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diff --git a/1910/CH5/EX5.3/Chapter53.sce b/1910/CH5/EX5.3/Chapter53.sce
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+// Display mode
+mode(0);
+// Display warning for floating point exception
+ieee(1);
+clear;
+clc;
+disp("Introduction to heat transfer by S.K.Som, Chapter 5, Example 3")
+//Air at temprature (T1=20°C) and 1 atmospheric pressure flows over a flat plate with a free stream velocity(Uinf) of 1m/s.
+Uinf=1;
+T1=20;
+//The length of plate is 1m and is heated over its entire length to a constant temprature of T2=100°C.
+T2=100;
+//For air at 20°C(The mean temprature of 100°C and 20°C),viscosity(mu=1.9*10^-5kg/(m*s)),density(rho=1.05kg/m^3),conductivity(k=0.03W/(m*K)),specific heat(cp=1.007kJ/(kg*K))
+//Prandtl number is Pr=0.7
+mu=1.9*10^-5;
+rho=1.05;
+k=0.03;
+cp=1.007;
+Pr=0.7;
+//For laminar flow over a plate Nusselt number is Nux=0.332*Rex^0.5*Pr^(1/3)
+//The boundary layer flow over a flat plate will be laminar if Reynolds number is Rex=(rho*Uinf*x)/mu<5*10^5
+//First of all we have to check whether the flow is laminar or not.
+//Let us check at x=1m
+x=1;
+disp("Reynolds number is")
+ReL=(rho*Uinf*x)/mu
+//There fore the flow is laminar and we can use the relationships of Nux,
+//Thus Rex=(1.05*1*x)/(1.9*10^-5)=0.5526*10^5*x
+//Therefore we can write Nux=(hx*x/k)=0.332*(0.5526*10^5*x)^0.5*Pr^(1/3)....or hx=2.08*x^(-1/2) W/(m^2*°C)
+//hbarL is the average heat transfer coefficient over a length(L)
+disp("The average heat transfer coefficient over a length(L)= 1m ,in W/m^2 is")
+L=1;
+hbarL=(1/L)*integrate("2.08*x^(-1/2)",'x',0,L)
+//Q is the rate of heat transfer
+disp("The rate of heat transfer in W/m of width is")
+Q=hbarL*L*(T2-T1)
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diff --git a/1910/CH5/EX5.4/Chapter54.sce b/1910/CH5/EX5.4/Chapter54.sce
new file mode 100755
index 000000000..5edd0e7b6
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+// Display mode
+mode(0);
+// Display warning for floating point exception
+ieee(1);
+clear;
+clc;
+disp("Introduction to heat transfer by S.K.Som, Chapter 5, Example 4")
+//Air at atmospheric pressure is required to flow over a circuit board to cool the electronics element mounted on it.
+//Chip has length (L)=3mm and width(B)=3mm located x=0.1m from the leading edge
+L=0.003;//in metre
+B=0.003;//in metre
+x=0.1;
+//The Nusselt no. is given by Nux=0.06*Rex^0.85*Pr^0.33
+//The chip has to dissipate E=50mW of energy while its surface temprature has to be kept below temprature,Ts=45°C and free strem Temptrature of air is Tinf=25°C
+Ts=45;
+Tinf=25;
+E=50*10^-3;//in watt
+//For air ,density(rho=1.2kg/m^3),viscosity(mu=1.8*10^5kg/(m*s)),conductivity(k=0.03W/(m*K)) and specific heat(cp=1000J/(kg*K))
+rho=1.2;
+mu=1.8*10^5;
+k=0.03;
+cp=1000;
+//Let the minimum flow velocity be U.
+//The local heat transfer coefficient hx where the chip is mounted is determined as hx=(k/x)*0.06*(rho*U*x/mu)^0.85*(mu*cp/k)^0.33
+disp("The local heat transfer coefficient hx is hx=27.063*U^0.85")
+//from an enrgy balance we can write 27.063*U^0.85*L*B*(Ts-Tinf)=E
+disp("The minimum flow velocity in m/s is")
+U=[E/(27.063*L*B*(Ts-Tinf))]^(1/0.85)
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diff --git a/1910/CH5/EX5.6/Chapter56.sce b/1910/CH5/EX5.6/Chapter56.sce
new file mode 100755
index 000000000..358bef46d
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+++ b/1910/CH5/EX5.6/Chapter56.sce
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+// Display mode
+mode(0);
+// Display warning for floating point exception
+ieee(1);
+clear;
+clc;
+disp("Introduction to heat transfer by S.K.Som, Chapter 5, Example 6")
+//Air at 1atm pressure and temprature(Tin)=30°C enters a tube of 25mm diameter(D) with a velocity(U) of 10m/s
+D=0.025;//in metre
+U=10;
+Tin=30;
+//Tube is heated so that a constant heat flux(q) of 2kW/m^2 is maintained at the wall whose temprature is deltaT=20°C above the bulk mean air temprature through the length of tube 
+//Let Tw-Tb=T
+q=2000;
+deltaT=20;
+//The length(L)= 2m
+L=2;
+//For air density(rho=1.2kg/m^3),specific heat(cp=1000J/(kg*K))
+rho=1.2;
+cp=1000;
+//From an energy balance of a control volume of air we get mdot*cp*(Tb+(dTb/dx)*deltax-Tb)=q*pi*D*deltax or (dTb/dx)=(q*pi*D)/(mdot*cp)
+//mass flow rate=mdot
+mdot=rho*%pi*D^2*U;
+//let (dTb/dx)=Y
+disp("(dTb/dx)in °C/m is")
+Y=(4*q*%pi*D)/(mdot*cp)
+//Tb2 is Exit bulk mean temprature
+disp("Therefore Exit bulk mean temprature Tb2 in °C is")
+Tb2=Tin+Y*2
+//Again we can write at any section of the tube hx*(Tw-Tb)=q or hx=q/(Tw-Tb)
+//hx is heat flux
+disp("Heat flux(hx) in W/(m^2*°C) is ")
+hx=q/(deltaT)
+//Since Tw-Tb remains the same,The heat transfer coefficient at all sections are the same
+//Now Overall Nusselt number,NuL=hx*D/k
+//The thermal conductivity of air at mean temprature of (30+83.4)/2=56.7°C is k=0.0285 W/(m*K)
+k=0.0285;
+disp("Overall Nusselt number is ")
+NuL=hx*D/k
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diff --git a/1910/CH5/EX5.7/Chapter57.sce b/1910/CH5/EX5.7/Chapter57.sce
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index 000000000..5ed652784
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+++ b/1910/CH5/EX5.7/Chapter57.sce
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+// Display mode
+mode(0);
+// Display warning for floating point exception
+ieee(1);
+clear;
+clc;
+disp("Introduction to heat transfer by S.K.Som, Chapter 5, Example 7")
+//A wall is exposed to nitrogen at one atmospheric pressure and temprature,Tinf=4°C.
+Tinf=4;
+//The wall is H=2.0m high and B=2.5m wide and is maintained at temprature,Ts=56°C
+Ts=56;
+H=2;
+B=2.5;
+A=H*B;//area is(A)
+//The average nusselt number NuHbar over the height of the plate is given by NuHbar=0.13*(Gr*Pr)^(1/3)
+//The properties of nitrogen at mean film temprature(Tf) is (56+4)/2=30°C are given as density(rho=1.142kg/m^3) ,conductivity(k=0.026W/(m*K)),
+//kinematic viscosity(nu=15.630*10^-6 m^2/s) ,Prandtl number(Pr=0.713)
+rho=1.142;
+k=0.026;
+nu=15.630*10^-6;
+Pr=0.713;
+Tf=30;
+//We first have to detrmine the value of Grashoff number,Gr.In consideration of nitrogen as an ideal gas,we can write
+//Beta(The volumetric coefficient of expansion)=1/T
+disp("Beta(The volumetric coefficient of expansion in K^-1 is")
+Beta=1/(273+Tf)
+//Now Gr=(g*Beta*(Ts-Tinf)*H^3)/nu^2
+g=9.81;//acceleration due to gravity
+disp("Grashoff number is")
+Gr=(g*Beta*(Ts-Tinf)*H^3)/nu^2
+disp("The average nusselt number is")
+NuHbar=0.13*(Gr*Pr)^(1/3)
+//hbar is the heat flux
+disp("Heat flux hbar in W/(m^2*°C)")
+hbar=NuHbar*k/H
+//Q is the heat loss from the plate
+disp("The heat loss from the plate in W is")
+Q=hbar*A*(Ts-Tinf)
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diff --git a/1910/CH5/EX5.8/Chapter58.sce b/1910/CH5/EX5.8/Chapter58.sce
new file mode 100755
index 000000000..1ef3d4879
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+++ b/1910/CH5/EX5.8/Chapter58.sce
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+// Display mode
+mode(0);
+// Display warning for floating point exception
+ieee(1);
+clear;
+clc;
+disp("Introduction to heat transfer by S.K.Som, Chapter 5, Example 8")
+//Eletric current passes through a L=0.5m long horizontal wire of D=0.1mm diameter.
+L=0.5;
+D=0.1*10^-3;
+//The wire is to be maintained at temprature,Twire=400K and the air is at temprature,Tair=300K.
+Twire=400;
+Tair=300;
+//The resistance of the wire(R) is 0.012ohm per meter.Nusselt number(NuL) over the length of wire to be 0.4.
+NuL=0.4;
+R=0.012;
+//At mean temprature of Tf=350K, The thermal conductivity of air is k=0.03W/(m*K)
+k=0.03;
+//Nusselt number is NuL=hbar*D/k
+//hbar is the heat flux
+disp("The heat flux in W/(m^2*K) is")
+hbar=NuL*k/D
+//Q is the heat loss from the wire
+disp("The heat loss from the wire is Q=hbar*pi*D*L*(Twire-Tair) in Watt")
+Q=hbar*%pi*D*L*(Twire-Tair)
+//At steady state the ohmic loss in the wire equals the heat loss from its surface Therfore I^2*R=Q
+//I is the current flow.
+disp("The current in Ampere is")
+I=(Q/(R*L))^0.5
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