diff options
Diffstat (limited to '2510/CH19')
| -rwxr-xr-x | 2510/CH19/EX19.1/Ex19_1.sce | 23 | ||||
| -rwxr-xr-x | 2510/CH19/EX19.10/Ex19_10.sce | 45 | ||||
| -rwxr-xr-x | 2510/CH19/EX19.11/Ex19_11.sce | 29 | ||||
| -rwxr-xr-x | 2510/CH19/EX19.12/Ex19_12.sce | 37 | ||||
| -rwxr-xr-x | 2510/CH19/EX19.16/Ex19_16.sce | 17 | ||||
| -rwxr-xr-x | 2510/CH19/EX19.18/Ex19_18.sce | 17 | ||||
| -rwxr-xr-x | 2510/CH19/EX19.19/Ex19_19.sce | 40 | ||||
| -rwxr-xr-x | 2510/CH19/EX19.2/Ex19_2.sce | 11 | ||||
| -rwxr-xr-x | 2510/CH19/EX19.20/Ex19_20.sce | 19 | ||||
| -rwxr-xr-x | 2510/CH19/EX19.3/Ex19_3.sce | 11 | ||||
| -rwxr-xr-x | 2510/CH19/EX19.4/Ex19_4.sce | 12 | ||||
| -rwxr-xr-x | 2510/CH19/EX19.5/Ex19_5.sce | 30 | ||||
| -rwxr-xr-x | 2510/CH19/EX19.6/Ex19_6.sce | 17 | ||||
| -rwxr-xr-x | 2510/CH19/EX19.7/Ex19_7.sce | 27 | ||||
| -rwxr-xr-x | 2510/CH19/EX19.9/Ex19_9.sce | 31 |
15 files changed, 366 insertions, 0 deletions
diff --git a/2510/CH19/EX19.1/Ex19_1.sce b/2510/CH19/EX19.1/Ex19_1.sce new file mode 100755 index 000000000..5fddaac8c --- /dev/null +++ b/2510/CH19/EX19.1/Ex19_1.sce @@ -0,0 +1,23 @@ +//Variable declaration: +H = 2.5 //Height of wall (m) +W = 4 //Width of wall (m) +h = 11 //Convective heat transfer coefficient (W/m^2.K) +T1 = 24 //Outside surface temperature ( C) +T3 = -15 //Outside air temperature ( C) +L = 7.62/10**3 //Insulation thickness (m) +k = 0.04 //Thermal conductivity of wool (W/m.K) + +//Calculation: +A = H*W //Heat transfer area (m^2) +Q = h*A*(T1-T3) //Heat transfer rate (W) +Ri = L/(k*A) //Insuation resistance (K/W) +Rc = 1/(h*A) //Convective resitance (K/W) +R = Ri+Rc //Total resistance (K/W) +Qt = (T1-T3)/R //Revised heat transfer rate (Btu/h) + +//Result: +printf("1. The heat transfer rate without insulation is : %.0f W .",Q) +printf("Or, the heat transfer rate without insulation is : %.0f Btu/h .",Q*3.412) +printf("2. The revised heat transfer rate with insulation is : %.0f W .",Qt) +printf("Or, the revised heat transfer rate with insulation is : %.0f Btu/h .",Qt*3.412) +printf("There is a calculation mistake in book.") diff --git a/2510/CH19/EX19.10/Ex19_10.sce b/2510/CH19/EX19.10/Ex19_10.sce new file mode 100755 index 000000000..fe576f6e6 --- /dev/null +++ b/2510/CH19/EX19.10/Ex19_10.sce @@ -0,0 +1,45 @@ +//Variable declaration: +r1 = 1.1/100 //Inside radius of %pipe (m) +r2 = 1.3/100 //Outside radius of %pipe (m) +r3 = 3.8/100 //Outside radius of asbestos insulation (m) +L = 1 //Length of tube (m) +h1 = 190 //Heat transfer coefficient from ethylene glycol to the stainless steel %pipe (W/m^2.K) +k2 = 19 //Thermal conductivity of %pipe (W/m.K) +h2 = 14 //Outside heat transfer coefficient from the air to the surface of the insulation (W/m^2.K) +k3 = 0.2 //Thermal conductivity of asbestos (W/m.K) +T1 = 124 //Hot ethylene glycol temperature ( C) +T5 = 2 //Surrounding air temperature ( C) +k4 = 0.0242 //Thermal conductivity of air (W/m.K) + +//Calculation: +A1 = 2*%pi*r1*L //Inside surface area of %pipe (m^2) (part1) +A2 = 2*%pi*r2*L //Outside surface area of %pipe (m^2) +A3 = 2*%pi*r3*L //Outside surface area of asbestos insulation (m^2) +R1 = 1/(h1*A1) //Inside convection resistance ( C/W) +R2 = log(r2/r1)/(2*%pi*k2*L) //Conduction resistance through the tube ( C/W) +R3 = 1/(h2*A2) //Outside convection resistance ( C/W) +Rt1 = R1+R2+R3 //Total resistance without insulation ( C/W) +Q1 = (T1 - T5)/Rt1 //Heat transfer rate without insulation (W) +R4 = log(r3/r2)/(2*%pi*k3*L) //Conduction resistance associated with the insulation ( C/W) (part 2) +R5 = 1/(h2*A3) //Outside convection resistance ( C/W) +Rt2 = R1+R2+R4+R5 //Total rsistance with the insulation ( C/W) +Q2 = (T1-T5)/Rt2 //Heat transfer rate with the insulation (W) +U1 = 1/(Rt2*A1) //Overall heat transfer coefficient based on the inside area (W/m^2.K) (part 3) +U3 = 1/(Rt2*A3) //Overall heat transfer coefficient based on the outside area (W/m^2.K) (part 4) +T3 = T1-(R1+R2)*Q2 //Temperature at the steelu2013insulation interface ( C) (part 5) +Bi1 = h2*(2*r3)/k3 //Outside Biot number (part 6) +Bi2 = h1*(2*r1)/k2 //Inside Biot number +Nu = h1*(2*r1)/k4 //Nusselt number of the air +rlm = (r3-r2)/log(r3/r2) //Log mean radius of the insulation (m) (part 7) + +//Result: +printf("1. The rate of heat transfer without insulation is : %.1f W.",Q1) +printf("2. The rate of heat transfer with insulation is : %.1f W.",Q2) +printf("3. The overall heat transfer coefficient based on the inside area of the tube is : %.2f W/m^2.K .",U1) +printf("4. The overall heat transfer coefficient based on the outside area of the insulation is : %.1f W/m^2.K .",U3) +printf("5. The temperature, T3, at the steelu2013insulation interface is : %.1f C.",T3) +printf("6. The inside Biot numbers is : %.2f",Bi2) +printf(" The outside Biot numbers is : %.2f",Bi1) +printf(" The Nusselt number is : %.1f",Nu) +printf("7. The log mean radius of insulation is : %.2f cm.",rlm*100) +printf("There is a printing mistake in book for unit in part 7.") diff --git a/2510/CH19/EX19.11/Ex19_11.sce b/2510/CH19/EX19.11/Ex19_11.sce new file mode 100755 index 000000000..2dc08e48c --- /dev/null +++ b/2510/CH19/EX19.11/Ex19_11.sce @@ -0,0 +1,29 @@ +//Variable declaration: +h1 = 800 //Heat transfer coefficient for steam condensing inside coil (Btu/h.ft^2. F) +h2 = 40 //Heat transfer coefficient for oil outside coil (Btu/h.ft^2. F) +h3 = 40 //Heat transfer coefficient for oil inside tank wal (Btu/h.ft^2. F) +h4 = 2 //Heat transfer coefficient for outer tank wall to ambient air (Btu/h.ft^2. F) +k1 = 0.039 //Thermal conductivity of insulation layer (Btu/h.ft. F) +l1 = 2/12 //Thickness of insulation layer (ft) +D = 10 //Diameter of tank (ft) +H = 30 //Height of tank (ft) +k2 = 224 //Thermal conductivity of copper tube (Btu/h.ft. F) +l2 = (3/4)/12 //Thickness of insulation layer (ft) +T1 = 120 //Temperature of tank ( F) +T2 = 5 //Outdoor temperature ( F) + +//Calculation: +Uo1 = 1/(1/h3+(l1/k1)+1/h4) //Overall heat transfer coefficient for tank (Btu/h.ft^2. F) +At = %pi*(D+2*l1)*H //Surface area of tank (ft^2) +Q = Uo1*At*(T1-T2) //Heat transfer rate lost from the tank (Btu/h) +//From table 6.3: +l2 = 0.049/12 //Thickness of coil (ft) +A = 0.1963 //Area of 18 guage, 3/4-inch copper tube (ft^2/ft) +Uo2 = 1/(1/h2+(l2/k2)+1/h1) //Overall heat transfer coefficient for coil (Btu/h.ft^2. F) +//From steam tables: +Tst = 240 //Temperature for 10 psia (24.7 psia) steam ( F) +Ac = Q/(Uo2*(Tst-T1)) //Area of tube (ft^2) +L = Ac/A //Lengt of tube (ft) + +//Result: +printf("The length ofcopper tubing required is : %.1f ft",L) diff --git a/2510/CH19/EX19.12/Ex19_12.sce b/2510/CH19/EX19.12/Ex19_12.sce new file mode 100755 index 000000000..85aef843f --- /dev/null +++ b/2510/CH19/EX19.12/Ex19_12.sce @@ -0,0 +1,37 @@ +//Variable declaration: +//For 1-inch %pipe schedule 40: +Di = 1.049/12 //Inside diameter (ft) +Do = 1.315/12 //Outside diameter (ft) +L = 8000 //Length of %pipe (ft) +hi = 2000 //Heat transfer coefficient inside of the %pipe (Btu/h.ft^2. F) +ho = 100 //Outside heat transfer coefficient (Btu/h.ft. F) +kl = 0.01 //Thermal conductivity of insulation (Btu/h.ft. F) +T1 = 240 //Steam temperature ( F) +T2 = 20 //Air temperature ( F) +k = 24.8 //Thermal conductivity for steel (Btu/h.ft. F) +Dxl = ([3/8,1/2,3/4,1])/12 //thickness(ft) +amt = ([1.51,3.54,5.54,8.36])/6 //Cost per feet($) + +//Calculation: +D_ = (Do-Di)/log(Do/Di) //log-mean diameter of the %pipe (ft) +Dl = Do+2*(Dxl) //Insulation thickness (ft) +D_l = [ 0.13849079 0.14734319 0.16423045 0.18025404] +//D_l = (Dl-Do)/log(Dl/Do) //log mean diameter of %pipe (ft) +Dxw = (Do-Di)/2 //%pipe thickness (ft) +Rw = Dxw/(k*%pi*D_*L) //Wall resistance ((Btu/h. F)^-1) +Ri = 1/(hi*%pi*Di*L) //Inside steam convection resistance ((Btu/h. F)^-1) +Rl = [ 0.00089782 0.00112517 0.00151421 0.00183947] //Dxl/(kl*%pi*D_l*L) //Insulation resistance ((Btu/h. F)^-1) +Ro = [ 2.31217835e-06 2.06248306e-06 1.69614504e-06 1.44031623e-06] //1/(ho*%pi*Dl*L) //Outside air convection resistance ((Btu/h. F)^-1) +R = [ 0.00090054, 0.00112764,0.00151632,0.00184132] //Total resistance ((Btu/h. F)^-1) +Uo = [ 0.25675435 0.18290211 0.11185958 0.07822176] //Overall outside heat transfer coefficient (Btu/h.ft^2. F) +Ui = [ 0.50543158 0.40364002 0.30017609 0.24719271] //Overall inside heat transfer coefficient (Btu/h.ft^2. F) +dT = T1-T2 +Ai = %pi*Di*L //Inside area (ft^2) +Q = Ui*Ai*dT //Energy loss (Btu/h) +function [a] =energyPerDollar(Q1,Q2,amt1,amt2) + a = ((Q1-Q2)/(8000*(amt2-amt1))) +endfunction +//Results: +printf("Energy saved per dollar ingoing from 3/8 to 1/2 inch is : %.1f Btu/h.$",energyPerDollar(Q(1),Q(2),amt(1),amt(2))) +printf("Energy saved per dollar ingoing from 1/2 to 3/4 inch is : %.1f Btu/h.$",energyPerDollar(Q(2),Q(3),amt(2),amt(3))) +printf("Energy saved per dollar ingoing from 3/4 to 1 inch is : %.1f Btu/h.$",energyPerDollar(Q(3),Q(4),amt(3),amt(4))) diff --git a/2510/CH19/EX19.16/Ex19_16.sce b/2510/CH19/EX19.16/Ex19_16.sce new file mode 100755 index 000000000..a4709155e --- /dev/null +++ b/2510/CH19/EX19.16/Ex19_16.sce @@ -0,0 +1,17 @@ +//Variable declaration: +ki = 0.44 //Thermal conductivity of insulation (Btu/h.ft. F) +ho = 1.32 //Air flow coefficient (Btu/h.ft^2. F) +OD = 2 //Outside diameter of pipe (in) + +//Calculation: +rc = (ki/ho)*12 //Outer critical radius of insulation (in) +ro = OD/2 //Outside radius of pipe (in) +L = rc-ro //Critical insulation thickness (in) + +//Result: +printf("The outer critical radius of insulation is : %.0f in .",rc) +if ro<rc then + printf("Since, ro<rc, the heat loss will increase as insulation is added.") +else + printf("Sice, ro>rc, the heat loss will decrease as insulation is added.") +end diff --git a/2510/CH19/EX19.18/Ex19_18.sce b/2510/CH19/EX19.18/Ex19_18.sce new file mode 100755 index 000000000..74fe78441 --- /dev/null +++ b/2510/CH19/EX19.18/Ex19_18.sce @@ -0,0 +1,17 @@ +//Variable declaration: +Lf = 6/12 //Length of firebrick (ft) +kf = 0.61 //Thermal conductivity of firebrick (Btu/h.ft. F) +A = 480 //Surface area of wall (ft^2) +Lw = 8/12 //Length of rock wool (ft) +kw = 0.023 //Thermal conductivity of rock wool (Btu/h.ft. F) +T1 = 1900 //Temperature of insulation of firebrick ( F) +T2 = 140 //Temperature of insulation of rock wool ( F) + +//Calculation: +Rf = Lf/(kf*A) //Resistance of firebrick (h. F/Btu) +Rw = Lw/(kw*A) //Resistance of rock wool (h. F/Btu) +R = Rf+Rw //Total resitance (h. F/Btu) +Q = (T1-T2)/R //Heat loss through the wall (Btu/h) + +//Result: +printf("The heat loss through the wall is : %.0f Btu/h .",Q) diff --git a/2510/CH19/EX19.19/Ex19_19.sce b/2510/CH19/EX19.19/Ex19_19.sce new file mode 100755 index 000000000..ccfca3b4a --- /dev/null +++ b/2510/CH19/EX19.19/Ex19_19.sce @@ -0,0 +1,40 @@ + +//Variable declaration: +h1 = 1700 //Steam heat-transfer coefficient (Btu/h.ft^2. F) +h2 = 2 //Air heat-transfer coefficient (Btu/h.ft^2. F) +A = 1 //Area of base (ft^2) (assumption) +k1 = 26 //Thermal conductivity of steel (Btu/h.ft. F) +k2 = 218 //Thermal conductivity of copper (Btu/h.ft. F) +t = 0.375 //Thickness of steel sheet (in) +h3 = 2500 //Increased steam heat-transfer coefficient (Btu/h.ft^2. F) +h4 = 12 //Increased air heat-transfer coefficient (Btu/h.ft^2. F) + +//Calculation: +R1 = 1/(h1*A) //Steam resistance (h. F/Btu) +R2 = 1/(h2*A) //Air resistance (h. F/Btu) +R3 = (t/12)/(k1*A) //Steel resistance (h. F/Btu) +Rt1 = R1+R2+R3 //Total resistance (with steel) (h. F/Btu) +R4 = (t/12)/(k2*A) //Copper resistance (h. F/Btu) (part 1) +Rt2 = R1+R2+R4 //Total resistance (with copper) (h. F/Btu) +R5 = 1/(h1*A) //New steam resistance (h. F/Btu) +Rt3 = R5+R2+R3 //Total resistance after increasing the steam coefficient (h. F/Btu) +R6 = 1/(h4*A) //Air resistance (h. F/Btu) +Rt4 = R1+R6+R3 //Total resistance after increasing the air coefficient (h. F/Btu) + +//Result: +if (Rt1==Rt2) then + printf("1.The rate of heat transfer is essentially unaffected.") +else + printf("1. The rate of heat transfer is essentially affected.") +end + +if (Rt1==Rt3) then + printf("2. The rate is again unaffected.") +else + printf("2. The rate is again affected.") +end +if (Rt1==Rt4) then + printf("3. The rate is unaffected for this case.") +else + printf("3. The rate is affected for this case.") +end diff --git a/2510/CH19/EX19.2/Ex19_2.sce b/2510/CH19/EX19.2/Ex19_2.sce new file mode 100755 index 000000000..768e8ce71 --- /dev/null +++ b/2510/CH19/EX19.2/Ex19_2.sce @@ -0,0 +1,11 @@ +//Variable declaration: +//From example 19.1: +T1 = 24 //Outside surface temperature ( C) +Ri = 0.0191 //Insulation resistance (K/W) +Q = 1383 //Revised heat transfer rate (Btu/h) + +//Calculation: +T2 = T1-Q*Ri //Temperature at outer surface of insulation ( C) + +//Result: +printf("The temperature at the outer surface of the insulation is : %.1f C .",T2) diff --git a/2510/CH19/EX19.20/Ex19_20.sce b/2510/CH19/EX19.20/Ex19_20.sce new file mode 100755 index 000000000..84af3eef1 --- /dev/null +++ b/2510/CH19/EX19.20/Ex19_20.sce @@ -0,0 +1,19 @@ +//Variable declaration: +rfo = 12/2 //Outside radius of firebrick (ft) +rfi = 5.167 //Inside radius of firebrick (ft) +rso = 6.479 //Outside radius of sil-o-cel (ft) +rsi = 6.063 //Inside radius of fsil-o-cel (ft) +L = 30 //Length of incinerator (ft) +kf = 0.608 //Thermal conductivity of firebrick (Btu/h.ft. F) +ks = 0.035 //Thermal conductivity of sil-o-cel (Btu/h.ft. F) + +//Calculation: +Rf= log(rfo/rfi)/(2*%pi*L*kf) //Resistance of firebrick (h.ft. F/Btu) +Rs= log(rso/rsi)/(2*%pi*L*ks) //Resistance of sil-o-cel (h.ft. F/Btu) +R = Rf+Rs //Total resistance (h.ft. F/Btu) +ro = exp(R*(2*%pi*L*ks))*rso //New outside radius of sil-o-cel (ft) +r= ro-rso //Extra thickness (ft) + +//Result: +printf("The extra thickness is : %.3f ft",r) +printf("Or, the extra thickness is : %.2f in.",r*12) diff --git a/2510/CH19/EX19.3/Ex19_3.sce b/2510/CH19/EX19.3/Ex19_3.sce new file mode 100755 index 000000000..c9b71d9d2 --- /dev/null +++ b/2510/CH19/EX19.3/Ex19_3.sce @@ -0,0 +1,11 @@ +//Variable declaration: +//From example 19.1: +h = 11 //Convective heat transfer coefficient (W/m^2.K) +L = 7.62/10**3 //Insulation thickness (m) +k = 0.04 //Thermal conductivity of wool (W/m.K) + +//Calculation: +Bi = h*L/k //Biot number + +//Result: +printf("The Biot nmuber is : %.1f ",Bi) diff --git a/2510/CH19/EX19.4/Ex19_4.sce b/2510/CH19/EX19.4/Ex19_4.sce new file mode 100755 index 000000000..0fd234072 --- /dev/null +++ b/2510/CH19/EX19.4/Ex19_4.sce @@ -0,0 +1,12 @@ +//Variable declaration: +k = 0.022 //Thermal conductivity of glass wool (Btu/h.ft. F) +T1 = 400 //Inside wall temperature ( F) +T2 = 25 //Outside wall temperature ( C) +L = 3/12 //Length of insulation cover (ft) + +//Calculation: +T_2 = T2*(9/5)+32 //Outside wall temperature in fahrenheit scale ( F) +QbyA = k*(T1-T_2)/L //Heat flux across the wall (Btu/h.ft^2) + +//Result: +printf("The heat flux across the wall is : %.1f Btu/h.ft^2 .",QbyA) diff --git a/2510/CH19/EX19.5/Ex19_5.sce b/2510/CH19/EX19.5/Ex19_5.sce new file mode 100755 index 000000000..01aa172ff --- /dev/null +++ b/2510/CH19/EX19.5/Ex19_5.sce @@ -0,0 +1,30 @@ +//Variable declaration: +w = 8 //Width of wall (m) +H = 3 //Height of wall (m) +h = 21 //Convective heat transfer coefficient between the air and the surface (W/m^2.K) +T1 = -18 //Outside surace of wall temperature ( C) +T3 = 26 //Surrounding air temperature ( C) +l1 = 80/100 //Reduction in cooling load +k = 0.0433 //Thermal conductivity of cork board insulation (W/m.K) +T = 12000 //Units Btu/h in 1 ton of refrigeration + +//Calculation: +A = w*H //Heat transfer area (m^2) (part 1) +Q1 = h*A*(T1-T3) //Rate of heat flow in the absence of insulation (W) +Q2 = Q1*3.4123/T //Rate of heat flow in the absence of insulation (ton of refrigeration) +l2 = 1-l1 //Reduced cooling load (part 2) +Q3 = l2*Q1 //Heat rate with insulation (W) +Rt = (T1-T3)/Q3 //Total thermal resistance ( C/W) +R2 = 1/(h*A) //Convection thermal resistance ( C/W) +R1 = Rt-R2 //Insulation conduction resistance ( C/W) +L = R1*k*A //Required insulation thickness (m) + +//Result: +printf("1. The rate of heat flow through the rectangular wall without insulation is : %.2f kW .",Q1/10**3) +printf("Or, the rate of heat flow through the rectangular wall without insulation in tons of refrigeration is : %.1f ton of refrigeration .",Q2) +if (Q1<0) then + printf("The negative sign indicates heat flow from the surrounding air into the cold room.") +else + printf(" The positive sign indicates heat flow from the surrounding air into the cold room.") +end +printf("2. The required thickness of the insulation board is : %.2f mm .",L*10**3) diff --git a/2510/CH19/EX19.6/Ex19_6.sce b/2510/CH19/EX19.6/Ex19_6.sce new file mode 100755 index 000000000..816774977 --- /dev/null +++ b/2510/CH19/EX19.6/Ex19_6.sce @@ -0,0 +1,17 @@ +//Variable declaration: +//From example 19.5: +Q = -4435.2 //Heat rate with insulation (W) +R2 = 0.00198 //Convection thermal resistance ( C/W) +T3 = 26 //Surrounding air temperature ( C) +h = 21 //Convective heat transfer coefficient between the air and the surface (W/m^2.K) +k = 0.0433 //Thermal conductivity of cork board insulation (W/m.K) +L = 0.00825 //Required insulation thickness (m) + +//Calculation: +T2 = T3+Q*R2 //Interface temperature ( C) (part 1) +Bi = h*L/k //Biot number (part 2) + +//Result: +printf("1. The interface temperature is : %.2f C .",T2) +printf("2. The Biot number is : %.0f ",Bi) +printf("3. Theoretical part.") diff --git a/2510/CH19/EX19.7/Ex19_7.sce b/2510/CH19/EX19.7/Ex19_7.sce new file mode 100755 index 000000000..4cc7b8abd --- /dev/null +++ b/2510/CH19/EX19.7/Ex19_7.sce @@ -0,0 +1,27 @@ +//Variable declaration: +D2 = 0.5/10**3 //External diameter of needle (m) +h3 = 12 //Heat transfer coefficient (W/m^2.K) +L = 1 //Insulation thickness (m) +T1 = 95 //Reactant temperature ( C) +T3 = 20 //Ambient air temperature ( C) +k1 = 16 //Thermal conductivity of needle (W/m.K) +k3 = 0.0242 //Thermal conductivity of air (W/m.K) +D3 = 2/10**3 //Diameter of rubber tube (m) + +//Calculation: +r2 = D2/2 //External radius of needle (m) +r3 = D3/2 //Radius of rubber tube (m) +Rt1 = 1/(h3*(2*%pi*r2*L)) //Thermal resistance ( C/W) +Q1 = (T1-T3)/Rt1 //Rate of heat flow in the absence of insulation (W) +Bi = h3*D2/k1 //Biot number +Nu = h3*D2/k3 //Nusselt number +R2 = log(r3/r2) //Thermal resistance of needle ( C/W) +R3 = 1/(h3*(2*%pi*r3*L)) //Thermal resistance of rubber tube ( C/W) +Rt2 = R2+R3 //Total thermal resistance ( C/W) +Q2 = (T1-T3)/Rt2 //Rate of heat loss (W) + +//Result: +printf("1. The rate of the heat loss from the hypodermic needle with the rubber insulation is : %.2f W .",Q1) +printf(" The rate of the heat loss from the hypodermic needle without the rubber insulation is : %.2f W .",Q2) +printf("2. The Biot number is : %f",Bi) +printf(" The nusselt number is : %.3f ",Nu) diff --git a/2510/CH19/EX19.9/Ex19_9.sce b/2510/CH19/EX19.9/Ex19_9.sce new file mode 100755 index 000000000..f5d3d31bb --- /dev/null +++ b/2510/CH19/EX19.9/Ex19_9.sce @@ -0,0 +1,31 @@ +//Variable declaration: +h = 140 //Convention heat transfer coefficient (W/m^2.K) +D1 = 10/10**3 //Rod diameter (m) +L = 2.5 //Rod length (m) +T1 = 200 //Surface temperature of rod ( C) +T2 = 25 //Fluid temperature ( C) +k = 1.4 //Thermal conductivity of bakellite (W/m.K) +l = 55/10**3 //Insulation thickness (m) + +//Calculation: +Q1 = h*%pi*D1*L*(T1-T2) //Rate of heat transfer for the bare rod (W) (part 1) +Bi = 2 //Critical Biot number (part 2) +D2 = Bi*k/h //Critical diameter associated with the bakelite coating (m) +r2 = D2/2 //Critical radius associated with the bakelite coating (m) +r1 = D1/2 //Rod radius (m) +R1 = log(r2/r1)/(2*%pi*k*L) //Insulation conduction resistance ( C/W) +R2 = 1/(h*(2*%pi*r2*L)) //Convection thermal resistance ( C/W) +Rt1 = R1+R2 //Total thermal resistance ( C/W) +Qc = (T1-T2)/Rt1 //Heat transfer rate at the critical radius (W) +r3 = r1+l //New radius associated with the bakelite coating after insulation (m) (part 3) +R3 = log(r3/r1)/(2*%pi*k*L) //Insulation conduction bakelite resistance ( C/W) +R4 = 1/(h*(2*%pi*r3*L)) //Convection bakelite thermal resistance ( C/W) +Rt2 = R3+R4 //Total bakelite thermal resistance ( C/W) +Q2 = (T1-T2)/Rt2 //Heat transfer rate at the bakelite critical radius (W) +Re = ((Q1-Q2)/Q1)*100 //Percent reduction in heat transfer rate relative to the case of a bare rod (%) + +//Result: +printf("1. The rate of heat transfer for the bare rod is : %0.f W .",Q1) +printf("2. The critical radius associated with the bakelite coating is : %.0f mm.",r2*10**3) +printf(" & the heat transfer rate at the critical radius is : %.0f W .",Qc) +printf("3. The fractional reduction in heat transfer rate relative to the case of a bare rod is : %.1f ",Re) |