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{
"metadata": {
"name": ""
},
"nbformat": 3,
"nbformat_minor": 0,
"worksheets": [
{
"cells": [
{
"cell_type": "heading",
"level": 1,
"metadata": {},
"source": [
"Chapter 1 : Measured thermodynamic Properties and Other Basic Concepts"
]
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 1.3 Page No : 28"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"# Variables\n",
"P = 1.4 ; \t\t\t# [MPa]\n",
"T = 333. \t\t\t#[K]\n",
"T1 = 320. \t\t\t#[K]\n",
"T2 = 360. \t\t\t#[K]\n",
"P_low = 1. \t\t\t#[MPa]\n",
"P_high = 1.5\n",
"V_cap_T1_P1 = 0.2678\n",
"V_cap_T2_P1 = 0.2873\n",
"V_cap_T1_P1_5 = 0.1765\n",
"V_cap_T2_P1_5 = 0.1899\n",
"\n",
"# Calculations\n",
"#At P = 1 MPa\n",
"V_cap_T333_P1 = V_cap_T1_P1 + (V_cap_T2_P1 - V_cap_T1_P1)*((T - T1)/(T2- T1))\n",
"\n",
"#Similarly at P=1.5 MPa\n",
"V_cap_T333_P1_5 = V_cap_T1_P1_5 + (V_cap_T2_P1_5 - V_cap_T1_P1_5)*((T - T1)/(T2 - T1))\n",
"\n",
"#At T=333*C\n",
"V_cap_P1_5 = V_cap_T333_P1_5 ;\n",
"V_cap_P1 = V_cap_T333_P1 ;\n",
"V_cap_P1_4 = V_cap_P1 + (V_cap_P1_5 - V_cap_P1)*((P - P_low)/(P_high - P_low)) ; \t\t\t#[m**3/kg]\n",
"\n",
"# Results\n",
"print 'Required specific volume = %.5f m**3/kg'%(V_cap_P1_4);\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Required specific volume = 0.19951 m**3/kg\n"
]
}
],
"prompt_number": 2
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 1.4 Page No : 29"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"# Variables\n",
"P = 1.4 ; \t\t\t#[MPa]\n",
"P_low = 1 ;\t\t\t#[MPa]\n",
"P_high = 1.5;\t\t\t#[MPa]\n",
"\n",
"#At T=333*C from interpolation we have\n",
"v_cap_P1_5 = 0.18086 ;\t\t\t#[m**3/kg]\n",
"v_cap_P1 = 0.27414 ;\t\t\t#[m**3/kg]\n",
"\n",
"# Calculations\n",
"#Molar volume is inversely proportional to pressure\n",
"v_cap_P1_4 = v_cap_P1 +(v_cap_P1_5 - v_cap_P1)*((1/P - 1/P_low)/(1/P_high - 1/P_low));\n",
"x=(0.19951-v_cap_P1_4)/v_cap_P1_4*100 ;\n",
"\n",
"# Results\n",
"print 'Specific volume m**3/kg) = %g m**3/kg'%(v_cap_P1_4);\n",
"print 'Percentage difference = %.1f %%'%(x);\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Specific volume m**3/kg) = 0.194186 m**3/kg\n",
"Percentage difference = 2.7 %\n"
]
}
],
"prompt_number": 1
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 1.5 page no : 29"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"# variables\n",
"V1 = 1 # volume\n",
"m1 = 2.5 # water\n",
"Vv = 5.042 # m**3/kg\n",
"Vt = .001023 # m**3/kg\n",
"T_44632 = 145 # C\n",
"T_39278 = 150 # C\n",
"\n",
"# Calculations\n",
"v1 = V1/m1 \n",
"x1 = (v1 - Vt)/(Vv - Vt)\n",
"\n",
"T2 = T_44632 + ( T_39278 - T_44632) * (( 0.4 - 0.44632)/(0.39278 - 0.44632))\n",
"\n",
"# Restuls\n",
"print \"Approximately %.f %% of the mass of water in the vapor\"%(round(x1,2)*100)\n",
"print \"Temperature T2 = %.1f C\"%T2\n",
"\n",
"# Note : Answer in book is wrong. Please calculate manually.\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Approximately 8 % of the mass of water in the vapor\n",
"Temperature T2 = 149.3 C\n"
]
}
],
"prompt_number": 8
}
],
"metadata": {}
}
]
}
|
