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-rw-r--r--Introduction_to_Heat_Transfer_by_S._K._Som/Chapter1.ipynb49
-rw-r--r--Introduction_to_Heat_Transfer_by_S._K._Som/Chapter10.ipynb138
-rw-r--r--Introduction_to_Heat_Transfer_by_S._K._Som/Chapter11.ipynb101
-rw-r--r--Introduction_to_Heat_Transfer_by_S._K._Som/Chapter2.ipynb185
-rw-r--r--Introduction_to_Heat_Transfer_by_S._K._Som/Chapter3.ipynb19
-rw-r--r--Introduction_to_Heat_Transfer_by_S._K._Som/Chapter4.ipynb114
-rw-r--r--Introduction_to_Heat_Transfer_by_S._K._Som/Chapter5.ipynb107
-rw-r--r--Introduction_to_Heat_Transfer_by_S._K._Som/Chapter6.ipynb102
-rw-r--r--Introduction_to_Heat_Transfer_by_S._K._Som/Chapter7.ipynb242
-rw-r--r--Introduction_to_Heat_Transfer_by_S._K._Som/Chapter8.ipynb172
-rw-r--r--Introduction_to_Heat_Transfer_by_S._K._Som/Chapter9.ipynb180
-rw-r--r--Introduction_to_Heat_Transfer_by_S._K._Som/chapter12.ipynb5
12 files changed, 128 insertions, 1286 deletions
diff --git a/Introduction_to_Heat_Transfer_by_S._K._Som/Chapter1.ipynb b/Introduction_to_Heat_Transfer_by_S._K._Som/Chapter1.ipynb
index b36f371b..d3b728b1 100644
--- a/Introduction_to_Heat_Transfer_by_S._K._Som/Chapter1.ipynb
+++ b/Introduction_to_Heat_Transfer_by_S._K._Som/Chapter1.ipynb
@@ -57,7 +57,7 @@
},
{
"cell_type": "code",
- "execution_count": 2,
+ "execution_count": 3,
"metadata": {
"collapsed": false
},
@@ -88,7 +88,7 @@
"#The thickness of masonry wall is Lm.\n",
"print\"The thickness of masonry wall is Lm in m\"\n",
"Lm=(km/kc)*(Lc/(0.8))\n",
- "print\"Lm=\",Lm\n"
+ "print\"Lm=\",Lm"
]
},
{
@@ -100,7 +100,7 @@
},
{
"cell_type": "code",
- "execution_count": 3,
+ "execution_count": 5,
"metadata": {
"collapsed": false
},
@@ -139,7 +139,7 @@
},
{
"cell_type": "code",
- "execution_count": 4,
+ "execution_count": 6,
"metadata": {
"collapsed": false
},
@@ -178,7 +178,7 @@
},
{
"cell_type": "code",
- "execution_count": 5,
+ "execution_count": 14,
"metadata": {
"collapsed": false
},
@@ -189,9 +189,9 @@
"text": [
"Introduction to heat transfer by S.K.Som, Chapter 1, Example 6\n",
"The rate of heat transfer from the plate is given by Q=hbr*A*(Ts-Tinf)\n",
- "Q= 16000.0\n",
+ "Q= 224.0\n",
"The rate of heat transfer can also be written in the form of Q=m*cp*|dT/dt| from an energy balance.\n",
- "Q= 16000.0\n",
+ "Q= 224.0\n",
"Equating the above two equations we get hbr=(m*cp*|dT/dt|)/(A*(Ts-Tinf)) in W/(m**2°C)\n",
"hbr= 11.2\n"
]
@@ -220,21 +220,7 @@
"print\"Q=\",Q\n",
"print\"Equating the above two equations we get hbr=(m*cp*|dT/dt|)/(A*(Ts-Tinf)) in W/(m**2°C)\"\n",
"hbr=(m*cp*10**3*X)/(A*(Ts-Tinf))\n",
- "print\"hbr=\",hbr\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n"
+ "print\"hbr=\",hbr"
]
},
{
@@ -246,7 +232,7 @@
},
{
"cell_type": "code",
- "execution_count": 6,
+ "execution_count": 2,
"metadata": {
"collapsed": false
},
@@ -286,7 +272,7 @@
},
{
"cell_type": "code",
- "execution_count": 7,
+ "execution_count": 4,
"metadata": {
"collapsed": false
},
@@ -312,7 +298,7 @@
"print\"The emitted radiant energy per unit surface area is given by Eb/A=sigma*T**4 in W/m**2\"\n",
"#Let Eb/A=F\n",
"F=sigma*(50+273.15)**4\n",
- "print\"F=\",F\n"
+ "print\"F=\",F"
]
},
{
@@ -324,7 +310,7 @@
},
{
"cell_type": "code",
- "execution_count": 8,
+ "execution_count": 4,
"metadata": {
"collapsed": false
},
@@ -381,7 +367,7 @@
},
{
"cell_type": "code",
- "execution_count": 10,
+ "execution_count": 19,
"metadata": {
"collapsed": false
},
@@ -390,9 +376,11 @@
"name": "stdout",
"output_type": "stream",
"text": [
- " Introduction to heat transfer by S.K.Som, Chapter 1, Example 10\n",
+ "Introduction to heat transfer by S.K.Som, Chapter 1, Example 10\n",
"Heat transfer from the outer surface takes place only by radiation is given by Q/A=F1=emi*sigma*(T2**4-T0**4)in W/m**2 for different values of tempratures in K\n",
+ "F1= 332.029390022\n",
"heat transfer from the outer surface can also be written as Q/A=F2=(Ti-To)/((1/hbri)+(L/k)+(1/hr)) in W/m**2 at different tempratures in K\n",
+ "F2= 332.132667923\n",
"The values of temprature that are considered are <298 K\n",
"Satisfactory solutions for Temprature in K is\n",
"T2= 292.5\n",
@@ -425,7 +413,9 @@
"#Radiation heat transfer coefficient(hr) is defined as Q/A=hr(T2-To)\n",
"#so hr=4.536*10**-8*T2**3\n",
"print\"Heat transfer from the outer surface takes place only by radiation is given by Q/A=F1=emi*sigma*(T2**4-T0**4)in W/m**2 for different values of tempratures in K\"\n",
+ "print\"F1=\",F1\n",
"print\"heat transfer from the outer surface can also be written as Q/A=F2=(Ti-To)/((1/hbri)+(L/k)+(1/hr)) in W/m**2 at different tempratures in K\"\n",
+ "print\"F2=\",F2\n",
"print\"The values of temprature that are considered are <298 K\"\n",
"for i in range(285,292):\n",
" T2=i\n",
@@ -498,8 +488,7 @@
"print\"The total heat loss by The pipe per unit length is given by Q/L=hbr*A*(T1-T2)+sigma*emi*A*(T1**4-T2**4) in W/m\"\n",
"#Let Q/L=F\n",
"F=hbr*A*((T1+273.15)-(T2+273.15))+sigma*emi*A*((T1+273.15)**4-(T2+273.15)**4)\n",
- "print\"F=\",F\n",
- "\n"
+ "print\"F=\",F"
]
}
],
diff --git a/Introduction_to_Heat_Transfer_by_S._K._Som/Chapter10.ipynb b/Introduction_to_Heat_Transfer_by_S._K._Som/Chapter10.ipynb
index 95f29d79..9eacc4ed 100644
--- a/Introduction_to_Heat_Transfer_by_S._K._Som/Chapter10.ipynb
+++ b/Introduction_to_Heat_Transfer_by_S._K._Som/Chapter10.ipynb
@@ -44,10 +44,6 @@
}
],
"source": [
- " \n",
- " \n",
- " \n",
- " \n",
"import math\n",
" \n",
"print\"Introduction to heat transfer by S.K.Som, Chapter 10, Example 1\"\n",
@@ -94,29 +90,7 @@
"print\"LMTD=\",LMTD\n",
"print\"Area(A)=Q/(U*LMTD) in m**2\"\n",
"A=Q/(U*LMTD)\n",
- "print\"A=\",A\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n"
+ "print\"A=\",A"
]
},
{
@@ -150,10 +124,6 @@
}
],
"source": [
- " \n",
- " \n",
- " \n",
- " \n",
"import math\n",
" \n",
"print\"Introduction to heat transfer by S.K.Som, Chapter 10, Example 2\"\n",
@@ -193,7 +163,7 @@
"#Area(A)=Q/(U*LMTD) in m**2\n",
"print\"Area(A)=Q/(U*LMTD) in m**2\"\n",
"A=Q/(U*LMTD)\n",
- "print\"A=\",A\n"
+ "print\"A=\",A"
]
},
{
@@ -233,10 +203,6 @@
}
],
"source": [
- " \n",
- " \n",
- " \n",
- " \n",
"import math\n",
" \n",
"print\"Introduction to heat transfer by S.K.Som, Chapter 10, Example 3\"\n",
@@ -284,13 +250,7 @@
"#overall heat transfer coefficient(U)=Q/(A*F*LMTD)\n",
"print\"overall heat transfer coefficient(U)=Q/(A*F*LMTD)in W/(m**2*K)\"\n",
"U=Q/(A*F*LMTD)\n",
- "print\"U=\",U\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n"
+ "print\"U=\",U"
]
},
{
@@ -342,10 +302,6 @@
}
],
"source": [
- " \n",
- " \n",
- " \n",
- " \n",
"import math\n",
" \n",
"print\"Introduction to heat transfer by S.K.Som, Chapter 10, Example 5\"\n",
@@ -424,33 +380,7 @@
"print\"To provide this surface area ,The length(L) of the tube required is given by L=A/(pi*D) in m\"\n",
"L=A/(math.pi*D)\n",
"print\"Hence same result is obtained for both methods\"\n",
- "print\"L=\",L\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n"
+ "print\"L=\",L"
]
},
{
@@ -482,10 +412,6 @@
}
],
"source": [
- " \n",
- " \n",
- " \n",
- " \n",
"import math\n",
" \n",
"print\"Introduction to heat transfer by S.K.Som, Chapter 10, Example 6\"\n",
@@ -520,13 +446,7 @@
"#Hence The total heat transfer rate (Q)=eff*Cmin*(Thi-Tci)in kW.\n",
"print\"The total heat transfer rate (Q)=eff*Cmin*(Thi-Tci) in kW\" \n",
"Q=eff*Cmin*(Thi-Tci)\n",
- "print\"Q=\",Q\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n"
+ "print\"Q=\",Q"
]
},
{
@@ -560,10 +480,6 @@
}
],
"source": [
- " \n",
- " \n",
- " \n",
- " \n",
"import math\n",
" \n",
"print\"Introduction to heat transfer by S.K.Som, Chapter 10, Example 7\"\n",
@@ -603,22 +519,7 @@
"#The exit temprature(Tho) of air is given by Thi-(Q/(mdota*cpa))\n",
"print\"The exit temprature of air in °C \"\n",
"Tho=Thi-(Q/(mdota*1000*cpa))#NOTE:-The answer slightly varies from the answer in book(i.e Tho=26°C) because the value of Q taken in book is approximated to 1*10**6W.\n",
- "print\"Tho=\",Tho\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n"
+ "print\"Tho=\",Tho"
]
},
{
@@ -656,10 +557,6 @@
}
],
"source": [
- " \n",
- " \n",
- " \n",
- " \n",
"import math\n",
" \n",
"print\"Introduction to heat transfer by S.K.Som, Chapter 10, Example 8\"\n",
@@ -702,28 +599,7 @@
"Tho=Tci;\n",
"print\"Effectiveness of heat exchanger is \"\n",
"eff=(mdoth*ch*(Thi-Tho))/(mdoth*ch*(Thi-Tci))\n",
- "print\"eff=\",eff\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n"
+ "print\"eff=\",eff"
]
}
],
diff --git a/Introduction_to_Heat_Transfer_by_S._K._Som/Chapter11.ipynb b/Introduction_to_Heat_Transfer_by_S._K._Som/Chapter11.ipynb
index a46cced0..c44923f7 100644
--- a/Introduction_to_Heat_Transfer_by_S._K._Som/Chapter11.ipynb
+++ b/Introduction_to_Heat_Transfer_by_S._K._Som/Chapter11.ipynb
@@ -35,10 +35,6 @@
}
],
"source": [
- " \n",
- " \n",
- " \n",
- " \n",
"import math\n",
" \n",
"print\"Introduction to heat transfer by S.K.Som, Chapter 11, Example 3\"\n",
@@ -57,7 +53,7 @@
"A1=2;\n",
"A3=2.5;\n",
"F31=(A1/A3)*F13\n",
- "print\"F31=\",F31\n"
+ "print\"F31=\",F31"
]
},
{
@@ -87,10 +83,6 @@
}
],
"source": [
- " \n",
- " \n",
- " \n",
- " \n",
"import math\n",
" \n",
"print\"Introduction to heat transfer by S.K.Som, Chapter 11, Example 4\"\n",
@@ -116,11 +108,7 @@
"#This implies F14=((F1,2-4*(A1+A2)))-A2*F24)/A2\n",
"print\"The view factor F14=((F1,2-4*(A1+A2)))-A2*F24)/A2\"\n",
"F14=((F124*(A1+A2))-(A2*F24))/A2\n",
- "print\"F14=\",F14\n",
- "\n",
- "\n",
- "\n",
- " \n"
+ "print\"F14=\",F14"
]
},
{
@@ -149,10 +137,6 @@
}
],
"source": [
- " \n",
- " \n",
- " \n",
- " \n",
"import math\n",
" \n",
"print\"Introduction to heat transfer by S.K.Som, Chapter 11, Example 5\"\n",
@@ -171,7 +155,7 @@
"#Let A1/A2=A\n",
"A=1/4;\n",
"F31=(A)*F13\n",
- "print\"F31=\",F31\n"
+ "print\"F31=\",F31"
]
},
{
@@ -206,10 +190,6 @@
}
],
"source": [
- " \n",
- " \n",
- " \n",
- " \n",
"import math\n",
" \n",
"print\"Introduction to heat transfer by S.K.Som, Chapter 11, Example 6\"\n",
@@ -252,27 +232,7 @@
"#Therefore we can write Q1=A1*sigma*(T1**4-F12*T2**4-F1s*Ts**4)\n",
"print\"The net rate of energy loss from the surface at 127°C if the surrounding other than the two surfaces act as black body at 300K in W \"\n",
"Q1=A1*sigma*(T1**4-F12*T2**4-F1s*Ts**4)\n",
- "print\"Q1=\",Q1\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n"
+ "print\"Q1=\",Q1"
]
},
{
@@ -302,10 +262,6 @@
}
],
"source": [
- " \n",
- " \n",
- " \n",
- " \n",
"import math\n",
" \n",
"print\"Introduction to heat transfer by S.K.Som, Chapter 11, Example 7\"\n",
@@ -327,17 +283,7 @@
"#So,The net rate of heat transfer when the two surfaces are black is Q/A=sigma*(T1**4-T2**4)\n",
"print\"The net rate of heat transfer when the two surfaces are black is Q/A=sigma*(T1**4-T2**4) in W\"\n",
"H=sigma*(T1**4-T2**4)\n",
- "print\"H=\",H\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n"
+ "print\"H=\",H"
]
},
{
@@ -371,10 +317,6 @@
}
],
"source": [
- " \n",
- " \n",
- " \n",
- " \n",
"import math\n",
" \n",
"print\"Introduction to heat transfer by S.K.Som, Chapter 11, Example 8\"\n",
@@ -411,20 +353,7 @@
"print\"Q2=\",Q2\n",
"print\"Error(E) is given By ((Q2-Q1)/Q1)*100 in percentage\"\n",
"E=((Q2-Q1)/Q1)*100\n",
- "print\"E=\",E\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n"
+ "print\"E=\",E"
]
},
{
@@ -455,10 +384,6 @@
}
],
"source": [
- " \n",
- " \n",
- " \n",
- " \n",
"import math\n",
" \n",
"print\"Introduction to heat transfer by S.K.Som, Chapter 11, Example 10\"\n",
@@ -503,19 +428,7 @@
"#So Q/A=(sigma*(T1**4-T2**4))/(R)\n",
"#Let Q/A=H\n",
"H=(sigma*(T1**4-T2**4))/(R)\n",
- "print\"H=\",H\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n"
+ "print\"H=\",H"
]
}
],
diff --git a/Introduction_to_Heat_Transfer_by_S._K._Som/Chapter2.ipynb b/Introduction_to_Heat_Transfer_by_S._K._Som/Chapter2.ipynb
index ba64d857..cda3a92c 100644
--- a/Introduction_to_Heat_Transfer_by_S._K._Som/Chapter2.ipynb
+++ b/Introduction_to_Heat_Transfer_by_S._K._Som/Chapter2.ipynb
@@ -18,7 +18,7 @@
},
{
"cell_type": "code",
- "execution_count": 2,
+ "execution_count": 1,
"metadata": {
"collapsed": false
},
@@ -119,31 +119,7 @@
"print\"Check for Ti(in °C)\"\n",
"Ti=T4-(Q*Ri)\n",
"print\"The value is same as given in the problem\"\n",
- "print\"Ti=\",Ti\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- " \n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n"
+ "print\"Ti=\",Ti"
]
},
{
@@ -171,10 +147,6 @@
}
],
"source": [
- " \n",
- "\n",
- " \n",
- " \n",
"import math\n",
" \n",
"print\"Introduction to heat transfer by S.K.Som, Chapter 2, Example 2\"\n",
@@ -194,7 +166,7 @@
"#Q=(Ti-To)/((Lb/kb)+(L/ki)) so L=ki*(((Ti-To)/Q)-(Lb/kb))\n",
"print\"The thickness of insulating material L=ki*(((Ti-To)/Q)-(Lb/kb)) in m\"\n",
"L=ki*(((Ti-To)/Q)-(Lb/kb))\n",
- "print\"L=\",L\n"
+ "print\"L=\",L"
]
},
{
@@ -206,7 +178,7 @@
},
{
"cell_type": "code",
- "execution_count": 1,
+ "execution_count": 3,
"metadata": {
"collapsed": false
},
@@ -235,7 +207,6 @@
}
],
"source": [
- " \n",
"import math\n",
" \n",
"print\"Introduction to heat transfer by S.K.Som, Chapter 2, Example 4\"\n",
@@ -281,17 +252,7 @@
"print\"q2=\",q2\n",
"print\"Heat flux qo=q1+q2 in W/m**2 \"\n",
"qo=q1+q2\n",
- "print\"qo=\",qo\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n"
+ "print\"qo=\",qo"
]
},
{
@@ -329,7 +290,6 @@
}
],
"source": [
- " \n",
"import math\n",
" \n",
"print\"Introduction to heat transfer by S.K.Som, Chapter 2, Example 6\"\n",
@@ -372,9 +332,7 @@
"X=(2*10**3)-(4*10**5*x);\n",
"Q=-k*X/10**6\n",
"#A check for the above results can be made from an energy balance of the plate as |(q/A)|@x=0+|(q/A)|@x=0.02=qG*0.02\n",
- "print\"Q=\",Q\n",
- "\n",
- "\n"
+ "print\"Q=\",Q"
]
},
{
@@ -386,7 +344,7 @@
},
{
"cell_type": "code",
- "execution_count": 2,
+ "execution_count": 16,
"metadata": {
"collapsed": false
},
@@ -410,7 +368,6 @@
}
],
"source": [
- " \n",
"import math\n",
" \n",
"print\"Introduction to heat transfer by S.K.Som, Chapter 2, Example 9\"\n",
@@ -458,16 +415,7 @@
"Q=(T1-Tinf)/(((X)/(2*math.pi*L*k))+(1/(h*2*math.pi*r2*L)))\n",
"#It is important to note that Q increases by 5.2% when the insulation thickness increases from 0.002m to critical thickness. \n",
"#Addition of insulation beyond the critical thickness decreases the value of Q (The heat loss).\n",
- "print\"Q=\",Q\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n"
+ "print\"Q=\",Q"
]
},
{
@@ -479,7 +427,7 @@
},
{
"cell_type": "code",
- "execution_count": 3,
+ "execution_count": 8,
"metadata": {
"collapsed": false
},
@@ -495,7 +443,6 @@
}
],
"source": [
- " \n",
"import math\n",
" \n",
"print\"Introduction to heat transfer by S.K.Som, Chapter 2, Example 10\"\n",
@@ -525,22 +472,7 @@
"#Therefore the thickness of insulation is given by t=r3-Do\n",
"print\"the thickness of insulation in metre is\"\n",
"t=r3-Do\n",
- "print\"t=\",t\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n"
+ "print\"t=\",t"
]
},
{
@@ -552,7 +484,7 @@
},
{
"cell_type": "code",
- "execution_count": 4,
+ "execution_count": 9,
"metadata": {
"collapsed": false
},
@@ -570,7 +502,6 @@
}
],
"source": [
- " \n",
"import math\n",
" \n",
"print\"Introduction to heat transfer by S.K.Som, Chapter 2, Example 11\"\n",
@@ -591,21 +522,7 @@
"print\"The temprature of wire at the centre in K is \"\n",
"To=Tw+((qG*ro**2)/(4*k))\n",
"#Note:The answer in the book is incorrect(value of D has been put instead of ro)\n",
- "print\"To=\",To\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n"
+ "print\"To=\",To"
]
},
{
@@ -655,10 +572,6 @@
}
],
"source": [
- "\n",
- " \n",
- " \n",
- " \n",
"import math\n",
" \n",
"print\"Introduction to heat transfer by S.K.Som, Chapter 2, Example 12\"\n",
@@ -730,12 +643,7 @@
"#the amount of ice in kG which melts during a 24 hour period is (mice)\n",
"print\"Therefore,the amount of ice(mice)in kG which melts during a 24 hour period is\"\n",
"mice=Qt/deltahf\n",
- "print\"mice=\",mice\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n"
+ "print\"mice=\",mice"
]
},
{
@@ -747,7 +655,7 @@
},
{
"cell_type": "code",
- "execution_count": 5,
+ "execution_count": 13,
"metadata": {
"collapsed": false
},
@@ -782,7 +690,6 @@
}
],
"source": [
- " \n",
"import math\n",
" \n",
"print\"Introduction to heat transfer by S.K.Som, Chapter 2, Example 13\"\n",
@@ -846,15 +753,7 @@
"Qinf=(h*P*k*A)**0.5*thetab\n",
"print\"We see that since k is large there is significant difference between the finite length and the infinte length cases\"\n",
"print\"However when the length of the rod approaches 1m,the result become almost same.\" \n",
- "print\"Qinf=\",Qinf\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n"
+ "print\"Qinf=\",Qinf"
]
},
{
@@ -866,7 +765,7 @@
},
{
"cell_type": "code",
- "execution_count": 6,
+ "execution_count": 14,
"metadata": {
"collapsed": false
},
@@ -882,7 +781,6 @@
}
],
"source": [
- " \n",
"import math\n",
" \n",
"print\"Introduction to heat transfer by S.K.Som, Chapter 2, Example 14\"\n",
@@ -899,41 +797,7 @@
"#The thermal conductivity of Rod B iskB\n",
"print\"The thermal conductivity of Rod B kB in W/(m*K) is \"\n",
"kB=kA*(xB/xA)**2\n",
- "print\"kB=\",kB\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n"
+ "print\"kB=\",kB"
]
},
{
@@ -945,7 +809,7 @@
},
{
"cell_type": "code",
- "execution_count": 7,
+ "execution_count": 15,
"metadata": {
"collapsed": false
},
@@ -971,7 +835,6 @@
}
],
"source": [
- " \n",
"import math\n",
" \n",
"print\"Introduction to heat transfer by S.K.Som, Chapter 2, Example 15\"\n",
@@ -1013,17 +876,7 @@
"#Heat loss from the plate is Qb\n",
"print\"Heat loss from the plate at 400K in W is\"\n",
"Qb=(N*(h*P*kal*A)**0.5*thetab*((math.cosh(m*L)-(thetaL/thetab))/(math.sinh(m*L))))+(((l*b)-(N*A))*h*thetab)+(l*b*h*thetab)\n",
- "print\"Qb=\",Qb\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n"
+ "print\"Qb=\",Qb"
]
}
],
diff --git a/Introduction_to_Heat_Transfer_by_S._K._Som/Chapter3.ipynb b/Introduction_to_Heat_Transfer_by_S._K._Som/Chapter3.ipynb
index 55a11dc9..6b51de29 100644
--- a/Introduction_to_Heat_Transfer_by_S._K._Som/Chapter3.ipynb
+++ b/Introduction_to_Heat_Transfer_by_S._K._Som/Chapter3.ipynb
@@ -34,9 +34,6 @@
}
],
"source": [
- "\n",
- "\n",
- " \n",
"import math\n",
" \n",
"print\"Introduction to heat transfer by S.K.Som, Chapter 3, Example 1\"\n",
@@ -59,7 +56,7 @@
"#Temperature in degree celcius\n",
"print\"Temperature at the centre in Degree C is\"\n",
"T = theta*100+100\n",
- "print\"T=\",T\n"
+ "print\"T=\",T"
]
},
{
@@ -90,9 +87,6 @@
}
],
"source": [
- " \n",
- " \n",
- " \n",
"import math\n",
" \n",
"print\"Introduction to heat transfer by S.K.Som, Chapter 3, Example 2\"\n",
@@ -117,7 +111,7 @@
"print\"theta=\",theta\n",
"print\"Temperature in K at centre point\"\n",
"T = theta*100+300\n",
- "print\"T=\",T\n"
+ "print\"T=\",T"
]
},
{
@@ -166,7 +160,6 @@
}
],
"source": [
- " \n",
"import math\n",
"import numpy\n",
" \n",
@@ -213,7 +206,7 @@
"print T7\n",
"print\"T8 in degree K\"\n",
"T8 = T[7]\n",
- "print T8\n"
+ "print T8"
]
},
{
@@ -255,7 +248,6 @@
}
],
"source": [
- " \n",
"import math\n",
"import numpy\n",
" \n",
@@ -300,7 +292,7 @@
"print T5\n",
"print\"T6 in degree C\"\n",
"T6 = T[4]\n",
- "print T6\n"
+ "print T6"
]
},
{
@@ -354,7 +346,6 @@
}
],
"source": [
- " \n",
"import math\n",
" \n",
"print\"Introduction to heat transfer by S.K.Som, Chapter 3, Example 6\"\n",
@@ -412,7 +403,7 @@
"print T8\n",
"print\"T9 in degree C\"\n",
"T9 = T[8]\n",
- "print T9\n"
+ "print T9"
]
}
],
diff --git a/Introduction_to_Heat_Transfer_by_S._K._Som/Chapter4.ipynb b/Introduction_to_Heat_Transfer_by_S._K._Som/Chapter4.ipynb
index c6207f4b..6c85f1a7 100644
--- a/Introduction_to_Heat_Transfer_by_S._K._Som/Chapter4.ipynb
+++ b/Introduction_to_Heat_Transfer_by_S._K._Som/Chapter4.ipynb
@@ -35,10 +35,6 @@
}
],
"source": [
- " \n",
- "\n",
- " \n",
- " \n",
"import math\n",
" \n",
"print\"Introduction to heat transfer by S.K.Som, Chapter 4, Example 1\"\n",
@@ -59,8 +55,7 @@
"if Bi<0.1:\n",
" print\"Problem is suitable for lumped parameter analysis\"\n",
"else:\n",
- " print\"Problem is not suitable for lumped parameter analysis\"\n",
- "\n"
+ " print\"Problem is not suitable for lumped parameter analysis\""
]
},
{
@@ -90,10 +85,6 @@
}
],
"source": [
- " \n",
- " \n",
- " \n",
- " \n",
"import math\n",
" \n",
"print\"Introduction to heat transfer by S.K.Som, Chapter 4, Example 2\"\n",
@@ -118,7 +109,7 @@
"#Required time in sec\n",
"t = (-8)*math.log(0.01);\n",
"print\"Time required in seconds\"\n",
- "print\"t=\",t\n"
+ "print\"t=\",t"
]
},
{
@@ -130,7 +121,7 @@
},
{
"cell_type": "code",
- "execution_count": 2,
+ "execution_count": 4,
"metadata": {
"collapsed": false
},
@@ -146,10 +137,6 @@
}
],
"source": [
- " \n",
- " \n",
- " \n",
- " \n",
"import math\n",
" \n",
"print\"Introduction to heat transfer by S.K.Som, Chapter 4, Example 3\"\n",
@@ -163,7 +150,7 @@
"#Maximum dimension in metre\n",
"a = ((6*k)*Bi)/h;\n",
"print\"Maximum dimension in metre for lumped parameter analysis\"\n",
- "print\"a=\",a\n"
+ "print\"a=\",a"
]
},
{
@@ -192,7 +179,6 @@
}
],
"source": [
- " \n",
"from scipy.integrate import quad\n",
"import math\n",
"print\"Introduction to heat transfer by S.K.Som, Chapter 4, Example 4\"\n",
@@ -228,7 +214,7 @@
"E = (((h*math.pi)*d)*H)*quad(lambda t:(80.0-25.0)*math.e*(-t/472.5),0,60.0*t)[0];\n",
"print\"Energy required for cooling in KJ\"\n",
"E = E/1000.0\n",
- "print \"E=\",E\n"
+ "print \"E=\",E"
]
},
{
@@ -258,10 +244,6 @@
}
],
"source": [
- " \n",
- " \n",
- " \n",
- " \n",
"import math\n",
" \n",
"print\"Introduction to heat transfer by S.K.Som, Chapter 4, Example 5\"\n",
@@ -297,7 +279,7 @@
"Q = ((((0.69*k)*2)*L)*(Tinfinity-Ti))/alpha;\n",
"print\"Heat transfer rate in MJ\"\n",
"Q = Q/(10**6)\n",
- "print\"Q=\",Q\n"
+ "print\"Q=\",Q"
]
},
{
@@ -327,10 +309,6 @@
}
],
"source": [
- " \n",
- " \n",
- " \n",
- " \n",
"import math\n",
" \n",
"print\"Introduction to heat transfer by S.K.Som, Chapter 4, Example 6\"\n",
@@ -368,7 +346,7 @@
"Q = (((0.4*k)*L)*(Ti-Tinfinity))/alpha;\n",
"print\"Heat transfer rate in MJ\"\n",
"Q = Q/(10**6)\n",
- "print\"Q=\",Q\n"
+ "print\"Q=\",Q"
]
},
{
@@ -398,10 +376,6 @@
}
],
"source": [
- " \n",
- " \n",
- " \n",
- " \n",
"import math\n",
" \n",
"print\"Introduction to heat transfer by S.K.Som, Chapter 4, Example 7\"\n",
@@ -439,7 +413,7 @@
"Q = (((((0.4*k)*math.pi)*ro)*ro)*(Ti-Tinfinity))/alpha;\n",
"print\"Heat transfer rate per unit length in MJ/m\"\n",
"Q = Q/(10**6)\n",
- "print\"Q=\",Q\n"
+ "print\"Q=\",Q"
]
},
{
@@ -467,10 +441,6 @@
}
],
"source": [
- " \n",
- " \n",
- " \n",
- " \n",
"import math\n",
"\n",
"print\"Introduction to heat transfer by S.K.Som, Chapter 4, Example 8\"\n",
@@ -499,7 +469,7 @@
"t = ((Fo*ro)*ro)/alpha;\n",
"print\"Time required in minutes\"\n",
"t = t/60\n",
- "print\"t=\",t\n"
+ "print\"t=\",t"
]
},
{
@@ -527,10 +497,6 @@
}
],
"source": [
- " \n",
- " \n",
- " \n",
- " \n",
"import math\n",
" \n",
"print\"Introduction to heat transfer by S.K.Som, Chapter 4, Example 9\"\n",
@@ -573,7 +539,7 @@
"#Temperature in °C\n",
"T = Tinfinity+z*(Ti-Tinfinity);\n",
"print\"Tempearture of bar in °C\"\n",
- "print\"T=\",T\n"
+ "print\"T=\",T"
]
},
{
@@ -610,10 +576,6 @@
}
],
"source": [
- " \n",
- " \n",
- " \n",
- " \n",
"import math\n",
" \n",
"print\"Introduction to heat transfer by S.K.Som, Chapter 4, Example 10\"\n",
@@ -683,17 +645,7 @@
"#Therefore ((To-Tinf)/(Ti-Tinf))plate1*((To-Tinf)/(Ti-Tinf))plate2=A*B\n",
"T=A*B\n",
"print\"The calculated value is very close to the required value of 0.6.Hence the time required for the centre of the beam to reach 310°C is nearly 1200s or 20 minutes.\" \n",
- "print\"T=\",T\n",
- " \n",
- " \n",
- " \n",
- " \n",
- " \n",
- " \n",
- " \n",
- " \n",
- " \n",
- " \n"
+ "print\"T=\",T"
]
},
{
@@ -721,10 +673,6 @@
}
],
"source": [
- " \n",
- " \n",
- " \n",
- " \n",
"import math\n",
" \n",
"print\"Introduction to heat transfer by S.K.Som, Chapter 4, Example 11\"\n",
@@ -746,22 +694,7 @@
"#Therefore 10/t**0.5=0.38...this implies t=(10/0.38)**2\n",
"print\"The time required for the temprature to reach 255°C at a depth of 80mm, in minutes is\"\n",
"t=(10/0.38)**2/60\n",
- "print\"T=\",T\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n"
+ "print\"T=\",T"
]
},
{
@@ -791,7 +724,6 @@
}
],
"source": [
- " \n",
"import math\n",
"import scipy \n",
"print\"Introduction to heat transfer by S.K.Som, Chapter 4, Example 12\"\n",
@@ -815,18 +747,7 @@
"print\"The temprature at a depth(x) of 100mm after a time(t) of 100 seconds,in °C is\"\n",
"T=Ti+((2*qo*(alpha*t/math.pi)**0.5)/(k))*math.e**((-x**2.0)/(4*alpha*t))-((qo*x)/(k))*scipy.special.erf(x/(2*(alpha*t)**0.5))\n",
"print\"T=\",T\n",
- "#NOTE:The answer in the book is incorrect(Calculation mistake)\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n"
+ "#NOTE:The answer in the book is incorrect(Calculation mistake)"
]
},
{
@@ -838,7 +759,7 @@
},
{
"cell_type": "code",
- "execution_count": 17,
+ "execution_count": 2,
"metadata": {
"collapsed": false
},
@@ -849,7 +770,7 @@
"text": [
"Introduction to heat transfer by S.K.Som, Chapter 4, Example 14\n",
"Temperature distribution after 25 mins in °C\n",
- "[[ 2.29192547e+02 2.91925466e+00 1.11801242e+00 4.34782609e-01\n",
+ "T= [[ 2.29192547e+02 2.91925466e+00 1.11801242e+00 4.34782609e-01\n",
" 1.86335404e-01 6.21118012e-02]\n",
" [ 8.75776398e+01 8.75776398e+00 3.35403727e+00 1.30434783e+00\n",
" 5.59006211e-01 1.86335404e-01]\n",
@@ -865,7 +786,6 @@
}
],
"source": [
- " \n",
"import math\n",
"import numpy\n",
" \n",
@@ -891,10 +811,8 @@
"#From Eq. 4.126\n",
"#Temperature distribution after one time step\n",
"T = numpy.linalg.inv(A)*B;\n",
- "\n",
- " \n",
"print\"Temperature distribution after 25 mins in °C\"\n",
- "print T\n"
+ "print\"T=\",T"
]
}
],
diff --git a/Introduction_to_Heat_Transfer_by_S._K._Som/Chapter5.ipynb b/Introduction_to_Heat_Transfer_by_S._K._Som/Chapter5.ipynb
index d3fc7380..5b3d46e7 100644
--- a/Introduction_to_Heat_Transfer_by_S._K._Som/Chapter5.ipynb
+++ b/Introduction_to_Heat_Transfer_by_S._K._Som/Chapter5.ipynb
@@ -38,7 +38,6 @@
}
],
"source": [
- " \n",
"import math \n",
"from scipy.integrate import quad\n",
" \n",
@@ -74,23 +73,7 @@
"#Q is the rate of heat transfer\n",
"print\"The rate of heat transfer in W/m of width is\"\n",
"Q=hbarL*L*(T2-T1)\n",
- "print\"Q=\",Q\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n"
+ "print\"Q=\",Q"
]
},
{
@@ -102,7 +85,7 @@
},
{
"cell_type": "code",
- "execution_count": 2,
+ "execution_count": 27,
"metadata": {
"collapsed": false
},
@@ -119,7 +102,6 @@
}
],
"source": [
- " \n",
"import math\n",
" \n",
"print\"Introduction to heat transfer by S.K.Som, Chapter 5, Example 4\"\n",
@@ -144,18 +126,7 @@
"#from an enrgy balance we can write as E=27.063*U**0.85*L*B*(Ts-Tinf)\n",
"print\"The minimum flow velocity in m/s is\"\n",
"U=(E/(27.063*L*B*(Ts-Tinf)))**(1/0.85)\n",
- "print\"U=\",U\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n"
+ "print\"U=\",U"
]
},
{
@@ -167,7 +138,7 @@
},
{
"cell_type": "code",
- "execution_count": 3,
+ "execution_count": 1,
"metadata": {
"collapsed": false
},
@@ -190,10 +161,8 @@
}
],
"source": [
- " \n",
"import math\n",
" \n",
- " \n",
"print\"Introduction to heat transfer by S.K.Som, Chapter 5, Example 6\"\n",
"#Air at 1atm pressure and temprature(Tin)=30°C enters a tube of 25mm diameter(D) with a velocity(U) of 10m/s\n",
"D=0.025;#in metre\n",
@@ -231,25 +200,7 @@
"k=0.0285;\n",
"print\"Overall Nusselt number is \"\n",
"NuL=hx*D/k\n",
- "print\"NuL=\",NuL\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n"
+ "print\"NuL=\",NuL"
]
},
{
@@ -261,7 +212,7 @@
},
{
"cell_type": "code",
- "execution_count": 4,
+ "execution_count": 3,
"metadata": {
"collapsed": false
},
@@ -285,7 +236,6 @@
}
],
"source": [
- " \n",
"import math\n",
" \n",
"print\"Introduction to heat transfer by S.K.Som, Chapter 5, Example 7\"\n",
@@ -324,21 +274,7 @@
"#Q is the heat loss from the plate\n",
"print\"The heat loss from the plate in W is\"\n",
"Q=hbar*A*(Ts-Tinf)\n",
- "print\"Q=\",Q\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n"
+ "print\"Q=\",Q"
]
},
{
@@ -370,11 +306,7 @@
}
],
"source": [
- " \n",
- " \n",
- " \n",
- " \n",
- " import math\n",
+ "import math\n",
" \n",
"print\"Introduction to heat transfer by S.K.Som, Chapter 5, Example 8\"\n",
"#Eletric current passes through a L=0.5m long horizontal wire of D=0.1mm diameter.\n",
@@ -401,28 +333,7 @@
"#I is the current flow.\n",
"print\"The current in Ampere is\"\n",
"I=(Q/(R*L))**0.5\n",
- "print\"I=\",I\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n"
+ "print\"I=\",I"
]
}
],
diff --git a/Introduction_to_Heat_Transfer_by_S._K._Som/Chapter6.ipynb b/Introduction_to_Heat_Transfer_by_S._K._Som/Chapter6.ipynb
index a4433df7..5a31da63 100644
--- a/Introduction_to_Heat_Transfer_by_S._K._Som/Chapter6.ipynb
+++ b/Introduction_to_Heat_Transfer_by_S._K._Som/Chapter6.ipynb
@@ -18,7 +18,7 @@
},
{
"cell_type": "code",
- "execution_count": 3,
+ "execution_count": 10,
"metadata": {
"collapsed": false
},
@@ -45,7 +45,6 @@
}
],
"source": [
- " \n",
"import math\n",
" \n",
"print\"Introduction to heat transfer by S.K.Som, Chapter 6, Example 1\"\n",
@@ -55,7 +54,7 @@
"mu=0.1;\n",
"b=0.005; #in metre\n",
" #Umax is maximum velocity\n",
- " Umax=(3.0/2)*Uav\n",
+ " Umax=(3/2)*Uav\n",
"print\"Umax in m/s is\"\n",
"Umax=(3/2)*Uav\n",
"print\"Umax=\",Umax\n",
@@ -82,31 +81,7 @@
" #Since pressure drop is considered at a distance of 2m so L=2m\n",
"L=2;\n",
"deltaP=(-X)*L\n",
- "print\"deltaP=\",deltaP\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n"
+ "print\"deltaP=\",deltaP"
]
},
{
@@ -118,7 +93,7 @@
},
{
"cell_type": "code",
- "execution_count": 2,
+ "execution_count": 14,
"metadata": {
"collapsed": false
},
@@ -138,7 +113,6 @@
}
],
"source": [
- " \n",
"import math\n",
" \n",
"print\"Introduction to heat transfer by S.K.Som, Chapter 6, Example 3\"\n",
@@ -159,28 +133,7 @@
" #The viscosity of oil is mu=(pi*D**4*X)/(128*Q*dz)\n",
"print\"The viscosity of oil(mu)in kg/(m*s)\"\n",
"mu=(math.pi*D**4*X)/(128*Q)\n",
- "print\"mu=\",mu\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n"
+ "print\"mu=\",mu"
]
},
{
@@ -212,10 +165,6 @@
}
],
"source": [
- " \n",
- " \n",
- " \n",
- " \n",
"import math\n",
" \n",
"print\"Introduction to heat transfer by S.K.Som, Chapter 6, Example 7\"\n",
@@ -239,26 +188,7 @@
" #Fd is drag force\n",
"print\"Drag force on one side of plate in N is\"\n",
"Fd=cfL*(rhoair*Uinf**2/2)*B*L\n",
- "print\"Fd=\",Fd\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n"
+ "print\"Fd=\",Fd"
]
},
{
@@ -294,10 +224,6 @@
}
],
"source": [
- " \n",
- " \n",
- " \n",
- " \n",
"import math\n",
" \n",
"print\"Introduction to heat transfer by S.K.Som, Chapter 6, Example 10\"\n",
@@ -326,21 +252,7 @@
" #The turbulent boundary layer thickness at the trailing edge is given by delta=L*(0.379/ReL**(1/5))\n",
"print\"The turbulent boundary layer thickness at the trailing edge in metre is \"\n",
"delta=L*(0.379/ReL**(1/5))\n",
- "print\"delta=\",delta\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n"
+ "print\"delta=\",delta"
]
}
],
diff --git a/Introduction_to_Heat_Transfer_by_S._K._Som/Chapter7.ipynb b/Introduction_to_Heat_Transfer_by_S._K._Som/Chapter7.ipynb
index 85b7eec5..bffd25f6 100644
--- a/Introduction_to_Heat_Transfer_by_S._K._Som/Chapter7.ipynb
+++ b/Introduction_to_Heat_Transfer_by_S._K._Som/Chapter7.ipynb
@@ -42,7 +42,6 @@
}
],
"source": [
- " \n",
"import math \n",
" \n",
"print\"Introduction to heat transfer by S.K.Som, Chapter 7, Example 1\"\n",
@@ -115,11 +114,7 @@
}
],
"source": [
- " \n",
- " \n",
- " \n",
- " \n",
- " import math\n",
+ "import math\n",
" \n",
"print\"Introduction to heat transfer by S.K.Som, Chapter 7, Example 2\"\n",
"#Atmospheric air at temprature,Tinf=300K and with a free stream Velocity Uinf=30m/s flows over a flat plate parallel to a side of length(L)=2m.\n",
@@ -161,24 +156,7 @@
"A=L*B;\n",
"print\"The rate of heat transfer per unit width in W is\"\n",
"Q=hbarL*A*(Tw-Tinf)\n",
- "print\"Q=\",Q\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n"
+ "print\"Q=\",Q"
]
},
{
@@ -218,11 +196,7 @@
}
],
"source": [
- " \n",
- " \n",
- " \n",
- " \n",
- " import math\n",
+ "import math\n",
" \n",
"print\"Introduction to heat transfer by S.K.Som, Chapter 7, Example 3\"\n",
"#Air at a pressure of 101kPa and temprature,Tinf=20°C flows with a velocity(Uinf) of 5m/s over a flat plate whose temprature is kept constant at Tw=140°C.\n",
@@ -272,33 +246,7 @@
"#Q is the rate of heat transfer\n",
"print\"The rate of heat transfer per unit width in W is\"\n",
"Q=h*A*(Tw-Tinf)\n",
- "print\"Q=\",Q\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n"
+ "print\"Q=\",Q"
]
},
{
@@ -340,11 +288,7 @@
}
],
"source": [
- " \n",
- " \n",
- " \n",
- " \n",
- " import math\n",
+ "import math\n",
" \n",
"print\"Introduction to heat transfer by S.K.Som, Chapter 7, Example 4\"\n",
"#Castor oil at temprature,Tinf=36°C flows over a heated plate of length,L=6m and breadth,B=1m at velocity,Uinf=0.06m/s\n",
@@ -392,29 +336,7 @@
"A=L*B;\n",
"print\"(c)The rate of heat transfer in W is\"\n",
"Q=hbarL*A*(Tw-Tinf)\n",
- "print\"Q=\",Q\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n"
+ "print\"Q=\",Q"
]
},
{
@@ -447,11 +369,7 @@
}
],
"source": [
- " \n",
- " \n",
- " \n",
- " \n",
- " import math\n",
+ "import math\n",
" \n",
"print\"Introduction to heat transfer by S.K.Som, Chapter 7, Example 5\"\n",
"#A flat plate of width B=1m is maintained at a uniform surface temprtaure(Tw)=225°C\n",
@@ -487,24 +405,7 @@
"#If qm be the power generation in W/m**2 within the module ,we can write from energy balance qm*(t/0.1000)*(l/0.1000)*(B)=hbarL*(t/0.1000)*(B)*(Tw-Tinf)\n",
"print\"The required power generation in W/m**3 is\"\n",
"qm=(hL*(l/0.1000)*(B)*(Tw-Tinf))/((t/0.1000)*(l/0.1000)*(B))\n",
- "print\"qm=\",qm\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n"
+ "print\"qm=\",qm"
]
},
{
@@ -540,11 +441,7 @@
}
],
"source": [
- " \n",
- " \n",
- " \n",
- " \n",
- " import math\n",
+ "import math\n",
"\n",
"print\"Introduction to heat transfer by S.K.Som, Chapter 7, Example 6\"\n",
"#An aircraft is moving at a velocity of Uinf=150m/s in air at an altitude where the pressure is 0.7bar and the temprature is Tinf=-5°C.\n",
@@ -579,21 +476,7 @@
"#Therefore we can write Surface temprature of wing, Tw=Tinf+(Qr/(2*hbarL))\n",
"print\"Surface temprature of wing in kelvin is\"\n",
"Tw=(273+Tinf)+(Qr/(2*hbarL))\n",
- "print\"Tw=\",Tw\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n"
+ "print\"Tw=\",Tw"
]
},
{
@@ -633,11 +516,7 @@
}
],
"source": [
- " \n",
- " \n",
- " \n",
- " \n",
- " import math\n",
+ "import math\n",
" \n",
"print\"Introduction to heat transfer by S.K.Som, Chapter 7, Example 7\"\n",
"#A fine wire having a diameter(D)=0.04mm is placed in an air stream at temprature,Tinf=25°C having a flow velocity of Uinf=60m/s perpendicular to wire.\n",
@@ -679,32 +558,7 @@
"#Heat transfer per unit length(qL) is given by pi*D*hbar*(Tw-Tinf)\n",
"print\"Heat transfer per unit length in W/m is\"\n",
"qL=math.pi*(D*10**-3)*hbar*(Tw-Tinf)\n",
- "print\"qL=\",qL\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n"
+ "print\"qL=\",qL"
]
},
{
@@ -742,11 +596,7 @@
}
],
"source": [
- " \n",
- " \n",
- " \n",
- " \n",
- " import math\n",
+ "import math\n",
"\n",
"print\"Introduction to heat transfer by S.K.Som, Chapter 7, Example 8\"\n",
"#Mercury and a light oil flowing at Uinf=4mm/s in a smooth tube having diameter(D)=25mm at a bulk temprature of 80°C.\n",
@@ -783,24 +633,7 @@
"#Ltoil is the thermal entry length for oil\n",
"print\"The thermal entry length for oil in m is\"\n",
"Ltoil=0.05*Reoil*Proil*D\n",
- "print\"Ltoil=\",Ltoil\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n"
+ "print\"Ltoil=\",Ltoil"
]
},
{
@@ -840,11 +673,7 @@
}
],
"source": [
- " \n",
- " \n",
- " \n",
- " \n",
- " import math\n",
+ "import math\n",
" \n",
"print\"Introduction to heat transfer by S.K.Som, Chapter 7, Example 9\"\n",
"#Air at one atmospheric pressure and temprature(Tbi=75°C) enters a tube of internal diameter(D)=4.0mm with average velocity(U)=2m/s\n",
@@ -895,21 +724,7 @@
"#Let Twe be the surface temprature at the exit plane.Then we can write hL*(Twe-Tbo)=qw\n",
"print\"The tube surface temprature at the exit plane in °C is \"\n",
"Twe=Tbo+(qw/hL)\n",
- "print\"Twe=\",Twe\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n"
+ "print\"Twe=\",Twe"
]
},
{
@@ -952,11 +767,7 @@
}
],
"source": [
- " \n",
- " \n",
- " \n",
- " \n",
- " import math\n",
+ "import math\n",
" \n",
"print\"Introduction to heat transfer by S.K.Som, Chapter 7, Example 10\"\n",
"#Air at one atmospheric pressure and temprature(Tbi=75°C) enters a tube of internal diameter(D)=4.0mm with average velocity(U)=2m/s\n",
@@ -1011,18 +822,7 @@
"#Let Twe be the surface temprature at the exit plane.Then we can write hL*(Twe-Tbo)=qw\n",
"print\"The tube surface temprature at the exit plane in °C is \"\n",
"Twe=Tbo+(qw/hL)\n",
- "print\"Twe=\",Twe\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n"
+ "print\"Twe=\",Twe"
]
},
{
@@ -1067,11 +867,7 @@
}
],
"source": [
- " \n",
- " \n",
- " \n",
- " \n",
- " import math\n",
+ "import math\n",
" \n",
"print\"Introduction to heat transfer by S.K.Som, Chapter 7, Example 11\"\n",
"#Liquid sulphur di oxide in a saturated state flows inside a L=5m long tube and D=25mm internal diameter with a mass flow rate(mdot) of 0.15 kg/s.\n",
diff --git a/Introduction_to_Heat_Transfer_by_S._K._Som/Chapter8.ipynb b/Introduction_to_Heat_Transfer_by_S._K._Som/Chapter8.ipynb
index a6c237e4..882c8bf9 100644
--- a/Introduction_to_Heat_Transfer_by_S._K._Som/Chapter8.ipynb
+++ b/Introduction_to_Heat_Transfer_by_S._K._Som/Chapter8.ipynb
@@ -79,20 +79,7 @@
"#The rate of heat transfer is given by q=hbarL*A*(Tw-Tinf)\n",
"print\"The rate of heat transfer in W is\"\n",
"q=hbarL*A*(Tw-Tinf)\n",
- "print\"q=\",q\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n"
+ "print\"q=\",q"
]
},
{
@@ -128,7 +115,6 @@
}
],
"source": [
- " \n",
"import math \n",
" \n",
"print\"Introduction to heat transfer by S.K.Som, Chapter 8, Example 2\"\n",
@@ -165,24 +151,7 @@
"#spac is the minimum spacing \n",
"print\"The minimum spacing in metre is\"\n",
"spac=2*delta\n",
- "print\"spac=\",spac\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n"
+ "print\"spac=\",spac"
]
},
{
@@ -219,7 +188,6 @@
}
],
"source": [
- "\n",
"from scipy.integrate import quad\n",
"print \"Introduction to heat transfer by S.K.Som, Chapter 8, Example 3\"\n",
"#Considering question 5.7\n",
@@ -275,14 +243,7 @@
"print \"Mass flow rate at x=0.8m,in kG is\"\n",
"I=quad(lambda y:465.9*(y-116*y*2+3341*y*3),0,delta)\n",
"mdot=rho*B*I[0]\n",
- "print\"mdot=\",mdot\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n"
+ "print\"mdot=\",mdot"
]
},
{
@@ -314,7 +275,6 @@
}
],
"source": [
- " \n",
"import math \n",
" \n",
"print\"Introduction to heat transfer by S.K.Som, Chapter 8, Example 4\"\n",
@@ -352,33 +312,7 @@
"hL=(2*k)/delta;\n",
"hbarL=(4.0/3)*(hL)#NOTE:The answer in the book is incorrect(calculation mistake)\n",
"print\"hL=\",hL\n",
- "print\"hbarL=\",hbarL\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n"
+ "print\"hbarL=\",hbarL"
]
},
{
@@ -432,7 +366,6 @@
}
],
"source": [
- " \n",
"import math \n",
" \n",
"print\"Introduction to heat transfer by S.K.Som, Chapter 8, Example 5\"\n",
@@ -509,51 +442,7 @@
"print hbarL\n",
"print\"The rate of heat transfer in W is \"\n",
"Q=hbarL*A*(Tw-Tinf)\n",
- "print Q\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n"
+ "print Q"
]
},
{
@@ -599,10 +488,6 @@
}
],
"source": [
- " \n",
- " \n",
- " \n",
- " \n",
"import math \n",
" \n",
"print\"Introduction to heat transfer by S.K.Som, Chapter 8, Example 6\"\n",
@@ -663,16 +548,7 @@
"#The current flowing in the wire I=(q/(R*L)**(1/2.0)\n",
"print\"The current flowing in the wire in Ampere is\"\n",
"I=(q/(R*L))**(1/2.0)\n",
- "print\"I=\",I\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n"
+ "print\"I=\",I"
]
},
{
@@ -709,7 +585,6 @@
}
],
"source": [
- " \n",
"import math \n",
" \n",
"print\"Introduction to heat transfer by S.K.Som, Chapter 8, Example 7\"\n",
@@ -750,26 +625,7 @@
"#The heat loss per meter length is given by q=hbar*A*(Tw-Tinf)\n",
"print\"The heat loss per meter length in W is\"\n",
"q=hbar*A*(Tw-Tinf)\n",
- "print\"q=\",q\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n"
+ "print\"q=\",q"
]
},
{
@@ -808,7 +664,6 @@
}
],
"source": [
- " \n",
"import math \n",
" \n",
"print\"Introduction to heat transfer by S.K.Som, Chapter 8, Example 8\"\n",
@@ -854,18 +709,7 @@
"print\"Hence,steady state Surface temprature in °C is\"\n",
"Tw=Tinf+(P/(hbarD*math.pi*D*L))\n",
"print\"Hence we see that our guess is in excellent agreement with the calculated value\"\n",
- "print\"Tw=\",Tw\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n"
+ "print\"Tw=\",Tw"
]
}
],
diff --git a/Introduction_to_Heat_Transfer_by_S._K._Som/Chapter9.ipynb b/Introduction_to_Heat_Transfer_by_S._K._Som/Chapter9.ipynb
index 7d76afa3..2944362a 100644
--- a/Introduction_to_Heat_Transfer_by_S._K._Som/Chapter9.ipynb
+++ b/Introduction_to_Heat_Transfer_by_S._K._Som/Chapter9.ipynb
@@ -46,10 +46,6 @@
}
],
"source": [
- " \n",
- " \n",
- " \n",
- " \n",
"import math\n",
" \n",
"print\"Introduction to heat transfer by S.K.Som, Chapter 9, Example 1\"\n",
@@ -90,18 +86,7 @@
"print\"Reynolds no. is\"\n",
"ReL=(4*mdotc)/(mu)\n",
"print\"Therefore the flow is laminar and hence the use of the equation is justified\"\n",
- "print\"ReL=\",ReL\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n"
+ "print\"ReL=\",ReL"
]
},
{
@@ -137,10 +122,6 @@
}
],
"source": [
- " \n",
- " \n",
- " \n",
- " \n",
"import math\n",
" \n",
"print\"Introduction to heat transfer by S.K.Som, Chapter 9, Example 2\"\n",
@@ -178,19 +159,7 @@
"#Re is reynolds number\n",
"print\"Reynolds number is\"\n",
"Re=(4*mdotc)/(mu*P)\n",
- "print\"Re=\",Re\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n"
+ "print\"Re=\",Re"
]
},
{
@@ -228,10 +197,6 @@
}
],
"source": [
- " \n",
- " \n",
- " \n",
- " \n",
"import math\n",
" \n",
"print\"Introduction to heat transfer by S.K.Som, Chapter 9, Example 3\"\n",
@@ -278,29 +243,7 @@
"#v is the average flow velocity\n",
"print\"Hence the average flow velocity at the trailing edge in m/s is\"\n",
"v=(mdotc)/(rho*delta*B)\n",
- "print\"v=\",v\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n"
+ "print\"v=\",v"
]
},
{
@@ -334,10 +277,6 @@
}
],
"source": [
- " \n",
- " \n",
- " \n",
- " \n",
"import math\n",
" \n",
"print\"Introduction to heat transfer by S.K.Som, Chapter 9, Example 4\"\n",
@@ -373,29 +312,7 @@
"#The rate of condensation is given by mdotc=(hbar*(pi*D*L)*(Tg-Tw))/hfg\n",
"print\"The total rate of condensation in kg/hr\"\n",
"mdotc=((hbar*(math.pi*D*L)*(Tg-Tw))/hfg)*3600\n",
- "print\"mdotc=\",mdotc\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n"
+ "print\"mdotc=\",mdotc"
]
},
{
@@ -423,10 +340,6 @@
}
],
"source": [
- " \n",
- " \n",
- " \n",
- " \n",
"import math\n",
" \n",
"print\"Introduction to heat transfer by S.K.Som, Chapter 9, Example 5\"\n",
@@ -445,15 +358,7 @@
"#h is heat transfer coefficient\n",
"print\"Heat transfer coefficient in W/m**2 is\"\n",
"h=(E*I)/(A*(T1-T2))\n",
- "print\"h=\",h\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n"
+ "print\"h=\",h"
]
},
{
@@ -483,10 +388,6 @@
}
],
"source": [
- " \n",
- " \n",
- " \n",
- " \n",
"import math\n",
" \n",
"print\"Introduction to heat transfer by S.K.Som, Chapter 9, Example 6\"\n",
@@ -513,23 +414,7 @@
"#E is the burn out voltage\n",
"print\"The burn out voltage in Volts is \"\n",
"E=(qc*A)/I\n",
- "print\"E=\",E\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n"
+ "print\"E=\",E"
]
},
{
@@ -558,10 +443,6 @@
}
],
"source": [
- " \n",
- " \n",
- " \n",
- " \n",
"import math\n",
" \n",
"print\"Introduction to heat transfer by S.K.Som, Chapter 9, Example 7\"\n",
@@ -587,20 +468,7 @@
"print\"Heat flux q in W/m**2 is\"\n",
"q=(mul*hfg)*(((rhol-rhov)*g)/sigma)**(1/2)*((cpl*(T1-T2))/(csf*hfg*Prl**n))**3 \n",
"print\"The peak heat flux for water at one atmospheric pressure is qc=1.24*10**6(found in example 9.6).Since q<qc,The regime of boiling is nucleate.\"\n",
- "print\"q=\",q\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n"
+ "print\"q=\",q"
]
},
{
@@ -612,7 +480,7 @@
},
{
"cell_type": "code",
- "execution_count": 1,
+ "execution_count": 21,
"metadata": {
"collapsed": false
},
@@ -627,15 +495,11 @@
"The surface temprature in °C is\n",
"Tw= 120.0\n",
"The value of the coefficient csf is \n",
- "csf= 0.0151329179422\n"
+ "csf= 0.0214423761571\n"
]
}
],
"source": [
- " \n",
- " \n",
- " \n",
- " \n",
"import math\n",
" \n",
"print\"Introduction to heat transfer by S.K.Som, Chapter 9, Example 8\"\n",
@@ -676,30 +540,8 @@
"#Now we use following equation to determine csf,q=(mul*hfg)*(((rhol-rhov)*g)/sigma1)**(1/2)*((cpl*(Tw-T))/(csf*hfg*Prl**n))**3 \n",
"#Manipulating above equation to find csf we get csf=((cpl*(Tw-T))/(((q/((mul*hfg)*(((rhol-rhov)*g)/sigma1)**(1/2))**(1/3))*hfg*Prl**n))\n",
"print\"The value of the coefficient csf is \"\n",
- "csf=((cpl*(Tw-T))/(((q/((mul*hfg)*(((rhol-rhov)*g)/sigma1)**(1.0/2)))**(1.0/3))*hfg*Prl**n))#[NOTE:The answer in the book is incorrect.(Calcultion mistake)]\n",
- "print\"csf=\",csf\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n"
+ "csf=((cpl*(Tw-T))/(((q/((mul*hfg)*(((rhol-rhov)*g)/sigma1)**(1/2)))**(1/3))*hfg*Prl**n))#[NOTE:The answer in the book is incorrect.(Calcultion mistake)]\n",
+ "print\"csf=\",csf"
]
}
],
diff --git a/Introduction_to_Heat_Transfer_by_S._K._Som/chapter12.ipynb b/Introduction_to_Heat_Transfer_by_S._K._Som/chapter12.ipynb
index 6fafa3fd..859ba636 100644
--- a/Introduction_to_Heat_Transfer_by_S._K._Som/chapter12.ipynb
+++ b/Introduction_to_Heat_Transfer_by_S._K._Som/chapter12.ipynb
@@ -39,7 +39,6 @@
}
],
"source": [
- "\n",
"import math\n",
"\n",
"print \"Introduction to heat transfer by S.K.Som, Chapter 12, Example 1\"\n",
@@ -78,7 +77,7 @@
"#mass flow rate of air is mair\n",
"print \"mass flow rate of air is given by m=Mair*Nair in kg/sec \"\n",
"mair=Mair*Nair\n",
- "print round(mair,11)\n"
+ "print round(mair,11)"
]
},
{
@@ -112,7 +111,6 @@
}
],
"source": [
- "\n",
"import math\n",
"print \"Introduction to heat transfer by S.K.Som, Chapter 12, Example 2\"\n",
"#The temprature of atmospheric air (T)=40°C which flows over a wet bulb thermometer.\n",
@@ -193,7 +191,6 @@
}
],
"source": [
- "\n",
"import math\n",
"\n",
"print \"Introduction to heat transfer by S.K.Som, Chapter 12, Example 3\"\n",