diff options
Diffstat (limited to '587')
143 files changed, 3483 insertions, 0 deletions
diff --git a/587/CH1/EX1.1/example1_1.sce b/587/CH1/EX1.1/example1_1.sce new file mode 100755 index 000000000..16991bac9 --- /dev/null +++ b/587/CH1/EX1.1/example1_1.sce @@ -0,0 +1,32 @@ +clear;
+clc;
+
+//Example 1.1(Heating of a copper ball)
+
+//(a)
+//density of the copper ball
+rho= 8950;//[kg/m^3]
+//Diameter of the copper ball
+D=0.1;//[m]
+//mass of the ball
+m=rho*(%pi/6)*(D^3);//[kg]
+//Specific Heat of copper
+Cp=0.395;//[kJ/Kg/m^3]
+//Initial Temperature
+T1=100;//[degree C]
+//Final Temperature
+T2=150;//[degree C]
+// The amount of heat transferred to the copper ball is simply the change in it's internal energy and is given by
+// Energy transfer to the system=Energy increase of the system
+Q=(m*Cp*(T2-T1));
+disp("kJ",Q,"Heat needs to be transferred to the copper ball to heat it from 100 to 150 degree celsius is ")
+//b
+//Time interval for which the ball is heated
+dT=1800;//[seconds]
+Qavg=(Q/dT)*1000;//[W]
+disp("W",Qavg,"Average Heat Transfer by the iron ball is ")
+
+//(c)
+//Heat Flux
+qavg=(Qavg/(%pi*(D^2)));//[W/m^2]
+disp("W/m^2",qavg,"Average flux is")
diff --git a/587/CH1/EX1.10/example1_10.sce b/587/CH1/EX1.10/example1_10.sce new file mode 100755 index 000000000..1341a6ed4 --- /dev/null +++ b/587/CH1/EX1.10/example1_10.sce @@ -0,0 +1,17 @@ +clear;
+clc;
+
+//Example1.10[Heat Loss from a Person]
+//Given:-
+T_room=20+273;//Temperature of breezy room[K]
+T_outr=29+273;//Average outer surface temperature of the person[K]
+As=1.6;//Exposed Surface Area[m^2]
+h=6;//Convection Heat transfer coefficient[W/m^2.K]
+e=0.95;//Emissivity of person
+sigma=5.67*(10^(-8));//Stephan's constant[W/m^2.degree Celcius]
+Q_conv=h*As*(T_outr-T_room);//[W]
+disp("W",Q_conv,"Rate of convection heat transfer from the person to the air in the room is")
+Q_rad=e*sigma*As*((T_outr^4)-(T_room^4));//[W]
+disp("W",Q_rad,"The rate of convection heat transfer from the person to the surrounding walls,cieling,fllor is")
+Q_total=Q_conv+Q_rad;//[W]
+disp("W",round(Q_total),"The rate of total heat transfer from the body is ")
\ No newline at end of file diff --git a/587/CH1/EX1.11/example1_11.sce b/587/CH1/EX1.11/example1_11.sce new file mode 100755 index 000000000..4e1a2ca97 --- /dev/null +++ b/587/CH1/EX1.11/example1_11.sce @@ -0,0 +1,36 @@ +clear;
+clc;
+
+//Example1.11[Heat transfer between two Isothermal Plates]
+//Given:-
+T1=300,T2=200;//Temperatures of two large parallel isothermal plates[K]
+L=0.01;//distance between both plates[m]
+e=1;//Emissivity of plates
+A=1;//Surface area of plates[m^2]
+T_avg=(T1+T2)/2;//Average temperature[K]
+sigma=5.67*(10^(-8));//Stefan's constant[W/m^2.K^4]
+//Solution (a)[space between plates is filled with air]
+k_air=0.0219;//The thermal conductivity of aair at the average temperature[W/m.K]
+Q_cond=k_air*A*(T1-T2)/L;//[W]
+Q_rad=e*sigma*A*((T1^4)-(T2^4));//[W]
+disp("W",round(Q_rad),"and",Q_cond,"The rates of conduction and radiation heat transfer between the plates through the air layer are respectively")
+Q_total_a=Q_cond+Q_rad;//[W]
+disp("W",round(Q_total_a),"Net rate of heat transfer is")
+disp("The heat transfer rate in reality will be higher because of the natural convection currents that are likely to occur in the air space between the plates")
+//Solution (b)[space between the plates is evacutaed]
+disp("when the air space b/w the plates is evacuted there is no conduction or convection,and the only heat transfer between the plates will be by radiation. ")
+disp("Therefore")
+Q_total_b=Q_rad;//[W]
+disp("W",round(Q_total_b),"Net rate of heat transfer is")
+//Solution (c)[space between the plates is filled with urethane insulation]
+k_insu=0.026;//At average temperature thermal conductivity of urethane insulation [W/m.K]
+disp("An opaque solid material placed b/w the two plates blocks direct radiation heat transfer between the plates")
+Q_cond_c=k_insu*A*(T1-T2)/L;//[W]
+Q_total_c=Q_cond_c;//[W]
+disp("W",round(Q_total_c),"The net rate of heat transfer through the urethane insulation is")
+//Solution (d)[the distance between the plates is filled with superinsulation]
+k_super=0.00002;//At average temperature thermal conductvity of superinsulation[W/m.K]
+disp("The layers of superinsulation prevent any direct radiation heat transfer between the plates")
+Q_cond_d=k_super*A*(T1-T2)/L;//[W]
+Q_total_d=Q_cond_d;//[W]
+disp("W",Q_total_d,"The net rate of heat transfer through the layer of superinsulation is")
diff --git a/587/CH1/EX1.13/example1_13.sce b/587/CH1/EX1.13/example1_13.sce new file mode 100755 index 000000000..88d6c4fd6 --- /dev/null +++ b/587/CH1/EX1.13/example1_13.sce @@ -0,0 +1,13 @@ +clear;
+clc;
+
+//Example1.13[Heating of a Plate by Solar Energy]
+//Given:-
+a=0.6;//absorptivity of exposed surface of plate
+q_incident=700;//Rate at which solar radiation incident on the plate [W/m^2]
+T_surr=25+273;//Surrounding air temperature[K]
+h=50;//Combined radiation and convection heat transfer coefficient[W/m^2.K]
+//Solution
+//Temperature keeps on increasing till a point comes at which the rate of heat loss from the plate equals the rate of solar energy absorbed, and the temperature of the plate no longer changes
+T_surface=T_surr+(a*(q_incident)/h);//[K]
+disp("degree Celcius",T_surface-273,"The plate surface temperature is")
\ No newline at end of file diff --git a/587/CH1/EX1.14/example1_14.sce b/587/CH1/EX1.14/example1_14.sce new file mode 100755 index 000000000..80992ae1b --- /dev/null +++ b/587/CH1/EX1.14/example1_14.sce @@ -0,0 +1,10 @@ +clear;
+clc;
+
+//Example1.14[Non-linear equation in two variable]
+//x1=x, x2=y
+function[f]=f2(x)
+f(1)=x(1)-x(2)-4;
+f(2)=x(1)^2+x(2)^2-x(1)-x(2)-20;
+deff('[f]=f2(x)',['f_1=x(1)-x(2)-4','f_2=x(1)^2+x(2)^2-x(1)-x(2)-20'])
+//To get the desired output assign an initial value such as x0=[1,1], [xs,fxs,m]=fsolve(x0',f2)
diff --git a/587/CH1/EX1.2/example1_2.sce b/587/CH1/EX1.2/example1_2.sce new file mode 100755 index 000000000..5bf339169 --- /dev/null +++ b/587/CH1/EX1.2/example1_2.sce @@ -0,0 +1,19 @@ +clear;
+clc;
+
+//Example 1.2(Heating of water in an Electric Teapot)
+//Mass of liquid water
+m1=1.2,m2=0.5;//[Kg]
+//Initial Temperature
+t1=15;//[Degree Celcius]
+//Final Temperature
+t2=95;//[Degree Celcius]
+//Specific heat of water
+cp1=4.186;//[kJ/kG.K]
+//Specific heat capacity of teapot
+cp2=.7;//[]
+Em=(m1*cp1*(t2-t1))+(m2*cp2*(t2-t1));//[kJ]
+//Rating of Electric Heating Equipment
+Em1=1.2;//[kJ/s]
+dt=(Em/Em1)/60;//[seconds]
+disp("minutes",round(dt),"of heat is","kJ",Em,"Time needed for this heater to supply ")
\ No newline at end of file diff --git a/587/CH1/EX1.3/example1_3.sce b/587/CH1/EX1.3/example1_3.sce new file mode 100755 index 000000000..6076aa96f --- /dev/null +++ b/587/CH1/EX1.3/example1_3.sce @@ -0,0 +1,24 @@ +clear; +clc; + +//Example1.3[Heat Loss from Heating Ducts in a Basement] +//Given:- +T_in=60+273;//Temperature of hot air while entering the duct[K] +T_out=54+273;//Temperature of hot air while leaving the duct[K] +T_avg=(T_in+T_out)/2;//Average temperature of air[K] +Cp=1.007;//[kJ/kg] +disp("kJ/kg",Cp,"K is",T_avg,"The constant pressure specific heat of air at the average temperature of") +P=100;//Pressure of air while entering the duct[kPa] +R=0.287;//Universal Gas Constant[kPa.(m^3/kg).K] +v=5;//Average velocity of flowing air[m/s] +neta=0.8;//Efficiency of natural gas furnace +ucost=1.60;//Cost of natural gas in that area[$/therm],where 1therm=105,500kJ +//Solution;- +rho=P/(R*T_in);//The density of air at the inlet conditions is[kg/m^3] +Ac=0.20*0.25;//Cross sectional area of the duct[m^2] +m_=rho*v*Ac;//[kg/s] +disp("kg/s",m_,"mass flow rate of air through the duct is") +Q_loss=m_*Cp*(T_in-T_out);//[kJ/s] +disp("kJ/s",Q_loss,"The rate of heat loss by the air is") +cost=(Q_loss*3600)*(ucost)*(1/105500)*(1/neta);//[$/h] +disp("per hour",cost,"$"," Cost of heat loss to the home owner is")
\ No newline at end of file diff --git a/587/CH1/EX1.4/example1_4.sce b/587/CH1/EX1.4/example1_4.sce new file mode 100755 index 000000000..1d7eb09a0 --- /dev/null +++ b/587/CH1/EX1.4/example1_4.sce @@ -0,0 +1,30 @@ +clear;
+clc;
+
+//Example1.4 (Electric Heating of a House at High Elevation)
+
+//(a)
+t1=10+273;//Initial temperature of house[K]
+t2=20+273;//Temperature after turning on heater[K]
+tavg=(t1+t2)/2;//Average temperature[K]
+cp=1.007;//[kJ/kg.K]
+cv=.720;//[kJ/kg.K]
+disp("kJ/kg.K",cp,"and",cv,"K",tavg,"at the average temperature of","The specific heat capacities of air")
+A=200;//The floor area[m^2]
+h=3;//Height of room[m]
+V=A*h;//Volume of the air in the house[m^3]
+P=84.6;//Pressure [kPa]
+R=0.287;//Universal gas constant[kPa.m^3/kg.K]
+m=(P*V)/(R*t1);//[kg]
+disp("kg",m,"Mass of air in the room is")
+Eincv=m*cv*(t2-t1);
+disp("kJ",Eincv,"The amount of energy transferred to air at constant volume is ")
+u_cost=0.075;//Unit cost of energy[$/kWh]
+Cost1=(Eincv*u_cost)/(3600);//[$]
+disp(Cost1,"Cost of Energy is $")
+
+//(b)
+Eincp=m*cp*(t2-t1);//[kJ]
+disp("kJ",Eincp,"The amount of energy transferred to air at constant is ")
+Cost2=(Eincp*u_cost)/3600;//[$]
+disp(Cost2,"Cost of Energy is $")
\ No newline at end of file diff --git a/587/CH1/EX1.5/example1_5.sce b/587/CH1/EX1.5/example1_5.sce new file mode 100755 index 000000000..23b1602a8 --- /dev/null +++ b/587/CH1/EX1.5/example1_5.sce @@ -0,0 +1,20 @@ +clear;
+clc;
+
+//Example1.5 (The cost of Heat loss through a Roof)
+
+//(a)
+k=0.8;//The thermal conductivity of the roof[W/m.degree.C]
+A=6*8;//Area of the roof[m^2]
+t1=15;//temperature of inner surface roof[degree C]
+t2=4;//temperature of outer surface roof[degree C]
+L=0.25;//thickness of roof[m]
+Q_=k*A*(t1-t2)/L;//[W]
+disp("W",Q_,"The steady rate of heat transfer through the roof is")
+
+//(b)
+dt=10;//time period[h]
+Q=Q_*dt/1000;//[kWh]
+u_cost=0.08;//Unit cost of energy[$/kWh]
+Cost=Q*u_cost;//[$]
+disp(Cost,"and its cost is $","kWh",Q,"The amount of heat lost through the roof")
diff --git a/587/CH1/EX1.6/example1_6.sce b/587/CH1/EX1.6/example1_6.sce new file mode 100755 index 000000000..95b0b39a0 --- /dev/null +++ b/587/CH1/EX1.6/example1_6.sce @@ -0,0 +1,16 @@ +clear;
+clc;
+
+//Example1.6 (Measuring the Thermal Conductivity of a Material)
+V=110;//Voltage diffrence b/w thermocouples[V]
+I=0.4;//Current drawn by thermocouples[A]
+We=V*I;//[W]
+disp("W",We,"The electrical power consumed by the resistance heater and converted to heat is")
+q_=We/2;//[W]
+disp("W",q_,"The rate of heat flow through each sample")
+dT=15;//Temperature drop in the direction of heat flow[degree C]
+l=.03;//length for which temperature change is measured[m]
+D=.05;//diameter of cylinder[m]
+a=(%pi*D^2)/4;//Cross-sectional area of the cylinder[m^2]
+K=(q_*l)/(a*dT);//[W/m.degreeC]
+disp("W/mC",K,"The thermal conductivity of the sample is")
diff --git a/587/CH1/EX1.7/example1_7.sce b/587/CH1/EX1.7/example1_7.sce new file mode 100755 index 000000000..c710e0b8c --- /dev/null +++ b/587/CH1/EX1.7/example1_7.sce @@ -0,0 +1,10 @@ +clear;
+clc;
+
+//Example1.7[Conversion between SI and English Units]
+W_to_btu_p_h=3.41214;//Conersion from Watt to btu per hour[btu/h]
+m_to_ft=3.2808;//Conversion from meter to english unit feet[ft]
+deg_C_to_deg_F=1.8;//Conversion from degree Celcius to degree Farenhiet
+W_per_m_deg_C=W_to_btu_p_h/(m_to_ft*deg_C_to_deg_F);//Conversion factor for 1W/m.degree Celcius[Btu/h.ft.degree Farenhiet]
+k_brick=0.72*W_per_m_deg_C;//[Btu/h.ft.degree Farenhiet]
+disp("Btu/h.ft.degree Farenhiet",k_brick,"The thermal conductivity of the brick in English units is")
diff --git a/587/CH1/EX1.8/example1_8.sce b/587/CH1/EX1.8/example1_8.sce new file mode 100755 index 000000000..56161439a --- /dev/null +++ b/587/CH1/EX1.8/example1_8.sce @@ -0,0 +1,20 @@ +clear;
+clc;
+
+//Example1.8[Measuring Convection Heat Transfer coefficient]
+//Given:-
+T_ambient=15;//Temperature of room[degree Celcius]
+T_surface=152;//Temperature of surface of wire[degree Celcius]
+L=2;//Length of wire[m]
+D=0.003;//Diameter of wire[m]
+V=60;//Voltage drop across the current wire[Volts]
+I=1.5;//Current flowing in the wire[amp]
+//Solution:-
+//When steady conditions are reached, the rate of heat loss from the wire equals the rate of heat generation in the wire as a result of resistance heating
+Q_=V*I;//[W]
+disp("W",Q_,"The rate of heat generated in the wire as a result of resistance heating is")
+As=%pi*D*L;//Surface Area of the wire[m^2]
+//Using Newton's Law of Cooling
+//and assuming all heat loss in wire to occur by convection
+h=Q_/(As*(T_surface-T_ambient));//[W/m^2.degree Celcius]
+disp("W/m^2.degree Celcius",h,"The convection Heat Transfer coefficient is" )
diff --git a/587/CH1/EX1.9/example1_9.sce b/587/CH1/EX1.9/example1_9.sce new file mode 100755 index 000000000..3fba653e8 --- /dev/null +++ b/587/CH1/EX1.9/example1_9.sce @@ -0,0 +1,15 @@ +clear;
+clc;
+
+//Example1.9[Radiation Effect on Thermal Comfort]
+//Given:-
+T_room=22+273;//Temperature fo room[K]
+T_wntr=10+273;//Average Temperature of inner surfaces of walls,floors and the cieling in winter[K]
+T_smmr=25+273;//Average Temperature of inner surfaces of walls,floors and the cieling in summer[K]
+T_outr=30+273;//Average outer surface temperature of the person[K]
+A=1.4;//The exposed surface area[m^2]
+e=0.95;//Emissivity of person
+sigma=5.67*(10^(-8));//Stefan's constant
+Q_rad_wntr=e*sigma*A*((T_outr^4)-(T_wntr^4));//[W]
+Q_rad_smmr=e*sigma*A*((T_outr^4)-(T_smmr^4));//[W]
+disp("W",Q_rad_smmr,"and",round(Q_rad_wntr),"The net rates of radiation heat transfer from the body to the surrounding walls,ceiling, and floor in winter and summer are respectively")
diff --git a/587/CH10/EX10.1/example10_1.sce b/587/CH10/EX10.1/example10_1.sce new file mode 100755 index 000000000..2c7b10384 --- /dev/null +++ b/587/CH10/EX10.1/example10_1.sce @@ -0,0 +1,25 @@ +clear;
+clc;
+
+//Example10.1[Nucleate Boiling of Water in a Pan]
+Ts=108;//Temp of surface of bottom of pan[degree Celcius]
+Tsat=100;//Saturation temp of water[degree Celcius]
+D=0.3;//Diameter[m]
+//Properties of water at the saturation temp
+rho_l=957.9;//Density of liquid[kg/m^3]
+rho_v=0.6;//Density of vapour[kg/m^3]
+Pr_l=1.75;//Prandtl no of liquid
+mu_l=0.282*10^(-3);//Viscosity of liquid[kg/m.s]
+Cp_l=4217;//Specific Heat of liquid[J/kg.degree Celcius]
+h_fg=2257*10^3;//[J/kg]
+sigma=0.0589;//[N/m]
+g=9.81;//Acc due to gravity[m/s^2]
+Csf=0.0130,n=1.0;
+//Solution(a):-
+q_nuc=mu_l*h_fg*((g*(rho_l-rho_v)/sigma)^(1/2))*((Cp_l*(Ts-Tsat)/(Csf*h_fg*(Pr_l^n)))^3);//[W/m^2]
+A=%pi*(D^2)/4;//Surface Area of bottom of the pan[m^2]
+Q_boiling=A*q_nuc;//[W]
+disp("W",Q_boiling,"(a) The rate of heat transfer during nucleate boiling becomes ")
+//Solution(b):-
+m=Q_boiling/h_fg;//[kg/s]
+disp("kg/s",m,"The rate of Evaporation of water is")
diff --git a/587/CH10/EX10.2/example10_2.sce b/587/CH10/EX10.2/example10_2.sce new file mode 100755 index 000000000..e949f861b --- /dev/null +++ b/587/CH10/EX10.2/example10_2.sce @@ -0,0 +1,24 @@ +clear;
+clc;
+
+//Example10.2[Peak Heat Flux in Nucleate Boiling]
+D=0.01;//[m]
+Tsat=100;//Saturation Temperature[degree Celcius]
+sigma=0.0589;//[N/m]
+//Properties of water at saturation temperature
+rho_l=957.9;//[kg/m^3]
+rho_v=0.6;//[kg/m^3]
+h_fg=2257*10^3;//[J/kg]
+mu_l=0.282*10^(-3);//[kg/m.s]
+Pr_l=1.75;//Prandtl number
+Cp_l=4217;//[J/kg.degree Celcius]
+Csf=0.0130,n=1.0;
+g=9.81;//[m/s^2]
+//Solution:-
+L_=(D/2)*((g*(rho_l-rho_v)/sigma)^(1/2));//dimensionless Parameter
+//For this value of L_ we have
+C_cr=0.12;//Constant
+q_max=C_cr*h_fg*((sigma*g*(rho_v^2)*(rho_l-rho_v))^(1/4));//[W/m^2]
+disp("W/m^2",q_max,"The maximum or critical heat flux is")
+Ts=(((q_max/(mu_l*h_fg*((g*(rho_l-rho_v)/sigma)^(1/2))))^(1/3))*(Csf*h_fg*Pr_l^n)/Cp_l)+Tsat;//[degree Celcius]
+disp("degree Celcius",round(Ts),"The surface temperature is")
\ No newline at end of file diff --git a/587/CH10/EX10.3/example10_3.sce b/587/CH10/EX10.3/example10_3.sce new file mode 100755 index 000000000..0d6693bde --- /dev/null +++ b/587/CH10/EX10.3/example10_3.sce @@ -0,0 +1,28 @@ +clear;
+clc;
+
+//Example10.3[Film Boiling of Water on a Heating Element]
+//Given:-
+D=0.005;//[m]
+e=0.05;//Emissivity
+Ts=350;//Surface temperature[degree Celcius]
+Tsat=100;//[degree Celcius]
+Tf=(Ts+Tsat)/2;//[degree Celcius]
+g=9.81;//[m/s^2]
+//Properties of water at Tsat
+rho_l=957.9;//[kg/m^3]
+h_fg=2257*10^3;//[J/kg]
+//Properties of vapor at film temp
+rho_v=0.444;//[kg/m^3]
+Cp_v=1951;//[J/kg.degree Celcius]
+mu_v=1.75*10^(-5);//[kg/m.s]
+k_v=0.0388;//[W/m.degree Celcius]
+//Solution:-
+q_film=0.62*(((g*(k_v^3)*rho_v*(rho_l-rho_v)*(h_fg+(0.4*Cp_v*(Ts-Tsat))))/(mu_v*D*(Ts-Tsat)))^(1/4))*(Ts-Tsat);//[W/m^2]
+disp("W/m^2)",q_film,"The film boiling heat flux is")
+q_rad=e*(5.67*10^(-8))*(((Ts+273)^4)-((Tsat+273)^4));//[W/m^2]
+disp("W/m^2",q_rad,"The radiation heat flux is")
+q_total=q_film+(3/4)*q_rad;//[W/m^2]
+disp("W/m^2",q_total,"The total heat flux is")
+Q_total=(%pi*D*1)*q_total;//[W]
+disp("W",Q_total,"The rate of heat transfer from the heating element to the water is")
diff --git a/587/CH10/EX10.4/example10_4.sce b/587/CH10/EX10.4/example10_4.sce new file mode 100755 index 000000000..c4b5fca61 --- /dev/null +++ b/587/CH10/EX10.4/example10_4.sce @@ -0,0 +1,31 @@ +clear;
+clc;
+
+//Example10.4[Condensation of steam on a Vertical Plate]
+//Given:-
+Tsat=100,Ts=80;//[degree Celcius]
+Tf=(Ts+Tsat)/2;//[degree Celcius]
+L=2,w=3;//Dimensions of Plate[m]
+g=9.81;//[m/s^2]
+//Properties of water at Tsat
+h_fg=2257*10^3;//[J/kg]
+rho_v=0.60;//[kg/m^3]
+//Properties of liquid water at Tf
+rho_l=965.3;//[kg/m^3]
+mu_l=0.315*10^(-3);//[kg/m.s
+Cp_l=4206;//[J/kg.degree Celcius]
+k_l=0.675;//[W/m.degree Celcius]
+nu_l=0.326*10^(-6);//[m^2/s]
+//Solution (a)
+h_fg_m=h_fg+0.68*Cp_l*(Tsat-Ts);//[J/kg]
+disp("J/kg",h_fg_m,"The modified latent heat of vapourization is")
+Re=((4.81+((3.70*L*k_l*(Tsat-Ts)*((g/nu_l^2)^(1/3)))/(mu_l*h_fg_m)))^(0.820));
+disp(ceil(Re),"For wavy laminar flow Reynolds number is")
+h=(Re*k_l*((g/nu_l^2)^(1/3)))/((1.08*(Re^(1.22)))-5.2);//[W/m^2.degree Celcius]
+disp("W/m^2.degree Celcius",h,"The conensation heat transfer coefficient is")
+As=w*L;//[m^2]
+Q=h*As*(Tsat-Ts);//[W]
+disp("W",Q,"The rate of heat transfer during condensation process is")
+//Solution (b)
+m=Q/h_fg_m;//[kg/s]
+disp("kg/s",m,"The rate of condensation of steam is")
diff --git a/587/CH10/EX10.5/example10_5.sce b/587/CH10/EX10.5/example10_5.sce new file mode 100755 index 000000000..28a973985 --- /dev/null +++ b/587/CH10/EX10.5/example10_5.sce @@ -0,0 +1,32 @@ +clear;
+clc;
+
+//Example10.5[Condensation of steam on a Vertical Tilted Plate]
+//Given:-
+Tsat=100,Ts=80;//[degree Celcius]
+Tf=(Ts+Tsat)/2;//[degree Celcius]
+L=2,w=3;//Dimensions of Plate[m]
+g=9.81;//[m/s^2]
+//Properties of water at Tsat
+h_fg=2257*10^3;//[J/kg]
+rho_v=0.60;//[kg/m^3]
+//Properties of liquid water at Tf
+rho_l=965.3;//[kg/m^3]
+mu_l=0.315*10^(-3);//[kg/m.s
+Cp_l=4206;//[J/kg.degree Celcius]
+k_l=0.675;//[W/m.degree Celcius]
+nu_l=0.326*10^(-6);//[m^2/s]
+theta=(%pi/6);//Angle at which plate is tilted[radians]
+//Solution (a)
+h_fg_m=h_fg+0.68*Cp_l*(Tsat-Ts);//[J/kg]
+disp("J/kg",h_fg_m,"The modified latent heat of vapourization is")
+Re=((4.81+((3.70*L*k_l*(Tsat-Ts)*((g/nu_l^2)^(1/3)))/(mu_l*h_fg_m)))^(0.820));
+disp(ceil(Re),"For wavy laminar flow Reynolds number is")
+h=((Re*k_l*((g/nu_l^2)^(1/3)))/((1.08*(Re^(1.22)))-5.2))*((cos(theta))^(1/4));//[W/m^2.degree Celcius]
+disp("W/m^2.degree Celcius",h,"The conensation heat transfer coefficient is")
+As=w*L;//[m^2]
+Q=h*As*(Tsat-Ts);//[W]
+disp("W",Q,"The rate of heat transfer during condensation process is")
+//Solution (b)
+m=Q/h_fg_m;//[kg/s]
+disp("kg/s",m,"The rate of condensation of steam is")
diff --git a/587/CH10/EX10.6/example10_6.sce b/587/CH10/EX10.6/example10_6.sce new file mode 100755 index 000000000..44924ec6b --- /dev/null +++ b/587/CH10/EX10.6/example10_6.sce @@ -0,0 +1,29 @@ +clear;
+clc;
+
+//Example10.6[Condensation of Steam on horizontal Tubes]
+//Given:-
+Tsat=40;//[degree Celcius]
+D=0.03;//[m]
+Ts=30;//Outer Surface temperature of tube[degree Celcius]
+Tf=(Ts+Tsat)/2;//Film Temperature[degree Celcius]
+g=9.81;//[m/s^2]
+//Properties of water at the saturation temp
+h_fg=2407*10^3;//[J/kg]
+rho_v=0.05;//[kg/m^3]
+//Properties of liquid water at the film temperature
+rho_l=994;//[kg/m^3]
+Cp_l=4178;//[J/kg.degree Celcius]
+mu_l=0.720*10^(-3);//[kg/m.s]
+k_l=0.623;//[W/m.degree Celcius]
+//Solution (a)
+h_fg_m=h_fg+0.68*Cp_l*(Tsat-Ts);//[J/kg]
+disp("J/kg",h_fg_m,"(a) The modified latent heat of vapourisation is")
+h_hori=0.729*(((g*(rho_l^2)*h_fg_m*(k_l^3))/(mu_l*D*(Tsat-Ts)))^(1/4));//[W/m^2.degree Celcius]
+disp("W/m^2.degree Celcius",h_hori,"The heat transfer coefficient for condensation on a single horizontal tube is")
+As=%pi*D*1;//[m^2]
+Q=h_hori*As*(Tsat-Ts);//[W]
+disp("W",Q,"The rate of heat transfer during condensation Process is")
+//Solution (b)
+m=Q/h_fg_m;//[kg/s]
+disp("kg/s",m,"(b) The rate of condensation of steam is")
\ No newline at end of file diff --git a/587/CH10/EX10.7/example10_7.sce b/587/CH10/EX10.7/example10_7.sce new file mode 100755 index 000000000..44fba2cff --- /dev/null +++ b/587/CH10/EX10.7/example10_7.sce @@ -0,0 +1,31 @@ +clear;
+clc;
+
+//Example10.7[Condensation of Steam on horizontal Tube Banks]
+//Given:-
+Tsat=40;//[degree Celcius]
+D=0.03;//[m]
+Ts=30;//Outer Surface temperature of tube[degree Celcius]
+Tf=(Ts+Tsat)/2;//Film Temperature[degree Celcius]
+g=9.81;//[m/s^2]
+N=3;//No of tubes in a vertical tier
+N_total=12;//Total number of tubes
+//Properties of water at the saturation temp
+h_fg=2407*10^3;//[J/kg]
+rho_v=0.05;//[kg/m^3]
+//Properties of liquid water at the film temperature
+rho_l=994;//[kg/m^3]
+Cp_l=4178;//[J/kg.degree Celcius]
+mu_l=0.720*10^(-3);//[kg/m.s]
+k_l=0.623;//[W/m.degree Celcius]
+//Solution (a)
+h_fg_m=h_fg+0.68*Cp_l*(Tsat-Ts);//[J/kg]
+disp("J/kg",h_fg_m,"(a) The modified latent heat of vapourisation is")
+h_hori_N=(0.729*(((g*(rho_l^2)*h_fg_m*(k_l^3))/(mu_l*D*(Tsat-Ts)))^(1/4)))*(1/(N^(1/4)));//[W/m^2.degree Celcius]
+disp("W/m^2.degree Celcius",h_hori_N,"The heat transfer coefficient for condensation 12 horizontal tube is")
+As=%pi*D*1*N_total;//[m^2]
+Q=h_hori_N*As*(Tsat-Ts);//[W]
+disp("W",Q,"The rate of heat transfer during condensation Process is")
+//Solution (b)
+m=Q/h_fg_m;//[kg/s]
+disp("kg/s",m,"(b) The rate of condensation of steam is")
\ No newline at end of file diff --git a/587/CH10/EX10.8/example10_8.sce b/587/CH10/EX10.8/example10_8.sce new file mode 100755 index 000000000..982c1fda8 --- /dev/null +++ b/587/CH10/EX10.8/example10_8.sce @@ -0,0 +1,18 @@ +clear;
+clc;
+
+//Example10.8[Replacing a Heat Pipe by a Copper Rod]
+//Given:-
+L=0.3;//[m]
+D=0.006;//[m]
+Q=180;//[W]
+del_T=3;//Temperature Difference [degree Celcius]
+//Properties of copper at room temperature
+rho=8933;//[kg/m^3]
+k=401;//[W/m.degree Celcius]
+//Solution:-
+A=Q*L/(k*del_T);//[m^2]
+d=sqrt(4*A/%pi);//[m]
+disp("cm",ceil(100*d),"The diameter of the copper pipe is")
+m=rho*A*L;//[kg]
+disp("kg",round(m),"Mass of the copper rod is")
\ No newline at end of file diff --git a/587/CH11/EX11.1/example11_1.sce b/587/CH11/EX11.1/example11_1.sce new file mode 100755 index 000000000..125e88ad8 --- /dev/null +++ b/587/CH11/EX11.1/example11_1.sce @@ -0,0 +1,38 @@ +clear;
+clc;
+
+//Example11.1[Overall Heat Transfer Coefficient of a Heat Exchanger]
+D_in=0.02;//Diameter of inner tubes[m]
+Di_out=0.03;//Inner Diameter of Outer tubes[m]
+mw=0.5;//Mass Flow Rate of water[kg/s]
+mo=0.8;//Mass Flow rate of oil[kg/s]
+Tw=45;//Average Temp of water[degree Celcius]
+To=80;//Average Temp of oil [degree Celcius]
+//Properties of water at Tw
+rho_w=990.1;//[kg/m^3]
+Pr_w=3.91;//Prandtl Number
+k_w=0.637;//[W/m.degree Celcius]
+nu_w=0.602*10^(-6);//[m^2/s]
+//Properties of oil at To
+rho_o=852;//[kg/m^3]
+Pr_o=499.3;//Prandtl Number
+k_o=0.138;//[W/m.degree Celcius]
+nu_o=3.794*10^(-5);//[m^2/s]
+//Solution:-
+Vw=mw/(rho_w*(%pi*(D_in^2)/4));//[m/s]
+disp("m/s",Vw,"The average velocity of water in the tube is")
+Re_w=Vw*D_in/nu_w;
+disp(Re_w,"The Reynolds number for flow of water in the tube is")
+Nu_w=0.023*(Re_w^(0.8))*(Pr_w^(0.4));
+disp(Nu_w,"The nusselt no for turbulent water flow")
+hi=k_w*Nu_w/D_in;//[W/m^2.degree Celcius]
+//For oil flow
+Dh=Di_out-D_in;//Hydraulic Diameter for the annular space[m]
+Vo=mo/(rho_o*(%pi*((Di_out^2)-(D_in^2))/4));//[m/s]
+disp("m/s",Vo,"The average velocity for flow of oil is")
+Re_o=Vo*Dh/nu_o;
+disp(Re_o,"The Reynolds number for flow of oil is")
+Nu_o=5.45;//Nusselt number for flow of oil usign the table 11.3 and interpolating for value corresponding to Di_out/D_in
+ho=Nu_o*k_o/Dh;//[W/m^2.degree Celcius]
+U=(1/((1/hi)+(1/ho)));//[W/m^2.degree Celcius]
+disp("W/m^2.degree Celcius",U,"The overall heat transfer Coefficient for the given heat exchanger is")
diff --git a/587/CH11/EX11.10/example11_10.sce b/587/CH11/EX11.10/example11_10.sce new file mode 100755 index 000000000..a6b471995 --- /dev/null +++ b/587/CH11/EX11.10/example11_10.sce @@ -0,0 +1,22 @@ +clear;
+clc;
+
+//Example11.10[Installing a Heat Exchanger to Save Energy and Money]
+//Given:-
+Cp=4.18;//[kJ/kg.degree Celcius]
+Th_in=80,Tc_in=15;//Inlet temperatures of hot and cold water[degree Celcius]
+m=15/60;//[kg/s]
+e=0.75;//Effectiveness
+t=24*365;//Operating Hours[hours/year]
+neta=0.8;//Eficiency
+cost=1.10;//[$/therm]
+//Solution:-
+Q_max=m*Cp*(Th_in-Tc_in);//[kJ/kg.degree Celcius]
+disp("kJ/kg.degree Celcius",Q_max,"Maximun Heat recover is")
+Q=e*Q_max;//[kJ/s]
+E_saved=Q*t*3600;//[kJ/year]
+disp("kJ/year",E_saved,"The energy saved during an entire year will be")
+F_saved=(E_saved/neta)*(1/105500);//[therms]
+disp("therms/year",F_saved,"Fuel savings will be")
+M_saved=F_saved*cost;//[$/year]
+disp("per year",M_saved,"The amount of money saved is $")
\ No newline at end of file diff --git a/587/CH11/EX11.2/Example11_2.sce b/587/CH11/EX11.2/Example11_2.sce new file mode 100755 index 000000000..bdbe9b33f --- /dev/null +++ b/587/CH11/EX11.2/Example11_2.sce @@ -0,0 +1,23 @@ +clear;
+clc;
+
+//Example11.2[Effect of Fouling on the Overall Heat Transfer Coefficient]
+//Given:-
+k=15.1;//[W/m^2.degree Celcius]
+Di=0.015;//Inner Diameter[m]
+Do=0.019;//Outer Diameter[m]
+Di_s=0.032;//Inner diameter of outer shell[m]
+L=1;//[m]
+hi=800;//W/m^2.degree Celcius
+ho=1200;//[W/m^2.degree Celcius]
+Rfi=0.0004;//[m^2.degree Celcius/W]
+Rfo=0.0001;//[m^2.degree Celcius/W]
+//Solution (a):-
+Ai=%pi*Di*L;//[m^2]
+Ao=%pi*Do*L;//[m^2]
+Ra=(1/(hi*Ai))+(Rfi/Ai)+((log(Do/Di))/(2*%pi*k*L))+(Rfo/Ao)+(1/(ho*Ao));//[m^2.degree Celcius/W]
+disp("m^2.degree Celcius/W",Ra,"The thermal Resistance for an unfinned shell and tube heat exchanger with fouling on both heat transfer surfaces is")
+//Solution (b):-
+Ui=1/(Ra*Ai);//[W/m^2.degree Celcius]
+Uo=1/(Ra*Ao);//[W/m^2.degree Celcius]
+disp("respectively","W/m^2.degree Celcius",Uo,"and",Ui,"The overall Heat transfer Coefficients based on the inner and outer surfaces of the tube are")
\ No newline at end of file diff --git a/587/CH11/EX11.3/example11_3.sce b/587/CH11/EX11.3/example11_3.sce new file mode 100755 index 000000000..e7f9b2e50 --- /dev/null +++ b/587/CH11/EX11.3/example11_3.sce @@ -0,0 +1,22 @@ +clear;
+clc;
+
+//Example11.3[The Condensation of Steam in a Condenser]
+//Given:-
+Th_in=30,Th_out=30,Tc_in=14,Tc_out=22;//Inlet and Outlet temperatures of hot and cold liquids [degree Celcius]
+A=45;//[m^2]
+U=2100;//[W/m^2.degree Celcius]
+h_fg=2431;//Heat of vapourisation of water at Th_i[kJ/kg]
+Cp=4184;//Specific heat of cold water [J/kg]
+//Solution:-
+del_T1=Th_in-Tc_out;//[degree Celcius]
+del_T2=Th_out-Tc_in;//[degree Celcius]
+del_T_lm=(del_T1-del_T2)/(log(del_T1/del_T2));//[degree Celcius]
+disp("degree Celcius",del_T_lm,"The logrithmic Mean temperature difference is")
+Q=U*A*del_T_lm;//[W]
+disp("W",Q,"The heat transfer rate in the condenser is")
+mw=Q/(Cp*(Tc_out-Tc_in));//[kg/s]
+disp("kg/s",mw,"The mass flow rate of the cooling water is")
+
+ms=(Q/(1000*h_fg));//[kg/s]
+disp("kg/s",ms,"The rate of condensation of steam is")
\ No newline at end of file diff --git a/587/CH11/EX11.4/example11_4.sce b/587/CH11/EX11.4/example11_4.sce new file mode 100755 index 000000000..2db731115 --- /dev/null +++ b/587/CH11/EX11.4/example11_4.sce @@ -0,0 +1,24 @@ +clear;
+clc;
+
+//Example11.4[Heating Water in a Counter Flow Heat Exchanger]
+//Given:-
+mw=1.2,mgw=2;//Mass Flow rate of water and geothermal fluid[kg/s]
+U=640;//Overall Heat transfer Coefficient[W/m^2.degree Celcius]
+Di=0.015;//[m]
+Tw_out=80,Tw_in=20;//Outlet and Inlet temp of water[degree Celcius]
+Tgw_in=160;//Inlet temp of geothermal fluid[degree Celcius]
+Cp_w=4.18,Cp_gw=4.31;//Specific Heats of water and geothermal fluid[kJ/kg.degree Celcius]
+//Solution:-
+Q=mw*Cp_w*(Tw_out-Tw_in);//[kW]
+disp("kW",ceil(Q),"The rate of heat transfer in the heat exchanger is")
+Tgw_out=(Tgw_in-(ceil(Q)/(mgw*Cp_gw)));//[degree Celcius]
+disp("degree Celcius",Tgw_out,"The outlet temp of geothermal fluid is")
+del_T1=Tgw_in-Tw_out;//[degree Celcius]
+del_T2=Tgw_out-Tw_in;//[degree Celcius]
+del_T_lm=(del_T1-del_T2)/(log(del_T1/del_T2));//[degree Celcius]
+disp("degree Celcius",del_T_lm,"The logrithmic Mean temperature difference is")
+As=1000*ceil(Q)/(U*del_T_lm);//[m^2]
+disp("m^2",As,"The surface area of the heat exchanger is")
+L=As/(%pi*Di);//[m]
+disp("m",round(L),"The length of the tube is")
diff --git a/587/CH11/EX11.5/example11_5.sce b/587/CH11/EX11.5/example11_5.sce new file mode 100755 index 000000000..187dede16 --- /dev/null +++ b/587/CH11/EX11.5/example11_5.sce @@ -0,0 +1,28 @@ +clear;
+clc;
+
+//Example11.5[Heating of Glycerine in a Multipass Heat Exchanger]
+//Given:-
+//A 2,4 shell and tube heat exchanger
+D=0.02;//Diameter[m]
+L=60;//Length of tube[m]
+Th_in=80,Th_out=40,Tc_in=20,Tc_out=50;//Inlet and Outlet temperatures water and glycerine[degree Celcius]
+hi=160,ho=25;//Convective Heat transfer coefficients on both side of tube[W/m^2.degree Celcius]
+Rf=0.0006;//Fouling Resistance[m^2.degree Celcius/W]
+//Solution:-
+As=%pi*D*L;//[m^2]
+del_T1=Th_in-Tc_out;//[degree Celcius]
+del_T2=Th_out-Tc_in;//[degree Celcius]
+del_T_lm=(del_T1-del_T2)/log(del_T1/del_T2);//[degree Celcius]
+disp("degree Celcius",del_T_lm,"The log mean temperature difference for the counter flow arrangement is")
+F=0.91;//Correction Factor
+//(a)
+Ua=1/((1/hi)+(1/ho));//[W/m^2.degree Celcius]
+disp("W/m^2.degree Celcius",Ua,"In case of no fouling, the over all heat transfer coefficient is")
+Qa=Ua*As*F*del_T_lm;//[W]
+disp("W",ceil(Qa),"And the rate of heat transfer is")
+//(b)
+Ub=1/((1/hi)+(1/ho)+(Rf));//[W/m^2.degree Celcius
+disp("W/m^2.degree Celcius",Ub,"When there is fouling on one of the surfaces, the overall heat transfer coefficient is")
+Qb=Ub*As*F*del_T_lm;//[W]
+disp("W",round(Qb),"And the rate of heat transfer is")
\ No newline at end of file diff --git a/587/CH11/EX11.6/example11_6.sce b/587/CH11/EX11.6/example11_6.sce new file mode 100755 index 000000000..2169148dc --- /dev/null +++ b/587/CH11/EX11.6/example11_6.sce @@ -0,0 +1,22 @@ +clear;
+clc;
+
+//Example11.6[Cooling of Water in an Automotive Radiator]
+//Given:-
+m=0.6;//Mass Flow rate of water[kg/s]
+Th_in=90,Th_out=65,Tc_in=20,Tc_out=40;//[degree Celcius]
+Di=0.005;//[m]
+L=0.65;//[m]
+n=40;//No of tubes
+Cp=4195;//[J/kg.degree Celcius]
+//Solution:-
+Q=m*Cp*(Th_in-Th_out);//[W]
+disp("W",Q,"The rate of heat transfer in the radiator from the hot water to the air is")
+Ai=n*%pi*Di*L;//[m^2]
+del_T1=Th_in-Tc_out;//[degree Celcius]
+del_T2=Th_out-Tc_in;//[degree Celcius]
+del_T_lm=(del_T1-del_T2)/log(del_T1/del_T2);//[degree Celcius]
+disp("degree Celcius",del_T_lm,"The log mean temperature difference for the counter flow arrangement is")
+F=0.97;//Correction Factor for this situation
+Ui=Q/(Ai*F*del_T_lm);//[W/m^2.degree Celcius]
+disp("W/m^2.degree Celcius",round(Ui),"the overall heat transfer coefficient is")
diff --git a/587/CH11/EX11.7/example11_6.sce b/587/CH11/EX11.7/example11_6.sce new file mode 100755 index 000000000..2169148dc --- /dev/null +++ b/587/CH11/EX11.7/example11_6.sce @@ -0,0 +1,22 @@ +clear;
+clc;
+
+//Example11.6[Cooling of Water in an Automotive Radiator]
+//Given:-
+m=0.6;//Mass Flow rate of water[kg/s]
+Th_in=90,Th_out=65,Tc_in=20,Tc_out=40;//[degree Celcius]
+Di=0.005;//[m]
+L=0.65;//[m]
+n=40;//No of tubes
+Cp=4195;//[J/kg.degree Celcius]
+//Solution:-
+Q=m*Cp*(Th_in-Th_out);//[W]
+disp("W",Q,"The rate of heat transfer in the radiator from the hot water to the air is")
+Ai=n*%pi*Di*L;//[m^2]
+del_T1=Th_in-Tc_out;//[degree Celcius]
+del_T2=Th_out-Tc_in;//[degree Celcius]
+del_T_lm=(del_T1-del_T2)/log(del_T1/del_T2);//[degree Celcius]
+disp("degree Celcius",del_T_lm,"The log mean temperature difference for the counter flow arrangement is")
+F=0.97;//Correction Factor for this situation
+Ui=Q/(Ai*F*del_T_lm);//[W/m^2.degree Celcius]
+disp("W/m^2.degree Celcius",round(Ui),"the overall heat transfer coefficient is")
diff --git a/587/CH11/EX11.8/example11_8.sce b/587/CH11/EX11.8/example11_8.sce new file mode 100755 index 000000000..ee1b405f8 --- /dev/null +++ b/587/CH11/EX11.8/example11_8.sce @@ -0,0 +1,32 @@ +clear;
+clc;
+
+//Example11.8[Using the Effectiveness- NTU Method]
+//Given:-
+mc=1.2,mh=2;//Mass Flow rate of water and geothermal fluid[kg/s]
+U=640;//Overall Heat transfer Coefficient[W/m^2.degree Celcius]
+Di=0.015;//[m]
+Tc_out=80,Tc_in=20;//Outlet and Inlet temp of water[degree Celcius]
+Th_in=160;//Inlet temp of geothermal fluid[degree Celcius]
+Cp_c=4.18,Cp_h=4.31;//Specific Heats of water and geothermal fluid[kJ/kg.degree Celcius]
+//Solution:-
+Ch=mh*Cp_h;//[kW/degree Celcius]
+Cc=mc*Cp_c;//[kW/degree Celcius]
+if(Ch>Cc) then,
+ Cmin=Cc;
+ c=Cmin/Ch;
+else
+ Cmin=Ch;
+ c=Cmin/Cc;
+end
+Q_max=Cmin*(Th_in-Tc_in);//[kW]
+disp("kW",Q_max,"The maximum heat transfer rate is")
+Q_ac=mc*Cp_c*(Tc_out-Tc_in);//[kW]
+e=Q_ac/Q_max;
+disp(e,"The effectiveness of the heat exchanger is")
+NTU=(1/(c-1))*log((e-1)/(e*c-1));
+disp(NTU,"The NTU of this counter flow heat exchanger is")
+As=NTU*Cmin*1000/U;//[m^2]
+disp("m^2",As,"The heat transfer surface area is")
+L=As/(%pi*Di);//[m]
+disp("m",round(L),"The length of the tube is")
\ No newline at end of file diff --git a/587/CH11/EX11.9/example11_9.sce b/587/CH11/EX11.9/example11_9.sce new file mode 100755 index 000000000..7d83ea1e5 --- /dev/null +++ b/587/CH11/EX11.9/example11_9.sce @@ -0,0 +1,34 @@ +clear;
+clc;
+
+//Example11.9[Cooling Hot Oil by Water in Multipass Heat Exchanger]
+//Given:-
+Cp_c=4.18,Cp_h=2.13;//Specific Heats of water and oil[kJ/kg]
+mc=0.2,mh=0.3;//Mass Flow rate of oil and water [kg/s]
+Th_in=150,Tc_in=20;//[degree Celcius]
+n=8;//No of tubes
+D=0.014;//[m]
+L=5;//[m]
+U=310;//Overall Heat transfer Coefficient[W/m^2.degree Celcius]
+//Solution:-
+Ch=mh*Cp_h;//[kW/degree Celcius]
+Cc=mc*Cp_c;//[kW/degree Celcius]
+if(Ch>Cc) then,
+ Cmin=Cc;
+ c=Cmin/Ch;
+else
+ Cmin=Ch;
+ c=Cmin/Cc;
+end
+Q_max=Cmin*(Th_in-Tc_in);//[kW]
+disp("kW",Q_max,"The maximum heat transfer rate is")
+As=n*%pi*D*L;//[m^2]
+disp("m^2",As,"Heat transfer Surface Area is")
+NTU=U*As/Cmin;
+disp(NTU,"The NTU of this heat exchanger is")
+e=0.47;//Determined from fig 11.26(c)using value of NTU and c
+Q=e*Q_max;//[kW]
+Tc_out=Tc_in+(Q/Cc);//[degree Celcius]
+Th_out=Th_in-(Q/Ch);//[degree Celcius]
+disp("degree Celcius",Tc_out,"to","degree Celcius",Tc_in,"The temperature of cooling water will rise from")
+disp("degree Celcius",Th_out,"to","degree Celcius",Th_in,"as it cools the hot oil from")
\ No newline at end of file diff --git a/587/CH12/EX12.1/example12_1.sce b/587/CH12/EX12.1/example12_1.sce new file mode 100755 index 000000000..d13342ebf --- /dev/null +++ b/587/CH12/EX12.1/example12_1.sce @@ -0,0 +1,22 @@ +clear;
+clc;
+
+//Example12.1[Radiation Emission from a Black Ball]
+//Given:-
+T=800;//Temperature of suspended ball[K]
+D=0.2;//Diameter[m]
+C1=3.74177*10^8;//[(micrometer^4)/m^2]
+C2=1.43878*10^4;//[micrometer.K]
+lambda=3;//[micrometer]
+//Solution (a):-
+Eb=(5.67*10^(-8))*(T^4);//[W/m^2]
+disp("of energy in the form of energy in the form of electromagnetic radiation per second per m^2","kJ",Eb/1000,"The ball emits")
+//Solution(b):-
+As=%pi*(D^2);//[m^2]
+disp("m^2",As,"The total Surface area of the ball is")
+del_t=5*60;//[seconds]
+Q_rad=Eb*As*del_t;//[J]
+disp("kJ",Q_rad/1000,"The total amount of radiation energy emitted from the entire ball is")
+//Solution (c)
+Eb_lambda=C1/((lambda^5)*((exp(C2/(lambda*T)))-1));//[W/m^2.micrometer]
+disp("W/m^2.micrometer",round(Eb_lambda),"The spectral blackbody emissive power")
diff --git a/587/CH12/EX12.2/example12_2.sce b/587/CH12/EX12.2/example12_2.sce new file mode 100755 index 000000000..408351a9c --- /dev/null +++ b/587/CH12/EX12.2/example12_2.sce @@ -0,0 +1,13 @@ +clear;
+clc;
+
+//Example12.2[Emission of Radiation from a Lightbulb]
+//Given:-
+T=2500;//Temp of the filament[K]
+lambda1=0.4,lambda2=0.76;//Visible ranfe[micrometer]
+f1=0.000321,f2=0.053035;//The black body radiation functions corresponding to lamda1*T and lambda2*T
+//Solution:-
+f3=f2-f1;
+disp(f3,"Fraction of radiation emitted between the two given wavelengths is")
+lambda_max=2897.8/T;//[micrometer]
+disp("micron",lambda_max,"The wavelength at which the emission of radiation from the filament peaks is")
\ No newline at end of file diff --git a/587/CH12/EX12.3/example12_3.sce b/587/CH12/EX12.3/example12_3.sce new file mode 100755 index 000000000..d0d06386a --- /dev/null +++ b/587/CH12/EX12.3/example12_3.sce @@ -0,0 +1,17 @@ +clear;
+clc;
+
+//Example12.3[Radiation Incident on a small surface]
+//Given:-
+A1=3^10^(-4);//[m^2]
+T1=600;//[k]
+A2=5*10^(-4);//[m^2]
+theta1=%pi*55/180,theta2=%pi*40/180;//[Radian]
+r=0.75;//[m]
+//Solution:-
+w_2_1=(A2*cos(theta2))/(r^2);//[Steradian]
+disp("sr",w_2_1,"The solid angle subtended by a2 when viewed from A1 is")
+I1=(5.67*10^(-8))*(T1^4)/(%pi);//[W/m^2.sr]
+disp("W/m^2.sr",I1,"The Intensity of radiation emitted by A1 is")
+Q1_2=I1*(A1*cos(theta1))*w_2_1;//[W]
+disp("W",Q1_2,"is ","Steradian",w_2_1,"through the solid angle","radians",theta1,"The rate of radiation energy emitted by A1 in the direction of")
\ No newline at end of file diff --git a/587/CH12/EX12.4/example12_4.sce b/587/CH12/EX12.4/example12_4.sce new file mode 100755 index 000000000..3d04f55e4 --- /dev/null +++ b/587/CH12/EX12.4/example12_4.sce @@ -0,0 +1,21 @@ +clear;
+clc;
+
+//Example12.4[Emissivity of a surface and emissive Power]
+e1=0.3;//For 0<= lambda <= 3micron
+e2=0.8;//3micron<=lambda<=7micron
+e3=0.1;//7micron<=lamda<infinity
+lambda1=3,lambda2=7;//[micron]
+T=800;//[K]
+//Solution:-
+p=lambda1*T;//[micron.K]
+q=lambda2*T;//[micron.K]
+//Hence blackbody radiation functions are
+f1=0.140256;
+f2=0.701046;
+f0_1=f1-0;
+f2_inf=1-f2;
+e_T=e1*f1+e2*(f2-f1)+e3*(1-f2);
+disp(e_T,"Average emissivity of the surface is")
+E=e_T*(5.67*10^(-8))*(T^4);//[W/m^2]
+disp("W/m^2",E,"The Emissive Power of the surface is")
diff --git a/587/CH12/EX12.5/example12_5.sce b/587/CH12/EX12.5/example12_5.sce new file mode 100755 index 000000000..48bdc4841 --- /dev/null +++ b/587/CH12/EX12.5/example12_5.sce @@ -0,0 +1,27 @@ +clear;
+clc;
+
+//Example12.6[Selective Absorber and Reflective Surfaces]
+//Given:-
+G_D=400,G_d=300;//Direct and diffuse components of solar radiation[W/m^2]
+Ts=320,T_sky=260;//[K]
+theta=20*%pi/180
+//Solution:-
+G_solar=(G_D*cos(theta))+G_d
+//(a)
+ab_a=0.9,e_a=0.9;//Grey absorber surface
+q_net_rad_a=ab_a*G_solar+e_a*(5.67*10^(-8))*((T_sky^4)-(Ts^4));//[W/m^2]
+disp("W/m^2",round(q_net_rad_a),"(a) The net radiation heat transfer is")
+//(b)
+ab_b=0.1,e_b=0.1;//Grey reflector surface
+q_net_rad_b=ab_b*G_solar+e_b*(5.67*10^(-8))*((T_sky^4)-(Ts^4));//[W/m^2]
+disp("W/m^2",round(q_net_rad_b),"The net radiation heat transfer is")
+//(c)
+ab_c=0.9,e_c=0.1;//Selective Absorber surface
+
+q_net_rad_c=ab_c*G_solar+e_c*(5.67*10^(-8))*((T_sky^4)-(Ts^4));//[W/m^2]
+disp("W/m^2",round(q_net_rad_c),"The net radiation heat transfer is")
+//(d)
+ab_d=0.1,e_d=0.9;//Selective reflector surface
+q_net_rad_d=ab_d*G_solar+e_d*(5.67*10^(-8))*((T_sky^4)-(Ts^4));//[W/m^2]
+disp("W/m^2",round(q_net_rad_d),"The net radiation heat transfer is")
\ No newline at end of file diff --git a/587/CH12/EX12.6/example12_6.sce b/587/CH12/EX12.6/example12_6.sce new file mode 100755 index 000000000..dde18ae6e --- /dev/null +++ b/587/CH12/EX12.6/example12_6.sce @@ -0,0 +1,28 @@ +clear;
+clc;
+
+//Example12.6[Installing Reflective Films on Windows]
+//Given:-
+A_glazing=40;//[m^2]
+SHGC_wof=0.766,SHGC_wf=0.261;//[kWh/year]
+unit_c_e=0.08;//[$/kWh]
+unit_c_f=0.5;//[$/therm]
+COP=2.5,neta=0.80;
+//Solution:-
+//For the months of June,July,August and Sepetember
+Q_summer=5.31*30+4.31*31+3.93*31+3.28*30;//[kWh/year]
+//For the months oct,Nov,Dec,Jan,Feb,Mar,Apr
+Q_winter=2.80*31+1.84*30+1.54*31+1.86*31+2.66*28+3.43*31+4.00*30;//[kWh/year]
+c_l_d=Q_summer*A_glazing*(SHGC_wof-SHGC_wf);//[kWh/year]
+disp("kWh/year",c_l_d,"The decrease in the annual cooling load is")
+h_l_i=Q_winter*A_glazing*(SHGC_wof-SHGC_wf);//[kWh/year]
+disp("kWh/year",h_l_i,"The increase in annual heating load is")
+d_c_c=c_l_d*(unit_c_e)/COP;//[$/year]
+i_h_c=h_l_i*(unit_c_f/29.31)/neta;//[$/year]
+disp("per year",i_h_c,"and $",d_c_c,"The corresponding decrease in cooling costs and the increase in heating costs are $")
+Cost_s=d_c_c-i_h_c;//[$/year]
+disp("per year",Cost_s,"The net annual cost savings due to the reflective film is $")
+I_cost=20*A_glazing;//[$]
+disp(I_cost,"The implementation Cost of installing films is $")
+pp=I_cost/Cost_s;//[years]
+disp("years",pp,"Payback Period is")
diff --git a/587/CH13/EX13.1/example13_1.sce b/587/CH13/EX13.1/example13_1.sce new file mode 100755 index 000000000..23dc9c93b --- /dev/null +++ b/587/CH13/EX13.1/example13_1.sce @@ -0,0 +1,18 @@ +clear;
+clc;
+
+//Example13.1[View Factors Associated with two Concentric Spheres]
+//Solution:-
+//The outer surface of the smaller sphere and inner surface of the larger sphere form a two surface enclosure
+N=2;
+disp("View Factors",N^2,"This enclosure involves ")
+x=(1/2)*N*(N-1);
+disp("view factor directly",x,"W need to determine only")
+F11=0;
+F12=1;
+disp("The Two view Factors")
+disp(F11,"Since no radiation leaving surface 1 strikes itself..... F11=")
+disp(F12,"Since all radiation leaving surface 1 strikes surface 2 F12=")
+disp("F12= ((r1/r2)^2)")
+disp("F22= 1-((r1/r2)^2)")
+disp("where r1 and r2 are radius of surface 1 and surface 2")
\ No newline at end of file diff --git a/587/CH13/EX13.10/example13_10.sce b/587/CH13/EX13.10/example13_10.sce new file mode 100755 index 000000000..5b2b8f9e9 --- /dev/null +++ b/587/CH13/EX13.10/example13_10.sce @@ -0,0 +1,42 @@ +clear;
+clc;
+
+//Example13.10[Heat Transfer through a Tubular Solar Collector]
+k=0.02588;//[W/m.degree Celcius]
+Pr1=0.7282,Pr2=0.7255;//Prandtl no
+nu1=1.608*(10^(-5)),nu2=1.702*10^(-5);//[m^2/s]
+T1=20,T2=40;//[degree Celcius]
+Tavg=((T1+T2)/2)+273;//[K]
+Do=0.1,L=1;//Dimensions of glass tube[m]
+Di=0.05;//Inner diameter of tube[m]
+Q_glass=30;//Rate of heat transfer from the outer surface of the glass cover[W]
+g=9.81;//[m^2/s]
+eo=0.9,ei=0.95;//Emissivity
+//Solution:-
+Ao=%pi*Do*L;//Heat transfer surface area of the glass cover[m^2]
+disp(Ao,Tavg)
+Ra_Do=g*Tavg*(T2-T1)*(Do^3)*Pr1/(nu1);
+disp(Ra_Do,"The Rayleigh number is")
+Nu=((0.6+((0.387*(Ra_Do^(1/6)))/((1+((0.559/Pr1)^(9/16)))^(8/27))))^2);
+disp(Nu,"The nusselt number is")
+ho=k*Nu/Do;//[W/m^2.degree Celcius]
+Qo_conv=ho*Ao*(T2-T1);//[W]
+Qo_rad=eo*5.67*10^(-8)*Ao*(((T2+273)^4)-((T1+273)^4));//[W]
+Qo_total=Qo_conv+Qo_rad;//[W]
+disp("W",Qo_total,"The total rate of heat loss from the glass cover
+Lc=(Do-Di)/2;//The characteristic length
+Ai=%pi*Di*L;//[m^2]
+//Assuming
+T_tube=54,T_cover=26;//Temperature of tube and glass cover[degree Celcius]
+T_avg=((T_tube+T_cover)/2)+273;//[K]
+Ra_L=g*T_avg*(T_tube-T_cover)*(Lc^3)*Pr2/(nu2);
+disp(Ra_L,"The Rayleigh number in this case is")
+F_cyl=((log(Do/Di))^4)/((Lc^3)*(((Di^(-3/5))+(Do^(-3/5)))^5));
+k_eff=0.386*k*((Pr2/(0.861+Pr2))^(1/4))*((F_cyl*Ra_L)^(1/4));
+disp("W/m.degree Celcius",k_eff,"The effective thermal conductivity is")
+QL_conv=2*%pi*k_eff*(T_tube-T_cover)/(log(Do/Di));
+disp("W",QL_conv,"The rate of heat transfer between the cylinders by convection is")
+QL_rad=((5.67*10^(-8))*Ai*(((T_tube+273)^4)-((T_cover+273)^4)))/((1/ei)+(((1-eo)/eo)*(Di/Do)));
+disp("W",QL_rad,"The radiation rate of heat transfer is")
+QL_total=QL_conv+QL_rad;//[W]
+disp("W",QL_total,"The total rate of heat loss from the glass cover is")
\ No newline at end of file diff --git a/587/CH13/EX13.11/example13_11.sce b/587/CH13/EX13.11/example13_11.sce new file mode 100755 index 000000000..16b12ad01 --- /dev/null +++ b/587/CH13/EX13.11/example13_11.sce @@ -0,0 +1,11 @@ +clear;
+clc;
+
+//Example13.11[Radiation Shields]
+//given:-
+e=0.1;//Emissivity of aluminium sheet
+T1=800,T2=500;//Temperatures of two parallel plates[K]
+e1=0.2,e2=0.7;//Emissivities of plates
+//So9lution:-
+q12=((5.67*10^(-8))*((T1^4)-(T2^4)))/((1/e1)+(1/e2)-1+(1/e)+(1/e)-1);//[W/m^2]
+disp("W/m^2",round(q12),"Radiation Heat Transfer")
diff --git a/587/CH13/EX13.12/example13_12.sce b/587/CH13/EX13.12/example13_12.sce new file mode 100755 index 000000000..f4bdc35c0 --- /dev/null +++ b/587/CH13/EX13.12/example13_12.sce @@ -0,0 +1,11 @@ +clear;
+clc;
+
+//Example13.12[Radiation Effect on Temperature Measurements]
+//Given:-
+Tw=400,Tth=650;//Temperature of duct wall and hota air flowing in it[K]
+e=0.6;//emissivity
+h=80;//Heat transfer coefficient[W/m^2.K]
+//Solution:-
+Tf=Tth+((e*5.67*10^(-8)*((Tth^4)-(Tw^4)))/h);//[K]
+disp("K",round(Tf),"The temperature of actual air is")
diff --git a/587/CH13/EX13.13/example13_13.sce b/587/CH13/EX13.13/example13_13.sce new file mode 100755 index 000000000..058aac981 --- /dev/null +++ b/587/CH13/EX13.13/example13_13.sce @@ -0,0 +1,19 @@ +clear;
+clc;
+
+//Example13.13[Effective Emissivity of Combustion Gases]
+d=5,H=5;//Diameter and height of cylindrical furnace[m]
+T=1200;//Temp of gases[K]
+P=2;//Pressure[atm]
+yN2=0.8,yH2O=0.08,yO2=0.07,yCO2=0.05;//Volumetric Composition
+//Solution:-
+Pc=yCO2*P;//[atm]
+Pw=yH2O*P;//[atm]
+disp("atm",Pw,"and","atm",Pc,"The partial pressures of CO2 and H2O are")
+L=0.6*d;//[m]
+x=Pc*L,y=Pw*L;//[m.atm]
+ec_1=0.16,ew_1=0.23;//Emissivity of CO2 and H2O at 1 atm pressure
+Cc=1.1,Cw=1.4;//Pressure Correction Factors are
+del_e=0.048;//Emissivity correction factor at T=1200K
+e_g=Cc*ec_1+Cw*ew_1-del_e;
+disp(e_g,"The effectivity of the combustion gases is")
\ No newline at end of file diff --git a/587/CH13/EX13.14/example13_14.sce b/587/CH13/EX13.14/example13_14.sce new file mode 100755 index 000000000..4333b1838 --- /dev/null +++ b/587/CH13/EX13.14/example13_14.sce @@ -0,0 +1,24 @@ +clear;
+clc;
+
+//Example13.14[Radiation Heat Transfer in a Cylindrical Furnace]
+//Given:-
+Ts=600;//Wall Temperature[K]
+d=5,H=5;//Diameter and Height of cylindrical furnace
+Tg=1200,eg=0.45;//Average gas temperature and average emissivity of the combustion gases
+Pc=0.10,L=3,Pw=0.16;//From Previous examples
+//Solution:-
+x=Pc*L*Ts/Tg;//[m.atm]
+y=Pw*L*Ts/Tg;//[m.atm]
+ec_1=0.11,ew_1=0.25;//Emissivities of CO2 and H2O corresponding to 600K and 1atm
+Cc=1.1,Cw=1.4;//Correction Factors
+a_c=Cc*((Tg/Ts)^(0.65))*(ec_1);
+a_w=Cw*((Tg/Ts)^(0.45))*ew_1;
+disp(a_w,"and",a_c,"The absorptivities of CO2 and H2O are")
+del_a=0.027;
+a_g=a_c+a_w-del_a;
+disp(a_g,"The absorptivity of the combustion gases is")
+As=(%pi*d*H)+(%pi*(d^2)/2);//[m^2]
+disp("m^2",round(As),"the surface area of the cylindrical surface is")
+Q_net=round(As)*(5.67*10^(-8))*((eg*(Tg^4))-(a_g*(Ts^4)));
+disp("W",Q_net,"The net rate of radiation heat transfer from the combustion gases to walls of the furnace is")
\ No newline at end of file diff --git a/587/CH13/EX13.15/example13_15.sce b/587/CH13/EX13.15/example13_15.sce new file mode 100755 index 000000000..8028653ea --- /dev/null +++ b/587/CH13/EX13.15/example13_15.sce @@ -0,0 +1,17 @@ +clear;
+clc;
+
+//Example13.15[Effect of Clothing on Thermal Comfort]
+h_rad=4.7,h_conv=4.0;//The radiation and convection heat transfer coefficient[W/m^2.degree Celcius]
+R_clo=0.6*0.155;//Thermal resistance of clothing[m^2.degree Celcius/W]
+T_skin=33,T_amb=22;//Skin and Ambient temperature[degree Celcius]
+As=1.8;//Surface area of an average man
+//Solution:-
+h_comb=h_conv+h_rad;//combined heat transfer coefficient[W/m^2.degree Celcius]
+Q_sen_clo=As*(T_skin-T_amb)/(R_clo+(1/h_comb));//[W]
+disp("W",Q_sen_clo,"The sensible heat loss from this person when clothed is")
+//On removing the clothes
+//R_clo=0 Clothing resistance on removing clothes
+//Setting both heat transfer rates equal to determine new ambient air temperature
+T_amb_new=T_skin-(Q_sen_clo*(1/h_comb)/As)//[degree Celcius]
+disp("degree Celcius",T_amb_new,"The ambient temperature now is")
\ No newline at end of file diff --git a/587/CH13/EX13.2/example13_2.sce b/587/CH13/EX13.2/example13_2.sce new file mode 100755 index 000000000..224004269 --- /dev/null +++ b/587/CH13/EX13.2/example13_2.sce @@ -0,0 +1,14 @@ +clear;
+clc;
+
+//Example13.2[Fraction of Radiation Leaving through an Opening]
+//Given:-
+r1=0.1;//Radius of enclosure[m]
+L=0.1;//Length of Enclosure[m]
+r2=0.05,r3=0.08;//Inner and outer radii of the ring[m]
+//Solution:-
+//Using Chart in Fig 13.7
+F12=0.11;
+F13=0.28;
+F1_ring=F13-F12;
+disp(F1_ring,"The fraction of the radiation leaving the base cyllinder enclosure that escapes through coaxial ring opening at its top surface is")
diff --git a/587/CH13/EX13.3/example13_3.sce b/587/CH13/EX13.3/example13_3.sce new file mode 100755 index 000000000..266347521 --- /dev/null +++ b/587/CH13/EX13.3/example13_3.sce @@ -0,0 +1,11 @@ +clear;
+clc;
+
+//Example13.3[View Factors Associated with a Tetragon]
+//Given:-
+//A pyramid with square base and it's sides being isoceles triangle
+//Solution:=
+F11=0;//Since base is a flat surface
+//F12=F13=F14=F15=x
+x=(1-F11)/4;
+disp("of total radiation",x,"Each side pf the four surfaces of the pyramid recieves")
\ No newline at end of file diff --git a/587/CH13/EX13.5/example13_5.sce b/587/CH13/EX13.5/example13_5.sce new file mode 100755 index 000000000..7c30bc4ca --- /dev/null +++ b/587/CH13/EX13.5/example13_5.sce @@ -0,0 +1,15 @@ +clear;
+clc;
+
+//Example13.5[The Crossed-Strings Method for View Factors]
+a=12,b=5;//With od long parallel plates[cm]
+c=6;//Distance between the plates
+L1=a,L2=b,L3=c;
+L4=sqrt((7^2)+(6^2));
+L5=sqrt((5^2)+(6^2));
+L6=sqrt((12^2)+(6^2));
+F12_1=((L5+L6)-(L3+L4))/(2*L1);
+F13=(L1+L3-L6)/(2*L1);
+F14=(L1+L4+L5)/(2*L1);
+F12_2=1-F13-F14;
+disp(F12_1,"Therefore from two different methods F12_1=F12_2=",F13,"F13=",F14,"F14=")
\ No newline at end of file diff --git a/587/CH13/EX13.6/example13_6.sce b/587/CH13/EX13.6/example13_6.sce new file mode 100755 index 000000000..457de6405 --- /dev/null +++ b/587/CH13/EX13.6/example13_6.sce @@ -0,0 +1,18 @@ +clear;
+clc;
+
+//Example13.6[Radiation Heat Transfer in a Black Furnace]
+//Given:-
+F12=0.2;
+A=5*5;//Area of 1 surface of cube[m^2]
+Tb=800,Tt=1500,Ts=500;//Temperature of base top and the side surfaces of the furbace[K]
+//Solution:-
+F11=0;
+Q11=0;
+F13=1-F11-F12;
+Q13=A*F13*(5.67*10^(-8))*((Tb^4)-(Ts^4));//[kW]
+disp("kW",round(Q13/1000),"The net rate of heat transfer from surface1 to surface3 is")
+Q12=A*F12*(5.67*10^(-8))*((Tb^4)-(Tt^4));//[kW]
+disp("kW",round(Q12/1000),"The net rate of radiation heat transfer from siurface1 to surface2 is")
+Q1=Q11+Q12+Q13;//[kW]
+disp("kW",round(Q1/1000),"Rhe net radiation heat transfer from the base surface is")
diff --git a/587/CH13/EX13.7/example13_7.sce b/587/CH13/EX13.7/example13_7.sce new file mode 100755 index 000000000..46c76e70e --- /dev/null +++ b/587/CH13/EX13.7/example13_7.sce @@ -0,0 +1,10 @@ +clear;
+clc;
+
+//Example13.7[Radiation Heat Transfer between Parallel Plates]
+//Given:-
+T1=800,T2=500;//Temp of parallel plates[K]
+e1=0.2,e2=0.7;//Emissivities
+//Solution:-
+q12=(5.67*10^(-8))*((T1^4)-(T2^4))/((1/e1)+(1/e2)-1);
+disp("is transferred from plate 1 to plate 2 by radiation per unit surface area of either plate","W",round(q12),"The net heat at the rate of")
\ No newline at end of file diff --git a/587/CH13/EX13.8/example13_8.sce b/587/CH13/EX13.8/example13_8.sce new file mode 100755 index 000000000..a21e403e3 --- /dev/null +++ b/587/CH13/EX13.8/example13_8.sce @@ -0,0 +1,30 @@ +clear;
+clc;
+
+//Example13.8[Radiation Heat Transfer in Cylindrical Furnace]
+//Given:-
+ro=1,H=1;//Radius amd height of cylinder[m]
+e1=0.8,e2=0.4;//Emissivities
+T1=700,T2=500;//Top and base temperatures of furnace[K]
+T3=400;//Side durface temperature[K]
+F11=0,F12=0.38;
+//Solution:-
+A1=%pi*(ro^2);//[m^2]
+A2=A1;//[m^2]
+A3=2*%pi*ro*H;//[m^2]
+F13=1-F11-F12;
+F21=F12;//Top and Bottom are symmetric
+F31=F13*(A1/A3);
+F23=F13;
+F32=F31;
+function[i]=rad(J)
+ i(1)=J(1)+(((1-e1)/e1)*((F12*(J(1)-(J(2))))+(F13*((J(1))-(J(3))))))-((T1^4)*(5.67*10^(-8)));
+ i(2)=J(2)+(((1-e2)/e2)*((F21*(J(2)-J(1)))+(F23*(J(2)-J(3)))))-((T2^4)*(5.67*10^(-8)));
+ i(3)=J(3)-((T3^4)*(5.67*10^(-8)));
+ deff('[i]=rad(J)',['i_1=J(1)+(((1-e1)/e1)*((F12*(J(1)-(J(2))))+(F13*((J(1))-(J(3))))))-((T1^4)*(5.67*10^(-8)))','i_2=J(2)+(((1-e2)/e2)*((F21*(J(2)-J(1)))+(F23*(J(2)-J(3)))))-((T2^4)*(5.67*10^(-8)))','i_3=J(3)-((T3^4)*(5.67*10^(-8)))'])
+ disp(J(3),J(2),J(1))
+ Q1=A1*((F12*(J(1)-J(2)))+(F13*(J(1)-J(3))));//[kW]
+ Q2=A2*((F21*(J(2)-J(1)))+(F13*(J(2)-J(3))));//[kW]
+ Q3=A3*((F31*(J(3)-J(1)))+(F32*(J(3)-J(2))));//[kW]
+ disp("kW",Q3/1000,Q2/1000,Q1/1000,"The net rates of radiation heat transfer at the three surfaces are")
+
\ No newline at end of file diff --git a/587/CH13/EX13.9/example13_9.sce b/587/CH13/EX13.9/example13_9.sce new file mode 100755 index 000000000..301966d2c --- /dev/null +++ b/587/CH13/EX13.9/example13_9.sce @@ -0,0 +1,15 @@ +clear;
+clc;
+
+//Example13.9[Radiation Heat Transfer in a Triangular Furnace]
+//Given:-
+A1=1,A2=1,A3=1;//Area of each side[m^2]
+T1=600,T2=1000;//[K]
+e=0.7;
+F12=0.5,F13=0.5,F23=0.5;//Symmetry
+//Solution:-
+Eb1=5.67*10^(-8)*(T1^4);//[W/m^2]
+Eb2=5.67*10^(-8)*(T2^4);//[W/m^2]
+Q=(Eb2-Eb1)/(((1-e)/(A1*e))+((((A1*F12)+(1/((1/(A1*F13))+(1/(A2*F23))))))^(-1)));//[kW]
+disp("kW",round(Q/1000),"Heat at the rate of")
+disp("must be supplied to the heated surface per unit lemgth of the duct to maintain steady operation in the furnace")
\ No newline at end of file diff --git a/587/CH14/EX14.1/example14_1.sce b/587/CH14/EX14.1/example14_1.sce new file mode 100755 index 000000000..03cef30c0 --- /dev/null +++ b/587/CH14/EX14.1/example14_1.sce @@ -0,0 +1,14 @@ +clear;
+clc;
+
+//Example14.1[Determining Mass Fractions from Mole Fractions]
+//Given:-
+yN2=0.781,yO2=0.209,yAr=0.01;//Mole fractions
+M_N2=28,M_O2=32,M_Ar=39.9;//Molar Masses
+//Solution:-
+M_air=yN2*M_N2+yO2*M_O2+yAr*M_Ar;//[kg/kmol]
+disp("kg/kmol",M_air,"The molar mass of air is")
+w_N2=yN2*M_N2/M_air;
+w_O2=yO2*M_O2/M_air;
+w_Ar=yAr*M_Ar/M_air;
+disp("respectively","percent",100*w_Ar,"and","percent",100*w_O2,",","percent",100*w_N2,"The mass fractions of N2, O2 and Ar in dry standard atmosphere are")
\ No newline at end of file diff --git a/587/CH14/EX14.10/example14_10.sce b/587/CH14/EX14.10/example14_10.sce new file mode 100755 index 000000000..d19eaba60 --- /dev/null +++ b/587/CH14/EX14.10/example14_10.sce @@ -0,0 +1,24 @@ +clear;
+clc;
+
+//Example14.10[Mass Convection inside a Circular Pipe]
+//Given:-
+D=0.015;//Inner Diameter[m]
+T=300;//Temp of air[K]
+P=1;//Pressure of air[atm]
+v=1.2;//Average velocity of air[m/s]
+nu=1.58*10^(-5);//Viscosity[m^2/s]
+//Solution:-
+//Water is Species Aand air is species B
+D_AB=(1.87*10^(-10))*(T^2.072)/P;//[m^2/s]
+disp("m^2/s",D_AB,"The mass diffusivity of water vapor in air at 300K is")
+Re=v*D/nu;
+disp(round(Re),"The Reynolds number for internal flow is")
+if(Re<2300) then
+ disp("laminar Flow")
+ Sh=3.66;//Sherwood number equals to Nusselt number
+ h_mass=Sh*D_AB/D;//[m/s]
+ disp("m/s",h_mass,"The mass transfer coefficient is")
+else
+ disp("Flow is not laminar")
+end
diff --git a/587/CH14/EX14.11/example14_11.sce b/587/CH14/EX14.11/example14_11.sce new file mode 100755 index 000000000..7d9678fe7 --- /dev/null +++ b/587/CH14/EX14.11/example14_11.sce @@ -0,0 +1,30 @@ +clear;
+clc;
+
+//Example14.11[Analogy between Heat and Mass Transfer]
+//Given:-
+//Napthalene is species A and air is species B
+M_A=128.2;//Molar Mass of A[kg/kmol]
+M_air=29;//Molar mass of B[kg/kmol]
+P=101325;//Pressure of Air[Pa]
+T=298;//Temperature[K]
+D_AB=0.61*10^(-5);//[m^2/s]
+v=2;//Stream velocity[m/s]
+rho=1.184;//Density of air[kg/m^3]
+Cp=1007;//Specific Heat[J/kg.K]
+a=2.141*10^(-5);//Absorptivity[m^2/s]
+w_inf=0;//Mass fraction of napthalene at free stream conditions
+P_As=11;//Vapor Pressure of Napthalene at surface[Pa]
+mA=12*10^(-3);//Mass of napthalene sublimated[kg]
+delta_t=15*60;//time of sublimation[s]
+As=0.3;//surface area of the body[m^2]
+//Solution:-
+w_As=(P_As/P)*(M_A/M_air);
+disp(w_As,"Mass fraction at the surface is")
+m_evap=mA/delta_t;//[kg/s]
+disp("kg/s",m_evap,"The rate of evaporation of napthalene is")
+h_mass=m_evap/(rho*As*(w_As-w_inf));
+disp("m/s",h_mass,"The mass convection coefficient is")
+//Using analogy between heat and mass transfer
+h_heat=rho*Cp*h_mass*((a/D_AB)^(2/3));//[W/m^2.degree Celcius]
+disp("W/m^2.degree Celcius",round(h_heat),"The average heat transfer coefficient is")
diff --git a/587/CH14/EX14.12/example14_12.sce b/587/CH14/EX14.12/example14_12.sce new file mode 100755 index 000000000..849b7bc93 --- /dev/null +++ b/587/CH14/EX14.12/example14_12.sce @@ -0,0 +1,26 @@ +clear;
+clc;
+
+//Example14.12[Evaporative Cooling of a Canned Drink]
+//Given:-
+//Water is species A and air is species B
+M_A=18,M_B=29;;//Molar Masses of water and air[kg/kmol]
+D_AB=2.5*10^(-5);//Diffusivity of water vapor in air[m^2/s]
+T_inf=30;//Ambient Temperature[degree Celcius]
+T_avg=(20+T_inf)/2;//Average temperature
+P=101.325;//Atmospheric Pressure[kPa]
+//Properties of A at 20 degree Celcius
+h_fg=2454;//[kJ/kg]
+Pv1=2.34;//Saturation vapor pressure[kPa]
+Pv2=4.25;//Vapor Pressure at 30 degree Celcius[kPa]
+//Properties of air at average temperature and 1 atm
+Cp=1.007;//[kJ/kg]
+a=2.141*10^(-5);//[m^2/s]
+phi=0.4;//Relative Humidity
+//Solution:-
+Le=a/D_AB;
+disp(Le,"The Lewis Number is")
+Pv_inf=phi*Pv2;//[kPa]
+disp("kPa",Pv_inf,"The vapor pressure of air away from the surface is")
+Ts=T_inf-(h_fg*M_A*(Pv1-Pv_inf)/(Cp*(Le^(2/3))*M_B*P));
+disp("degree Celcius",Ts,"The temperature of the drink can be lowered to")
\ No newline at end of file diff --git a/587/CH14/EX14.13/example14_13.sce b/587/CH14/EX14.13/example14_13.sce new file mode 100755 index 000000000..8ca942520 --- /dev/null +++ b/587/CH14/EX14.13/example14_13.sce @@ -0,0 +1,72 @@ +clear;
+clc;
+
+//Example14.13[Heat Loss from Uncovered Hot Water Baths]
+//Given:-
+Ts=50+273;//Uniform temperature of water[K]
+T_surr=20+273;//Average temperature of surrounding surfaces[K]
+T_inf=25+273;//Ambient temperature[K]
+As=3.5*1;//Surface area of water bath[m^2]
+p=2*(3.5+1);//Perimeter of top surface of water bath[m]
+e=0.95;//Emissivity of liquid water
+phi=0.52;//Relative Humidity
+Rv=0.4615;//Universal Gas Constant[kPa.m^3/kg.K]
+Ra=0.287;//Universal Gas Constant[kPa.m^3/kg.K]
+g=9.81;//[m^2/s]
+//solution:-
+//(a)
+Q_rad=e*As*(5.67*10^(-8))*((Ts^4)-(T_surr^4));
+disp("W",round(Q_rad),"The radiation heat loss from the water to the surrounding surface is")
+//(b)
+Tavg=(Ts+T_inf)/2;//Average temperature[degree Celcius]
+P=92;//Atmospheric pressure[kPa]
+//At average temperature Tavg and Pressure P,Properties of dry air:-
+k=0.02644;//[W/m.degree Celcius]
+Pr=0.7262;//Prandtl number, independent of pressure
+a=(2.312*10^(-5))/P;//Absorptivity[m^2/s]
+nu=(1.849*10^(-5));//Kinematic viscosity[m^2/s]
+//At T_surr properties of water are:-
+h_fg=2383;//[kJ/kg]
+Pvs=12.35;//[kPa]
+Psat=3.17;//Saturation Pressure of water at surface temp[kPa]
+//The air at surface is saturated therefore vapor pressure at surface is simple the saturation pressure of water at the surface temperature
+Pv_inf=phi*Psat;//[kPa]
+//At the surface
+rho_vs=Pvs/(Rv*Ts);
+disp("kg/m^3",rho_vs,"Density of water vapor at the surface is")
+rho_as=(P-Pvs)/(Ra*Ts);
+disp("kg/m^3",rho_as,"Density of air at the surface is")
+rho_s=rho_vs+rho_as;
+disp("kg/m^3",rho_s,"Density of mixture at the surface is")
+//Away from the surface
+rho_vinf=Pv_inf/(Rv*T_inf);
+disp("kg/m^3",rho_vinf,"Density of vapor away from the surface is")
+rho_ainf=(P-Pv_inf)/(Ra*T_inf);
+disp("kg/m^3",rho_ainf,"Density of air away from the surface is")
+rho_inf=rho_ainf+rho_vinf;
+disp("kg/m^3",rho_inf,"The density of mixture away from the surface is")
+Lc=As/p;
+disp("m",Lc,"The characteristic length is")
+Gr=g*(rho_inf-rho_s)*(Lc^3)/(((rho_inf+rho_s)/2)*(nu^2));
+disp(Gr,"The Grashof number is")
+Nu=0.15*((Gr*Pr)^(1/3));
+disp(Nu,"The Nusselt number is")
+h_conv=Nu*k/Lc;
+disp("W/m^2.degree Celcius",h_conv,"The convection heat transfer coefficient is")
+Q_conv=h_conv*As*(Ts-T_inf);
+disp("W",ceil(Q_conv),"The natural convection heat transfer rate is")
+//(c)
+D_AB=(1.87*10^(-10))*(Tavg^2.072)/(P/101.325);
+disp("m^2/s",D_AB,"The mass diffusivity of water vapor in air at the average temperature is")
+Sc=nu/D_AB;
+disp(Sc,"The Schmidt Number is")
+Sh=0.15*((Gr*Sc)^(1/3));
+disp(Sh,"The Sherwood Number is")
+h_mass=Sh*D_AB/Lc;
+disp("m/s",h_mass,"The mass transfer coefficient is")
+mv=h_mass*As*(rho_vs-rho_vinf);
+disp("kg/s",mv,"The evaporation rate is")
+Q_evap=mv*h_fg;
+disp("kW",Q_evap,"The rate of heat transfer by evaporation is")
+Q_total=Q_rad+Q_conv+1000*Q_evap;
+disp("W",Q_total,"The total rate of heat transfer from the water to the surrounding air and surfaces is")
\ No newline at end of file diff --git a/587/CH14/EX14.2/example14_2.sce b/587/CH14/EX14.2/example14_2.sce new file mode 100755 index 000000000..2f40be9a4 --- /dev/null +++ b/587/CH14/EX14.2/example14_2.sce @@ -0,0 +1,13 @@ +clear;
+clc;
+
+//Example14.2[Mole Fraction of Water Vapor at the surface of a Lake]
+//Given:-
+P_vapor=1.705;//Partial Pressure of water vapor in the air at the lake surface is saturation pressure of watre at 15 degree Celcius[kPa]
+T_lake=15;//[degree Celcius]
+P=92;//Atmospheric pressure at lake level [kPa]
+//Solution:-
+y_vapor=P_vapor/P;
+disp(y_vapor,"The mole fraction of water vapor in the air at the surface of lake is")
+y_water=1-y_vapor;//Since water contains dissolved air
+disp(y_water,"Mole fraction of liquid water in lake")
\ No newline at end of file diff --git a/587/CH14/EX14.3/example14_3.sce b/587/CH14/EX14.3/example14_3.sce new file mode 100755 index 000000000..efbfed81a --- /dev/null +++ b/587/CH14/EX14.3/example14_3.sce @@ -0,0 +1,13 @@ +clear;
+clc;
+
+//Example14.3[Mole Fraction of Dissolved Air in Water]
+//Given:-
+P_vapor=1.96;//The partial presure of water vapor in the air at the lake surface is the saturation pressure of water at 17 degree Celcius
+H=62000;//Henry's constant for air dissolved in water at 290K[bar]
+P=92;//Atmospheric Pressure at lake level[kPa]
+//Solution:-
+P_dryair=P-P_vapor;//[kPa]
+disp("bar",P_dryair/100,"The partial pressure of dry air is")
+y_dryair=(P_dryair/100)/H;
+disp(y_dryair,"The mole fraction of air in the water is")
\ No newline at end of file diff --git a/587/CH14/EX14.4/example14_4.sce b/587/CH14/EX14.4/example14_4.sce new file mode 100755 index 000000000..ba2884f90 --- /dev/null +++ b/587/CH14/EX14.4/example14_4.sce @@ -0,0 +1,13 @@ +clear;
+clc;
+
+//Example14.4[Diffusion of Hydrogen Gas into a Nickel Plate]
+//Given:-
+s=0.00901;//Solubility of hydrogen in nickel at 358K[kmol/m^3.bar]
+P_H2=300/100;//[bar]
+M_H2=2;//Molar Mass of hydrogen[kg/kmol]
+//Solution:-
+C_H2=s*P_H2;//[kmol/m^3]
+disp("kmol/m^3",C_H2,"The molar density of hydrogen in the nickel at the interface is")
+rho_H2=C_H2*M_H2;//[kg/m^3]
+disp("kg/m^3",rho_H2,"Mass Density of hydrogen is")
diff --git a/587/CH14/EX14.5/example14_5.sce b/587/CH14/EX14.5/example14_5.sce new file mode 100755 index 000000000..b2b56eb34 --- /dev/null +++ b/587/CH14/EX14.5/example14_5.sce @@ -0,0 +1,16 @@ +clear;
+clc;
+
+//Example14.5[Diffusion of Hydrogen through a Spherical Container]
+//Given:-
+CA1=0.087,CA2=0;//Molar concentration of hydrogen in the nickel at inner and outer surfaces[kmol/m^3]
+r2=4.8/2;//Outer radius[m]
+t=0.06;//Thickness of shell[m]
+D_AB=1.2*(10^(-12));//Diffusion coefficient for hydrogen in the nickel at the specified temperature is[m^2/s]
+M_H2=2;//Molar Mass of H2[kg/kmol]
+//Solution:-
+r1=((2*r2)-(2*t))/2;//Inner radius[m]
+N_diff=4*%pi*r1*r2*D_AB*(CA1-CA2)/(r2-r1);
+disp("kmol/s",N_diff,"The molar flow rate of hydrogen through the shell by diffusion is")
+m_diff=M_H2*N_diff;
+disp("kg/s",m_diff,"The mass flow rate of hydrogen is")
\ No newline at end of file diff --git a/587/CH14/EX14.6/example14_6.sce b/587/CH14/EX14.6/example14_6.sce new file mode 100755 index 000000000..d10d4fa76 --- /dev/null +++ b/587/CH14/EX14.6/example14_6.sce @@ -0,0 +1,25 @@ +clear;
+clc;
+
+//Example14.6[Condensation and Freezing of Moisture in Walls]
+Ti=20,To=-16;//Indoor and outdoor temperatures[degree Celcius]
+R_wall=3.05;//Total thermal resistance of the wall[m^2.degree Celcius/W]
+A=1;//Heat transfer area[m^2]
+R_ext=0.40;//The thermal resistance of the exterior part of the wall beyond the insulation[m^2.degree Celcius/W]
+Rv_int=0.012+0.0004,Rv_ext=0.0138+0.019;//Indoor and outdoor vapor resistances[Pa.m^2.s/ng]
+phi1=0.6,phi2=0.7;//Indoor and outdoor Relative Humidity
+Psat1=2340,Psat2=151;//Indoor and outdoor saturation pressures[Pa]
+//Solution:-
+Q_wall=A*(Ti-To)/R_wall;//[W]
+disp("W",Q_wall,"The rate of heat transfer through unit area of wall is")
+T_I=To+(Q_wall*R_ext);
+disp("degree Celcius",T_I,"The temperature of outer sheathing interface is")
+P=234;//The saturation pressure of water at temp T_I[Pa]
+Pv1=phi1*Psat1;
+Pv2=phi2*Psat2;
+disp("Pa",round(Pv2),"and","Pa",Pv1,"The vapor pressure at the indoor and the outdoor is")
+mv_int=A*(Pv1-P)/Rv_int;
+mv_ext=A*(P-Pv2)/Rv_ext;
+disp("ng/s",mv_ext,"and","ng/s",mv_int,"The rate of moisture flow through the interior and exterior parts of the wall is")
+mv_freezing=mv_int-mv_ext;
+disp("ng/s",mv_freezing,"The moisture is freezing in the insulation at the rate of")
\ No newline at end of file diff --git a/587/CH14/EX14.7/example14_7.sce b/587/CH14/EX14.7/example14_7.sce new file mode 100755 index 000000000..56027e119 --- /dev/null +++ b/587/CH14/EX14.7/example14_7.sce @@ -0,0 +1,15 @@ +clear;
+clc;
+
+//Example14.7[Hardening of Steel by the diffusion of carbon]
+//Given:-
+D_AB=4.8*10^(-10);//Diffusion coefficient of carbon in steel at the furnace temperature [m^2/s]
+wA_i=0.0015;//Initial carbon concentration
+wA_e=0.012;//Equilibrium concentration of carbon
+wA_t=0.01;//Concentration of carbon after desired time
+x=0.0005;//Diffusion distance[m]
+//Solution:
+a=(wA_t-wA_i)/(wA_e-wA_i);
+//Tthe argument whose complimentary error function is a=0.81 is 0.17
+t=(x^2)/(4*D_AB*(0.17^2));//[seconds]
+disp("seconds",round(t),"Time taken to reach desired level of hardening is")
\ No newline at end of file diff --git a/587/CH14/EX14.8/example14_8.sce b/587/CH14/EX14.8/example14_8.sce new file mode 100755 index 000000000..4b2affd77 --- /dev/null +++ b/587/CH14/EX14.8/example14_8.sce @@ -0,0 +1,29 @@ +clear;
+clc;
+
+//Example14.8[Venting of Helium into the Atmosphere by Diffusion]
+//Given:-
+D_AB=7.2*10^(-5);//Diffusion coefficient of air in helium[m^2/s]
+M_He=4,M_air=29;//Molar masses of helium and air[kg/kmol]
+D=0.005;//Internal diameter of tube[m]
+L=15;//Length of tube[m]
+R1=8.314;//Universal Gas Constant[kPa.m^3/kmol.K]
+R2=2.0769;//Universal Gas Constant[kPa.m^3/kg.K]
+T=298;///Ambient temperature[K]
+//Solution:-
+A=%pi*(D^2)/4;//Flow area[m^2]
+P_He0=1,P_HeL=0;//Pressure of helium at x=0 i.e. bottom of tube and at x=L i.e. at the top of the tube [atm]
+N_He=D_AB*A*(P_He0-P_HeL)*(101.3)/(R1*T*L);
+disp("kmol/s",N_He,"The molar flow rate of Helium is")
+m_He=N_He*M_He;
+disp("kg/s",m_He,"Mass flow rate of helium is")
+N_air=-N_He;//Equimolar counter diffusion process
+m_air=N_air*M_air;
+disp("kg/s",m_air,"The flow rate of air into the pipeline is")
+w_air=m_air/(m_air+m_He);
+disp("which is negligible",w_air,"Mass fraction of air in the helium pipeline is")
+m_net=m_He+m_air;//[kg/s]
+//Taking density of mixture at x=0 to be the density of helium as the mass fraction of air at the bottom is very small
+rho=P_He0*101.325/(R2*T);//[kg/m^3]
+V=m_net/(rho*A);//[m/s]
+disp("m/s",V,"The average flow velocity at the bottom of the tube is")
diff --git a/587/CH14/EX14.9/example14_9.sce b/587/CH14/EX14.9/example14_9.sce new file mode 100755 index 000000000..285e448aa --- /dev/null +++ b/587/CH14/EX14.9/example14_9.sce @@ -0,0 +1,25 @@ +clear;
+clc;
+
+//Example14.9[Measuring Diffusion Coefficient by the Stefan tube]
+//Given:-
+D=0.03;//Diameter of tube[m]
+P=83.5;//Atmospheric Pressure at an elevation of 1600m[kPa]
+T=20+273;//Ambient temperature[K]
+R=8.314;//Universal Gas Constant[kPa.m^3/kmol.K]
+P_vapor0=2.34;//The saturation pressure of water at 20 degree Celcius[kPa]
+M_vapor=18;//Molar mass of water vapor[kg/kmol]
+x=0.4;//Distance from water surface to the open end of the tube[m]
+//Solution:-
+//water vapor is species A
+yA0=P_vapor0/P;
+disp(yA0,"The mole fraction of water vapor (species A) at the Interface is")
+yAL=0;//mole fraction of water vapor on the top of the tube
+C=P/(R*T);//[kmol/m^3]
+A=%pi*(D^2)/4;//[m^2]
+disp("m^2",A,"The cross sectional area of tube")
+m_vapor=(1.23*10^(-3))/(15*24*3600);//Rate of evaporation [kg/s]
+N_vapor=m_vapor/M_vapor;
+disp("kmol/s",N_vapor,"The molar flow rate of vapor is")
+D_AB=(N_vapor/A)*(x/C)/log((1-yAL)/(1-yA0));
+disp("m^2/s",D_AB,"Binary diffusion coefficient of water vapor in air at 20 degree Celcius and 83.5kPa")
\ No newline at end of file diff --git a/587/CH2/EX2.10/example2_10.sce b/587/CH2/EX2.10/example2_10.sce new file mode 100755 index 000000000..ac5566c81 --- /dev/null +++ b/587/CH2/EX2.10/example2_10.sce @@ -0,0 +1,20 @@ +clear;
+clc;
+
+//Example2.10[Combined Convection, Radiation and Heat Flux]
+//Given:-
+T_surf1=20;//Ambient temperature in the interior of the house[degree Celcius]
+T_surf2=5;// Ambient temperature outside the house[degree Celcius]
+L=0.2;// Thickness of the wall[m]
+a=0.5;// absorptivity of outer surface of wall
+h_in=6;//Convection heat transfer coefficient for inner surface of wall[W/m^2.degree Celcius]
+h_out=25;//Convection heat transfer coefficient for outer surface of wall[W/m^2.degree Celcius]
+k=0.7;//The thermal conductivity of wall material[W/m.degree Celcius]
+e=0.9;//Emissivity of outer surface of wall
+//Solution:-
+//The heat transfer though the wall is given to be steady and one dimensional and thus temperature depends on x only i.e. T=T(x)
+disp("The boundary condition ont the inner surface of the wall at x=0 can be expressed as -k(dT(0)/dx)=h_in[T_surf1-T(0)]")
+disp("degree Celcius",T_surf1,"and","W/m^2.degree Celcius",h_in,"where h_in and T_surf are respectively ")
+disp("The boundary condition on the outer surface at x=L can be expressed as ")
+disp("-kdT(L)/dx=h_out[T(L)-T_surf2]+e*sigma[(T(L)^4)-(T_sky^4)]-a*q_solar")
+disp("where T_sky is temperature of the sky and q_solar is the incident solar heat flux")
\ No newline at end of file diff --git a/587/CH2/EX2.11/example2_11.sce b/587/CH2/EX2.11/example2_11.sce new file mode 100755 index 000000000..ce3b24229 --- /dev/null +++ b/587/CH2/EX2.11/example2_11.sce @@ -0,0 +1,22 @@ +clear;
+clc;
+
+//Example2.11[Heat Conduction in a Plane Wall]
+//Given:-
+k_wall=1.2;//Thermal conductivity of wall[W/m.degree Celcius]
+L=0.2;//Thickness of wall[m]
+As=15;//Surface area[m^2]
+T1=120,T2=50;//The two sides of the wall are maintained at these constant temperatures[degree Celcius]
+//Solution (a)
+disp("Differential equation can be expressed as d^2(T)/(dx^2)=0")
+disp("with boundary conditions ")
+disp("degree Celcius",T1,"T(0)=T1=")
+disp("degree Celcius",T2,"T(L)=T2=")
+disp("integrating this we get,")
+disp("dT/dx=C1", ,"where C1 is an arbitrary constant")
+disp("integrating we obtain temperature to follow following relation :-")
+disp("and substituting values in above equation","T(x)=((T2-T1)/L)*x+T1 ")
+T3=(((T2-T1)/L)*(0.1))+T1;
+disp("degree Celcius",T3,"The value of temperature at x=0.1m is")
+Q_wall=-k_wall*As*((T2-T1)/L);//[W]
+disp("W",Q_wall,"The rate of heat conduction through the wall is")
\ No newline at end of file diff --git a/587/CH2/EX2.13/example2_13.sce b/587/CH2/EX2.13/example2_13.sce new file mode 100755 index 000000000..8328d0729 --- /dev/null +++ b/587/CH2/EX2.13/example2_13.sce @@ -0,0 +1,20 @@ +clear;
+clc;
+
+//Example2.13[Heat Conduction in the Base Plate of an Iron]
+//Given:-
+k=15;//[W/m.degree Celcius]
+A=300*10^(-4);//Base Area[m^2]
+L=0.005;//Thickness[m]
+T_surr=20;//Temp of surrounding[degree Celcius]
+h=80;//Convection het transfer coefficient[W/m^2.degree Celcius]
+Q=1200;//[W]
+//Solution:-
+q=Q/A;//[W/m^2]
+disp("W/m^2",q,"Uniform Heat Flux to whicj inner surface of the base plate is subjected")
+//Integration Constants
+C1=-q/k;
+C2=T_surr+(q/h)+(q*L/k);
+T_0=T_surr+q*((L/k)+(1/h));//[degree Celcius]
+T_L=T_surr+q*(1/h);//[degree Celcius]
+disp("respectively","degree Celcius",T_L,"and","degree Celcius",round(T_0),"The temperatures at the inner and outer surfaces of the plate i.e. at x=0 and x=L are")
\ No newline at end of file diff --git a/587/CH2/EX2.14/example2_14.sce b/587/CH2/EX2.14/example2_14.sce new file mode 100755 index 000000000..d49bf3ec8 --- /dev/null +++ b/587/CH2/EX2.14/example2_14.sce @@ -0,0 +1,23 @@ +clear;
+clc;
+
+//Example2.14[Heat Conduction in a Solar Heated Wall]
+//Given:-
+L=0.06;//Thickness of wall[m]
+k=1.2;//Thermal Conductivity[W/m.degree Celcius]
+e=0.85;//Emissivity
+a=0.26;//Solar absorptivity
+T1=300;//Temp of Inner surface of Wall[K]
+q_solar=800;//Incident rate of solar radiation[W/m^2]
+T_space=0;//Temp of outer space[K]
+//Solution:-
+//Integrating results into
+function[f]=temp(T)
+ f(1)=(((a*q_solar)-(e*5.67*10^(-8)*T(1)^4))*(L/k))+T1-T(1);
+ deff('[f]=temp(T)',['f_1(T)=(((a*q_solar)-(e*5.67*10^(-8)*T(1)^4))*(L/k))+T1-T(1)'])
+endfunction
+
+ disp("K",xs,"The outer surface temperature is ")
+ //First execute the program with x0=[1] then [xs,fxs,m]=fsolve(x0',temp) then re-execute to obtain correct output as for 1st exeution 'xs' is undefined
+ q=k*(T1-xs)/L;
+ disp("W/m^2",round(q),"The steady rate of heat conduction through the wall is")
\ No newline at end of file diff --git a/587/CH2/EX2.15/example2_15.sce b/587/CH2/EX2.15/example2_15.sce new file mode 100755 index 000000000..c21bcd5e1 --- /dev/null +++ b/587/CH2/EX2.15/example2_15.sce @@ -0,0 +1,17 @@ +clear;
+clc;
+
+//Example2.15[Heat Loss through a Steam Pipe]
+//Given:-
+L=20;//Pipe Length[m]
+k=20;//[W/m.degree Celcius]
+r1=0.06;//Inner Radius[m]
+r2=0.08;//Outer Radius[m]
+T1=150;//Temp of inner surface[degree Celcius]
+T2=60;//Temp of outer surface[degree Celcius]
+//Solution:-
+//Integrating differential equation we get T(r)=C1logr+C2, where C1 and C2 are
+C1=(T2-T1)/log(r2/r1);
+C2=T1-((T2-T1/log(r2/r1))*log(r1));
+Q_cyl=2*%pi*k*L*(T1-T2)/(log(r2/r1));
+disp("kW",round(Q_cyl/1000),"The rate of heat conduction through the pipe is")
diff --git a/587/CH2/EX2.16/example2_16.sce b/587/CH2/EX2.16/example2_16.sce new file mode 100755 index 000000000..058c758f3 --- /dev/null +++ b/587/CH2/EX2.16/example2_16.sce @@ -0,0 +1,16 @@ +clear;
+clc;
+
+//Example2.16[Heat Conduction through a Spherical Shell]
+//Given:-
+r1=0.08;//Inner Radius[m]
+r2=0.1;//Outer radius[m]
+k=45;//Thermal conductivity[W/m.degree Celcius]
+T1=200;//Temperature of inner surface[degree Celcius]
+T2=80;//Temperarure of outer surface[degree Celcius]
+//Solution:-
+//Integrating the differential equation twicw we get T(r)=-C1/r+C2, where
+C1=r1*r2*(T1-T2)/(r2-r1);
+C2=((r2*T2)-(r1*T1))/(r2-r1);
+Q_sphere=4*%pi*k*r1*r2*(T1-T2)/(r2-r1);
+disp("kW",Q_sphere/1000,"The rate of heat conduction through the container wall is")
diff --git a/587/CH2/EX2.17/example2_17.sce b/587/CH2/EX2.17/example2_17.sce new file mode 100755 index 000000000..f6f6f63bb --- /dev/null +++ b/587/CH2/EX2.17/example2_17.sce @@ -0,0 +1,16 @@ +clear;
+clc;
+
+//Example2.17[Centerline Temperature of a Resistance Heater]
+//Given:-
+k=15;//Thermal conductivity of heater wire[W/m.K]
+E_gen=2000;//Total heat generation[W]
+l=0.5;//Length of resistance heater wire[m]
+D=0.004;//Diameter of wire[m]
+Ts=105;//Outer sorface Temperarure[degree Celcius]
+//Solution:-
+V_wire=%pi*(D^2)*l/4;//Volume of wire[m^3]
+e_gen=E_gen/V_wire;//[W/m^3]
+disp("W/m^3",e_gen,"The heat generation per unit volume of the wire is")
+Tc=Ts+(e_gen*(D^2)/(4*4*k));//[degree Celcius]
+disp("degree Celcius",round(Tc),"The center temperature of the wire is ")
diff --git a/587/CH2/EX2.18/example2_18.sce b/587/CH2/EX2.18/example2_18.sce new file mode 100755 index 000000000..72636f37d --- /dev/null +++ b/587/CH2/EX2.18/example2_18.sce @@ -0,0 +1,14 @@ +clear;
+clc;
+
+//Example2.18[Variation of Temperature in a Resistance Heater]
+//Given:-
+k=13.55;//[W/m.degree Celcius]
+ro=0.005;//[m]
+e_gen=4.3*10^7;//rate of resistance heating[W/m^3]
+Ts=108;//Surface temperature[degree Celcius]
+//Solution:-
+//Integrating we get
+//T(r)=Ts+((e_gen*(ro^2-r^2)/4k))
+T_0=Ts+((e_gen*ro^2)/(4*k));
+disp("degree Celcius",round(T_0),"The temperature at the centreline,r=0 is")
diff --git a/587/CH2/EX2.19/example2_19.sce b/587/CH2/EX2.19/example2_19.sce new file mode 100755 index 000000000..cdc440bfa --- /dev/null +++ b/587/CH2/EX2.19/example2_19.sce @@ -0,0 +1,15 @@ +clear;
+clc;
+
+//Example2.19[Heat Conduction in a two layer medium]
+//Given:-
+k_wire=15,k_ceramic=1.2;//[W/m.degree Celcius]
+r1=0.002,r2=0.007;//[m]
+e_gen=50*10^6;//[W/m^3]
+Ts=45;//[degree Celcius]
+//Solution:-
+T1=(((e_gen*(r1^2)*log(r2/r1))/(2*k_ceramic))+Ts);//[degree Celcius]
+disp("degree Celcius",T1,"The Interface temperature is")
+T_wire=T1+((e_gen*(r1^2))/(4*k_wire));//[degree Celcius]
+disp("degree Celcius",T_wire,"The temperature at the centreline(r=0) is")
+disp("Thus the temperature of the centreline is slightly above the interface temperature")
\ No newline at end of file diff --git a/587/CH2/EX2.2/example2_2.sce b/587/CH2/EX2.2/example2_2.sce new file mode 100755 index 000000000..17185e937 --- /dev/null +++ b/587/CH2/EX2.2/example2_2.sce @@ -0,0 +1,14 @@ +clear;
+clc;
+
+//Example2.2[Heat Generation in a Hair Dryer]
+//Given:-
+E_gen=1200;//[Total rate of heat generation]
+L=80;//Length of wire[cm]
+D=0.3;//Diameter of wire[cm]
+//Solution:-
+V_wire=%pi*(D^2)*L/4;//Volume of the wire[cm^3]
+e_gen=E_gen/V_wire;//[W/cm^3]
+As=%pi*D*L;//Suface Area of wire[m^2]
+Q_=E_gen/As;//[W/cm^2]
+disp("W/cm^2",Q_,"and","W/cm^3",round(e_gen),"The rate of heat generation in the wire per unit volume and heat flux on the outer surface of the wire as a result of this heat generation are respectively")
\ No newline at end of file diff --git a/587/CH2/EX2.21/example2_21.sce b/587/CH2/EX2.21/example2_21.sce new file mode 100755 index 000000000..274d65dc9 --- /dev/null +++ b/587/CH2/EX2.21/example2_21.sce @@ -0,0 +1,17 @@ +clear;
+clc;
+
+//Example2.21[Heat Conduction through a Wall with k(T)]
+//Given:-
+//k varies with temperature as k=k0(1+bT)
+k0=38;//[W/m]
+b=9.21*(10^(-4));//[k^(-1)]
+h=2,w=0.7,t=0.1;//Height,width and thickness of plates respectively[m]
+T1=600,T2=400;//Temperature maintained on the two sides of the plate[K]
+//Solution:-
+A=h*w;//Surface area of plate[m^2]
+Tavg=(T1+T2)/2;//Average temperature of plate[K]
+kavg=k0*(1+(b*Tavg));//[W/m.K]
+disp("W/m.K",kavg,"The average thermal conductivity of the medium is")
+Q_=kavg*A*(T1-T2)/t;//[W]
+disp("kW",round(Q_/1000),"The rate of heat conduction through the plate is")
\ No newline at end of file diff --git a/587/CH2/EX2.4/example2_4.sce b/587/CH2/EX2.4/example2_4.sce new file mode 100755 index 000000000..17836636d --- /dev/null +++ b/587/CH2/EX2.4/example2_4.sce @@ -0,0 +1,16 @@ +clear;
+clc;
+
+//Example2.3[Heat Conduction in a Resistance Heater]
+//Given:-
+E_gen=2000;//Total rate of heat generation in the wire[W]
+L=0.5;//Length of cyllindrical shaped wire[m]
+D=0.004;//Diameter of wire[m]
+k_heater=15;//Thermal conductivity of wire[W/m.K]
+//Solution:-
+//The resistance wire is considered to be a very long cylinder since its length is more than 100 times its diameter.Heat is generated uniformly in the wire and the conditions on the outer surface of the wire are uniform.Hence it is reasonable to expect the temperature int he wire to vary in radial r direction only and thus heat transfer to be one dimensional,T=T(r)
+V_wire=%pi*(D^2)*L/4;//Volume of the wire[m^3]
+e_gen=E_gen/V_wire;//[W/m^3]
+disp("W/m^3",e_gen,"The rate of heat generation in the wire per unit volume is")
+const=e_gen/k_heater;
+disp("= 0",const,"The equation governing the variation of temperature int he wire is simply (1/r)d/dr(r.dT/dr)+")
\ No newline at end of file diff --git a/587/CH2/EX2.5/example2_5.sce b/587/CH2/EX2.5/example2_5.sce new file mode 100755 index 000000000..7ef00a219 --- /dev/null +++ b/587/CH2/EX2.5/example2_5.sce @@ -0,0 +1,11 @@ +clear;
+clc;
+
+//Example2.5[Cooling of a Hot Metal Ball in Air]
+//Given:-
+T_ball=300;//Temeprature of ball[degree Celcius]
+T_surr=25;//Temperature of ambient air[degree Celcius]
+//Solution:-
+//The ball in initially at a uniform temperature and is cooled uniformly from the entire outer surface.Also, the temperature within the ball changes with the radial distance r and the time t. T=T(r,t)
+disp("The thermal conductivity is given to be variable,and there is no heat generation in the ball, therefore the differential equation governing the variation of temperature in the ball is")
+disp("(1/(r^2)d/dr((r^2)k(dT/dr))=rho*c(dT/dt)")
\ No newline at end of file diff --git a/587/CH2/EX2.6/example2_6.sce b/587/CH2/EX2.6/example2_6.sce new file mode 100755 index 000000000..fcf80bc41 --- /dev/null +++ b/587/CH2/EX2.6/example2_6.sce @@ -0,0 +1,14 @@ +clear;
+clc;
+
+//Example2.6[Heat Conduction in a Short Cylinder]
+//Given:-
+//Radius R and height h of the small cylinder
+T=300;//Temperature of cylinder[degree Celcius]
+T_ambient=20;//Temperature of ambient air[degree Celcius]
+//Variation is thermal conductivity is negligible
+//The cylinder is cooled unifromly from the top and bottom surfaces in the z-direction as well as the lateral surface in the radial r-direction.Also Temperature at any point in the ball changes with time during cooling. Therefore this is a two dimensional transient heat conduction problem i.e. T=T(r,z,t)
+disp("The differential equation governing the variation of temperature in the billet is ")
+disp("((1/r)(d/dr)(k*r(dT/dr)))+((d/dz)(k(dT/dz)))=rho*c(dT/dt)")
+disp("In case of constant thermal conductivity it reduces to")
+disp("((1/r)(d/dr)(r(dT/dr)))+(d^2)T/(dz^2)=(1/a)(dT/dt)")
\ No newline at end of file diff --git a/587/CH2/EX2.7/example2_7.sce b/587/CH2/EX2.7/example2_7.sce new file mode 100755 index 000000000..05ecabfad --- /dev/null +++ b/587/CH2/EX2.7/example2_7.sce @@ -0,0 +1,20 @@ +clear;
+clc;
+
+//Example2.7[Heat Flux boundary Condition]
+//Given:-
+Q=800;//Heat transfer rate[W]
+D=0.2;//Diameter of pan[m]
+L=0.003;//Thickness of pan[m]
+T_in=110;//T(L) Temperature of the inner surface of the pan[degree Celcius]
+neta=0.9;//Percent of total heat transferred to the pan
+//Solution;-
+//The inner and outer surfaces of the bottom section of the pan can be represented by x=0 and x=L,respectively. During steady operation the temperature will depend on x only and thus T=T(x).
+//Solution:-
+actual_Q=neta*Q;//90 percent of the 800W is transferred to the pan at that surface
+A=%pi*(D^2)/4;//Bottom Surface Area[m^2]
+disp("-k*dT(0)/dx=q_")
+q_=actual_Q/(1000*A);//[kW/m^2]
+//The boundary condition on this surface can be expressed as
+disp("degree Celcius",T_in,"T(L)=")
+disp("m",L,"where L=")
diff --git a/587/CH2/EX2.8/example2_8.sce b/587/CH2/EX2.8/example2_8.sce new file mode 100755 index 000000000..a231f9002 --- /dev/null +++ b/587/CH2/EX2.8/example2_8.sce @@ -0,0 +1,15 @@ +clear;
+clc;
+
+//Example2.8[Convection and Insulation Boundary Conditions]
+//Given:-
+T_steam=200;//Temperature of steam[degree Celcius]
+r_in=0.08;//Inner radii of pipe[m]
+r_out=0.085;//Outer radii of pipe[m]
+h=65;//convection heat transfer coefficient on the inner surface of the pipe[W/m^2.K]
+//Heat transfer through the pipe material predominantly is in the radial direction and thus can be approximated as being one-dimensional
+disp("Taling the direction of heat transfer to be the positive r direction, the boundary condition on that surface can be expressed as")
+disp("-k(dT(r_in,t)/dr)=h(T_steam-T(r1))")
+//The pipe is said to be well insulated on the outside, and thus heat loss through the outer surface of the pipe can be assumed to be negligible.
+disp("Then the boundary at the outer surface can be expressed as")
+disp("dT(r_out,t)/dr=0")
\ No newline at end of file diff --git a/587/CH2/EX2.9/example2_9.sce b/587/CH2/EX2.9/example2_9.sce new file mode 100755 index 000000000..2373d66ff --- /dev/null +++ b/587/CH2/EX2.9/example2_9.sce @@ -0,0 +1,21 @@ +clear;
+clc;
+
+//Example2.9[Combined Convection and Radiation Condition]
+//Given:-
+T_ball=300;//Temperature of spherical metal ball[degree Celcius]
+T_ambient=27;//Temperature of ambient air[degree Celcius]
+k=14.4;//Thermal conductivity of the ball material[W/m.K]
+h=25;//average convection heat transfer coefficient on the outer surface of the ball[W/m^2.K]
+e=0.6;//Emissivity of outer surface of the ball
+T_surr=290;//
+//This is one-dimensional transient heat transfer problem since the temperature within the ball changes with the radial distance r and the time t i.e. T=T(r,t)
+//Taking the moment the ball is removed from the oven to be t=0
+disp("The initial condition can be expressed as")
+disp("T(r,0)=T_ball")
+disp("degree Celcius",T_ball)
+//The problem possesses symmetry about the mid point(r=0) since the isotherms in this case are concentric spheres, and thus no heat is crossing the mid point of the ball.
+disp("The boundary condition at the midpoint i.e. r=0 can be expressed as dT(0,t)/dr=0")
+//The heat conducted to the outer surface of the ball is lost to the environment by convection and radiation.
+disp("Taking the direction of heat transfer to be the positive r direction, the boundary condition on the outer surface can be expressed as")
+disp("-k(dT(r_out,t)/dr)=h[T(r_out)-T_ambient]+e*sigma[(T(r_out)^4)-(T_ambient^4)]")
\ No newline at end of file diff --git a/587/CH3/EX3.1/example3_1.sce b/587/CH3/EX3.1/example3_1.sce new file mode 100755 index 000000000..12226fa40 --- /dev/null +++ b/587/CH3/EX3.1/example3_1.sce @@ -0,0 +1,21 @@ +clear;
+clc;
+//Example 3.1 [Heat Loss through a Wall]
+
+//assumptions:-
+//1)Heat transfer through the wall is steady
+//2)Heat transfer is one-imensional
+//Properties:
+k=0.9;//[W/m.K]
+disp("W/m.K",k,"The thermal conductivity is given to be")
+//Heat transfer through the wall is by conduction
+A=(3*5);//[m^2]
+disp("m^2",A,"The area of the wall is")
+T1=16;//temperature of inner wall[degree Celcius]
+T2=2;//Temperature of Outer wall[degree Celcius]
+delta_T=T1-T2;//Temperature Gradient[degree Celcius]
+L=0.3;//Length of wall along which heat is being transferred[m]
+R_wall=L/(k*A);//[degree Celcius/W]
+disp("degree Celicus/W",R_wall,"Thermal Resistnace offered by the wall is")
+Q_=(delta_T/R_wall);//[W]
+disp("W",Q_,"The steady rate of heat transfer through the wall is ")
diff --git a/587/CH3/EX3.10/example3_10.sce b/587/CH3/EX3.10/example3_10.sce new file mode 100755 index 000000000..c9e4cda75 --- /dev/null +++ b/587/CH3/EX3.10/example3_10.sce @@ -0,0 +1,12 @@ +clear;
+clc;
+
+//Example3.10[Maximum Power dissipation of a Transistor]
+//Given:-
+T_ambient=25;//Ambient temperature[degree Celcius]
+T_case=85;//Maximum temperature of the case[degree Celcius]
+R_case_ambient=20;//Resistance for convection b/w case and ambient [degree Celcius/W]
+//Solution:-
+Q_=(T_case-T_ambient)/R_case_ambient;//[W]
+disp("W",Q_,"The given power transistor should not be operated at power levels above")
+disp("if is its case temperature is not to exceed 85 degree Celcius")
\ No newline at end of file diff --git a/587/CH3/EX3.11/example3_11.sce b/587/CH3/EX3.11/example3_11.sce new file mode 100755 index 000000000..1c2de75f5 --- /dev/null +++ b/587/CH3/EX3.11/example3_11.sce @@ -0,0 +1,11 @@ +clear;
+clc;
+
+//Example3.11[Selecting a Heat Sink for a Transistor]
+//Given:-
+Q_=60;//Rate of heat transfer from given transistor at at full power[W]
+T_ambient=30;//Temperature of ambient air[degree Celcius]
+T_case=90;//Maximum temperature of case[degree Celcius]
+//Solution:-
+R_sink=(T_case-T_ambient)/Q_;//[degree Celcius/W]
+disp("degree Celcius/W",R_sink,"The thermal resistance b/w the transistor attached to the heat sink and the ambient air for the specified temperature difference is ")
\ No newline at end of file diff --git a/587/CH3/EX3.12/example3_12.sce b/587/CH3/EX3.12/example3_12.sce new file mode 100755 index 000000000..3ff875a66 --- /dev/null +++ b/587/CH3/EX3.12/example3_12.sce @@ -0,0 +1,48 @@ +clear;
+clc;
+
+//Example3.12[Effect of fins on Heat transfer from steam pipes]
+//Given:-
+k_fin=180;//thermal conductivity of aluminium alloy fins[W/m.degree Celcius]
+D_tout=0.03;//Outer diameter of tube[m]
+D_fout=0.06;//Outer diameter of circular fins[m]
+t=0.002;//thickness of fin[m]
+s=0.003;//distance between fins attached to the tube[m]
+n=200;//number of fins per meter of tube
+L=1;//length of tube[m]
+T_surr=25;//Surrounding temperature[degree Celcius]
+T_wall=120;//Temperature of wall of the tube[degree Celcius]
+h=60;//Combined heat transfer coefficient[W/m^2.degree Celcius]
+//Solution:-
+disp("In case of no fins")
+A_nf=%pi*D_tout*L;//Area of tube with no fins attached[m^2]
+//Using Newton's Law of cooling
+Q_nf=h*A_nf*(T_wall-T_surr);//[W]
+disp("W",Q_nf,"Rate of heat transfer when no finis attached")
+//The efficiency of the circular fins attached to a circular tube is plotted in Fig 3.43
+L_fin=(D_fout-D_tout)/2;//[m]
+//In this case we have following corrected parameters
+r2c=(D_fout+t)/2;//[m]
+Lc=L_fin+(t/2);//[m]
+Ap=Lc*t;//[m^2]
+r=r2c/(D_tout/2);
+alpha=(Lc*sqrt(Lc))*sqrt(h/(k_fin*Ap));//efficiency
+disp(alpha)
+//for above value of alpha efficiency is found out from the plot in fig 3.43
+neta=0.96;
+A_f=2*%pi*((r2c^2)-((D_tout/2)^2));//Area of tube with fins attached to it[m^2]
+Q_f_max=h*A_f*(T_wall-T_surr);//maximum rate of heat transfer[W]
+Q_f=neta*Q_f_max;//Heat transfer through tube with fins is efficiency times the maximum rate of heat transfer[W]
+disp("W",Q_f,"Heat transfer due to the finned tube")
+//From unfinned portion
+A_uf=%pi*D_tout*s;//Unfinned area between two consecutive fins[m^2]
+Q_uf=h*A_uf*(T_wall-T_surr);//[W]
+disp("W",Q_uf,"Heat transfer from the unfinned portion of the tube is")
+//Since there are 200 fins per meter of the tube hence 200 interfin spacing
+Q_tf=n*(Q_f+Q_uf);//[W]
+disp("W",Q_tf,"The total Heat transfer from the finned tube is")
+Q_increase=Q_tf-Q_nf;//[W]
+disp("W",Q_increase,"The increase in heat transfer from the tube per meter of length as a result of the addition of fins is")
+eff=Q_tf/Q_nf;//Effectiveness
+disp(eff,"The rate of heat transfer from the steam tube increases by a factor of")
+disp("as a result of adding fins")
diff --git a/587/CH3/EX3.13/example3_13.sce b/587/CH3/EX3.13/example3_13.sce new file mode 100755 index 000000000..c5605827b --- /dev/null +++ b/587/CH3/EX3.13/example3_13.sce @@ -0,0 +1,18 @@ +clear;
+clc;
+
+//Example3.13[Heat Loss from Buried Steam Pipes]
+//Given:-
+T_esurf=10;//Surface temperatur of earth[degree Celcius]
+T_psurf=80;//Outer surface temperature of pipe[degree Celcius]
+k_soil=0.9//Thermal Conductivity of soil[W/m.degree Celcius]
+L=30;//Length of pipe[m]
+D=0.1;//Diameter of pipe[m]
+z=0.5;//Depth at which pipe is kept[m]
+//Solution:-
+//Calculating shape factor
+if(z>(1.5*D))then
+ S=(2*%pi*L)/(log((4*z)/D)), end;//[m]
+ disp(S,"Shape factor is")
+ Q_=S*k_soil*(T_psurf-T_esurf);//[W]
+ disp("W",Q_,"The steady rate of heat transfer from the pipe is")
\ No newline at end of file diff --git a/587/CH3/EX3.14/example3_14.sce b/587/CH3/EX3.14/example3_14.sce new file mode 100755 index 000000000..9fe3be6a1 --- /dev/null +++ b/587/CH3/EX3.14/example3_14.sce @@ -0,0 +1,17 @@ +clear;
+clc;
+
+//Example3.14[Heat Transfer between Hot and Cold Water pipes]
+//Given:-
+T_hot=70;//Surface Temperature of hot pipe[degree Celcius]
+T_cold=15;//Surface Temperature of cold pipe[degree Celcius]
+L=5;//Length of both pipes[m]
+D=0.05;//Diameter of both the pipes[m]
+z=0.3;//Distance between centreline of both the pipes[m]
+k=0.75;//Thermal Conductivity of the concerte[W/m.degree Celcius]
+//Solution:-
+//Calculating Shape Factor
+S=(2*%pi*L)/(acosh(((4*(z^2))-(D^2)-(D^2))/(2*D*D)));//[m]
+disp("m",S,"Shave factor for given configuration is")
+Q_=S*k*(T_hot-T_cold);//[W]
+disp("W",Q_,"The steady rate of heat transfer between the pipes becomes")
diff --git a/587/CH3/EX3.15/example3_15.sce b/587/CH3/EX3.15/example3_15.sce new file mode 100755 index 000000000..ca05c8108 --- /dev/null +++ b/587/CH3/EX3.15/example3_15.sce @@ -0,0 +1,32 @@ +clear;
+clc;
+
+//Example3.15[Cost of Heat Loss through walls in winter]
+//Given:-
+R_va_insu=2.3;//thickness to thermal conductivity ratio[m^2.degreeCelcius/W]
+L1=12;//length of first wall of house[m]
+L2=12;//length of second wall of house[m]
+L3=9;//length of third wall of house[m]
+L4=9;//length of fourth wall of house[m]
+H=3;//height of all the walls[m]
+T_in=25;//Temperature inside house[degree Celcius]
+T_out=7;;//average temperature of outdoors on a certain day[degree Celcius]
+ucost=0.075;//Unit Cost of elctricity[$/kWh]
+h_in=8.29,h_out=34.0;//Heat transfer coefficients for inner and outer surface of the walls respectively[W/m^2.degree Celcius]
+v=24*(3600/1000);//velocity of wind[m/s]
+//Solution:-
+//Heat transfer Area of walls=(Perimeter*Height)
+A=(L1+L2+L3+L4)*H;//[m^2]
+//Individual Resistances
+R_conv_in=1/(h_in*A);//Convection Resistance on inner surface of wall[degree Celcius/W]
+R_conv_out=1/(h_out*A);//Convection Resistance on outer surface of wall[degree Celcius/W]
+R_wall=R_va_insu/A;//Conduction resistance to wall[degree Celcius/W]
+//All resistances are in series
+R_total=R_conv_in+R_wall+R_conv_out;//[degree Celcius/W]
+Q_=(T_in-T_out)/R_total;//[W]
+disp("W",Q_,"The steady rate of heat transfer through the walls of the house is")
+delta_t=24;//Time period[h]
+Q=(Q_/1000)*delta_t;//[kWh/day]
+disp("kWh/day",Q,"The total amount of heat lost through the walss during a 24 hour period ")
+cost=Q*ucost;//[$/day]
+disp("per day",cost,"Cost of heat consumption is $")
diff --git a/587/CH3/EX3.16/example3_16.sce b/587/CH3/EX3.16/example3_16.sce new file mode 100755 index 000000000..728a8eae6 --- /dev/null +++ b/587/CH3/EX3.16/example3_16.sce @@ -0,0 +1,24 @@ +clear;
+clc;
+
+//Example3.16[The R-value of a Wood Frame Wall]
+//Given:-
+f_area_insu=0.75;//area fraction for the insulation section
+f_area_stud=0.25;//area fraction for the stud
+R_bstud=3.05;//Total unit thermal resistance of section between studs[m^.degree Celcius/W]
+R_atstud=1.23;//Total unit thermal resistance of section at studs[m^.degree Celcius/W]
+P=50;//Perimeter of the building[m]
+H=2.5;//height of the walls[m]
+T_in=22;//Temperature inside the walls[degree Celcius]
+T_out=-2;//Temperature outside the walls[degree Celcius]
+//Solution:-
+U_bstud=1/R_bstud;//[W/m^2.degree Celcius]
+U_atstud=1/R_atstud;//[W/m^2.degree Celcius]
+Total_U=(f_area_insu*U_bstud)+(f_area_stud*R_atstud);//[W/m^2.degree Celcius]
+disp("W/m^",Total_U,"Overall U factor is")
+disp("degree Celcius.m^2/W",(1/Total_U),"Overall unit thermal Resistance is")
+///Since glazing constitutes 20% of the walls,
+A_wall=(0.80)*P*H;//[m^2]
+Q_=Total_U*A_wall*(T_in-T_out);//[W]
+disp("W",Q_,"The rate of heat loss through the walls under design conditions is")
+//Answer is slighthly different from book because of no of digits after decimal pont used here is quite large
\ No newline at end of file diff --git a/587/CH3/EX3.17/example3_17.sce b/587/CH3/EX3.17/example3_17.sce new file mode 100755 index 000000000..80dd7d4f5 --- /dev/null +++ b/587/CH3/EX3.17/example3_17.sce @@ -0,0 +1,16 @@ +clear;
+clc;
+
+//Example13.17[The R value of a Wall with Rigid Foam]
+//Given:-
+//using values from previous example
+R_old=2.23;//AS written in book[m^2.degree Celcius/W]
+//R value of of the fibreboard and the foam insulation, respectively
+R_removed=0.23;//[m^2.degree Celcius/W]
+R_added=0.98;//[m^2.degree Celcius/W]
+//Solution:-
+R_new=R_old-R_removed+R_added;//[m^2.degree Celcius/W]
+increase=((R_new-R_old)/R_old)*100;
+disp("m^2.degree Celcius/W",R_old,"Old R value is")
+disp("m^2.degree Celcius/W",R_new,"New R value is")
+disp(increase,"Percent increase in R-value")
\ No newline at end of file diff --git a/587/CH3/EX3.2/example3_2.sce b/587/CH3/EX3.2/example3_2.sce new file mode 100755 index 000000000..32bf6e3f2 --- /dev/null +++ b/587/CH3/EX3.2/example3_2.sce @@ -0,0 +1,28 @@ +clear;
+clc;
+
+//Example 3.2[Heat Loss through a Single Pane Window]
+//Assumptions :-
+//1)Heat transfer through the window is steady
+//2)Heat transfer through the wall is one dimensional
+k=0.78;//[W/m.K]
+disp("W/m.K",k,"The thermal conductivity is given to be")
+L=0.008;//Thickness of glass window[m]
+A=(0.8*1.5);//Area of the window[m^2]
+T_1=20;//Temeprature of inner surface of glass window[dgree Celcius]
+T_2=-10;//Temeprature of outer surface of glass window[dgree Celcius]
+h_in=10;//Heat transfer coefficient on the inner surface of the window[W/m^2]
+h_out=40;//Heat transfer coefficient on the outer surface of the window[W/m^2]
+//Convection Resistance
+R_conv1=1/(h_in*A);//[degree Celcius/W]
+R_conv2=1/(h_out*A);//[degree Celcius/W]
+//Conduction Resistance
+R_cond=L/(k*A);//[degree Celcius/W]
+//Net Resistance are in series
+R_total=R_conv1+R_conv2+R_cond;//[degree Celcius/W]
+disp("degree Celcius/W",R_total,"The total Resistance offered by glass window")
+Q_=(T_1-T_2)/R_total;//[W]
+disp("W",Q_,"Steady rate of Heat Transfer through the window is")
+//Knowing the rate of Heat Transfer
+T1=T_1-(Q_*R_conv1);//[degree Celciusthe inner surface temperature of the window glass can be determined from]
+disp("degree Celcius",T1,"Inner Surface Temperature of the window glass")
diff --git a/587/CH3/EX3.3/example3_3.sce b/587/CH3/EX3.3/example3_3.sce new file mode 100755 index 000000000..fffe46cb8 --- /dev/null +++ b/587/CH3/EX3.3/example3_3.sce @@ -0,0 +1,30 @@ +clear;
+clc;
+
+//Example3.3[:Heat Loss through double pane windows]
+//Given:-
+k_g=0.78;//Thermal conductitvity of glass [W/m.K]
+k_a=0.026;//Thermal conductivity of air space[W/m.K]
+L_g=.004;//Thickness of glass layer[m]
+L_a=0.01;//Thickness of air space[m]
+h_in=10;//ConvectionHeat transfer coefficient on the inner surface of the window[W/m^2]
+h_out=40;//ConvectionHeat transfer coefficient on the outer surface of the window[W/m^2]
+T_1=20;//Outer wall Temperature [degree Celcius]
+T_2=-10;//Inner wall Temperature [degree Celcius]
+//Solution:-
+A=(0.8*1.5);//Area of glass window[m^2]
+//Convection Resistances
+R_conv1=1/(h_in*A);//Due to convection heat transfer between inner atmosphere and glass[degree Celcius/W]
+R_conv2=1/(h_out*A);//Due to convection heat transfer between outer atmosphere and glass[degree Celcius/W]
+//Conduction Resistances
+R_cond1=L_g/(k_g*A);//Due to conduction heat transfer through the glass[degree Celcius/W]
+R_cond2=R_cond1;//Glass Medium is seperated by air spac hence two glass mediums are created[degree Celcius/W]
+R_cond3=L_a/(k_a*A);//Due to conduction heat transfer through the air space[degree Celcius/W]
+//Net Resistance offered by window is the sum of all the individual resistances written in the oreder of their occurence
+R_total=R_conv1+R_cond1+R_cond2+R_cond3+R_conv2;//[degree Celcius/W]
+disp("degree Celcius/W",R_total,"The net resistance offered is")
+Q_=(T_1-T_2)/R_total;//[W]
+disp("W",Q_,"The steady rate of Heat transfer through the window is")
+//Inner surface temperature of the window is given by
+T1=T_1-(Q_*R_conv1);//[degree Celcius]
+disp("degree Celcius",T1,"Inner Surface Temperature of the window is")
\ No newline at end of file diff --git a/587/CH3/EX3.4/example3_4.sce b/587/CH3/EX3.4/example3_4.sce new file mode 100755 index 000000000..defd9782d --- /dev/null +++ b/587/CH3/EX3.4/example3_4.sce @@ -0,0 +1,17 @@ +clear;
+clc;
+
+//Example3.4[Equivalent Thickness for Contact Resistance]
+//Given:-
+k=237;//Thermal conductivity of aluminium[W/m.K]
+L=0.01;//Thickness of aluminium plate[m]
+hc=11000;//Thermal contact conductance[W/m^2.K]
+//Solution:-
+Rc=1/hc;//[m^2.K/W]
+disp("Since thermal contact resistance is the inverse of thermal contact conductance")
+disp("m^2.K/W",Rc,"Hence Therml contact Resistance is")
+//For a unit surface area, the thermal resistance of a flat plate is defined as
+R=L/k;
+//Equivalent thickness for R=Rc
+L=k*Rc;//[m]
+disp("cm",(100*L),"Equivalent thickness is")
\ No newline at end of file diff --git a/587/CH3/EX3.5/example3_5.sce b/587/CH3/EX3.5/example3_5.sce new file mode 100755 index 000000000..4503d02ba --- /dev/null +++ b/587/CH3/EX3.5/example3_5.sce @@ -0,0 +1,24 @@ +clear;
+clc;
+
+//Example3.5[Contact Reistance of Transistors]
+//Given:-
+k=386;//Thermal Conductivity of Copper[W/m.K]
+hc=42000;//Contact Conductance coreesponding to copper-aluminium interface for the case of 1.17-1.4 micron roughness and 5MPa[pressure, which is close to given to what we have[W/m^2.K]
+Ac=.0008;//Contact area b/w the case and the plate[m^2]
+A=0.01;//Plate area for each resistor[m^2]
+L=0.01;//Thickness of plate[m]
+ho=25;//Heat tranfer coefficient for back surface
+T_1=20;//Ambient Temperature[degree Celcius]
+T_2=70;//Maximum temperature of case[degree Celcius]
+//Solution:-
+//Resistances Offered
+R_interface=1/(hc*Ac);//Resistance offered at the copper aluminium interface[degree Cecius/W]
+R_plate=L/(k*A);//conduction resistance offered by coppr plate[degree Cecius/W]
+R_conv=1/(ho*A);//Convection resistance offerd by back surface of casing[degree Cecius/W]
+R_total=R_interface+R_plate+R_conv;//[degree Cecius/W]
+disp("degree Cecius/W",R_total,"The total thermal Tesistance is")
+Q_=(T_2-T_1)/R_total;//[W]
+disp("W",Q_,"The rate of heat transferred is")
+delta_T=Q_*R_interface;//[degree Celcius]
+disp("degree Celcius",delta_T,"The temperature jump at the interface is given by")
\ No newline at end of file diff --git a/587/CH3/EX3.6/example3_6.sce b/587/CH3/EX3.6/example3_6.sce new file mode 100755 index 000000000..268855b73 --- /dev/null +++ b/587/CH3/EX3.6/example3_6.sce @@ -0,0 +1,42 @@ +clear;
+clc;
+
+//Example3.6[Heat Loss through a Composite Wall]
+//Given:-
+//We consider a 1m deep and 0.25 m high portion of the wall since it is representative of the entire wall
+//Assuming any cross-section of the wall normal to the x-direction to be isothermal
+k_b=0.72;//thermal conductivity of bricks[W/m.K]
+k_p=0.22;//thermal conductivity of plaster layers[W/m.K]
+k_f=0.026;//thermal conductivity of foam layers[W/m.K]
+T_in=20;//Indoor Temperature[dgeree Celcius]
+T_out=-10;//Outdoor Temperature[dgeree Celcius]
+h_in=10;//Inner heat transfer coefficient[W/m^2.K]
+h_out=25;//Outer heat transfer coefficient[W/m^2.K]
+L_f=0.03;//Thickness of foam layer[m]
+L_p=0.02;//Thickness of plaster[m]
+L_b=0.16;//Thickness of brick wall[m]
+L_c=0.16;//Thickness of central plaster layer[m]
+A1=(0.25*1);//[m^2]
+A2=(0.015*1);//[m^2]
+A3=(0.22*1);//[m^2]
+//Resistances offered:-
+R_in=1/(h_in*A1);//Resistance to conevction heat transfer from inner surface[degree Celcius/W]
+R1=L_f/(k_f*A1);//Conduction Resistance offered by outer foam layer[degree Celcius/W]
+R2=L_p/(k_p*A1);//Conduction Resistance offered by Outer side Plaster Wall[degree Celcius/W]
+R6=R2;//Conduction Resistance offered by Inner side Plaster Wall[degree Celcius/W]
+R3=L_c/(k_p*A2);//Conduction Resistance offered by one side central Plaster wall[degree Celcius/W]
+R5=R3;//Conduction Resistance offered by other side central Plaster wall[degree Celcius/W]
+R4=L_b/(k_b*A3);//Conduction Resistance offered by Brick Wall[degree Celcius/W]
+R_out=1/(h_out*A1);//Convection Resistance from outer surface[degree Celcius/W]
+//R_in,R1,R2,R6,R_out are connected in series
+//R3,R4,R5 are connected in parallel
+R_mid=1/((1/R3)+(1/R4)+(1/R5));//Effective Parrallel Resistance
+R_total=(R_in+R1+R2+R_mid+R6+R_out);//[degree Celcius/W]
+disp("degree Celcius/W",R_total,"Net Resistance offered is")
+Q_=(T_in-T_out)/R_total;//[W]
+disp("W",Q_,"The steady rate of heat transfer through the wall is")
+Q_p=Q_/A1;//[W/m^2]
+disp("W/m^2",Q_p,"Heat Transfer per unit area is")
+A_total=3*5;//Total Area of wall[m^2]
+Q_total=Q_p*A_total;//[W]
+disp("W",Q_total,"Thr rate of heat transfer through the entire wall")
diff --git a/587/CH3/EX3.7/example3_7.sce b/587/CH3/EX3.7/example3_7.sce new file mode 100755 index 000000000..373298fd7 --- /dev/null +++ b/587/CH3/EX3.7/example3_7.sce @@ -0,0 +1,50 @@ +clear;
+clc;
+
+//Example3.7[Heat Transfer to a Spherical Container]
+//Radiation effect is being considered. For the black tank emissivity=1
+//Given:-
+k=15;//thermal conductivity of stainless steel[W/m.degree Celcius]
+T_ice=0+273;//temeperature of iced water[K]
+T_tank=22+273;//temperature of tank stored at room temperature[K]
+h_in=80;//Heat Transfer Coefficient at the inner surface of the tank[W/m^2.degree Celcius]
+h_out=10;//Heat Transfer Coefficient at the outer surface of the tank[W/m^2.degree Celcius]
+heat_f=333.7;//Heat of fusion of water at atmospheric pressure[kJ/kg]
+e=1;//emissivity of tank
+sigma=5.67*(10^(-8));//Stefan's [W/m^2.K^4]
+D1=3;//inner diameter[m]
+D2=3.04;//Outer diameter[m]
+//Solution:-
+//a)
+A1=(%pi)*(D1^2);//Inner Surface area of the tank[m^2]
+A2=(%pi)*(D2^2);//outer Surface area of the tank[m^2]
+disp("The radiation heat transfer coefficient is given by ")
+disp("h_rad=e*sigma*((T2^2)+(T_tank^2))*(T2+T_tank)")
+disp("But we dont know the outer surface temperature T2 of the tank. hence we assume a T2 value")
+disp("since heat transfer inside the tank is larger ")
+T2=5+273;//[K]
+disp("K",T2,"Therefore taking T2 =")
+h_rad=e*sigma*((T2^2)+(T_tank^2))*(T2+T_tank);//[W/m^2.K]
+disp("W/m^2.degree Celcius",h_rad,"The radiation heat transfer coefficient is determined to be")
+//Individual Thermal Resistances Offered
+R_in=1/(h_in*A1);//Resistance to convetion from inner side of tank[degree Celcius/W]
+R_sphere=((D2-D1)/2)/(4*%pi*k*(D1/2)*(D2/2));//Resistance to conduction due to ice sphere[degree Celcius/W]
+R_out=1/(h_out*A2);//Resistance to convection from outer side of tank[degree Celcius/W]
+R_rad=1/(h_rad*A2);//Resistance to radiation heat transfer[degree Celcius/W]
+//R_out and R_rad are in parallel connection,
+R_eq=(1/((1/R_out)+(1/R_rad)));//[degree Celcius/W]
+//R_in,R_sphere and R_eq are connected in series
+R_total=R_in+R_sphere+R_eq;//[degree Celcius/W]
+Q_=(T_tank-T_ice)/R_total;//[W]
+disp("W",Q_,"The steady rate of heat transfer to the iced water is")
+disp("We determine outer surface temperature to check the validity of assumption")
+T2=T_tank-(Q_*R_eq);//[K]
+disp("K",T2)
+disp("which is sufficiently close to 278 K")
+//b)
+delta_t=24;//Time duration[h]
+Q=Q_*delta_t*(3600/1000);//[kJ]
+disp("kJ",Q,"The total amount of heat transfer during a 24 hour period is")
+//It takes 333.7 kJ of energy to melt 1kg of ice at 0 degree Celcius
+m_ice=Q/heat_f;//[kg]
+disp("kg",m_ice,"The amount of ice that will melt during 24h period is")
diff --git a/587/CH3/EX3.8/example3_8.sce b/587/CH3/EX3.8/example3_8.sce new file mode 100755 index 000000000..f272c3242 --- /dev/null +++ b/587/CH3/EX3.8/example3_8.sce @@ -0,0 +1,32 @@ +clear;
+clc;
+
+//Example3.8[Heat Loss through an Insulated Steam Pipe]
+//Given:-
+T_steam=320;//[degree Celcius]
+T_surr=5;//[degree Celcius]
+k_iron=80;//Thermal conductivity of cast iron[W/m.degree Celcius]
+k_insu=0.05;//Thermal conductivity of glass wool insulation[W/m.degree Celcius]
+h_out=18;//Covection heat transfer coefficient outside the pipe[w/m^2.degree Celcius]
+h_in=60;//Covection heat transfer coefficient insideside the pipe[w/m^2.degree Celcius]
+D_in=0.05;//Inner diameter of pipe[m]
+D_out=0.055;//Outer diameter of pipe[m]
+t=0.03;//Thickness of insulation[m]
+r=(D_out/2)+t;//Effective outer radius[m]
+L=1;//Length of pipe[m]
+//Solution:-
+//Areas of surfaces exposed to convection
+A1=2*%pi*(D_in/2)*L;//Inner Area of pipe[m^2]
+A2=2*%pi*(r)*L;//Outer Area of pipe[m^2
+//Individual Thermal Resistances
+R_conv_in=1/(h_in*A1);//Resistance to convetion from inner surface of pipe[degree Celcius/W]
+R_pipe=(log(D_out/D_in))/(2*%pi*k_iron*L);//Resitance to conduction through iron pipe[degree Celcius/W]
+R_insu=(log(r/(D_out/2)))/(2*%pi*k_insu*L);//Resistance to conduction through insulation[degree Celcius/W]
+R_conv_out=1/(h_out*A2);//Resistance to convetion from outer surface of insulation on pipe[degree Celcius/W]
+//All resistances are in series
+R_total=R_conv_in+R_pipe+R_insu+R_conv_out;//Total Resistance[degree Celcius]
+Q_=(T_steam-T_surr)/R_total;//[W]
+disp("W",Q_,"The Steady rate of heat loss from the steam per m length of pipe is")
+delta_T_pipe=Q_*R_pipe;//[degree Celcius]
+delta_T_insu=Q_*R_insu;//[degree Celcius]
+disp("degree Celcius",delta_T_insu,"and",delta_T_pipe,,"The temperature drop across the pipe and the insulation is respectively")
\ No newline at end of file diff --git a/587/CH3/EX3.9/example3_9.sce b/587/CH3/EX3.9/example3_9.sce new file mode 100755 index 000000000..65dec257a --- /dev/null +++ b/587/CH3/EX3.9/example3_9.sce @@ -0,0 +1,31 @@ +clear;
+clc;
+
+//Example3.9[Heat Loss from an Insulated Electric Wire]
+//Given:-
+k_insu=0.15;//[W/m.degree Celcius]
+V=8;//Voltage drop across wire[Volts]
+I=10;//Current flowimg through the wire[Amperes]
+T_atm=30;//Temperature of atmosphere to which wire is exposed[degree Celcius]
+h=12;//heat transfer coefficient[W/m^2.degree Celcius]
+L=5;//length of wire[m]
+D=0.003;//diameter of wire[m]
+t=0.002;//thickness of insulation[m]
+r=(D/2)+t;//Effective radius[m]
+//Solution:-
+//Rate of heat generated in the wire becomes equal to the rate of heat transfer
+Q_=V*I;//[W]
+disp("W",Q_,"Heat generated in the wire is")
+A2=2*%pi*r*L;//Outer surface area[m^2]
+//Resistances offered
+R_conv=1/(h*A2);//Convection resistance for the outer sueface of insulation[degree Celcius/W]
+R_insu=(log(r/(D/2)))/(2*%pi*k_insu*L);//Conduction resitance for the plastic insulation[degree Celcius/W]
+//Effective Resistance
+R_total=R_conv+R_insu;//[degree Celcius/W]
+//Interface Temperature can be determined from
+T1=T_atm+(Q_*R_total);//[degree Celcius]
+disp("degree Celcius",T1,"The interface temperature is")
+//Critical radius
+r_cr=k_insu/h;//[m]
+disp("mm",r_cr*1000,"The critical radius of insulation of the plastic cover is")
+//Larger value of critical radius ensures that increasing the thickness of insulation upto critical radius will increase the rate of heat transfer
\ No newline at end of file diff --git a/587/CH4/EX4.1/example4_1.sce b/587/CH4/EX4.1/example4_1.sce new file mode 100755 index 000000000..ddfcdfcbc --- /dev/null +++ b/587/CH4/EX4.1/example4_1.sce @@ -0,0 +1,25 @@ +clear;
+clc;
+
+//Example4.1[Temperature Measurement by Thermocouples]
+//Given:-
+//Temperature of a gas stream is to be measured by a thermocouple whose junction can be approximated as a sphere
+D=0.001;//Diameter of junction sphere[m]
+//Properties of the junction
+k=35;//Thermal conductivity[W/m.degree Celcius]
+rho=8500;//desity[kg/m^3]
+Cp=320;//Specific heat[J/kg.degree Celcius]
+h=210;//Convection heat transfer coefficient between junction and the gas[W/m^2.degree Celcius]
+//Solution:-
+//V=(%pi/6)*(D^3)
+Lc=(((%pi/6)*(D^3))/(%pi*(D^2)));//The characteristic length of the junction[m]
+Bi=h*Lc/k;//Biot Number
+if(Bi<0.1) then,
+disp(Bi,"Since Bi=")
+disp("is less than 0.1 hence lumped system is applicable and the error involved in this approximation is negligible")
+end;
+b=h/(rho*Cp*Lc);//Exponent time constant[s^(-1)]
+disp("s^(-1)",b,"Time constant for given lumped heat capacity system")
+//In order to read 99% of intial temperature difference between the junction and the gas we must have ((T(t)-T_end)/(Ti-T_end))=0.01
+t=-1*(log(0.01))/b;
+disp("seconds",round(t),"Required time is")
\ No newline at end of file diff --git a/587/CH4/EX4.10/example4_10.sce b/587/CH4/EX4.10/example4_10.sce new file mode 100755 index 000000000..1313996ed --- /dev/null +++ b/587/CH4/EX4.10/example4_10.sce @@ -0,0 +1,28 @@ +clear;
+clc;
+
+//Example4.10[Cooling of a Long Cylinder by Water]
+Ti=200;//Initial Temperature of aluminium cylinder[degree Celcius]
+Tf=15;//Temperature of water in which cylinder is kept[degree Celcius]
+h=120;//Heat transfer Coefficent[W/m^2.degree Celcius]
+t=5*60;//[seconds]
+//Properties of aluminium at room temperature
+k=237;//Thermal conductivity[W/m.degree Celcius]
+a=9.71*(10^(-5));//Thermal diffusivity[m^/s]
+r=0.1;//Radius of cylinder[m]
+x=0.15;//[m]
+//Solution:-
+Bi=(h*r)/k;//Biot number
+//Corresponding to this biot no coefficients for a cylinder
+lambda=0.3126,A=1.0124;
+tau=(a*t)/(r^2);
+//Using one term approximation
+theta0=A*exp(-(lambda^2)*tau);
+neta=x/(2*sqrt(a*t));
+u=(h*sqrt(a*t))/k;
+v=(h*x)/k;
+w=(u^2);
+theta_semiinfinite=1-erfc(neta)+(exp(v+w)*erfc(neta+u));
+theta=theta_semiinfinite*theta0;
+T_x_t=Tf+(theta*(Ti-Tf));//[degree Celcius]
+disp("degree Celcius",ceil (T_x_t),"the temperature at the center of the cylinder 15cm from the exposed bottom surface")
\ No newline at end of file diff --git a/587/CH4/EX4.11/example4_11.sce b/587/CH4/EX4.11/example4_11.sce new file mode 100755 index 000000000..6719706ee --- /dev/null +++ b/587/CH4/EX4.11/example4_11.sce @@ -0,0 +1,23 @@ +clear;
+clc;
+
+//Example4.11[Refrigerating Steaks while Avoiding Frostbite]
+//Given:-
+Ti=25;//Initial temperature of steaks[degree Celcius]
+Tf=-15;//Temperature of refrigerator[degree Celcius]
+L=0.015;//Thickness of steaks[m]
+//Properties of steaks
+k=0.45;//[W/m.degree Celcius]
+rho=1200;//density[kg/m^3]
+a=9.03*(10^(-8));//Thermal diffusivity[m^2/s]
+Cp=4.10;//Specific Heat [kJ/kg]
+T_L=2,T_0=8;//[degree Celcius]
+//Solution:-
+//In the limiting case the surface temperature at x=L from the centre will be 2 degree C,while midplane temperature is 8 degree C in an environment at -15 degree C we have
+x=L;
+p=(T_L-Tf)/(T_0-Tf);
+//For this value of p we have
+Bi=1/1.5;//Biot number
+h=(Bi*k)/L;//[W/m^2.degree Celcius]
+disp("W/m^2.degree Celcius",h,"The convection heat transfer coefficient should be kept below the value")
+disp("to satisfy the constraints on the temperature of the steak during refrigeration")
\ No newline at end of file diff --git a/587/CH4/EX4.2/example4_2.sce b/587/CH4/EX4.2/example4_2.sce new file mode 100755 index 000000000..52408fbd1 --- /dev/null +++ b/587/CH4/EX4.2/example4_2.sce @@ -0,0 +1,26 @@ +clear;
+clc;
+
+//Example4.2[Predicting the time of Death]
+//Given:-
+T_room=20;//Temperature of room[degree Celcius]
+T_body_f=25;//Temperature of dead body after some time[degree Celcius]
+T_body_i=37;//Temperature of dead body just after death[degree Celcius]
+h=8;//Heat transfer Coefficient[W/m^2.degree Celcius]
+L=1.7;//Length of body which is assumed to be cylindrical in shape[m]
+r=0.15;//Radius of cylindrical body
+//Average human body is 72% water by mass, thus we assumne body to have properties of water
+rho=996;//Density[kg/m^3]
+k=0.617;//Thermal conductivity[W/m.degree Celcius]
+Cp=4178;//Specific Heat[J/kg.degree Celcius]
+//Solution:-
+Lc=(%pi*(r^2)*L)/((2*%pi*r*L)+(2*%pi*(r^2)));//Characteristic length of body[m]
+Bi=(h*Lc)/k;//Biot no
+if(Bi>0.1) then,
+ disp("lumped system analysis is not applicable, but we can still use it to get a rough estimate of time of death")
+ b=h/(rho*Cp*Lc);//[s^(-1)]
+ x=(T_body_i-T_room)/(T_body_f-T_room);
+//exp(-b*t)=x;
+t=(1/b)*log(x);//time elapsed[seconds]
+disp("hour",t/3600,"As a rough estimate the person dies about")
+disp("before the body was found")
diff --git a/587/CH4/EX4.3/example4_3.sce b/587/CH4/EX4.3/example4_3.sce new file mode 100755 index 000000000..8de8e73d6 --- /dev/null +++ b/587/CH4/EX4.3/example4_3.sce @@ -0,0 +1,27 @@ +clear;
+clc;
+
+//Example4.3[Boiling Eggs]
+//Given:-
+T1=5;//Initial temperature of egg[degree Celcius]
+T2=95;//Temperature of Boiling Water[degree Celcius]
+h=1200;//Convection heat transfer coefficient of egg[W/m^2.degree Celcius]
+r=0.025;//Radius of egg[m]
+T3=70;//Final temperature attained by centre of egg[degree Celcius]
+k=0.627;//Thermal conductivity[W/m.degree Celcius]
+a=0.151*(10^(-6));//Thermal diffusivity[m^2/s]
+//Solution:-
+Bi=(h*r)/k;//Biot Number
+if(Bi>0.1) then,
+disp("the lumped system analysis is not applicable")
+//Findinf coefficient for a sphere corresponding to this bi are,
+lambda1=3.0754,A1=1.9959;
+x=(T3-T2)/(T1-T2);
+tau=(-1/(lambda1^2))*log(x/A1);
+disp(tau,"Fourier no is")
+//Since fourier no is greater than 0.2, cooking time is determined from the definition of fourier no to be
+t=(tau*(r^2))/a;//[seconds]
+disp("minutes",(t/60),"The time taken for center of egg to reach 70 degree Celcius temperature")
+else,
+ disp("the lumped system is not applicable")
+end
\ No newline at end of file diff --git a/587/CH4/EX4.4/example4_4.sce b/587/CH4/EX4.4/example4_4.sce new file mode 100755 index 000000000..7696f226b --- /dev/null +++ b/587/CH4/EX4.4/example4_4.sce @@ -0,0 +1,25 @@ +clear;
+clc;
+
+//Example4.4[Heating of Brass Plates in an Oven]
+T_in=20;//Initial uniform temperature of brass plate[degree Celcius]
+T_f=500;//Temperature of the oven[degree Celcius]
+t=7*60;//[seconds]
+h=120;//combined convection and radiation heat transfer coefficient[W/m^2.degree Celcius]
+L=0.04/2;//Thickness of plate 2L=0.004[m]
+//Properties of brass at room temperature are:-
+k=110;//Thermal conductivity[W/m.degree Celcius]
+rho=8530;//density[kg/m^3]
+Cp=380;//Specific Heat Capacity[J.kg.degree Celcius]
+a=33.9*(10^(-6));//Thermal Diffusivity[m^2/s]
+//Solution:-
+Bi1=1/(k/(h*L));
+tau1=(a*t)/(L^2);
+//For above values of biot no and fourier no we have
+p=0.46;// p=(T0-T_f)/(T_in-T_f),where T0 is temperature of inner surface of plate at time t
+x=L;
+Bi2=Bi1;
+//For above condition of x/L ratio and Biot number we have
+q=0.99;//q=(T-T_f)/(T_in-T_f), where T is temperature of outer surface of plate after time t
+T=T_f+((p*q)*(T_in-T_f));//[degree Celcius]
+disp("degree Celcius",ceil (T),"The surface temperature of the plates when they leave the oven will be")
diff --git a/587/CH4/EX4.5/example4_5.sce b/587/CH4/EX4.5/example4_5.sce new file mode 100755 index 000000000..d103b09f6 --- /dev/null +++ b/587/CH4/EX4.5/example4_5.sce @@ -0,0 +1,31 @@ +clear;
+clc;
+
+//Example4.5[Cooling of a long Stainless Steel Cylindrical Shaft]
+//Given:-
+Ti=600;//Temperature of cylinder just after taking it out of the oven[degree Celcius]
+h=80;//average heat transfer coefficient[W/m^2.degree Celcius]
+t=45*60;//Time for cooling[seconds]
+r=0.1;//Radius of cylinder[m]
+l=1;//Length of cylinder[m]
+//Properties of stainless steel cylinder
+k=14.9;//Thermal conductivity[W/m.degree Celcius]
+rho=7900;//Density[kg/m^3]
+Cp=477;//Specific Heat Capacity[J/kg.degree Celcius]
+a=3.95*(10^(-6));//Thermal diffusivity[m^2/s]
+T_f=200;//Ambient temperature[degree Celcius]
+//Solution:-
+Bi1=(h*r)/k;
+tau1=(a*t)/(r^2);
+//For biot no=Bi1 and fourier no=tau1,we have
+p=0.40;//p=(T(0)-T_f)/(Ti-T_f)
+T_0=T_f+(p*(Ti-T_f));//[degree Celcius]
+disp("degree Celcius",T_0,"The center temperature of the shaft after 45 minutes is")
+//Determining actual heat transfer
+m=rho*%pi*(r^2)*l;//[kg]
+Q_max=m*Cp*(Ti-T_f)*(1/1000);//[kJ]
+x=(Bi1^2)*tau1;
+//For biot no= Bi1 and (h^2)at/(k^2)=x, we have
+y=0.62;//y=Q/Q_max
+Q=y*Q_max;//[kJ]
+disp("kJ",round(Q),"The total heat transfer from the shaft during 45 minutes of cooling is")
diff --git a/587/CH4/EX4.6/example4_6.sce b/587/CH4/EX4.6/example4_6.sce new file mode 100755 index 000000000..8b01c98e9 --- /dev/null +++ b/587/CH4/EX4.6/example4_6.sce @@ -0,0 +1,23 @@ +clear;
+clc;
+
+//Example4.6[Minimum Burial Depth of Water Pipes to avoid Freezing]
+//Given:-
+//Soil properties:-
+k=0.4;//Thermal conductivity[W/m.degree Celcius]
+a=0.15*(10^(-6));//Thermal diffusivity[m^2/s]
+T_in=15;//Initial uniform temperature of ground[degree Celcius]
+T_x=0;//Temperature after 3 months[degree Celcius]
+Ts=-10;//Temperature of surface[degree Celcius]
+//Solution:-
+//The temperature of the soil surrounding the pipes wil be 0 degree Celcius after three months in the case of minimum burial depth, therefore we have
+x=(h/k)*(sqrt(a*t));
+//Since h tends to infinty
+x=%inf;
+y=(T_x-T_in)/(Ts-T_in);
+//For values of x and y we have
+neta=0.36;
+t=90*24*60*60;//[seconds]
+x=2*neta*sqrt(a*t);//[m]
+disp("m",x,"Water pipes must be burried to a depth of at least ")
+disp("so as to avoid freezing under the specified harsh winter conditions")
diff --git a/587/CH4/EX4.7/example4_7.sce b/587/CH4/EX4.7/example4_7.sce new file mode 100755 index 000000000..0c62b61c3 --- /dev/null +++ b/587/CH4/EX4.7/example4_7.sce @@ -0,0 +1,16 @@ +clear;
+clc;
+
+//Example4.7[Surface Temperature Rise of Heated Blocks]
+//Given:-
+flux=1250;//Constant solar heat flux[W/m^2]
+T=20;//Temperature of black painted wood block[degree Celcius]
+k_wood=1.26;//Thermal conductivity of wood at room temperature[W/m.K]
+a_wood=1.1*(10^(-5));//Diffusivity of wood at room temperature[m^2/s]
+k_aluminium=237;//Thermal conductivity of aluminium at room temperature[W/m.K]
+a_aluminium=9.71*(10^(-5));//Diffusivity of aluminium at room temperature[m^2/s]
+t=20*60;//[seconds]
+//Solution:-
+Ts_wood=T+((flux/k_wood)*(sqrt((4*a_wood*t)/%pi)));//[degree Celcius]
+Ts_aluminium=T+((flux/k_aluminium)*(sqrt((4*a_aluminium*t)/%pi)));//[degree Celcius]
+disp("respectively","degree Celcius",round (Ts_aluminium),"and",ceil (Ts_wood),"The surface temperature fro both the wood and aluminium blocks are ")
\ No newline at end of file diff --git a/587/CH4/EX4.8/example4_8.sce b/587/CH4/EX4.8/example4_8.sce new file mode 100755 index 000000000..b8d935961 --- /dev/null +++ b/587/CH4/EX4.8/example4_8.sce @@ -0,0 +1,34 @@ +clear;
+clc;
+
+//Example4.8[Cooling of a Short Brass Cylinder]
+//Given:-
+Ti=120;//Initial Temperature[degree Celcius]
+T_ambient=25;//Temperature of atmospheric air[degree Celcius]
+h=60;//convetcion heat transfer coefficient[W/m^2.degree Celcius]
+r=0.05;//radius of cylinder[m]
+L=0.06;//thickness[m]
+a=3.39*(10^(-5));//Diffusivity of brass[m^2/s]
+k=110;//Thermal conductivity of brass[W/m.degree Celcius]
+t=900;//[seconds]
+//Solution (a):-
+disp("At the center of the plane wall")
+tau1=(a*t)/(L^2);
+Bi1=(h*L)/k;
+disp("respectively",Bi1,"and",tau1,"Fourier no and Biot no are")
+disp("At the center of the cylinder")
+tau2=(a*t)/(r^2);
+Bi2=(h*r)/k;
+disp("respectively",Bi2,"and",tau2,"Fourier no and Biot no are")
+theta_wall_c=0.8;//(T(0,t)-T_ambient)/(Ti-T_ambient)
+theta_cyl_c=0.5;//(T(0,t)-T_ambient)/(Ti-T_ambient)
+T_center=T_ambient+((theta_wall_c*theta_cyl_c)*(Ti-T_ambient));//[degree Celcius]
+disp("degree Celcius",round (T_center),"The temperature at the center of the short cylinder is")
+//Solution (b):-
+//The centre of the top surface of the cylinder is still at the center of the lonf cylinder(r=0),but at the outer surface of the plane wall(x=L).
+x=L;//[m]
+y=x/L;
+//For Bi=Bi1 and x=1
+theta_wall_L=0.98*theta_wall_c;//(T(L,t)-T_ambient)/(Ti-T_ambient)
+T_surface=T_ambient+((theta_wall_L*theta_cyl_c)*(Ti-T_ambient));//[degree Celcius]
+disp("degree Celcius",round (T_surface),"The temperature at the top surface of the cylinder")
\ No newline at end of file diff --git a/587/CH4/EX4.9/example4_9.sce b/587/CH4/EX4.9/example4_9.sce new file mode 100755 index 000000000..8cd7a2a99 --- /dev/null +++ b/587/CH4/EX4.9/example4_9.sce @@ -0,0 +1,33 @@ +clear;
+clc;
+
+//Example4.9[Heat transfer from a Short Cylinder]
+//Given:-
+Ti=120;//Initial Temperature[degree Celcius]
+T_ambient=25;//Temperature of atmospheric air[degree Celcius]
+rho=8530;//density of brass cyliner[kg/m^3]
+Cp=0.380;//Specific heat of brass cylinder[kJ/kg.degree Celcius]
+r=0.05;//radius[m]
+H=0.12;//Height of cylinder[m]
+h=60;//convetcion heat transfer coefficient[W/m^2.degree Celcius]
+a=3.39*(10^(-5));//Diffusivity of brass [m^2/s]
+k=110;//Thermal conductivity of brass[W/m.degree Celcius]
+L=0.06;//[m]
+t=900;//[seconds]
+//Solution:-
+m=rho*(%pi*(r^2)*H);//mass of cylinder[kg]
+Q_max=m*Cp*(Ti-T_ambient);//[kJ]
+disp("At the center of the plane wall")
+tau1=(a*t)/(L^2);
+Bi1=(h*L)/k;
+x=(Bi1^2)*tau1;
+//For given x and Bi1
+p=0.23;//(Q/Qmax) for plane wall
+disp("At the center of the cylinder")
+tau2=(a*t)/(r^2);
+Bi2=(h*r)/k;
+y=(Bi2^2)*tau2;
+//For given y and Bi2
+q=0.47;//(Q/Qmax) for infinite cylinder
+Q=Q_max*(p+(q*(1-p)));//[kJ]
+disp("kJ",ceil (Q),"The total heat transfer from the cylinder during the first 15 minutes of cooling is")
\ No newline at end of file diff --git a/587/CH5/EX5.1/example5_1.sce b/587/CH5/EX5.1/example5_1.sce new file mode 100755 index 000000000..d9ab2f9b9 --- /dev/null +++ b/587/CH5/EX5.1/example5_1.sce @@ -0,0 +1,26 @@ +clear;
+clc;
+
+//Example5.1[Steady Heat Conduction in a Large Uranium Plate]
+//Given:-
+L=0.04;//Thickness of plate[m]
+k=28;//Thermal conductivity[W/m.degree Celcius]
+e_gen=5*(10^6);//Rate of heat generation per unit volume[W/m^3]
+h=45;//Heat transfer coefficient[W/m^2]
+T_ambient=30;//Ambient temperature[degree Celcius]
+//Solutio:-
+M=3;//No of nodes
+//These nodes are chosen to be at the two surfaces of the plate and the mid point
+del_x=L/(M-1);//Nodal Spacing[m]
+//Let the nodes be 0,1 and 2. and temperatures at these nodes are
+T0=0;//Temperature at node 0[degree Celcius]
+//Finding temperatures at other two nodes using finite difference method
+c1=e_gen*(del_x^2)/k;
+c2=(-h*del_x*T_ambient/k)-(c1/2);
+function[temp]=f1(T)
+temp(1)=2*T(1)-T(2)-c1;
+temp(2)=T(1)-1.032*T(2)-c2;
+deff('[temp]=f1(T)',['temp_1=2*T(1)-T(2)-c1','temp_2=T(1)-1.032*T(2)-c2'])
+//To find the solution assume an initial value x0=[a,b]
+//then equate [xs,fxs,m]=fsolve(x0',f1)
+
\ No newline at end of file diff --git a/587/CH5/EX5.2/example5_2.sce b/587/CH5/EX5.2/example5_2.sce new file mode 100755 index 000000000..0d64682fb --- /dev/null +++ b/587/CH5/EX5.2/example5_2.sce @@ -0,0 +1,36 @@ +clear;
+clc;
+
+//Example5.2[Heat transfer from triangular fins]
+//Given:-
+k=180;//Thermal conductivity of aluminium alloy[W/m.degree Celcius]
+L=0.05;//length of fin[m]
+b=0.01;//Base thickness of fin[m]
+T_surr=25;//Temperature of surrounding[degree Celcius
+h=15;//heat transfer coefficient[W/m^2.degree Celcius]
+M=6;//No of equally spaced nodes along the fin
+//Solution (a)
+del_x=L/(M-1);//Nodal Spacing[m]
+T0=200;//Temperature at node 0[degree Celcius]
+theta=atan(b/2*L);
+//sigmaQ_all_sides=kA_left((T_(m-1)-T_m)/del_X)+((T_(m+1)-T_m)/del_x)+(hA_conv(T_surr-T_m))=0
+//Simplifying above equation we get
+disp("((5.5-m)T_(m-1))-((10.008-2m)Tm)+((4.5-m)T_m+1)=-0.29")
+//Putting m=1,2,3,4 we get five equations in five unknowns
+//Solving these five equations we get temperatures at node 1,2,3,4 and 5 respectively
+function[node]=f5(T)
+ node(1)=-8.008*T(1)+3.5*T(2)+0*T(3)+0*T(4)+0*T(5)+900.209;
+ node(2)=3.5*T(1)-6.008*T(2)+2.5*T(3)+0*T(4)+0*T(5)+0.209;
+ node(3)=0*T(1)+2.5*T(2)-4.008*T(3)+1.5*T(4)+0*T(5)+0.209;
+ node(4)=0*T(1)+0*T(2)+1.5*T(3)-2.008*T(4)+0.5*T(5)+0.209;
+ node(5)=0*T(1)+0*T(2)+0*T(3)+1*T(4)-1.008*T(5)+0.209;
+ deff('[node]=f5(T)',['f_1=-8.008*T(1)+3.5*T(2)+0*T(3)+0*T(4)+0*T(5)+900.209','f_2=3.5*T(1)-6.008*T(2)+2.5*T(3)+0*T(4)+0*T(5)+0.209','f_3=0*T(1)+2.5*T(2)-4.008*T(3)+1.5*T(4)+0*T(5)+0.209','f_4=0*T(1)+0*T(2)+1.5*T(3)-2.008*T(4)+0.5*T(5)+0.209','f_5=0*T(1)+0*T(2)+0*T(3)+1*T(4)-1.008*T(5)+0.209'])
+ //Solution(b)
+ T1=T(1),T2=T(2),T3=T(3),T4=T(4),T5=T(5);
+ w=1;//width[m]
+ Q_fin=(h*w*del_x/cos(theta))*[(T0+2*(T1+T2+T3+T4)+T5-10*T_surr)];//[W]
+ disp("W",Q_fin,"The total rate of heat transfer from the fin is")
+ //Solution(c)
+ Q_max=(h*2*w*L/cos(theta)*(T0-T_surr));//[W]
+neta=Q_fin/Q_max;
+disp(neta,"Efficiency of the fin is")
diff --git a/587/CH5/EX5.3/example5_3.sce b/587/CH5/EX5.3/example5_3.sce new file mode 100755 index 000000000..ea1f8c015 --- /dev/null +++ b/587/CH5/EX5.3/example5_3.sce @@ -0,0 +1,25 @@ +clear;
+clc;
+
+//Example5.3[SteadLy Two-Dimensional Heat Conduction in L-Bars]
+//Given:-
+e_gen=2*(10^6);//Heat generated per unit volume[W/m^3]
+k=15;//Thermal heat conductivity[W/m.degree Celcius]
+T_ambient=25;//Temperature of ambient air[degree Celcius]
+T_surface=90;//Temperature of the bottom surface[degree Celcius]
+h=80//convection coefficient[W/m^2]
+q_R=5000;//Heat flux to which right surface is subjected[W/m^2]
+del_x=0.012,del_y=0.012;//Distance between equally spaced nodes[m]
+//Solution:-
+//After substituing values in equations of all nodal points finally we have nine equation and nine unknowns
+function[temp]=f9(T)
+ temp(1)=-2.064*T(1)+1*T(2)+0*T(3)+1*T(4)+0*T(5)+0*T(6)+0*T(7)+0*T(8)+0*T(9)+11.2;
+ temp(2)=1*T(1)-4.128*T(2)+1*T(3)+0*T(4)+2*T(5)+0*T(6)+0*T(7)+0*T(8)+0*T(9)+22.4;
+ temp(3)=0*T(1)+1*T(2)-2.128*T(3)+0*T(4)+0*T(5)+1*T(6)+0*T(7)+0*T(8)+0*T(9)+12.8;
+ temp(4)=1*T(1)+0*T(2)+0*T(3)-4*T(4)+2*T(5)+109.2;
+ temp(5)=0*T(1)+1*T(2)+0*T(3)+1*T(4)-4*T(5)+1*T(6)+0*T(7)+0*T(8)+0*T(9)+109.2;
+ temp(6)=0*T(1)+0*T(2)+1*T(3)+0*T(4)+2*T(5)-6.128*T(6)+1*T(7)+0*T(8)+0*T(9)+212.0;
+ temp(7)=0*T(1)+0*T(2)+0*T(3)+0*T(4)+0*T(5)+1*T(6)-4.128*T(7)+1*T(8)+0*T(9)+202.4;
+ temp(8)=0*T(1)+0*T(2)+0*T(3)+0*T(4)+0*T(5)+0*T(6)+1*T(7)-4.128*T(8)+T(9)+202.4;
+ temp(9)=0*T(1)+0*T(2)+0*T(3)+0*T(4)+0*T(5)+0*T(6)+0*T(7)+1*T(8)-2.064*T(9)+105.2;
+ deff('[temp]=f9(T)',['f_1= -2.064*T(1)+1*T(2)+0*T(3)+1*T(4)+0*T(5)+0*T(6)+0*T(7)+0*T(8)+0*T(9)+11.2','f_2=1*T(1)-4.128*T(2)+T(3)+0*T(4)+2*T(5)+0*T(6)+0*T(7)+0*T(8)+0*T(9)+22.4','f_3=0*T(1)+T(2)-2.128*T(3)+0*T(4)+0*T(5)+T(6)+0*T(7)+0*T(8)+0*T(9)+12.8','f_4=T(1)+0*T(2)+0*T(3)-4*T(4)+2*T(5)+109.2','f_5=0*T(1)+T(2)+0*T(3)+T(4)-4*T(5)+T(6)+0*T(7)+0*T(8)+0*T(9)+109.2','f_6=0*T(1)+0*T(2)+T(3)+0*T(4)+2*T(5)-6.128*T(6)+T(7)+0*T(8)+0*T(9)+212.0','f_7=0*T(1)+0*T(2)+0*T(3)+0*T(4)+0*T(5)+T(6)-4.128*T(7)+T(8)+0*T(9)+202.4','f_8=0*T(1)+0*T(2)+0*T(3)+0*T(4)+0*T(5)+0*T(6)+T(7)-4.128*T(8)+T(9)+202.4','f_9=0*T(1)+0*T(2)+0*T(3)+0*T(4)+0*T(5)+0*T(6)+0*T(7)+T(8)-2.064*T(9)+105.2'])
\ No newline at end of file diff --git a/587/CH5/EX5.4/example5_4.sce b/587/CH5/EX5.4/example5_4.sce new file mode 100755 index 000000000..ec3459878 --- /dev/null +++ b/587/CH5/EX5.4/example5_4.sce @@ -0,0 +1,32 @@ +clear;
+clc;
+
+//Example5.4[Heat Loss through Chimneys]
+//Given:-
+k=1.4;//Thermal conductivity of concrete[W/m.degree Celcius]
+A=0.2*0.2;//Area of flow section[m^2]
+t=0.2;//Thickness of the wall[m]
+Ti=300+273;//Average temperature of gases[K]
+hi=70;//Convection heat transfer coefficient inside the chimney[W/m^2]
+ho=21;//Convection heat transfer coefficient outside the chimney[W/m^2]
+To=20+273;//Temperature od outer air[Kelvin]
+e=0.9;//Emissivity
+delx=0.1,dely=0.1;//Nodal spacing [m]
+//Solution:-
+//Substituing values in all nodal equations and and solving these equations we get temperature at all nodes
+function[temp]=fu9(T)
+ temp(1)=7*T(1)-T(2)-T(3)-2865;
+ temp(2)=-T(1)+8*T(2)-2*T(4)-2865;
+ temp(3)=-T(1)+4*T(3)-2*T(4)-T(6);
+ temp(4)=-T(2)-T(3)+4*T(4)-T(5)-T(7);
+ temp(5)=-2*T(4)+4*T(5)-2*T(8);
+ temp(6)=-T(2)-T(3)+3.5*T(6)+(0.3645*(10^(-9))*(T(6)^4))-456.2;
+ temp(7)=-2*T(4)-T(6)+7*T(7)+(0.729*(10^(-9))*(T(7)^4))-T(8)-912.4;
+ temp(8)=-2*T(5)-T(7)+7*T(8)+(0.729*(10^(-9))*(T(8)^4))-912.4;
+ temp(9)=-T(8)+2.5*T(9)+(0.3645*(10^(-9))*(T(9)^4))-456.2;
+ deff('[temp]=fu9(T)',['f_1=7*T(1)-T(2)-T(3)-2865','f_2=-T(1)+8*T(2)-2*T(4)-2865','f_3=-T(1)+4*T(3)-2*T(4)-T(6)','f_4=-T(2)-T(3)+4*T(4)-T(5)-T(7)','f_5=-2*T(4)+4*T(5)-2*T(8)','f_6=-T(2)-T(3)+3.5*T(6)+(0.3645*(10^(-9))*(T(6)^4))-456.2','f_7=-2*T(4)-T(6)+7*T(7)+(0.729*(10^(-9))*(T(7)^4))-T(8)-912.4','f_8=-2*T(5)-T(7)+7*T(8)+(0.729*(10^(-9))*(T(8)^4))-912.4','f_9=-T(8)+2.5*T(9)+(0.3645*(10^(-9))*(T(9)^4))-456.2'])
+T1=T(1),T2=T(2),T3=T(3),T4=T(4),T5=T(5),T6=T(6),T7=T(7),T8=T(8),T9=T(9);
+T_wall=(0.5*T6+T7+T8+0.5*T9)/(0.5+1+1+0.5);
+disp("Kelvin",T_wall,"The average temperature at the outer surface of the chimney weighed by the surface area is")
+Q_chimney=(ho*4*0.6*1*(T_wall-To))+(e*5.67*(10^-8)*0.6*1*((T_wall^4)-((260^4))));//[W]
+disp("W",Q_chimney,"The heat transfer is")
\ No newline at end of file diff --git a/587/CH5/EX5.5/example5_5.sce b/587/CH5/EX5.5/example5_5.sce new file mode 100755 index 000000000..8c0669966 --- /dev/null +++ b/587/CH5/EX5.5/example5_5.sce @@ -0,0 +1,26 @@ +clear;
+clc;
+
+//Example5.5[Transient Heat Conduction in a Large Uranium Plate]
+//Given:-
+k=28;//[W/m.degree Celcius]
+a=12.5*10^(-6);//Thermal diffusivity[m^2/s]
+T1_0=200,T2_0=200;//Initial Temperature[degree Celcius]
+e_gen=5*10^6;//Heat generated per unit volume[W/m^3]
+h=45;//heat transfer coefficient[W/m^2.degree Celcius]
+T0=0;//Temperature at node 0[degree Celcius]
+L=0.04;//[m]
+M=3;//No of nodes
+t=15;//[seconds]
+//Solution (a):-
+delx=L/(M-1);//[m]
+//The nodes are 0,1 and 2
+tau=(a*t)/(delx^2);//Fourier no
+//Substituing this value of tau in nodal equations
+//The nodal temperatures T1_1 and T2_1 at t=15sec
+T1_1=0.0625*T1_0+0.46875*T2_0+33.482;//[degree Celcius]
+T2_1=0.9375*T1_0+0.032366*T2_0+34.386;//[degree Celcius]
+//Similarly the nodal themperatures T1_2,T2_2 at t1=2*t=30sec are
+T1_2=0.0625*T1_1+0.46875*T2_1+33.482;//[degree Celcius]
+T2_2=0.9375*T1_1+0.032366*T2_1+34.386;//[degree Celcius]
+disp("degree Celcius",T2_2,T1_2,"and",T2_1,T1_1,"Temperatures at node 1 and 2 are respectively")
diff --git a/587/CH5/EX5.6/example5_6.sce b/587/CH5/EX5.6/example5_6.sce new file mode 100755 index 000000000..3d27a8876 --- /dev/null +++ b/587/CH5/EX5.6/example5_6.sce @@ -0,0 +1,49 @@ +clear;
+clc;
+
+//Example5.6[Solar Energy Storage in Trombe Walls]
+//Given:-
+hin=10;//[W/m^2]
+A=3*75;//[m^2]
+Tin=21;//[degree Celcius]
+k=0.69;//[W/m.degree Celcius]
+a=4.44*10^(-7);//diffusivity[m^2/s]
+kappa=0.77;
+delx=0.06;//The nodal spacing[m]
+L=0.3;//Length of wall[m]
+Tout=0.6,q_solar=360;//Ambient temperature in degree Celcius and Solar Radiation between 7am to 10 am
+//Solution:-
+M=(L/delx)+1;
+disp(M,"No of nodes are")
+//Stability Criterion
+del_t=(delx^2)/(3.74*a);//[seconds]
+disp("s",del_t,"The maximum allowable value of the time step is")
+//Therefore any step less than del_t can be used to solve this problem,for convinience let's choose
+delt=900;//[seconds]
+tao=a*delt/(delx^2);
+disp(tao,"The mesh Fourier number is")
+//Initially at 7am or t=0,the temperature of the wall is said to vary linearly between 21 degree Celcius at node 0 and -1 at node 5
+//Temp between two neighbouring nodes is
+temp=(21-(-1))/5;//[degree Celcius]
+T0_0=Tin;
+T1_0=T0_0-temp;
+T2_0=T1_0-temp;
+T3_0=T2_0-temp;
+T4_0=T3_0-temp;
+T5_0=T4_0-temp;
+T0_1=((1-3.74*tao)*T0_0)+(tao*(2*T1_0+36.5));
+T1_1=(tao*(T0_0+T2_0))+(T1_0*(1-(2*tao)));
+T2_1=(tao*(T1_0+T3_0))+(T2_0*(1-(2*tao)));
+T3_1=(tao*(T2_0+T4_0))+(T3_0*(1-(2*tao)));
+T4_1=(tao*(T3_0+T5_0))+(T4_0*(1-(2*tao)));
+T5_1=(T5_0*(1-(2.70*tao)))+(tao*((2*T4_0)+(0.70*Tout)+(0.134*q_solar)));
+disp("Nodal temperatures at 7:15am are")
+disp("degree Celcius",T0_1,"Node0:")
+disp("degree Celcius",T1_1,"Node1:")
+disp("degree Celcius",T2_1,"Node2:")
+disp("degree Celcius",T3_1,"Node3:")
+disp("degree Celcius",T4_1,"Node4:")
+disp("degree Celcius",T5_1,"Node5:")
+Q_wall=hin*A*delt*(((round(T0_1)+T0_0)/2)-Tin);//[J]
+disp("J",Q_wall,"The amount of heat transfer during the first time step or during the first 15 min period is")
+//Similarly using values from the table given we can find temperature at various nodes after required time interval
diff --git a/587/CH6/EX6.1/example6_1.sce b/587/CH6/EX6.1/example6_1.sce new file mode 100755 index 000000000..04ba74322 --- /dev/null +++ b/587/CH6/EX6.1/example6_1.sce @@ -0,0 +1,25 @@ +clear;
+clc;
+
+//Example6.1[Temperature Rise of Oil is a Journal Bearing]
+//Given:-
+k=0.145;//[W/m.K]
+mu=0.8374;//[kg/m.s]or[N.s/m^2]
+T1=20;//Temperature of both the plates[degree Celcius]
+t=0.002;//Thickness of oil film between the plates[m]
+v=12;//Velocity with which plates move[m/s]
+//Solution (a):-
+//Relation between velocity and temperature variation
+disp("T(y)=T0+(mu*(v^2)/(2*k))[(y/L)-((y/L)^2)]")
+//Solution(b):-
+//The location of maximum temperature is determined by setting dT/dy=0 and solving for y
+//(mu*(v^2)/(2*k*L))*(1-(2*y/L))=0
+L=1;//Random initialisation of variable L, where L is length of plates
+y=L/2;
+//T_max=T(L/2)
+T_max=T1+((mu*(v^2)/(2*k))*(((L/2)/L)-(((L/2)^2)/(L^2))));
+disp("degree Celcius",ceil(T_max),"Maximum temperature occurs at mid plane and its value is")
+//heat flux q0=-kdt/dy|y=0;=-kmu*v^2/(2*k*L)
+q0=-(mu*k*(v^2)/(2*k*t))/1000;//Heat flux from one plate [kW/m^2]
+qL=-((k*mu*(v^2))*(1-2)/(2*k*t*1000));//Heat flux from another plate[kW/m^2]
+disp("kW/m^2",qL,"Heat fluxes at the two plates are equal in magnitude but opposite in sign and the value of magnitude is")
diff --git a/587/CH6/EX6.2/example6_2.sce b/587/CH6/EX6.2/example6_2.sce new file mode 100755 index 000000000..a705ad377 --- /dev/null +++ b/587/CH6/EX6.2/example6_2.sce @@ -0,0 +1,19 @@ +clear;
+clc;
+
+//Example6.2[Finding Convection Coefficient from Drag Measurement]
+//Given:-
+//Properties of air
+rho=1.204;//[kg/m^3]
+Cp=1007;//[J/kg.K]
+Pr=0.7309;//Prandtl number
+w=2;//Width of plate[m]
+L=3;//Characteristic length of plate[m]
+v=7;//velocity of air[m/s]
+Fd=0.86;//Total grag force[N]
+//Solution:-
+As=2*w*L;//Since both sides of plate are exposed to air flow[m^2]
+//For flat plates drag force is equivalent to friction coefficient Cf
+Cf=Fd/(rho*As*(v^2)/2);
+h=(Cf*rho*v*Cp)/(2*(Pr^(2/3)));//[W/m^2.degree Celcius]
+disp("respectively","W/m^2.degree Celcius",h,"and",Cf,"Friction Factor and average heat transfer coefficient are")
diff --git a/587/CH7/EX7.1/example7_1.sce b/587/CH7/EX7.1/example7_1.sce new file mode 100755 index 000000000..b8a237e46 --- /dev/null +++ b/587/CH7/EX7.1/example7_1.sce @@ -0,0 +1,33 @@ +clear;
+clc;
+
+//Example7.1[Flow of hot oil over a Flat Plate]
+//Given:-
+T_oil=60;//Temp of engine oil[degree Celcius]
+T_plate=20;//Temp of flat plate[degree Celcius]
+Rec=5*10^5;//Critical reynolds number for laminar flow
+Tf=(T_oil+T_plate)/2;//Film temperature[degree Celcius]
+v=2;//[m/s]
+//Properties of engine oil at film temperature
+rho=876;//[kg/m^3]
+Pr=2962;//Prandtl number
+k=0.1444;//[W/m.degree Celcius]
+nu=2.485*10^(-4);//dynamic viscosity[m^2/s]
+L=5;//Length of plate[m]
+ReL=(v*L)/nu;
+if(ReL<Rec) then,
+ disp("We have laminar flow over the entire plate")
+ Cf=1.33*(ReL^(-0.5));
+ disp(Cf,"The average friction coefficient is")
+ //Pressure Drag is zero and thus Cd=Cf for parallel floe over a flat plate
+ Fd=Cf*5*1*rho*(v^2)/2;//[N]
+ disp("N",Fd,"The drag force acting on the plate per unit width is")
+else,
+ disp("flow is turbulent")
+end
+Nu=0.664*(ReL^(0.5))*(Pr^(1/3));//Nusselt Number
+disp(ceil(Nu),"Nusselt Number is")
+h=k*Nu/L;//[W/m^2.degree Celcius]
+disp("W/m^.degree Celcius",h,"Convective heat transfer coefficient is")
+Q=h*(5*1)*(T_oil-T_plate);//[W]
+disp("W",round(Q),"Heat flow rate is")
\ No newline at end of file diff --git a/587/CH7/EX7.2/example7_2.sce b/587/CH7/EX7.2/example7_2.sce new file mode 100755 index 000000000..780a50018 --- /dev/null +++ b/587/CH7/EX7.2/example7_2.sce @@ -0,0 +1,44 @@ +clear;
+clc;
+
+//Example7.2[Cooling of a Hot Block by Forced Air at High Elevation]
+//Given:-
+ReC=5*10^5;//critical Reynolds number
+v=8;//Velocity of air[m/s]
+T_air=20;//Initial Temperature of air[degree Celcius]
+T_plate=140;//Temperature of flat plate[degree Celcius]
+T_film=(T_air+T_plate)/2;//Film Temperature of air[degree Celcius]
+//Properties of air at film temperature[degree Celcius]
+k=0.02953;//[W/m.degree Celcius]
+Pr=0.7154;//Prandtl Number
+nu=2.097*10^(-5);//Kinematic Viscosity at 1 atm Pressure [m^2/s]
+nu_ac=nu/0.823;//Kinematic viscosity at pressure 0.823 atm[m^2/s]
+//Solution(a)
+L1=6;//Characteristic length of plate along the flow of air[m]
+w1=1.5;//width[m]
+ReL1=(v*L1)/nu_ac;//Reynolds number
+if(ReL1>ReC) then,
+ disp("Flow is not laminar")
+ //We have average Nusselt Number
+ Nu1=((0.037*(ReL1^(0.8)))-871)*(Pr^(1/3));
+ disp(ceil(Nu1),"Nusselt Number is")
+ h1=k*Nu1/L1;//[W/m^2.degree Celcius]
+ As1=w1*L1;//Flow Area of plate[m^2]
+ Q1=h1*As1*(T_plate-T_air);
+ disp("W",Q1,"Heat Flow Rate is")
+else,
+ disp("Flow is laminar")
+end
+//Solution(b)
+L2=1.5;//Characteristic length of plate along flow of air[m]
+ReL2=v*L2/nu_ac;//Reynolds Number
+if(ReL2<Rec) then,
+ disp("Flow is laminar")
+ Nu2=0.664*(ReL2^(0.5))*(Pr^(1/3));
+ disp(ceil(Nu2),"Nusselt Number is")
+ h2=k*Nu2/L2;//[W/m^2.degree Celcius]
+ Q2=h2*As1*(T_plate-T_air);
+ disp("W",ceil(Q2),"The heat transfer rate is")
+else,
+ disp("Flow is turbulent")
+end
diff --git a/587/CH7/EX7.3/example7_3.sce b/587/CH7/EX7.3/example7_3.sce new file mode 100755 index 000000000..647f50b8a --- /dev/null +++ b/587/CH7/EX7.3/example7_3.sce @@ -0,0 +1,42 @@ +clear;
+clc;
+
+//Example7.3[Cooling of Plastic Sheets by Forced Air]
+//Given:-
+T_p=95;//Temp of plastic Sheet[degree Celcius]
+T_air=25;//Temp of air[degree Celcius]
+v=3;//Velocity of flowing air[m/s]
+L=0.6;//Length of plastic sheet[m]
+w=1.2;//width[m]
+k=0.02808;//[W/m.degree Celcius]
+Pr=0.7202;//Prandtl Number
+nu=1.896*10^(-5);//[m^2/s]
+rho=1200;//[kg/m^3]
+Cp=1700;//[J/kg.degree Celcius]
+vp=(9/60);//Velocity of moving plastic[m/s]
+tp=0.001;//Thickness of plastic[m]
+ReC=5*10^5;//Crictical Reynolds Number
+e=0.9;//emissivity
+//Solution(a)
+L1=2*L;//Considering both sides of plastic sheet[m]
+ReL1=v*L1/nu;//Reynolds number
+if(ReL1<ReC) then,
+ disp("(a) Flow is laminar")
+ Nu1=0.664*(ReL1^0.5)*(Pr^(1/3));
+ disp(Nu1,"The nusselt number is")
+ h=k*Nu1/L1;//[W/m^2.degree Celcius]
+ As=L1*w;//[m^2]
+ Q_conv=h*As*(T_p-T_air);//[W]
+ disp("W",ceil(Q_conv),"The covection heat flow rate is")
+ Q_rad=e*(5.67*10^(-8))*As*(((T_p+273)^4)-((T_air+273)^4));//[W]
+ disp("W",(Q_rad),"Radiation heat transfer rate is")
+ Q_total=Q_conv+Q_rad;//[W]
+ disp("W",ceil(Q_total),"The rate of cooling of the plastic sheet by combined convection and radiation is")
+else
+ disp("(a) The Flow is turbulent")
+end
+//Solution(b)
+At=w*tp;//[m^2]
+m=rho*At*vp;//ass of th plastic rolling out per unit time[kg/s]
+T2=T_p+(-Q_total/(m*Cp));//[degree Celcius]
+disp("degree Celcius",T2,"(b) The temperature of the plastic sheet as it leaves the cooling section is")
diff --git a/587/CH7/EX7.4/example7_4.sce b/587/CH7/EX7.4/example7_4.sce new file mode 100755 index 000000000..d392eb295 --- /dev/null +++ b/587/CH7/EX7.4/example7_4.sce @@ -0,0 +1,17 @@ +clear;
+clc;
+
+//Example7.4[Drag Force Acting on a Pipe in a River]
+//Given:-
+T_water=15;//[degree Celcius]
+vw=4;//Velocity of water[m/s]
+od=0.022;//Outer diameter of pipe[m]
+w=30;//width of river[m]
+//At 15 degree C properties of water
+rho=999.1;//[kg/m^3]
+mu=1.138*10^(-3);//viscosity[kg/m.s]
+Re=(rho*vw*od)/mu;//Reynolds number
+Cd=1.0;//Dreag coefficient
+A=w*od;//Frontal area for flow past a cylinder[m^2]
+Fd=Cd*A*rho*(vw^2)/2;//[N]
+disp("kN",Fd/1000,"The drag force acting on the pipe is")
\ No newline at end of file diff --git a/587/CH7/EX7.5/example7_5.sce b/587/CH7/EX7.5/example7_5.sce new file mode 100755 index 000000000..981db8e7f --- /dev/null +++ b/587/CH7/EX7.5/example7_5.sce @@ -0,0 +1,20 @@ +clear;
+clc;
+
+//Example7.5[Heat Loss from a Steam Pipe in Windy Air]
+d=0.1//diameter of pipe[m]
+Ts=110;//Temp of ecternal surface of pipe[degree Celcius]
+Ta=10;//Temp of air[degree Celcius]
+va=8;//Velocity of air[m/s]
+Tf=(Ts+Ta)/2;//Film temperature[degree Celcius]
+k=0.02808;//[W/m.degree Celcius]
+Pr=0.7202;//Prandtl Number
+nu=1.896*10^(-5);//Kinematic viscosity[m^2/s]
+//Solution:-
+Re=(va*d)/nu;//Reynolds Number
+Nu=0.3+((0.62*(Re^(0.5))*(Pr^(1/3)))/((1+((0.4/Pr)^(2/3)))^(1/4))*[(1+((Re/282000)^(5/8)))^(4/5)]);
+disp(round(Nu),"The nusselt number is")
+h=k*Nu/d;//[W/m^2.degree Celcius]
+As=%pi*d*1;//Area of pipe per unit length[m^2]
+Q=h*As*(Ts-Ta);//[W]
+disp("W",ceil(Q),"The rate of heat loss from the pipe per unit of its length is")
\ No newline at end of file diff --git a/587/CH7/EX7.6/example7_6.sce b/587/CH7/EX7.6/example7_6.sce new file mode 100755 index 000000000..eab4b82ae --- /dev/null +++ b/587/CH7/EX7.6/example7_6.sce @@ -0,0 +1,31 @@ +clear;
+clc;
+
+//Example7.6[Cooling of a Steel Ball by Forced Air]
+//Given:-
+rho=8055;//[kg/m^3]
+Cp=480;//[J/kg.degree Celcius]
+To=300;//Temp of oven[degree Celcius]
+Ta=25;//Temp of air[degree Celcius]
+va=3;//Velocity of air[m/s]
+Ts=200;//Dropped temp of surface of ball[degree Celcius]
+Ts_avg=(Ts+To)/2;//[degree Celcius]
+d=0.25;//[m]
+mu_s=2.76*10^(-5);//Dynamic Viscosity at average surface temperature[kg/m.s]
+//Properties of air at 25 degree Celcius
+k=0.02551;//[W/m.degree Celcius]
+nu=1.562*10^(-5);//kinematic viscosity[m^2/s]
+mu=1.849*10^(-5);//Dynamic viscosity of air at 25 degree C[kg/m.s]
+//Solution:-
+Re=va*d/nu;//[Reynolds Number]
+Nu=2+[(0.4*(Re^(1/2)))+(0.06*(Re^(2/3)))]*(Pr^(0.4))*((mu/mu_s)^(1/4));
+disp(ceil(Nu),"The Nusselt number is")
+h=k*Nu/d;//[W/m^2.degree Celcius]
+As=%pi*(d^2);//[m^2]
+Q_avg=h*As*(Ts_avg-Ta);//[W]
+disp("W",ceil(Q_avg),"The average rate of heat transfer from Newtons Law of cooling is")
+m=rho*%pi*(d^3)/6;//[kg]
+Q_total=m*Cp*(To-Ts);//[J]
+disp("J",Q_total,"The total heat transferred from the ball is")
+delta_t=Q_total/Q_avg;//[s]
+disp("hour",delta_t/3600,"The time of cooling is")
\ No newline at end of file diff --git a/587/CH7/EX7.7/example7_7.sce b/587/CH7/EX7.7/example7_7.sce new file mode 100755 index 000000000..f8004016b --- /dev/null +++ b/587/CH7/EX7.7/example7_7.sce @@ -0,0 +1,44 @@ +clear;
+clc;
+
+//Example7.7[Preheating Air by Geothermal Water in a Tube Bank]
+//Given:-
+Ta_in=20;//Temp of air whileentering the duct[degree Celcius]
+v=4.5;//mean velocity[m/s]
+T_tw=120;//Temp of geothermal water[degree Celcius]
+od=0.015;//Outer Diameter of tubes[m]
+SL=0.05,ST=0.05;//Longitudinal and transverse pitches[m]
+//Properties of air at mean temp
+k=0.02808;//[W/m.K]
+rho=1.059;//[kg/m^3]
+Cp=1007;//[J/kg.K]
+Pr=0.7202;//Prandtl no
+Pr_s=0.7073;//Prandtl no at temp = 120 degree C
+mu=2.008*10^(-5);//Viscosity[kg/m.s]
+rho_in=1.204;//density of air at inlet conditions[kg/m^3]
+//Solution:-
+v_max=(ST*v)/(ST-od);//maximu velocity[m/s]
+Re=rho*v_max*od/mu;//Reynolds Number
+disp(Re,"Reynolds number is")
+Nu=0.27*(Re^(0.63))*(Pr^(0.36))*((Pr/Pr_s)^0.25);
+disp(Nu,"The nusselt number is")
+Nl=6;//No of rows of tubes
+Nt=10;//No of tubes in each row
+F=0.945;//For Nl=6, correction factor
+Nu_Nl=F*Nu;
+h=Nu_Nl*k/od;//[W/m^2.degree Celcius]
+N=Nl*Nt;//Total no of tubes
+//For unit tube length
+As=N*%pi*od*1;//[m^2]
+m=rho_in*v*(Nt*ST*1);//[kg/s]
+disp("kg/s",m,"Mass flow rate of air is")
+Te=T_tw-((T_tw-Ta_in)*exp((-As*h)/(m*Cp)));//[degree Celcius]
+disp("degree Celcius",Te,"Fluid exit temperature is")
+T_ln=(((T_tw-Te)-(T_tw-Ta_in))/(log((T_tw-Te)/(T_tw-Ta_in))));//[degree Celcius]
+disp("degree Celcius",T_ln,"Log mean temperature difference is")
+Q=h*As*T_ln;//[W]
+disp("W",Q,"Rate of heat transfer is")
+//For given Re and SL/od ratio friction coefficient is
+f=0.16;
+delta_P=Nl*f*rho*(v_max^2)/2;//[Pa]
+disp("Pa",delta_P,"The pressure drop across the tube bank is")
diff --git a/587/CH7/EX7.8/example7_8.sce b/587/CH7/EX7.8/example7_8.sce new file mode 100755 index 000000000..736b5f6c2 --- /dev/null +++ b/587/CH7/EX7.8/example7_8.sce @@ -0,0 +1,35 @@ +clear;
+clc;
+
+//Example7.8[Effect of insulation on Surface Temperature]
+//Given:-
+Ti=120;//Initial temp of hot water[degree Celcius]
+k_pipe=15;//W/m.degree Celcius
+ri=0.008,ro=0.01;//Inner and outer radii[m]
+t=0.002;//Thickness of pipe[m]
+To=25;//Ambient temperature[degree Celcius]
+Ts=40;//Maximum Temp of outer surface of insulation[degree Celcius]
+hi=70,ho=20;//Heat transfer coefficients inside and outside of the pipe[W/m^2.degree Celcius]
+k_insu=0.038;//[W/m.degree Celcius]
+L=1;//section of pipe[m]
+//Solution:-
+//Areas of surfaces exposed to convection
+A1=2*%pi*ri*L;//[m^2]
+//Individual Thermal Resistances
+R_conv1=1/(hi*A1);//[degree Celcius/W]
+R_pipe=(log(ro/ri))/(2*%pi*k_pipe*L);//[degree Celcius/W]
+//R_insu=(log(r3/ri))/(2*%pi*k_insu*L)
+//R_conv2=1/(ho*2*%pi*r3*L)
+//R_total=R_conv1+R_conv2+R_pipe+R_insu
+//Q=(Ti-To)/R_total;
+//Q=(Ts-To)/R_conv2;
+//Equating both Q we get
+function[r]=radius(r3)
+ r(1)=1884*r3(1)*(0.284+0.0024+4.188*log((r3(1))/0.01)+(1/(125.6*r3(1))))-95;
+ deff('[r]=radius(r3)',['radius_3=1884*r3(1)*(0.284+0.0024+4.188*log((r3(1))/0.01)+(1/(125.6*r3(1))))-95'])
+endfunction
+ disp("m",xs,"The outer radius of the insulation is")
+ t=xs-ro;//[m]
+ disp("cm",100*t,"The minimum thickness of fibreglass insulation required is")
+ ///Correct output will be displayed after executing the codes once and then re-executin them
+
\ No newline at end of file diff --git a/587/CH7/EX7.9/example7_9.sce b/587/CH7/EX7.9/example7_9.sce new file mode 100755 index 000000000..bfbd1ec75 --- /dev/null +++ b/587/CH7/EX7.9/example7_9.sce @@ -0,0 +1,32 @@ +clear;
+clc;
+
+//Example7.9[Optimum Thickness of insulation]
+//Given:-
+k_insu=0.024;//[Btu/h.ft^2.degree Farenhiet]
+Ts=180;//temp of exposed surface of oven[degree F]
+Ta=75;//temp of ambient air[degree F]
+L=12;//length[ft]
+d=8;//Diameter[m]
+time=5840;//[h/year]
+ho=3.58;//Heat transfer coefficient on the outer surface[tu/h.ft^2.degree F]
+unit_c1=0.75;//[$/therm]
+unit_c2=2.70;//Unit cost of insulation[4/ft^2]
+neta=0.8;//Efficiency
+//Solution:-
+As=(2*%pi*((d/2)^2))+(2*%pi*L*d/2);//Exposed surface area[ft^2]
+disp(As)
+Q=ho*ceil(As)*(Ts-Ta);//[Btu/h]
+Q_total=(1/neta)*Q*time/(100000);//[therms]
+disp("Therms",Q_total,"The total amount of heat loss from the surrounding is")
+annual_c1=Q_total*unit_c1;//[$/year]
+disp("per year",annual_c1,"The annual fuel cost of the oven before insulation is $")
+R_conv=1/(ho*ceil(As));
+R_insu=(1/12)/(k_insu*ceil(As));//Thickness id 1inch or 1/12 ft
+Q_insu=(Ts-Ta)/(R_conv+R_insu);//[Btu/hr]
+Q_insu_total=(1/neta)*Q_insu*time*(1/100000);//[therms]
+disp("therms",Q_insu_total,"Total energy consumption by oven on being insulated")
+annual_c2=Q_insu_total*unit_c1;//[$/yr]
+insu_cost=(unit_c2*ceil(As));//Insulation cost[$]
+Total_c=annual_c2+insu_cost;//[$]
+disp(Total_c,"The sum of insulation cost and heat loss costs is $")
diff --git a/587/CH8/EX8.1/example8_1.sce b/587/CH8/EX8.1/example8_1.sce new file mode 100755 index 000000000..c0bbb2086 --- /dev/null +++ b/587/CH8/EX8.1/example8_1.sce @@ -0,0 +1,25 @@ +clear;
+clc;
+
+//Example8.1[Heating of water in a tube by Steam]
+//Given:-
+id=0.025;//Internal diameter[m]
+Tin=15;//Initial temp[degree Celcius]
+m_=0.3;//Flow rate[kg/s]
+h=800/1000;//avg heat transfer coefficient[W/m^2.degree Celcius]
+Tf=115;//Final temp of water[degree Celcius]
+Ts=120;//[degree Celcius]
+Hs=2203;//Heat of condensation of steam at 120 degree Celcius[kJ/kg]
+Tavg=(Tin+Tf)/2;//[degree Celcius]
+Cp=4187;//Sp Heat of water at Tavg[J/kg.degree Celcius]
+//Solution:-
+Q_=m_*Cp*(Tf-Tin)/1000;//[kW]
+disp("kW",Q_,"The rate of heat transfer is")
+del_Tf=Ts-Tf;//[degree Celcius]
+del_Tin=Ts-Tin;//[degree Celcius]
+ln_del_T=(del_Tf-del_Tin)/(log(del_Tf/del_Tin));//[degree Celcius]
+disp("degree Celcius",ln_del_T,"Logrithmic Mean temperature difference is")
+A=Q_/(h*ln_del_T);//[m^2]
+disp("m^2",A,"Heat Transfer surface area is")
+l=A/(%pi*id);//[m]
+disp("m",round(l),"Required tube length is")
\ No newline at end of file diff --git a/587/CH8/EX8.2/example8_2.sce b/587/CH8/EX8.2/example8_2.sce new file mode 100755 index 000000000..9914bb538 --- /dev/null +++ b/587/CH8/EX8.2/example8_2.sce @@ -0,0 +1,24 @@ +clear;
+clc;
+
+//Example8.2[Pressure Drop in a tube]
+//Given:-
+Tw=5;//Temperature of water[degree Celcius]
+//Properties of water at Tw
+rho=999.9;//[kg/m^3]
+mu=1.519*10^(-3);//Viscosity[kg/m.s]
+d=0.003;//diameter[m]
+l=10;//length[m]
+v_avg=0.9;//Average flow velocity[m/s]
+//Solution:-
+Re=(rho*v_avg*d)/mu;
+disp(Re,"The reynolds number is ")
+f=64/ceil(Re);
+disp(f,"Friction factor is")
+del_P=f*l*rho*(v_avg^2)/(2*d);//[N/m^2]
+disp("kPa",del_P/1000,"The Pressure drop is ")
+V=v_avg*(%pi*(d^2))/4;//[m^3/s]
+disp("m^3/s",V,"Volumetric flow rate is")
+W_pump=V*del_P;//[W]
+disp("W",W_pump,"Mechanical Power Input of")
+disp("is needed to overcome the frictional losses in the flow due to viscosity")
diff --git a/587/CH8/EX8.3/example8_3.sce b/587/CH8/EX8.3/example8_3.sce new file mode 100755 index 000000000..f1b4e8cd0 --- /dev/null +++ b/587/CH8/EX8.3/example8_3.sce @@ -0,0 +1,38 @@ +clear;
+clc;
+
+//Example8.3[Flow of Oil in a Pipeline through a Lake]
+//Given:-
+Ts=0;//Temp of lake[degree Celcius]
+Ti=20;//Temp of oil[degree Celcius]
+d=0.3;//Diameter[m]
+l=200;//length of pipe[m]
+//At 20 degree Celcius
+rho=888.1;//[kg/m^3]
+nu=9.429*10^(-4);//Kinematic viscosity[m^2/s]
+k=0.145;//[W/m.degree Celcius]
+Cp=1880;//[J/kg.degree Celcius]
+Pr=10863;//Prandtl Number
+v_avg=2;//[m/s]
+//Solution(a)
+Re=v_avg*d/nu;
+disp(ceil(Re),"The Reynolds number is")
+Lt=0.05*Re*Pr*d;//[m]
+disp("m",Lt,"The thermal entry length is")
+Nu=3.66+((0.065*(d/l)*Re*Pr)/(1+(0.04*(((d/l)*Re*Pr)^(2/3)))));
+h=(k*Nu)/d;//[W/m^2.degree Celcius]
+As=%pi*d*l;//[m^2]
+m_=rho*%pi*((d/2)^2)*v_avg;//[kg/s]
+Te=Ts-((Ts-Ti)*exp((-h*As)/(m_*Cp)));//[degree Celcius]
+disp("degree Celcius",Te,"Exit temperature of oil is")
+//Solution(b):-
+ln_del_T=(Ti-Te)/(log((Ts-Te)/(Ts-Ti)));//[degree Celcius]
+disp("degree Celcius",ln_del_T,"The logrithmic mean temperature difference is")
+Q=h*As*ln_del_T;//[W]
+disp("W",Q,"The rate of heat loss from the oil are")
+//Solution(c)
+f=64/Re;//Friction factor is
+del_P=l*rho*(v_avg^2)/(2*d);//[N/m^2]
+disp(del_P);
+W_pump=m_*del_P/rho;//[kW]
+disp("pump just to overcome the friction in the pipe as the oil flows","kW",W_pump/1000,"We need a")
\ No newline at end of file diff --git a/587/CH8/EX8.4/example8_4.sce b/587/CH8/EX8.4/example8_4.sce new file mode 100755 index 000000000..9c91fedd1 --- /dev/null +++ b/587/CH8/EX8.4/example8_4.sce @@ -0,0 +1,26 @@ +clear;
+clc;
+
+//Example8.4[Pressure Drop in a Water tube]
+Tw=15;//temp of water while entering[degree Celcius]
+rho=999.1;//[kg/m^3]
+mu=1.138*10^(-3);//Viscosity[kg/m.s]
+id=0.05;//Internal diameter[m]
+V=5.5*10^(-3);//Flow rate[m^3/s]
+l=60;//length of tube[m]
+e=0.002*10^(-3);//[m]
+//Solution:-
+v=V/(%pi*(id^2)*(1/4));//Mean Velocity[m/s]
+Re=rho*v*id/mu;
+disp(Re,"Reynolds Number is")
+//Flow is turbulent
+r=e/id;//Relative roughness of the tube
+function[Func]=fric(fac)
+ Func(1)=(1/(fac(1)^(1/2)))+(2*log((0.00004/3.7)+(2.51/(122900*fac(1)^(1/2)))));
+ deff('[Func]=fric(fac)',['fric_1=(1/(fac(1)^(1/2)))+(2*log((0.00004/3.7)+(2.51/(122900*fac(1)^(1/2)))))'])
+endfunction
+disp(xs,"Friction Factor is")
+del_P=xs*l*rho*(v^2)/(2*id);//[kPa]
+disp("Pa",del_P,"The pressure drop is")
+W_pump=V*del_P;//[W]
+disp("W",W_pump,"The required poer input tp overcome the frictional losses in the tube is")
\ No newline at end of file diff --git a/587/CH8/EX8.5/example8_5.sce b/587/CH8/EX8.5/example8_5.sce new file mode 100755 index 000000000..ae73686a7 --- /dev/null +++ b/587/CH8/EX8.5/example8_5.sce @@ -0,0 +1,32 @@ +clear;
+clc;
+
+//Example8.5[Heating of water by Resistance Heaters in a tube]
+Ti=15;//Initial Temp[degree Celcius]
+Tf=65;//Final Temp[degree Celcius]
+d=0.03;//Internal diameter[m]
+l=5;//length[m]
+V=10*10^(-3);//flow rate of water[m^3/s]
+Tavg=(Ti+Tf)/2;//[degree Celcius]
+//Properties of water at Tavg
+rho=992.1;//[kg/m^3]
+Cp=4170;//[J/kg.degree Celcius]
+k=0.631;//[W/m.degree Celcius]
+nu=0.658*10^(-6);//[m^2/s]
+Pr=4.32;//Prandtl Number
+//Solution:-
+Ac=%pi*(d^2)*(1/4);//[m^2]
+As=%pi*d*l;//[m^2]
+m_=rho*V*(1/60);//[kg/s]
+Q_=m_*Cp*(Tf-Ti)/1000;//[kW]
+disp("kW",Q_,"The power rating of the heater is")
+qs=Q_/As;//[kW/m^2]
+disp("kW/m^2",qs,"Heat flux is")
+v_avg=V/(Ac*60);//[m/s]
+Re=v_avg*d/nu;//[Reynolds Number]
+Lt=10*d;//Entry length [m]
+Nu=0.023*(Re^(0.8))*(Pr^(0.4));
+disp(Nu,"The nussel number is")
+h=k*Nu/d;//[W/m^2]
+Ts=Tf+(qs*1000/h);//[degree Celcius]
+disp("degree Celcius",round(Ts),"The surface temperature of the pipe at the exit becomes")
diff --git a/587/CH8/EX8.6/example8_6.sce b/587/CH8/EX8.6/example8_6.sce new file mode 100755 index 000000000..cfa192bde --- /dev/null +++ b/587/CH8/EX8.6/example8_6.sce @@ -0,0 +1,30 @@ +clear;
+clc;
+
+//Example8.6[Heat Loss from the ducts of a Heating System]
+Ti=80;//Inlet temp[degree Celcius]
+A=0.2*0.2;//Area of cross section[m^2]
+l=8;//Length of tube[m]
+V=0.15;//[m^3/s]
+Td=60;//Temperature of duct[degree Celcius]
+//Properties of air at inlet conditions
+rho=0.9994;//[kg/m^3]
+Cp=1008;//[J/kg.degree Celcius]
+k=0.02953;//[W/m.degree Celcius]
+nu=2.097*10^(-5);//[m^2/s]
+Pr=0.7154;//Prandtl number
+//Solution:-
+Dh=4*A/(4*0.2);//Hydraulic Diameter[m]
+v_avg=V/A;//[m/s]
+Re=v_avg*Dh/nu;
+disp(Re,"Reynolds number is")
+Lt=10*Dh;//Entry length
+Nu=0.023*(Re^(0.8))*(Pr^(0.3));
+h=Nu*k/Dh;//[W/m^2.degree Celcius]
+As=4*0.2*l;//[m^2]
+m_=rho*V;//[kg/s]
+Te=Td-((Td-Ti)*exp((-h*As)/(m_*Cp)));//[degree Celcius]
+disp("degree Celcius",Te,"The exit temperature of air is")
+ln_delT=(Ti-Te)/(log((Td-Te)/(Td-Ti)));//[degree Celcius]
+Q=h*As*ln_delT;//[W]
+disp("respectively","W",round(Q),"and","degree Celcius",ln_delT,"The logrithmic mean temperature difference and the rate of heat loss from the air are")
\ No newline at end of file diff --git a/587/CH8/EX8.7/example8_7.sce b/587/CH8/EX8.7/example8_7.sce new file mode 100755 index 000000000..76be5f097 --- /dev/null +++ b/587/CH8/EX8.7/example8_7.sce @@ -0,0 +1,23 @@ +clear;
+clc;
+
+//Example8.7[Non-isothermal fully developed Friction in the Transition Region]
+//Given:-
+q=8;//Wall heat flux[kW/m^2]
+xm=0.34;//Mass fraction
+d=0.0158;//Inside diameter[m]
+V=1.32*10^(-4);//Flow rate[m^3/s]
+Pr=11.6;//Prandtl Number
+nu=1.39*10^(-6);//[m^2/s]
+p=1.14;//(mu_b/mu_s)i.e. ratio of viscosities of two substances
+Gr=60800;//Grashof number
+//Solution:-
+Ac=%pi*(d^2)*(1/4);//[m^2]
+Re=(V/Ac)*d/nu;
+disp(Re,"Reynolds number is")
+//For bell mouth inlet shape
+Cf1=((1+((round(Re)/5340)^(-0.099)))^(-6.32))*(p^(-2.58-0.42*(60.800^(-0.41))*(11.6^0.265)));
+disp(Cf1,"For bell mouth inlet friction coefficient is")
+//For square edged inlet Case
+Cf2=(0.0791/(Re^(0.25)))*(p^(-0.25));
+disp(Cf2,"For square edged inlet case coefficient of friction is")
\ No newline at end of file diff --git a/587/CH8/EX8.8/example8_8.sce b/587/CH8/EX8.8/example8_8.sce new file mode 100755 index 000000000..21ad7b2dd --- /dev/null +++ b/587/CH8/EX8.8/example8_8.sce @@ -0,0 +1,27 @@ +clear;
+clc;
+
+//Example8.8[Heat transfer in the Transition Region]
+//Given:-
+xm=0.6;//mass fraction of glycol
+V=2.6*10^(-4);//Flow rate[m^3/s]
+d=0.0158;//inside diameter[m]
+Gr=51770;//grashof number
+Pr=29.2;//Prandtl number
+nu=3.12*10^(-6);//[m^2/s]
+p=1.77;//mu_t/mu_s
+q=90;//A particular loctaion x with x/d=q
+//Solution:-
+Ac=%pi*(d^2)/4;
+Re=(V/Ac)*d/nu;
+disp(Re,"Reynolds Number is")
+//Value of Re lies in transition Region
+Nu_lam=1.24*(((Re*Pr/q)+(0.025*((Gr*Pr)^(0.75))))^(1/3))*(p^(0.14));
+Nu_tur=0.023*(Re^(0.8))*(Pr^0.385)*(q^(-0.0054))*(p^(0.14));
+//(a)
+Nu_tran_a=Nu_lam+((exp((1766-Re)/276)+(Nu_tur^(-0.955)))^(-0.955));
+disp(Nu_tran_a,"(a) Nusselt number for re-entrant inlet is")
+Nu_tran_b=Nu_lam+((exp((2617-Re)/207)+(Nu_tur^(-0.950)))^(-0.950));
+disp(Nu_tran_b,"(b) Nusselt number for square edged inlet is")
+Nu_tran_c=Nu_lam+((exp((6628-Re)/237)+(Nu_tur^(-0.980)))^(-0.980));
+disp(Nu_tran_c,"(c) Nusselt number for bell mouth inlet is")
\ No newline at end of file diff --git a/587/CH9/EX9.1/example9_1.sce b/587/CH9/EX9.1/example9_1.sce new file mode 100755 index 000000000..79d18474b --- /dev/null +++ b/587/CH9/EX9.1/example9_1.sce @@ -0,0 +1,31 @@ +clear;
+clc;
+
+//Example9.1[Heat Loss from Hot Water Pipes]
+//Given:-
+l=6;//Length[m]
+d=0.08;//diameter[m]
+T_room=20;//[degree Celcius]
+Ts=70;//Surface temperature of pipe[degree Celcius]
+Tf=(Ts+T_room)/2;//Film temperature[degree Celcius]
+//Properties of air at Tf
+k=0.02699;//[W/m.degree Celcius]
+Pr=0.7241;//Prandtl number
+nu=1.750*10^(-5);//[m^2/s]
+b=(1/(Tf+273));//[K^-1]
+g=9.81;//Acc dur to gravity[m/s^2]
+e=1;//Emissivity
+//Solution:-
+Lc=d;//Characteristic length[m]
+Ra_d=g*b*(Ts-T_room)*(d^3)*Pr/(nu^2);
+disp(Ra_d,"The Rayleigh number is")
+Nu=((0.6+((0.387*(Ra_d^(1/6)))/((1+((0.559/Pr)^(9/16)))^(8/27))))^2);
+disp(Nu," The natural convection Nusselt number is")
+h=k*Nu/d;//[W/m^2.degree Celcius]
+As=%pi*d*l;//[m^2]
+Q=h*As*(Ts-T_room);//[W]
+disp("by natural convection","W",round(Q),"The pipe loses heat to the air in the room at a rate of")
+Q_rad=e*As*(5.76*10^(-8))*(((Ts+273)^4)-((T_room+273)^4));//[W]
+disp("W",ceil(Q_rad),"The radiation heat transfer is")
+
+
diff --git a/587/CH9/EX9.2/example9_2.sce b/587/CH9/EX9.2/example9_2.sce new file mode 100755 index 000000000..15bc120dd --- /dev/null +++ b/587/CH9/EX9.2/example9_2.sce @@ -0,0 +1,42 @@ +clear;
+clc;
+
+//Example9.2[Cooling of a Plate in different orientaions]
+L=0.6;//side of square plate[m]
+T_surr=30;//[degree Celcius]
+Tp=90;//Temp of plate[degree Celcius]
+Tf=(Tp+T_surr)/2;//Film temperature[degree Celcius]
+//Properties of air at Tf
+k=0.02808;//[W/m.degree Celcius]
+Pr=0.7202;//Prandtl number
+nu=1.896*10^(-5);//Kinematic viscosity[m^2/s]
+b=1/(Tf+273);//[K^-1]
+g=9.81;//Acc due to gravity[m/s^2]
+//Solution (a)
+Lc_a=L;//Characteristic length
+Ra_1=g*b*(Tp-T_surr)*(L^3)*Pr/(nu^2);
+disp(Ra_1,"(a) The Rayleigh no is")
+Nu_a=((0.825+(0.387*(Ra_1^(1/6)))/((1+((0.492/Pr)^(9/16)))^(8/27)))^2);
+disp(Nu_a,"The natural convection Nusselt number is")
+h_a=k*Nu_a/L;//[W/m^2.degree Celcius]
+As=L^2;//[m^2]
+Q_a=h_a*As*(Tp-T_surr);//[W]
+disp("W",ceil(Q_a),"Heat loss to the surrounding is")
+//Solution (b)
+Lc_b=As/(4*L);//[m]
+Ra_2=g*b*(Tp-T_surr)*(Lc_b^3)*Pr/(nu^2);
+disp(Ra_2,"(b) The Rayleigh number is")
+Nu_b=0.54*(Ra_2^(1/4));
+disp(Nu_b,"The natural convection Nusselt number is")
+h_b=k*Nu_b/Lc_b;//[W/m^2.degree Celcius]
+Q_b=h_b*As*(Tp-T_surr);//[W]
+disp("W",round(Q_b),"Heat Loss is")
+//Solution (c)
+Lc_c=Lc_b
+Nu_c=(0.27*Ra_2^(1/4));
+disp(Nu_c,"(c) Natural convection Nusselt number")
+h_c=k*Nu_c/Lc_c;//[W/m^2.degree Celcius]
+Q_c=h_c*As*(Tp-T_surr);//[W]
+disp("W",Q_c,"Heat Loss is")
+Q_rad=e*(5.67*10^(-8))*As*(((Tp+273)^4)-((T_surr+273)^4));//[W]
+disp("W",round(Q_rad),"Radiation heat loss is")
diff --git a/587/CH9/EX9.3/example9_3.sce b/587/CH9/EX9.3/example9_3.sce new file mode 100755 index 000000000..dbfa763ad --- /dev/null +++ b/587/CH9/EX9.3/example9_3.sce @@ -0,0 +1,29 @@ +clear;
+clc;
+
+//Example9.3[Optimum Fin Spacing of a Heat Sink]
+//Given:-
+w=0.12;//width[m]
+l=0.18;//length[m]
+t=0.001;//thickness[m]
+H=0.024;//height[m]
+Ts=80;//Bast temperature[degree Celcius]
+T_surr=30;//[degree Celcius]
+Tf=(Ts+T_surr)/2;//[degree Celcius]
+//Properties of air at film temperature
+k=0.02772;//[W/m.degree Celcius]
+Pr=0.7215;//Prandtl number
+nu=1.847*10^(-5);//[m^2/s]
+b=1/(Tf+273);//[K^-1]
+g=9.81;//[m/s^2]
+//Solution:-
+Ra_l=g*b*(Ts-T_surr)*(l^3)*Pr/(nu^2);
+disp(Ra_l,"The Rayleigh number is")
+S_opt=2.714*l/(Ra_l^(0.25));//[m]
+disp("mm",S_opt*100,"The optimum spacing is")
+n=w/(S_opt+t);
+disp(round(n),"The no of for this optimum fin spacing are")
+Nu_opt=1.307;//Optimum Nusselt number
+h=Nu_opt*k/S_opt;//[W/m^2.degree Celcius]
+Q=h*2*round(n)*l*H*(Ts-T_surr);//[W]
+disp("W",Q,"The rate of natural convection heat transfer")
diff --git a/587/CH9/EX9.4/example9_4.sce b/587/CH9/EX9.4/example9_4.sce new file mode 100755 index 000000000..7113df50b --- /dev/null +++ b/587/CH9/EX9.4/example9_4.sce @@ -0,0 +1,25 @@ +clear;
+clc;
+
+//Example9.4[Heat Loss through a Double Pane Window]
+//Given:-
+H=0.8;//Height[m]
+L=0.02;//Air gap[m]
+w=2;//Width[m]
+T1=12,T2=2;//Glass Surface temperatures across the air gap
+Tavg=(T1+T2)/2;//[degree Celcius]
+k=0.02416;//[W/m.degree Celcius]
+Pr=0.7344;//Prandtl Number
+nu=1.4*10^(-5);//Kinematic Viscosity[m^2/s]
+g=9.81;//[m/s^2]
+//Solution:-
+Lc=L;//Characteristic length
+b=1/(Tavg+273);//[K^-1]
+Ra_L=g*b*(T1-T2)*Pr*(Lc^3)/(nu^2);
+disp(Ra_L,"The Rayleigh Number is")
+Nu=0.42*(Ra_L^(1/4))*(Pr^(0.012))*((H/L)^(-0.3));
+disp(Nu,"The Nusselt Number is")
+As=H*w;//[m^2]
+h=k*Nu/L;//[W/m^2.degree Celcius]
+Q=h*As*(T1-T2);
+disp("W",Q,"Rate at which Heat is Lost through the window is")
\ No newline at end of file diff --git a/587/CH9/EX9.5/example9_5.sce b/587/CH9/EX9.5/example9_5.sce new file mode 100755 index 000000000..fe82ef8a7 --- /dev/null +++ b/587/CH9/EX9.5/example9_5.sce @@ -0,0 +1,24 @@ +clear;
+clc;
+
+//Example9.5[Heat Transfer through a Spherical Enclosure]
+//Given:-
+Di=0.2;//Inner Diameter[m]
+Do=0.3;//Outer Diameter[m]
+Ti=320,To=280;//The surface temperatures of two spheres enclosing the air[K]
+Tavg=(Ti+To)/2;//[K]
+//Properties at Tavg
+k=0.02566;//[W/m.K]
+Pr=0.7290;//Prandtl Number
+nu=1.58*10^(-5);//[m^2/s]
+b=(1/Tavg);
+g=9.81;//[m/s^2]
+//Solution:-
+Lc=(Do-Di)/2;//Characteristic length[m]
+Ra_L=g*b*(Ti-To)*(Lc^3)*Pr/(nu^2);
+disp(Ra_L,"The Rayleigh Number is")
+Fsph=Lc/(((Di*Do)^4)*((((Di^(-7/5))+(Do^(-7/5))))^5));
+keff=0.74*k*((Pr/(0.861+Pr))^(1/4))*((Fsph*Ra_L)^(1/4));//[W/m.K]
+disp(Fsph,keff)
+Q=keff*(%pi*Di*Do/Lc)*(Ti-To);//[W]
+disp("W",Q,"The rate of heat transfer between the spheres is")
\ No newline at end of file diff --git a/587/CH9/EX9.6/example9_6.sce b/587/CH9/EX9.6/example9_6.sce new file mode 100755 index 000000000..cefe008a6 --- /dev/null +++ b/587/CH9/EX9.6/example9_6.sce @@ -0,0 +1,41 @@ +clear;
+clc;
+//Example9.6[Heating Water in a Tube by Solar Enegy]
+//Given:-
+Ts=40;//Glass Temp[degree Celcius]
+T_surr=20;//Surrounding temperature[degree Celcius]
+Tavg=(Ts+T_surr)/2;//[degree Celcius]
+Do=0.1;//[m]
+Di=0.05;//[m]
+L=1;//[m]
+//Properties of glass at Tavg
+k=0.02588;//[W/m.degree Celcius]
+Pr=0.7282;//Prandtl Number
+nu=1.608*10^(-5);//[m^2/s]
+b=1/(Tavg+273);//[K^-1]
+
+Q=30;//Rate pof absorpto\ion of solar radiation[W]
+g=9.81;//[m/s^2]
+//Solution:-
+Ao=%pi*Do*L;//Heat transfer surface area of the glass cover[m^2]
+Ra_D=g*b*(Ts-T_surr)*(Do^3)*Pr/(nu^2);
+disp(Ra_D,"The Rayleigh Number is")
+Nu=((0.6+((0.387*(Ra_D^(1/6)))/((1+((0.550/Pr)^(9/16)))^(8/27))))^2);
+disp(Nu,"The Nusselt number is")
+ho=k*Nu/Do;//[W/m^2.degree Celcius]
+Qo=ho*Ao*(Ts-T_surr);//[W]
+disp("W",Qo,"The rate of natural convection heat transfer from the glass cover to the ambient air is")
+//Value of Qo is less than 30W so assuming a higher temp of glass cover
+T_surr1=41;//[degree Celcius]
+Ts1=90;//[degree Celcius]
+Tavg1=(T_surr1+Ts1)/2;//[degree Celcius]
+b1=1/(Tavg1+273);//[K^-1]
+Lc=(Do-Di)/2;//Characteristic length [m]
+Ra_L1=g*b1*(Ts1-T_surr1)*(Lc^3)*Pr/(nu^2);
+disp(Ra_L1,"The Rayleigh number on assuming higher temperatures")
+Fcyl=((log(Do/Di))^4)/((Lc^3)*(((Di^(-3/5))+(Do^(-3/5)))^5));
+keff=0.386*k*((Pr/(0.861+Pr))^(1/4))*((Fcyl*Ra_L1)^(1/4));//[W/m.degree Celcius]
+Q1=2*%pi*keff*(Ts1-T_surr1)/(log(Do/Di));//[W]
+disp("W",Q1,"The rate of heat transfer between the cylinders is")
+//Obtained value of Q1 is more than 30 W, so using hit and trial aand suuming more values we get the tube temperature to be 82 degree Celcius,
+disp("Therefore tube will reach an equilibrium temperature of 82 degree Celcius when the pump fails")
\ No newline at end of file diff --git a/587/CH9/EX9.7/example9_7.sce b/587/CH9/EX9.7/example9_7.sce new file mode 100755 index 000000000..4ec529542 --- /dev/null +++ b/587/CH9/EX9.7/example9_7.sce @@ -0,0 +1,15 @@ +clear;
+clc;
+
+//Example9.7[U factor for Center of glass Section of Windows]
+//Given:-
+e=0.84;//Emissivity
+//For winter season
+hi=8.29;//[W/m^2.degree Celcius]
+ho=34.0;//[W/m^2.degree Celcius]
+//Solution:-
+e_eff=1/((1/e)+(1/e)-1);//Effective emissivity of air space
+//the effective emissivity and an average air space temperature of 0 degree Celcius read
+h_space=7.2;//[W/m^2.degree Celcius]
+U_center=1/((1/hi)+(1/ho)+(1/h_space));//[W/m^s.degree Celcius]
+disp("W/m^2.degree Celcius",U_center,"The center of glass U-factor value is")
diff --git a/587/CH9/EX9.8/example9_8.sce b/587/CH9/EX9.8/example9_8.sce new file mode 100755 index 000000000..9600aeffc --- /dev/null +++ b/587/CH9/EX9.8/example9_8.sce @@ -0,0 +1,21 @@ +clear;
+clc;
+
+//Example9.8[Heat Loss through Aluminium Framed Windows]
+//Given:-
+H=1.2;//Height[m]
+w=1.8;//Width[m]
+Ti=22;//Inside temp[degree Celcius]
+To=-10;//Outside temp[degree Celcius]
+U_a=6.63,U_b=3.51,U_c=1.92,hi=8.3;//[W/m^.degree Celcius]
+//Solution:-
+A_win=h*w;//[m^2]
+Q_win_a=U_a*A_win*(Ti-To);//[W]
+T_glass_a=Ti-(Q_win_a/(hi*A_win));//[degree Celcius]
+disp("degree Celcius",T_glass_a,"(a) The Inner surface temperature of the window glass is")
+Q_win_b=U_b*A_win*(Ti-To);//[W]
+T_glass_b=Ti-(Q_win_b/(hi*A_win));//[degree Celcius
+disp("degree Celcius",T_glass_b,"(b) The Inner surface temperature of the window glass is")
+Q_win_c=U_c*A_win*(Ti-To);//[W]
+T_glass_c=Ti-(Q_win_c/(hi*A_win));//[degree Celcius]
+disp("degree Celcius",T_glass_c,"(c) The Inner surface temperature of the window glass is")
\ No newline at end of file diff --git a/587/CH9/EX9.9/example9_9.sce b/587/CH9/EX9.9/example9_9.sce new file mode 100755 index 000000000..23c1a4ae2 --- /dev/null +++ b/587/CH9/EX9.9/example9_9.sce @@ -0,0 +1,14 @@ +clear;
+clc;
+
+//Example9.9[U-Factor of a Double-Door Window]
+//Given:-
+A_win=1.8*2.0;//[m^2]
+A_glazing=2*1.72*0.94;//[m^2]
+U_c=3.24,U_e=3.71,U_f=2.8;//U factors for the center edge and frame sections respectively [W/m^2.degree Celcius]
+//Solution:-
+A_frame=A_win-A_glazing;//[m^2]
+A_center=2*(1.72-0.13)*(0.94-0.13);//[m^2]
+A_edge=A_glazing-A_center;//[m^2]
+U_win=((U_c*A_center)+(U_e*A_edge)+(U_f*A_frame))/A_win;//[W/m^2.degree Celcius]
+disp("W/m^2.degree Celcius",U_win,"The overall U factor of the entire window is")
\ No newline at end of file |