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{
"cells": [
{
"cell_type": "markdown",
"metadata": {},
"source": [
"# Chapter 2: Zeroth Law Of Thermodynamics"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Example 2.1: Engineering_Thermodynamics_by_Onkar_Singh_Chapter_2_Example_1.sce"
]
},
{
"cell_type": "code",
"execution_count": null,
"metadata": {
"collapsed": true
},
"outputs": [],
"source": [
"// Display mode\n",
"mode(0);\n",
"// Display warning for floating point exception\n",
"ieee(1);\n",
"clear;\n",
"clc;\n",
"disp('Engineering Thermodynamics by Onkar Singh,Chapter 2,Example 1')\n",
"Tf=98.6;//temperature of body in farenheit\n",
"disp('degree celcius and farenheit are related as follows')\n",
"disp('Tc=(Tf-32)/1.8')\n",
"disp('so temperature of body in degree celcius')\n",
"Tc=(Tf-32)/1.8"
]
}
,
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Example 2.2: Engineering_Thermodynamics_by_Onkar_Singh_Chapter_2_Example_2.sce"
]
},
{
"cell_type": "code",
"execution_count": null,
"metadata": {
"collapsed": true
},
"outputs": [],
"source": [
"// Display mode\n",
"mode(0);\n",
"// Display warning for floating point exception\n",
"ieee(1);\n",
"clear;\n",
"clc;\n",
"disp('Engineering Thermodynamics by Onkar Singh,Chapter 2,Example 2')\n",
"t1=0;//ice point temperature in degree celcius\n",
"p1=3;//thermometric property for ice point\n",
"t2=100;//steam point temperature in degree celcius\n",
"p2=8;//thermometric property for steam point\n",
"p3=6.5;//thermometric property for any temperature\n",
"disp('using thermometric relation')\n",
"disp('t=a*log(p)+(b/2)')\n",
"disp('for ice point,b/a=')\n",
"b=2*log(p1)\n",
"disp('so b=2.1972*a')\n",
"disp('for steam point')\n",
"a=t2/(log(p2)-(2.1972/2))\n",
"disp('and b=')\n",
"b=2.1972*a\n",
"disp('thus, t=a*log(p3)+(b/2) in degree celcius')\n",
"t=a*log(p3)+(b/2)\n",
"disp('so for thermodynamic property of 6.5,t=302.83 degree celcius')"
]
}
,
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Example 2.3: Engineering_Thermodynamics_by_Onkar_Singh_Chapter_2_Example_3.sce"
]
},
{
"cell_type": "code",
"execution_count": null,
"metadata": {
"collapsed": true
},
"outputs": [],
"source": [
"// Display mode\n",
"mode(0);\n",
"// Display warning for floating point exception\n",
"ieee(1);\n",
"clear;\n",
"clc;\n",
"disp('Engineering Thermodynamics by Onkar Singh,Chapter 2,Example 3')\n",
"disp('emf equation')\n",
"disp('E=(0.003*t)-((5*10^-7)*t^2))+(0.5*10^-3)')\n",
"disp('using emf equation at ice point,E_0 in volts')\n",
"t=0;//ice point temperature in degree celcius\n",
"disp('E_0=(0.003*t)-((5*10^-7)*t^2)+(0.5*10^-3)')\n",
"E_0=(0.003*t)-((5*10^-7)*t^2)+(0.5*10^-3)\n",
"disp('using emf equation at steam point,E_100 in volts')\n",
"t=100;//steam point temperature in degree celcius\n",
"disp('E_100=(0.003*t)-((5*10^-7)*t^2)+(0.5*10^-3)')\n",
"E_100=(0.003*t)-((5*10^-7)*t^2)+(0.5*10^-3)\n",
"disp('now emf at 30 degree celcius using emf equation(E_30)in volts')\n",
"t=30;//temperature of substance in degree celcius\n",
"E_30=(0.003*t)-((5*10^-7)*t^2)+(0.5*10^-3)\n",
"disp('now the temperature(T) shown by this thermometer')\n",
"disp('T=((E_30-E_0)/(E_100-E_0))*(T_100-T_0) in degree celcius')\n",
"T_100=100;//steam point temperature in degree celcius\n",
"T_0=0;//ice point temperature in degree celcius\n",
"T=((E_30-E_0)/(E_100-E_0))*(T_100-T_0)\n",
"disp('NOTE=>In this question,values of emf at 100 and 30 degree celcius is calculated wrong in book so it is corrected above so the answers may vary.')"
]
}
,
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Example 2.4: Engineering_Thermodynamics_by_Onkar_Singh_Chapter_2_Example_4.sce"
]
},
{
"cell_type": "code",
"execution_count": null,
"metadata": {
"collapsed": true
},
"outputs": [],
"source": [
"// Display mode\n",
"mode(0);\n",
"// Display warning for floating point exception\n",
"ieee(1);\n",
"clear;\n",
"clc;\n",
"disp('Engineering Thermodynamics by Onkar Singh,Chapter 2,Example 4')\n",
"t1=0;//temperature at ice point\n",
"t2=100;//temperature at steam point\n",
"t3=50;//temperature of gas\n",
"disp('emf equation,e=0.18*t-5.2*10^-4*t^2 in millivolts')\n",
"disp('as ice point and steam points are two reference points,so')\n",
"disp('at ice point,emf(e1)in mV')\n",
"e1=0.18*t1-5.2*10^-4*t1^2\n",
"disp('at steam point,emf(e2)in mV')\n",
"e2=0.18*t2-5.2*10^-4*t2^2\n",
"disp('at gas temperature,emf(e3)in mV')\n",
"e3=0.18*t3-5.2*10^-4*t3^2\n",
"disp('since emf variation is linear so,temperature(t)in degree celcius at emf of 7.7 mV')\n",
"t=((t2-t1)/(e2-e1))*e3\n",
"disp('temperature of gas using thermocouple=60.16 degree celcius')\n",
"disp('% variation in temperature reading with respect to gas thermometer reading of 50 degree celcius')\n",
"variation=((t-t3)/t3)*100"
]
}
,
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Example 2.5: Engineering_Thermodynamics_by_Onkar_Singh_Chapter_2_Example_5.sce"
]
},
{
"cell_type": "code",
"execution_count": null,
"metadata": {
"collapsed": true
},
"outputs": [],
"source": [
"// Display mode\n",
"mode(0);\n",
"// Display warning for floating point exception\n",
"ieee(1);\n",
"clear;\n",
"clc;\n",
"disp('Engineering Thermodynamics by Onkar Singh,Chapter 2,Example 5')\n",
"disp('let the conversion relation be X=aC+b')\n",
"disp('where C is temperature in degree celcius,a&b are constants and X is temperature in X degree ')\n",
"disp('at freezing point,temperature=0 degree celcius,0 degree X')\n",
"disp('so by equation X=aC+b')\n",
"X=0;//temperature in degree X\n",
"C=0;//temperature in degree celcius\n",
"disp('we get b=0')\n",
"b=0;\n",
"disp('at boiling point,temperature=100 degree celcius,1000 degree X')\n",
"X=1000;//temperature in degree X\n",
"C=100;//temperature in degree celcius\n",
"a=(X-b)/C\n",
"disp('conversion relation')\n",
"disp('X=10*C')\n",
"disp('absolute zero temperature in degree celcius=-273.15')\n",
"disp('absolute zero temperature in degree X=')\n",
"10*-273.15"
]
}
],
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{
"text": "MetaKernel Magics",
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