′ ( x ) = q ′′j exp ( − x / L j ) where q ′′j =1x106 W/m2 is the maximum heat flux. (at the joint, x = 0) and Lj = 2.0 cm is a measure of the extent of the heat flux. The back side of the plates are exposed to aggressive liquid cooling by a jet of fluid at Tf = -35°C with h = 5000 W/m2-K. A half-symmetry model of the problem is shown in Figure P2.2-1. The thickness of the plate is b = 3.5 cm and the width of a single plate is W = 8.5 cm. You may assume that the welding process is steady-state and 2-D. You may neglect convection from the top of the plate. The conductivity of the plate material is k = 38 W/m-K. both edges are clamped and held at fixed temperature. heat flux joint. impingement cooling with liquid jets qm′′. k = 38 W/m-K. W = 8.5 cm b = 3.5 cm. y. Ts = 25°C. x T f = −35°C 2 h = 500 W/m -K. Figure P2-2-1: Welding process and half-symmetry model of the welding process.. a.) Develop a separation of variables solution to the problem. Implement the solution in EES and prepare a plot of the temperature as a function of x at y = 0, 1.0, 2.0, 3.0, and 3.5 cm. The problem shown in Figure P2.2-1 is governed by the partial differential equation: ∂ 2T ∂ 2T + =0 ∂x 2 ∂y 2. (1). and has the boundary conditions: ∂T ∂x. =0 x =0. (2).

(19) Tx =W = Ts k. ∂T ∂y. = h (Ty =0 − T f. (3). ). (4). y =0. k. ∂T ∂y. = qm′′. (5). y =b. Notice that the problem has 3 non-homogeneous boundary conditions and cannot, as it is written, be solved using separation of variables. However, the simple transformation:. θ = T − Ts. (6). transforms the boundary condition at x = W to a homogeneous boundary condition and provides a homogeneous direction (x). The transformed differential equation and boundary conditions are: ∂ 2θ ∂ 2θ + =0 ∂x 2 ∂y 2. (7). and has the boundary conditions: dθ dx. =0. (8). x =0. θ x =W = 0 k. ∂θ ∂y. = h (θ y =0 − θ f. (9). ). (10). y =0. k. ∂θ ∂y. = qm′′. (11). y =b. where. θ f = T f − Ts. (12). T = TX TY. (13). The separated solution is assumed:.

(20) and substituted into Eq. (7); the process leads to two ordinary differential equations:. d 2TX + λ 2 TX = 0 2 dx. (14). d 2TY − λ 2 TY = 0 2 dy. (15). TX = C1 sin ( λ x ) + C2 cos ( λ x ). (16). TY = C3 sinh ( λ y ) + C4 cosh ( λ y ). (17). The solutions to Eqs. (14) and (15) are:. We will solve the eigenproblem first (i.e., the problem in the homogeneous direction). The boundary condition at x = 0, Eq. (8), leads to: dTX dx. x =0. = ⎡⎣C1 λ cos ( λ x ) − C2 λ sin ( λ x ) ⎤⎦ x =0 = C1 λ = 0. (18). which indicates that C1 = 0 and therefore: TX = C2 cos ( λ x ). (19). The boundary condition at x = W, Eq. (9), leads to: TX x =W = C2 cos ( λ W ) = 0. (20). Equation (20) provides our eigenvalues:. λi =. (1 + 2 i ) π 2W. for i = 1..∞. (21). and the eigenfunctions are: TX i = C2,i cos ( λi x ). (22). The solution for TX is the sum of the eigenfunctions: ∞. TX = ∑ C2,i cos ( λi x ) i =0. (23).

(21) and the solution for the temperature is: ∞. TX = ∑ C2,i cos ( λi x ) ⎡⎣C3,i sinh ( λi y ) + C4,i cosh ( λi y ) ⎤⎦. (24). i =0. Note that C2,i can be absorbed in the other constants: ∞. θ = ∑ cos ( λi x ) ⎡⎣C3,i sinh ( λi y ) + C4,i cosh ( λi y ) ⎤⎦. (25). i =0. Equation (25) is substituted into the boundary condition at y = 0, Eq. (10): ∞. k ∑ cos ( λi x ) ⎣⎡C3,i λi cosh ( λi 0 ) + C4,i λi sinh ( λi 0 ) ⎦⎤ = i =0. (26). ⎧∞ ⎫ h ⎨∑ cos ( λi x ) ⎡⎣C3,i sinh ( λi 0 ) + C4,i cosh ( λi 0 ) ⎤⎦ − θ f ⎬ ⎩ i =0 ⎭ or ∞ ⎡∞ ⎤ k ∑ cos ( λi x ) C3,i λi = h ⎢ ∑ cos ( λi x ) C4,i − θ f ⎥ i =0 ⎣ i =0 ⎦. (27). We take advantage of the orthogonality of the eigenfunctions by multiplying Eq. (27) by an arbitrary eigenfunction and integrating from x = 0 to x = L in order to achieve: W W ⎡ k C3,i λi ∫ cos 2 ( λi x ) dx = h ⎢C4,i ∫ cos 2 ( λi x ) dx − θ f 0 0 ⎣. The integrals in Eq. (28) are evaluated using Maple: > restart; > assume(i,integer); > lambda:=(1+2*i)*Pi/(2*W);. λ :=. ( 1 + 2 i~ ) π 2W. > int((cos(lambda*x))^2,x=0..W);. W 2 > int(cos(lambda*x),x=0..W);. 2 ( -1 ) i~ W ( 1 + 2 i~ ) π. ⎤. W. ∫ cos ( λ x ) dx ⎥⎦ i. 0. (28).

(22) which leads to: k C3,i λi W 2. i ⎡ C4,i W 2 ( −1) W ⎤ = h⎢ −θ f ⎥ (1 + 2 i ) π ⎥⎦ ⎢⎣ 2. (29). or. 4 ( −1) kλ θf = C4,i − i C3,i h (1 + 2 i ) π i. (30). which is one equation for each of the two pairs of unknown coefficients. Equations (25) and Error! Reference source not found. are substituted into the final boundary condition at y = b, Eq. (11): ∞ ⎛ x k ∑ cos ( λi x ) ⎡⎣C3,i λi cosh ( λi b ) + C4,i λi sinh ( λi b ) ⎤⎦ = q ′′j exp ⎜ − ⎜ Lj i =0 ⎝. ⎞ ⎟⎟ ⎠. (31). Again, the orthogonality of the eigenfunctions is used to obtain: W W ⎛ x k ⎡⎣C3,i λi cosh ( λi b ) + C4,i λi sinh ( λi b ) ⎤⎦ ∫ cos 2 ( λi x ) dx = q ′′j ∫ exp ⎜ − ⎜ Lj 0 0 ⎝. ⎞ ⎟⎟ cos ( λi x ) dx ⎠. (32). or C3,i λi cosh ( λi b ) + C4,i λi sinh ( λi b ) =. 2 q ′′ Wkj. W. ⎛. x ⎞ ⎟⎟ cos ( λi x ) dx j ⎠. ∫ exp ⎜⎜ − L 0. ⎝. (33). which is the 2nd relationship between the coefficient pairs. The integral on the right hand side of Eq. (32) is accomplished using Maple: > int(exp(-x/L_j)*cos(lambda*x),x=0..W); ⎛ ⎞ ⎜⎜ ⎟⎟ ⎛ ⎞ ⎛⎜⎜ − L_j ⎞⎟⎟ ⎜⎜ ⎟ ⎝ ⎝ L_j ⎠ ⎠ i~ i~ 2 L_j W ⎝ 2 W e + π L_j ( -1 ) + 2 π L_j ( -1 ) i~ ⎟⎠ e 4 W2 + π 2 L_j2 + 4 π 2 L_j2 i~ + 4 π 2 L_j2 i~ 2 W. and pasted directly into EES in order to accomplish the implementation. The inputs are entered in EES: $UnitSystem SI MASS RAD PA K J. W.

(23) $TABSTOPS 0.2 0.4 0.6 0.8 3.5 in "Inputs" b=3.5 [cm]*convert(cm,m) W=8.5 [cm]*convert(cm,m) k=38 [W/m-K] T_s=converttemp(C,K,25) T_f=converttemp(C,K,-35) h=5000 [W/m^2-K] q``_j=1e6 [W/m^2] L_j=2.0 [cm]*convert(cm,m). "plate thickness" "plate width" "plate conductivity" "edge temperature" "fluid temperature" "heat transfer coefficient" "heat flux at x=0" "extent of the heat flux". and a position is specified: "Position" x_bar=0.5 x=x_bar*W y_bar=0.5 y=y_bar*b. The solution is implemented:: N=10 [-] "number of terms" theta_f=(T_f-T_s) "transformed fluid temperature" duplicate i=0,N lambda[i]=(1+2*i)*pi/(2*W) theta_f*4*(-1)î/((1+2*i)*pi)=C_4[i]-k*lambda[i]*C_3[i]/h C_3[i]*lambda[i]*cosh(lambda[i]*b)+C_4[i]*lambda[i]*sinh(lambda[i]*b)=(2*q``_j/(W*k))* (2*W*L_j*(2*W*exp(W/L_j)+Pi*L_j*(-1)î+2*Pi*L_j*(-1)î*i)*exp(-W/L_j)/(4*W^2+Pi^2*L_j^2 +4*Pi^2*L_j^2*i+4*Pi^2*L_j^2*i^2)) theta[i]=cos(lambda[i]*x)*(C_3[i]*sinh(lambda[i]*y)+C_4[i]*cosh(lambda[i]*y)) end T=T_s+sum(theta[0..N]) T_C=converttemp(K,C,T). Note that the bold portion of the code was pasted from Maple. Figure 2 illustrates the temperature as a function of dimensionless axial position (x/W) for various values of y..

(24) Figure 2: Temperature as a function of x/W for various values of y.. b.) Prepare a contour plot of the temperature distribution. Figure 3 illustrates a contour plot of the temperature distribution in the plate.. Figure 3: Contour plot of the temperature distribution in the plate..

(25) Problem 2.2-2: Derive a Partial Differential Equation Figure P2.2-2 illustrates a thin plate that is exposed to air on upper and lower surfaces. The heat transfer coefficient between the top and bottom surfaces is h and the air temperature is Tf. The thickness of the plate is th and its width and height are a and b, respectively. The conductivity of the plate is k. The top edge is fixed at a uniform temperature, T1. The right edge is fixed at a different, uniform temperature, T2. The left edge of the plate is insulated. The bottom edge of the plate is exposed to a heat flux, q ′′ . This problem should be done on paper. T1. a T2 k y. b x q ′′. th. h ,Tf. h ,T f. Figure P.2.2-2: Thin plate exposed to air.. a.) The temperature distribution within the plate can be considered 2-D (i.e., temperature variations in the z-direction can be neglected) if the plate is thin and conductive. How would you determine if this approximation is valid? The Biot number characterizes the ratio of conduction in the z direction (which I want to neglect) to convection from the plate surface (which I will consider): Bi =. Rcond , z Rconv. =. th h a b th h = 2k ab 1 2k. (1). b.) Derive the partial differential equation and boundary conditions that would need to be solved in order to obtain an analytical solution to this problem. A differential control volume has extend dx x dy and extends through the thickness of the plate, as shown in Figure 2..

(26) Figure 2: Differential control volume used to derive the governing partial differential equation.. The first law balance suggested by Figure 2 is:. q x + q y = q x + dx + q y + dy + qconv. (2). or 0=. ∂q y ∂q x dx + dy + qconv ∂x ∂y. (3). Substituting the rate equations into the energy balance leads to: 0=. ∂ ⎡ ∂T ⎤ ∂ ⎡ ∂T ⎤ dx + ⎢ − k th dx ⎥ dy + 2 dx dy h (T − T f ) −k th dy ⎢ ⎥ ∂x ⎣ ∂x ⎦ ∂y ⎣ ∂y ⎦. (4). or. ∂ 2T ∂ 2T 2 h + − (T − T f ) = 0 ∂x 2 ∂y 2 k th. (5). The boundary conditions at the upper and right edges are:. Tx = a = T2. (6). Ty =b = T1. (7).

(27) The boundary condition at the left edge is obtained from an interface balance which requires that the conduction heat flux in the x-direction at this insulated edge be zero:. q ′′x , x =0 = 0. (8). or −k. ∂T ∂x. =0. (9). x =0. which leads to:. ∂T ∂x. =0. (10). x =0. The boundary condition at the bottom edge is also obtained from an interface energy balance: q ′′ = −k. ∂T ∂y. (11) y =0. or ∂T ∂y. =− y =0. q ′′ k. (12).

(28) Problem 2.2-3 (2-3 in text): Heat Transfer Coefficient Measurement Device You are the engineer responsible for a simple device that is used to measure heat transfer coefficient as a function of position within a tank of liquid (Figure P2.2-3). The heat transfer coefficient can be correlated against vapor quality, fluid composition, and other useful quantities. The measurement device is composed of many thin plates of low conductivity material that are interspersed with large, copper interconnects. Heater bars run along both edges of the thin plates. The heater bars are insulated and can only transfer energy to the plate; the heater bars are conductive and can therefore be assumed to come to a uniform temperature as a current is applied. This uniform temperature is assumed to be applied to the top and bottom edges of the plates. The copper interconnects are thermally well-connected to the fluid; therefore, the temperature of the left and right edges of each plate are equal to the fluid temperature. This is convenient because it isolates the effect of adjacent plates from one another which allows each plate to measure the local heat transfer coefficient. Both surfaces of the plate are exposed to the fluid temperature via a heat transfer coefficient. It is possible to infer the heat transfer coefficient by measuring heat transfer required to elevate the heater bar temperature a specified temperature above the fluid temperature. top and bottom surfaces exposed to fluid 2 T∞ = 20°C, h = 50 W/m -K copper interconnet, T∞ = 20°C. a = 20 mm b = 15 mm. plate: k = 20 W/m-K th = 0.5 mm. heater bar, Th = 40°C. Figure P2.2-3: Device to measure heat transfer coefficient as a function of position.. The nominal design of an individual heater plate utilizes metal with k = 20 W/m-K, th = 0.5 mm, a = 20 mm, and b = 15 mm (note that a and b are defined as the half-width and half-height of the heater plate, respectively, and th is the thickness as shown in Figure P2-3). The heater bar temperature is maintained at Th = 40ºC and the fluid temperature is T∞ = 20ºC. The nominal value of the average heat transfer coefficient is h = 50 W/m2-K. a.) Develop an analytical model that can predict the temperature distribution in the plate under these nominal conditions. The problem can be simplified and tackled using the quarter-symmetry model of a single plate, shown in Figure P2.2-3, is specified mathematically in Figure 2(a)..

(29) (a). (b) Figure 2: Quarter symmetry model of a plate.. The lines of symmetry (x =0 and y = 0) are adiabatic, the top edge (y = b) is held at the heater temperature (Th) and the right edge (x = a) is held at the fluid temperature (T∞). These boundary conditions are expressed below: ∂T ∂x. =0 x =0. Tx =a = T∞.

(30) ∂T ∂y. =0 y =0. Ty =b = Th The differential control volume and associated first law balance are shown in Figure 2(a):. q x + q y = q x + dx + q y + dy + qconv or, after expanding the x + dx and y + dy terms: 0=. ∂q y ∂q x dx + dy + qconv ∂x ∂y. Substituting the rate equations into the energy balance leads to: 0=. ∂ ⎡ ∂T ⎤ ∂ ⎡ ∂T ⎤ −k th dy dx + ⎢ − k th dx dy + 2 dx dy h ( T − T∞ ) ⎢ ⎥ ∂x ⎣ ∂x ⎦ ∂y ⎣ ∂y ⎥⎦. or. ∂ 2T ∂ 2T 2 h + − T − T∞ = 0 ∂x 2 ∂y 2 k th. (. ). (1). Equation (1) is not homogeneous; however, a transformation variable can be identified by inspection:. θ = T − T∞. (2). Substituting Eq. (2) into Eq. (1) transforms the governing partial differential equation into a homogeneous equation; i.e., any multiple of θ will satisfy transformed partial differential equation: ∂ 2θ ∂ 2θ + 2 − m2 θ = 0 2 ∂x ∂y. where. m=. 2h k th. (3).

(31) Substituting Eq. (2) into the boundary conditions leads to: ∂θ ∂x. =0. θ x =a = 0 ∂θ ∂y. (4). x =0. =0. (5) (6). y =0. θ y =b = θ h. (7). where. θ h = Th − T∞ The transformed problem is shown in Figure 2(b); note that the transformed problem is linear and consists of a homogeneous partial differential equation and homogeneous boundary conditions in the x-direction. Therefore, we can apply a separation of variables solution to the problem and x is our homogeneous direction. The solution is assumed to be the product of θX which is a function of x and θY which is a function of y:. θ ( x, y ) = θ X ( x ) θ Y ( y ) Substituting the product solution into the differential equation leads to:. θY. d 2θ X d 2θ Y θ + X − m2 θ X θY = 0 2 2 dx dy. Dividing by the product θX θY: 2 d 2θ X d θ Y 2 dx 2 + dy − m 2 = 0 θ Y . X θ. −λ 2. λ2. Recall that x is our homogeneous direction; therefore, we need the θX group in the equation to equal a negative constant (-λ2); the remaining part of the equation must equal the positive value.

(32) of the same constant (λ2). Therefore, the two ordinary differential equations that result from the separation process are:. d 2θ X + λ2 θ X = 0 2 dx. (8). d 2θ Y − ( λ 2 + m2 )θY = 0 dy 2. (9). and. The next step is to solve the eigenproblem; the solution to the ordinary differential equation for θX is:. θ X = C1 cos ( λ x ) + C2 sin ( λ x ). (10). The 1st boundary conditions in the homogeneous direction, Eq. (4), leads to: dθ X dx. =0. (11). x =0. Substituting Eq. (10) into Eq. (11) leads to:. dθ X dx. x =0. = −C1 λ sin ( λ 0 ) + C2 λ cos ( λ 0 ) = 0 . . =0. =1. which can only be true if C2 = 0. the 2nd boundary condition in the x-direction, Eq. (5), leads to:. θ x =a = C1 cos ( λ a ) = 0 which can only be true if the argument of the cosine is 0. Therefore, the eigenfunctions and eigenvalues for the problem are:. θ X i = C1,i cos ( λi x ) where λi =. ( 2 i − 1) π 2. a. The solution to the ordinary differential equation in the non-homogeneous direction, Eq. (9) can be obtained using Maple: > restart; > assume((lambda^2+m^2),positive); > ODEy:=diff(diff(thetaY(y),y),y)-(lambda^2+m^2)*thetaY(y)=0;.

(33) 2 ⎞ ⎛d ODEy := ⎜ 2 thetaY( y ) ⎟⎟ − ( λ∼ 2 + m~ 2 ) thetaY( y ) = 0 ⎜ dy ⎝ ⎠. > qYs:=dsolve(ODEy);. qYs := thetaY( y ) = _C1 e. λ∼ 2 + m~2 y ). (. + _C2 e. ( − λ∼2 + m~2 y ). Note that Maple can convert this exponential form to an equivalent form involving hyperbolic sines and cosines with the convert command. > convert(qYs,'trigh');. thetaY( y ) = ( _C1 + _C2 ) cosh( λ∼ 2 + m~ 2 y ) + ( _C1 − _C2 ) sinh( λ∼ 2 + m~ 2 y ). θ Yi = C3,i cosh. (. ). λi2 + m 2 y + C4,i sinh. (. λi2 + m 2 y. ). The solution associated with the ith eigenvalue is therefore:. θi = θ X i θ Yi = cos ( λi x ) ⎡⎢C3,i cosh ⎣. (. ). λi2 + m 2 y + C4,i sinh. (. ). λi2 + m 2 y ⎤⎥ ⎦. (12). where the constant C1,i is absorbed into the constants C3,i and C4,i. It is good practice to verify that Eq. (12) satisfies both boundary conditions in the x-direction and the partial differential equation for any value of i: > restart; > assume(i,integer); > lambda:=(2*i-1)*Pi/(2*a);. λ :=. ( 2 i~ − 1 ) π 2a. > theta:=(x,y)->cos(lambda*x)*(C_3*cosh(sqrt(lambda^2+m^2)*y)+C_4*sinh(sqrt(lambda^2+m^2)*y));. θ := ( x, y ) → cos( λ x ) ( C_3 cosh( λ 2 + m 2 y ) + C_4 sinh( λ 2 + m 2 y ) ) > eval(diff(theta(x,y),x),x=0);. 0 > theta(a,y);. 0 > diff(diff(theta(x,y),x),x)+diff(diff(theta(x,y),y),y)-m^2*theta(x,y);. 1 ( 2 i~ − 1 ) π x ⎞ 2 2 − cos⎛⎜⎜ ⎟⎟ ( 2 i~ − 1 ) π 2a 4 ⎝ ⎠ ⎛ ⎛ ⎛ ⎞ ( 2 i~ − 1 ) 2 π 2 ⎜ ⎜ ⎜ + 4 m 2 y ⎟⎟ ⎜ ⎜ ⎜ 2 a ⎜ ⎜ ⎟ ⎜ ⎜⎜ C_3 cosh⎜⎜ ⎟⎟ + C_4 sinh⎜⎜ 2 ⎝ ⎝ ⎠ ⎝. ⎞⎞ ( 2 i~ − 1 ) 2 π 2 + 4 m 2 y ⎟⎟ ⎟⎟ 2 a ⎟⎟ ⎟⎟ ⎟⎟ 2 ⎠⎠.

(34) ⎛ ⎜ ⎜ ( 2 i~ − 1 ) π x ⎞ ⎜⎜ a 2 + cos⎛⎜⎜ ⎟⎟ ⎜ 2a ⎝ ⎠⎝ 2 2 ⎛ ⎞ ( 2 i~ − 1 ) π ⎜ + 4 m 2 y ⎟⎟ ⎜ 2 1 a ⎜ ⎟ ⎛ ( 2 i~ − 1 ) 2 π 2 ⎞ ⎜ ⎟⎟ ⎜ + 4 m 2 ⎟⎟ C_3 cosh⎜ 2 ⎜ 2 4 a ⎝ ⎠⎝ ⎠ 2 2 ⎞ ⎛ ⎞ ( 2 i~ − 1 ) π 2 ⎟ ⎜ ⎟ + 4 m y ⎟ ⎜ ⎟ 1 a2 ⎟ ⎜ ⎟ ⎛ ( 2 i~ − 1 ) 2 π 2 ⎞ 2 ⎟⎟ ⎜ + C_4 sinh⎜⎜ + 4 m ⎟⎟ ⎟⎟ − m 2 2 ⎜ 2 4 a ⎝ ⎠⎝ ⎠⎠ ( 2 i~ − 1 ) π x ⎞ cos⎛⎜⎜ ⎟⎟ 2 a ⎝ ⎠ ⎛ ⎞ ⎞⎞ ⎛ ⎛ ( 2 i~ − 1 )2 π2 ( 2 i~ − 1 ) 2 π 2 2 ⎜ ⎜ ⎟ ⎜ + 4 m y + 4 m 2 y ⎟⎟ ⎟⎟ ⎜ ⎜ ⎟ ⎜ 2 2 a a ⎜ ⎜ ⎟ ⎜ ⎟⎟ ⎜⎜ C_3 cosh⎜⎜ ⎟⎟ + C_4 sinh⎜⎜ ⎟⎟ ⎟⎟ 2 2 ⎝ ⎝ ⎠ ⎝ ⎠⎠ > simplify(%);. 0. The general solution is the sum of the solution for each eigenvalue: ∞. ∞. θ = ∑θi = ∑ cos ( λi x ) ⎡⎢C3,i cosh i =1. ⎣. i =1. (. ). λi2 + m 2 y + C4,i sinh. (. ). λi2 + m 2 y ⎤⎥ ⎦. (13). The general solution must satisfy the boundary conditions in the non-homogeneous direction. Substituting Eq. (13) into Eq. (6) leads to: ∂θ ∂y. ⎡ ⎤ 2 2 2 2 2 2 2 2 ⎢ = ∑ cos ( λi x ) C3,i λi + m sinh λi + m 0 + C4,i λi + m cosh λi + m 0 ⎥ = 0 ⎢ i =1 . . ⎥ =0 =1 ⎣⎢ ⎦⎥. y =0. ). (. ∞. (. ). or ∞. ∑C i =1. 4,i. λi2 + m 2 cos ( λi x ) = 0. which can only be true if C4,i = 0: ∞. θ = ∑ Ci cosh i =1. (. ). λi2 + m 2 y cos ( λi x ). (14).

(35) where the subscript 3 has been removed from C3,i because it is the only remaining undetermined constant. Substituting Eq. (14) into Eq. (7) leads to: ∞. θ y =b = ∑ Ci cosh i =1. (. ). λi2 + m 2 b cos ( λi x ) = θ h. This equation is multiplied by cos(λj x) and integrated from 0 to a: ∞. ∑ Ci cosh i =1. (. ). a. a. 0. 0. λi2 + m 2 b ∫ cos ( λi x ) cos ( λ j x ) dx = θ h ∫ cos ( λ j x ) dx. Orthogonality causes all of the terms in the summation to integrate to zero except for the one in which i = j: Ci cosh. (. λ +m b 2 i. 2. ) ∫ cos (λ x ) dx = θ ∫ cos (λ x ) dx a. a. 2. i. h. 0. i. 0. Maple is used to carry out the integrations: > restart; > assume(i,integer); > lambda:=(2*i-1)*Pi/(2*a);. λ := > int((cos(lambda*x))^2,x=0..a);. ( 2 i~ − 1 ) π 2a a 2. > int(cos(lambda*x),x=0..a);. ( 1 + i~ ). 2 ( -1 ) a ( 2 i~ − 1 ) π. which leads to an equation for each of the constants.. θ h 4 ( −1). 1+ i. Ci =. The inputs are entered in EES: $UnitSystem SI MASS RAD PA C J $Tabstops 0.2 0.4 0.6 0.8 3.5 "Inputs". cosh. (. ). λi2 + m 2 b ( 2 i − 1) π.

(36) th_mm=0.5 [mm] th=th_mm * convert(mm,m) k=20 [W/m-K] a_mm=20 [mm] a= a_mm * convert(mm,m) b= 15 [mm] *convert(mm,m) T_h = converttemp(C,K,40 [C]) T_infinity=converttemp(C,K,20 [C]) h_bar=50 [W/m^2-K]. "thickness of plate in mm" "thickness of plate" "conductivity of plate" "half-width of plate in mm" "half-width of plate" "half-height of plate" "heater temperature" "fluid temperature" "heat transfer coefficient". The position is specified using dimensionless variables x_bar and y_bar: "position" x_bar=0.5 y_bar=0.5 x=x_bar*a y=y_bar*b. The solution is evaluated: theta_h=T_h-T_infinity "heater temperature difference" m=sqrt(2*h_bar/(k*th)) "solution constant" N=100 "number of terms" duplicate i=1,N lambda[i]=(2*i-1)*pi/(2*a) "eigenvalues" C[i]=theta_h*(4*(-1)^(1+i)/(2*i-1)/Pi)/(cosh(sqrt(lambda[i]^2+m^2)*b)) theta[i]=C[i]*cos(lambda[i]*x)*cosh(sqrt(lambda[i]^2+m^2)*y) end T=sum(theta[1..N])+T_infinity T_C=converttemp(K,C,T). A parametric table is created and used to generate the contour plot shown in Figure 3..

(37) Figure 3: Contour plot of temperature on the plate.. b.) The measured quantity is the rate of heat transfer to the plate from the heater ( qh ) and therefore the relationship between qh and h (the quantity that is inferred from the heater power) determines how useful the instrument is. Determine the heater power. The heater power can be computed by integrating the conduction heat transfer along the top surface according to: a. qh = 4 ∫ k 0. ∂θ ∂y. th dx. (15). y =b. where the factor of 4 comes from the quarter symmetry of the model. substituted into Eq. (15): ∞. qh = 4 k th ∑ Ci λi2 + m 2 sinh i =1. (. λi2 + m 2 b. Equation (14) is. ) ∫ cos (λ x ) dx a. i. 0. which leads to: ∞. qh = 2 k th a ∑ Ci λi2 + m 2 sinh i =1. Equation (16) is evaluated in EES: "heater power". (. λi2 + m 2 b. ). (16).

(38) duplicate i=1,N q_dot_h[i]=2*k*th*a*C[i]*sqrt(lambda[i]^2+m^2)*sinh(sqrt(lambda[i]^2+m^2)*b) end q_dot_h=sum(q_dot_h[1..N]). c.) If the uncertainty in the measurement of the heater power is δ qh = 0.01 W, estimate the uncertainty in the measured heat transfer coefficient ( δ h ). A parametric table is used to explore the relationship between h and qhtr ; the two columns in the table are the variables h_bar and q_dot_h. Figure 4 illustrates the heat transfer coefficient as a function of the heater power, all else held constant.. Figure 4: Heat transfer coefficient as a function of the heater power.. Note that a good measuring device would show a strong relationship between the physical quantity being measured ( h ) and the measurement ( qh ); this would be indicated by a large partial derivative of heat transfer coefficient with respect to heater power. Given the uncertainty in the heater power measurement, δ qh (related to, for example, the resolution of the data acquisition system, noise, etc.), it is possible to estimate the uncertainty in the measurement of the heat transfer coefficient, δ h , according to:. δh =. ∂h δ qh ∂qh. For example, if the uncertainty in the heater power is 0.01 W then Fig. 3 suggests that the partial derivative of heat transfer coefficient with respect to heater power is approximately 125 W/m2-.

(39) K-W and the uncertainty in the heat transfer coefficient measurement would be δ h = 1.8 W/m2K. As an engineer designing this measurement device, you would like to calculate not the heater power but rather the partial derivative in the heat transfer coefficient with respect to heater power in order to automate the process of computing δ h and therefore evaluate how design changes affect the instrument. This would be difficult to accomplish analytically; note that both m and the constants Ci in Eq. (16) are functions of h . It is more convenient to determine the partial derivative numerically. That is, set the heat transfer coefficient at its nominal value plus a small amount ( h + Δh ) and evaluate the heater power ( qh , h +Δh ); note that Δh should be small relative to the nominal value of the heat transfer coefficient. Then set the heat transfer coefficient at its nominal value less a small amount ( h − Δh ) and evaluate the heater power ( qh , h −Δh ). The partial derivative is approximately: ∂h 2 Δh ≈ ∂qh qh ,h +Δh − qh ,h −Δh. Since qh needs to be evaluated at several values of the heat transfer coefficient, it is convenient to have the computation of the heater power occur within a function that is called twice within the equation window. Because the solution is in the form of an EES code, this is an ideal problem to use a MODULE. A MODULE is a stand-alone EES program that can be called from the main EES equation window. The MODULE is provided with inputs and it calculates outputs. The protocol of a call to a MODULE involves the name of the MODULE followed by a series of inputs separated by a colon from a series of outputs. For example, we will create a MODULE Heaterpower that calculates the value of the variable q_dot_h. It is important to note that all of the variables from the main equation window that are required by the MODULE must be passed to the MODULE; the MODULE can only access variables that are passed to it as parameters or using the $COMMON directive. The MODULE Heaterpower must be placed at the top of the EES code and is a small selfcontained EES program; we create the MODULE by copying those lines of the main EES code that are required to calculate the variable q_dot_h. MODULE Heaterpower(th,k,a,b,T_h,T_infinity,h_bar:q_dot_h) theta_h=T_h-T_infinity "heater temperature difference" m=sqrt(2*h_bar/(k*th)) "solution constant" N=100 "number of terms" duplicate i=1,N lambda[i]=(2*i-1)*pi/(2*a) "eigenvalues" C[i]=theta_h*(4*(-1)^(1+i)/(2*i-1)/Pi)/(cosh(sqrt(lambda[i]^2+m^2)*b)) q_dot_h[i]=2*k*th*a*C[i]*sqrt(lambda[i]^2+m^2)*sinh(sqrt(lambda[i]^2+m^2)*b) end q_dot_h=sum(q_dot_h[1..N]) end.

(40) The calling protocol for the MODULE consists of a series of inputs (the variables d, k, a, b, T_htr, T_f, and h) that are separated by a series outputs (in this case only the variable q_dot_htr) by a colon. MODULES are most useful where a certain sequence of code must be executed multiple times; in this case, the MODULE Heaterpower enables the partial derivative to be easily computed. delta_q_dot_htr=0.01 [W] dh=1 [W/m^2-K]. "heater power resolution" "perturbation of heat transfer coefficient". CALL Q_dot_heater(d,k,a,b,T_htr,T_f,h-dh:q_dot_htr_minus) CALL Q_dot_heater(d,k,a,b,T_htr,T_f,h+dh:q_dot_htr_plus) dhdqdot=2*dh/(q_dot_htr_plus-q_dot_htr_minus) delta_h=dhdqdot*delta_q_dot_htr. The modifications to the EES code verifies that the uncertainty in heat transfer coefficient, delta_h, is 1.26 W/m2-K and provides a convenient tool for assessing the impact of the various design parameters on the performance of the measurement system. For example, Figure 5 illustrates the uncertainty in the heat transfer coefficient as a function of the plate thickness (th) for various values of its half-width (a).. Figure 5: Uncertainty in the measured heat transfer coefficient as a function of the plate thickness for various values of the plate half-width..

(41) Problem 2.2-4: Local Heat Transfer Coefficient Measurement Figure P2.2-4(a) illustrates a proposed device to measure the local heat transfer coefficient from a surface undergoing a boiling heat transfer process. Micro-scale heaters and temperature sensors are embedded in a substrate in a regularly spaced array, as shown. evaporating fluid. heaters computational temperature domain sensors Figure P2.2-4(a): An array of micro-scale heaters and temperature sensors embedded in a substrate.. The heaters are activated, producing a heat flux that is removed primarily from the surface of the substrate exposed to evaporating fluid. The temperature sensors are embedded in sets of two located very near the surface. Each set of thermocouples are used to infer the local heat flux to the surface ( qs′′ ) and the surface temperature (Ts); these quantities are sufficient to measure the heat transfer coefficient. A half-symmetry model of the region of the substrate between two adjacent heaters (see Figure 2.2-4(a)) is shown in Figure P2.2-4(b). T∞ = 20°C 2 h = 2500 W/m -K. y1 = 1.9 mm. y. 5 2 qh′′ = 1x10 W/m. b = 2 mm W = 5 mm x. y2 = 1.8 mm. k = 3.5 W/m-K Figure P2.2-4(b): A half-symmetry model of the region of the substrate between two adjacent heaters.. The thickness of the substrate is b = 2 mm and the half-width between adjacent heaters is W = 5 mm. The substrate has conductivity k = 3.5 W/m-K and you may assume that the presence of the temperature sensors does not affect the temperature distribution. The heat flux exposed to the computational domain at x = W is qh′′ = 1x105 W/m2. The heat transfer coefficient between the evaporating fluid at T∞ = 20ºC and the surface is h = 2500 W/m2-K. a.) Develop a separation of variables solution based on the computational domain shown in Figure P2.2-4(b). Implement your solution in EES. The known information is entered in EES: $UnitSystem SI MASS RAD PA C J.

(42) $Tabstops 0.2 0.4 0.6 0.8 3.5 "Inputs" k=3.5 [W/m-K] T_infinity=converttemp(C,K,20 [C]) h=2500 [W/m^2-K] W=5 [mm]*convert(mm,m) b=2 [mm]*convert(mm,m) y_1=1.9 [mm]*convert(mm,m) y_2=1.8 [mm]*convert(mm,m) q_h=1e5 [W/m^2]. "conductivity" "fluid temperature" "heat transfer coefficient" "half-width" "substrate thickness" "temperature sensor #1 position" "temperature sensor #2 position" "edge heat flux". The problem shown in Figure P2.2-4(b) is governed by the partial differential equation: ∂ 2T ∂ 2T + =0 ∂x 2 ∂y 2. (1). and has the boundary conditions:. k. ∂T ∂x. x =0. ∂T ∂x. x =W. ∂T ∂y. −k. ∂T ∂y. =0. (2). = qh′′. (3). =0. (4). y =0. = h (Ty =b − T∞ ). (5). y =b. Neither the x- or y-directions are homogeneous. However, by defining the temperature difference relative to T∞, both boundary conditions in the y-direction can be made homogeneous:. θ = T − T∞. (6). The transformed differential equation and boundary conditions are: ∂ 2θ ∂ 2θ + =0 ∂x 2 ∂y 2. and has the boundary conditions:. (7).

(43) k. ∂θ ∂x. x =0. ∂θ ∂x. x =W. ∂θ ∂y −k. ∂θ ∂y. =0. (8). = qh′′. (9). =0. (10). = h θ y =b. (11). y =0. y =b. The separated solution is assumed:. θ = θ X θY. (12). and substituted into Eq. (7); the process leads to two ordinary differential equations: d 2θ X − λ2 θ X = 0 2 dx. (13). d 2θ Y + λ 2 θY = 0 dy 2. (14). Notice that the sign of the constant is selected so that the solution in the homogeneous direction (y) is sines and cosines. The solution to Eq. (14) is:. θ Y = C1 sin ( λ y ) + C2 cos ( λ y ). (15). The boundary condition at y = 0, Eq. (10), leads to: dθ Y dy. = C1 λ cos ( λ 0 ) − C2 λ sin ( λ 0 ) = 0. (16). y =0. which indicates that C1 = 0 and therefore:. θ Y = C2 cos ( λ y ). (17). The boundary condition at y = b, Eq. (11), leads to: k C2 λ sin ( λ b ) = h C2 cos ( λ b ). (18).

(44) Equation (18) provides the eigencondition for the problem: tan ( λi b ) =. hb k ( λi b ). (19). The roots of Eq. (19) occur in regular intervals of the argument of the tangent function. A residual is defined by rearranging Eq. (19): Res = tan ( λi b ) −. hb k ( λi b ). (20). Figure P2.2-4-2 illustrates the residual defined in Eq. (20); the eigenvalues correspond to the points where the residual is zero. 1 0.75 0.5. Residual. 0.25 0 -0.25 -0.5 -0.75 -1 0. eigenvalues 1.57 3.14 4.71 6.28 7.85 9.42 10.99 12.56 14.13 15.7. λb. Figure P2.2-4-2: Residual as a function of λb.. The eigenvalues lie in regular intervals of λb, as shown in Figure 2.2-4-2 and are identified automatically in EES by defining arrays and using them as the lower limit, upper limit, and guess values for the entries in the eigenvalue array in the Variable Information window (Figure P2.2-43).. Figure P2.2-4-3: Variable Information window. {Res=tan(lambdab)-h*b/(k*lambdab)}. "residual of the eigencondition".

(45) N=20 [-] duplicate i=1,N upperlimit[i]=(i-1/2)*pi lowerlimit[i]=(i-1)*pi guess[i]=(i-3/4)*pi tan(lambdab[i])-h*b/(k*lambdab[i])=0 lambda[i]=lambdab[i]/b end. The eigenfunctions are:. θ Yi = C2,i cos ( λi y ). (21). The solution to Eq. (13) for each eigenvalue is:. θ X i = C3,i sinh ( λi x ) + C4,i cosh ( λi x ). (22). The general solution for for each eigenvalue is:. θi = θ X i θ Yi = C2,i cos ( λi y ) ⎡⎣C3,i sinh ( λi x ) + C4,i cosh ( λi x ) ⎤⎦. (23). The sum of these solutions is, itself, a solution (note that the constant C2,i is absorbed into the other constants): ∞. ∞. i =1. i =1. θ = ∑ θi = ∑ cos ( λi y ) ⎡⎣C3,i sinh ( λi x ) + C4,i cosh ( λi x ) ⎤⎦. (24). Equation (24) is substituted into the boundary condition at x= 0, Eq. (8): ∂θ ∂x. ∞. x =0. = ∑ cos ( λi y ) ⎡⎣C3,i λi cosh ( λi 0 ) + C4,i λi sinh ( λi 0 ) ⎤⎦. (25). i =1. or ∞. ∑ cos ( λ y ) C i =1. i. 3,i. λi = 0. (26). which can only be true if C3,i = 0; therefore: ∞. θ = ∑ Ci cos ( λi y ) cosh ( λi x ) i =1. (27).

(46) where the constant C4,i is referred to simply as Ci since it is the only remaining undetermined constant associated with each eigenvalue. Equation (27) is substituted into the boundary condition at x = W, Eq. (9): ∞. k ∑ Ci cos ( λi y ) λi cosh ( λi W ) = qh′′. (28). i =1. The eigenfunctions are orthogonal; therefore, Eq. (28) is multiplied by an arbitary eigenfunction and integrated from y = 0 to y = b. The result is: Ci k λi cosh ( λi W ). b. b. ∫ cos ( λ y ) dy = q′′ ∫ cos ( λ y ) dy 2. i. y =0. i. (29). y =0. The two integrals in Eq. (29) are evaluated in Maple: > restart; > int((cos(lambda*y))^2,y=0..b);. 1 cos( λ b ) sin( λ b ) + λ b 2 λ. > int(cos(lambda*y),y=0..b);. sin( λ b ) λ. Therefore, the constants can be evaluated according to: Ci = q ′′. sin ( λi b ) k λ cosh ( λi W ) Integrali 2 i. (30). where ⎡cos ( λi b ) sin ( λi b ) + λi b ⎤⎦ Integrali = ⎣ 2 λi duplicate i=1,N Integral[i]=(cos(lambdab[i])*sin(lambdab[i])+lambdab[i])/(2*lambda[i]) C[i]=q_h*sin(lambdab[i])/(k*lambda[i]^2*cosh(lambdab[i])*Integral[i]) end. The solution is obtained according to Eq. (27) at a particular position: x_bar=0.5 [-] y_bar=0.5 [-] x=x_bar*W y=y_bar*b. (31).

(47) duplicate i=1,N theta[i]=C[i]*cos(lambda[i]*y)*cosh(lambda[i]*x) end T=T_infinity+sum(theta[i],i=1,N) T_C=converttemp(K,C,T). b.) Prepare a contour plot of the temperature distribution in the substrate. Figure P2.2-4-4 illustrates a contour plot of temperature in the substrate. 1 0.9. contours of constant temperature (°C). 0.8 0.7 55.72. y/b. 0.6 49.9. 0.5. 61.54. 44.08. 0.4. 32.45 38.27. 0.3. 67.36. 0.2. 73.17. 0.1 0 0. 78.99. 0.1. 0.2. 0.3. 0.4. 0.5. 0.6. 0.7. 0.8. 0.9. 1. x/W. Figure P2.2-4-4: Contour plot of temperature.. The position of temperature sensors #1 and #2 at a particular value of x are y1 = 1.9 mm and y2 = 1.8 mm, respectively (see Figure P2.2-4(b)). The surface temperature measurement extracted from these measured temperatures is associated with a linear extrapolation to the surface at y = 0: Ts ,m = T2 + (T1 − T2 ). ( b − y2 ) ( y1 − y2 ). (32). The heat flux measurement extracted from these measured temperatures is obtained from Fourier's law according to:. qs′′,m = k. (T2 − T1 ) ( y1 − y2 ). (33). c.) What is the heat transfer coefficient measured by the device at x/W = 0.5? That is, based on the temperatures T1 and T2 predicted by your model at x/W = 0.5, calculate the measured heat transfer coefficient according to:. hm =. (T. qs′′,m. s ,m. − T∞ ). (34). and determine the discrepancy of your measurement relative to the actual heat transfer coefficient..

(48) The temperatures measured by the sensor set at x/W = 0.5 are evaluated: duplicate i=1,N theta_1[i]=C[i]*cos(lambda[i]*y_1)*cosh(lambda[i]*x) theta_2[i]=C[i]*cos(lambda[i]*y_2)*cosh(lambda[i]*x) end T_1=T_infinity+sum(theta_1[i],i=1,N) T_2=T_infinity+sum(theta_2[i],i=1,N). The surface temperature, heat flux, and heat transfer coefficient are evaluated according to Eqs. (32) through (34): T_s_m=T_2+(T_1-T_2)*(b-y_2)/(y_1-y_2) q_s_m=k*(T_2-T_1)/(y_1-y_2) h_m=q_s_m/(T_s_m-T_infinity). which leads to hm = 2358 W/m2-K. This is 5.6% in error relative to the actual heat transfer coefficient. d.) Plot the % error associated with the device configuration (i.e., the discrepancy between the measured and actual heat transfer coefficient from part (c)) as a function of axial position, x. The % error is calculated according to: deltah=100*abs(h-h_m)/h. and shown in Figure P2.2-4-5 as a function of x/W. Error in heat transfer coefficient (%). 20 18 16 14 12 10 8 6 4 2 0 0. 0.1. 0.2. 0.3. 0.4. 0.5. 0.6. 0.7. 0.8. 0.9. 1. x/W. Figure P2.2-4-5: Error as a function of axial location..

(49) Problem 2.2-5 Three heater blocks provide heat to the back-side of a fin array that must be tested, as shown in Figure P2.2-5. 4 2 q1′′ = 3x10 W/m 4 2 q2′′ = 3x10 W/m. k = 30 W/m-K c = 1 cm. y th = 1 cm. L = 7 cm. d = 4 cm. x 2 h f = 35 W/m -K T f = 35°C N fins = 150 2 As, fin = 8 cm 2 Ab, fin = 0.8 cm η fin = 0.85. 2 h = 20 W/m -K T∞ = 20°C. Figure P2.2-5: An array of fins installed on a base plate energized by three heater blocks.. The fins are installed on a base plate that has half-width of L= 7 cm, thickness of th = 1 cm, and width W = 20 cm (into the page). The base plate material has conductivity k = 30 W/m-K. The edge of the base plate (at x = L) is exposed to air at T∞ = 20°C with heat transfer coefficient h = 20 W/m2-K. The middle of the plate (at x = 0) is a line of symmetry and can be modeled as being adiabatic. The bottom of the plate (at y = 0) has an array of Nfin = 150 fins installed. Each fin has surface area As,fin= 8 cm2, base area Ab,fin = 0.8 cm2, and efficiency ηfin = 0.85. The fin and the base material are exposed to fluid at Tf = 35°C with heat transfer coefficient h f = 20 W/m2-K. The top of the plate (at y = th) is exposed to the heat flux from the heater blocks. The heat flux is distributed according to:. q ′′y =th. ⎧q1′′ if x < c ⎪ = ⎨0 if c < x < ( d + c ) ⎪q ′′ if x > ( d + c ) ⎩ 2. where q1′′ = 3x104 W/m2, q2′′ = 3x104 W/m2, c = 1 cm and d = 4 cm. a.) Determine an effective heat transfer coefficient that can be applied to the surface at y = th in order to capture the combined effect of the fins and the unfinned base area. The known information is entered in EES: "PROBLEM 2.2-5" $UnitSystem SI MASS RAD PA K J $TABSTOPS 0.2 0.4 0.6 0.8 3.5 in "Inputs" k=30 [W/m-K] h_bar=20 [W/m^2-K]. "thermal conductivity" "heat transfer coefficient to ambient air".

(50) W=20 [cm]*convert(cm,m) L=7 [cm]*convert(cm,m) th=1 [cm]*convert(cm,m) h_bar_f=35 [W/m^2-K] A_s_fin=8 [cm^2]*convert(cm^2,m^2) A_b_fin=0.8 [cm^2]*convert(cm^2,m^2) N_fin=150 [-] eta_fin=0.85 [-] T_infinity=converttemp(C,K,20[C]) T_f=converttemp(C,K,35 [C]) c=1 [cm]*convert(cm,m) d=4 [cm]*convert(cm,m) q``_1=3e4 [W/m^2] q``_2=3e4 [W/m^2]. "width of base" "half-length of base" "thickness of base" "heat transfer coefficient to fluid" "surface area of each fin" "base area of each fin" "number of fins" "fin efficiency" "ambient air temperature" "fluid temperature" "width of first heating zone" "space between heating zones" "heat flux in 1st heating zone" "heat flux in 2nd heating zone". The effective heat transfer coefficient is defined so that it provides the same rate of heat transfer as the finned surface:. W L heff = (W L − N fins Ab , fin ) h f + N fins As , fin η fin h f. (1). W*L*h_bar_eff=(W*L-N_fin*A_b_fin)*h_bar_f+eta_fin*N_fin*A_s_fin*h_bar_f "effective heat transfer coefficient". which leads to heff = 260 W/m2-K. b.) Develop a separation of variables solution for the temperature distribution within the fin base material. The problem shown in Figure P2.2-5 is governed by the transformed partial differential equation: ∂ 2θ ∂ 2θ + =0 ∂x 2 ∂y 2. (2). and has the boundary conditions: ∂θ ∂x. =0. (3). = h θ x=L. (4). ⎧q1′′ if x < c ⎪⎪ = ⎨0 if c < x < ( d + c ) ⎪ ⎪⎩q2′′ if x > ( d + c ). (5). −k. ∂θ k ∂y. y =th. ∂θ ∂y. x =0. x=L.

(51) ∂θ heff ⎡⎣θ f − θ y =0 ⎤⎦ = − k ∂y. (6) y =0. where. θ = T − T∞. (7). θ f = T f − T∞. (8). θ = θ X θY. (9). and. The separated solution is assumed:. and substituted into Eq. (2); the process leads to two ordinary differential equations:. d 2θ X + λ2 θ X = 0 dx 2. (10). d 2θ Y − λ 2 θY = 0 2 dy. (11). Notice that the sign of the constant is selected so that the solution in the homogeneous direction (x) is sines and cosines. The solution to Eq. (10) is:. θ X = C1 sin ( λ x ) + C2 cos ( λ x ). (12). The boundary condition at x = 0, Eq. (3) requires that C1 = 0 and therefore:. θ X = C2 cos ( λ x ). (13). The boundary condition at x = L, Eq. (4), leads to the eigencondition for the problem:. tan ( λi L ) =. hL k ( λi L ). (14). The eigenvalues lie in regular intervals of λL and are identified automatically in EES by defining arrays and using them as the lower limit, upper limit, and guess values for the entries in the eigenvalue array in the Variable Information window. Nterm=11 [-] "number of terms to use in the solution" "Setup guess values and lower and upper bounds for eigenvalues".

(52) duplicate i=1,Nterm lowerlimit[i]=(i-1)*pi upperlimit[i]=lowerlimit[i]+pi/2 guess[i]=lowerlimit[i]+pi/4 end "Identify eigenvalues" duplicate i=1,Nterm tan(lambdaL[i])=h_bar*L/(k*lambdaL[i]) lambda[i]=lambdaL[i]/L end. "eigencondition" "eigenvalue". The eigenfunctions are:. θ X i = C2,i cos ( λi x ). (15). The solution to Eq. (11) for each eigenvalue is:. θ Yi = C3,i sinh ( λi y ) + C4,i cosh ( λi y ). (16). The general solution for for each eigenvalue is:. θi = θ X i θ Yi = C2,i cos ( λi x ) ⎡⎣C3,i sinh ( λi y ) + C4,i cosh ( λi y ) ⎤⎦. (17). The sum of these solutions is, itself, a solution (note that the constant C2,i is absorbed into the other constants): ∞. ∞. i =1. i =1. θ = ∑ θi = ∑ cos ( λi x ) ⎡⎣C3,i sinh ( λi y ) + C4,i cosh ( λi y ) ⎤⎦. (18). Equation (18) is substituted into the boundary condition at y = 0, Eq. (6): ∞ ∞ ⎡ ⎤ heff ⎢θ f − ∑ C4,i cos ( λi x ) ⎥ = − k ∑ C3,i λi cos ( λi x ) i =1 i =1 ⎣ ⎦. (19). Equation (19) is multiplied by an eigenfunction and integrated from x = 0 to x = L: L. L. L. 0. 0. 0. heff θ f ∫ cos ( λi x ) dx − heff C4,i ∫ cos 2 ( λi x ) dx = − k C3,i λi ∫ cos 2 ( λi x ) dx. (20). or L. L. heff θ f ∫ cos ( λi x ) dx = ⎡⎣ heff C4,i − k C3,i λi ⎤⎦ ∫ cos 2 ( λi x ) dx 0 0 . Integral1. (21).

(53) The interals in Eq. (21) are evaluated in Maple: > restart; > int(cos(lambda[i]*x),x=0..L);. sin( λ i L ) λi. > int((cos(lambda[i]*x))^2,x=0..L);. 1 cos( λ i L ) sin( λ i L ) + λ i L λi 2. The integrals are put into EES, leading to a relationship between C3,i and C4,i : duplicate i=1,Nterm Int1[i]=1/2*(cos(lambda[i]*L)*sin(lambda[i]*L)+lambda[i]*L)/lambda[i] h_bar_eff*(T_f-T_infinity)*sin(lambda[i]*L)/lambda[i]=(h_bar_eff*C4[i]-k*C3[i]*lambda[i])*Int1[i] end. Equation (18) is substituted into the boundary condition at y = th, Eq. (5): ⎧q1′′ if x < c ⎪⎪ λi k ∑ cos ( λi x ) ⎣⎡C3,i cosh ( λi th ) + C4,i sinh ( λi th ) ⎦⎤ = ⎨0 if c < x < ( d + c ) i =1 ⎪ ⎪⎩q2′′ if x > ( d + c ) ∞. (22). Again, the orthogonality of the eigenfunctions are used to obtain: L. c. λi k ⎡⎣C3,i cosh ( λi th ) + C4,i sinh ( λi th ) ⎤⎦ ∫ cos 2 ( λi x ) dx = q1′′ ∫ cos 2 ( λi x ) dx + q2′′ 0. 0. The integrals are carried out in Maple: > int(cos(lambda[i]*x),x=0..c);. sin( λ i c ) λi. > int(cos(lambda[i]*x),x=(d+e)..L);. −sin( λ i ( d + e ) ) + sin( λ i L ) λi. and used to complete the computation of the constants in EES: duplicate i=1,Nterm. L. ∫ cos ( λ x ) dx (23) 2. i. d +c.

(54) Int2[i]=(-sin(lambda[i]*(d+c))+sin(lambda[i]*L))/lambda[i] k*(C3[i]*lambda[i]*cosh(lambda[i]*L)+C4[i]*lambda[i]*sinh(lambda[i]*L))*Int1[i]=& q``_1*sin(lambda[i]*c)/lambda[i]+q``_2*Int2[i] end. The solution is evaluated at an arbitary position: x_bar=0.5 [-] y_bar=0 [-] x=x_bar*L y=y_bar*th duplicate i=1,Nterm theta[i]=cos(lambda[i]*x)*(C3[i]*sinh(lambda[i]*y)+C4[i]*cosh(lambda[i]*y)) end T=sum(theta[1..Nterm])+T_infinity T_C=converttemp(K,C,T). c.) Prepare a plot showing the temperature as a function of x at various values of y. Figure 2 illlustrates the temperature as a function of x/L at various values of y/th. 83 y/th = 1. Temperature (°C). 82 y/th = 0.75 81 y/th = 0.5 80. y/th = 0.25. 79. 78 0. y/th = 0. 0.1. 0.2. 0.3. 0.4. 0.5. 0.6. 0.7. 0.8. 0.9. 1. Dimensionless axial position, x/L Figure 2: Temperature as a function of x/L for various values of y/th.. d.) The goal of the base plate is to provide an uniform heat flow to each fin. Assess the performance of the base plate by plotting the rate of heat flux transferred to the fluid as a function of x at y = 0. The heat flux to the fluid is computed according to:. q ′′f = heff (Ty =0 − T f. ). (24).

(55) q``_f=h_bar_eff*(T-T_f). Figure 3 shows the heat flux to the fluid as a function of axial location. 19000 3 W/m-K. 2. Heat flux to fluid (W/m ). 18000 17000 16000 15000 14000. 10 W/m-K. 13000 30 W/m-K 100 W/m-K. 12000 11000 0. 0.1. 0.2. 0.3. 0.4. 0.5. 0.6. 0.7. 0.8. 0.9. 1. Dimensionless axial location, x/L Figure 3: Heat flux to fluid as a function of x/L.. e.) Overlay on your plot for (d) the rate of heat flux transferred to the fluid for various values of the base plate conductivity. Figure 3 includes various values of conductivity and shows that as conductivity increases, more energy is lost to the edge and the heat flux along the bottom is more uniform..

(56) Problem 2.2-6 (2-4 in text) A laminated composite structure is shown in Figure P2.2-6. H = 3 cm. 2 q ′′ = 10000 W/m. Tset = 20°C W = 6 cm. Tset = 20°C. kx = 50 W/m-K ky = 4 W/m-K Figure P2.2-6: Composite structure exposed to a heat flux.. The structure is anisotropic. The effective conductivity of the composite in the x-direction is kx = 50 W/m-K and in the y-direction it is ky = 4 W/m-K. The top of the structure is exposed to a heat flux of q ′′ = 10,000 W/m2. The other edges are maintained at Tset = 20°C. The height of the structure is H = 3 cm and the half-width is W = 6 cm. a.) Develop a separation of variables solution for the 2-D steady-state temperature distribution in the composite. The known information is entered in EES: $UnitSystem SI MASS RAD PA K J $TABSTOPS 0.2 0.4 0.6 0.8 3.5 in "Inputs" W=6 [cm]*convert(cm,m) H=3 [cm]*convert(cm,m) q``=10000 [W/m^2] k_x=50 [W/m-K] k_y=4 [W/m-K] T_set=converttemp(C,K,20[C]). "width of laminate" "height of laminate" "heat flux" "conductivity in the x-direction" "conductivity in the y-direction" "specified edge temperatures". A half-symmetry model of the problem shown in Figure P2.2-6 is governed by the transformed partial differential equation: kx. ∂ 2θ ∂ 2θ + k =0 y ∂x 2 ∂y 2. (1). with the boundary conditions: ∂θ ∂x. =0. (2). x =0. θ x =W = 0. (3).

(57) ky. ∂θ ∂y. = q ′′. (4). y=H. θ y =0 = 0. (5). θ = T − Tset. (6). θ = θ X θY. (7). where. The separated solution is assumed:. and substituted into Eq. (1):. kx θY. d 2θ X d 2θ Y k θ X + y =0 dx 2 dx 2. (8). Dividing through by θY θX leads to:. d 2θ X d 2θ Y dx 2 + k y dx 2 = 0 k θY θX . x . −λ2. (9). λ2. The process leads to two ordinary differential equations:. d 2θ X + λ2 θ X = 0 2 dx. (10). d 2θ Y − β 2 λ 2 θY = 0 dy 2. (11). where. β2 =. kx ky. (12). beta=sqrt(k_x/k_y). Notice that the sign of the constant is selected so that the solution in the homogeneous direction (x) is sines and cosines. The solution to Eq. (10) is:.

(58) θ X = C1 sin ( λ x ) + C2 cos ( λ x ). (13). The boundary condition at x = 0, Eq. (2) requires that C1 = 0 and therefore:. θ X = C2 cos ( λ x ). (14). The boundary condition at x = L, Eq. (3), leads to the eigencondition for the problem: cos ( λi W ) = 0. (15). which requires that:. λi =. (1 + 2 i ) π 2W. for i = 0,1,...∞. N_term=11 [-] duplicate i=0,N_term lambda[i]=(1+2*i)*pi/(2*W) end. (16). "number of terms" "eigenvalues". The eigenfunctions are:. θ X i = C2,i cos ( λi x ). (17). The solution to Eq. (11) for each eigenvalue is:. θ Yi = C3,i sinh ( β λi y ) + C4,i cosh ( β λi y ). (18). The general solution for for each eigenvalue is:. θi = θ X i θ Yi = C2,i cos ( λi x ) ⎡⎣C3,i sinh ( β λi y ) + C4,i cosh ( β λi y ) ⎤⎦. (19). The sum of these solutions is, itself, a solution (note that the constant C2,i is absorbed into the other constants): ∞. ∞. i =1. i =1. θ = ∑ θi = ∑ cos ( λi x ) ⎡⎣C3,i sinh ( β λi y ) + C4,i cosh ( β λi y ) ⎤⎦. (20). The boundary condition at y = 0, Eq. (5), leads to C4,i = 0: ∞. ∞. i =1. i =1. θ = ∑ θi = ∑ Ci cos ( λi x ) sinh ( β λi y ). (21).

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