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{
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"worksheets": [
{
"cells": [
{
"cell_type": "heading",
"level": 1,
"metadata": {},
"source": [
"\n",
"Chapter 2 : Basic Thermodynamics"
]
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 2.1 Page: 27"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
" \n",
"import math \n",
"\n",
"# Variables\n",
"m = 1. #[lbm] Mass of the steam\n",
"T_1 = 300. #[F] Initial temperature\n",
"P_1 = 14.7 #[psia] Initial pressure\n",
"P_sorronding = 14.7 #[psia]\n",
"Q = 50. #[Btu] Amount of the energy added to the system as heat\n",
"\n",
"# This is a closed system and we can apply the following equations\n",
"# delta_U_system = sum(dQ_in_minus_out) + sum(dW_in_minus_out) (A)\n",
"# dS_system = (m*ds)_system = sum((dQ)/T)_in_minus_out + dS_reversible (B)\n",
"\n",
"# From the steam tables, we look up the properties of steam at temperature 300F and pressure 14.7 psia and find \n",
"u_initial = 1109.6 #[Btu/lbm] Internal energy of the steam\n",
"h_initial = 1192.6 #[Btu/lbm] Enthalpy of the steam\n",
"s_initial = 1.8157 #[Btu/(lbm*R)] Entropy of the steam\n",
"\n",
"# The work here is done by the system, equal to\n",
"# -delta_w = P*A_piston*delta_x = P*m*delta_v\n",
"\n",
"# Calculations\n",
"# Substituting this in the equation (A) and rearranging, we have\n",
"# m*delta_(u + P*v) = m*delta_h = delta_Q\n",
"# From which we can solve for the final specific enthalpy\n",
"h_final = h_initial + Q #[Btu/lbm]\n",
"\n",
"# Now, by the linear interpolation we find that at h = 1242.6 Btu/lbm and P = 1 atm, temperature of the steam is given \n",
"T_2 = 405.7 #[F] Final temperature\n",
"\n",
"# At this final temperature and pressure we have the steam properties \n",
"u_final = 1147.7 #[Btu/lbm]\n",
"s_final = 1.8772 #[Btu/(lbm*R)]\n",
"\n",
"# Thus, increase in the internal energy, enthalpy and entropy are \n",
"delta_u = u_final - u_initial #[Btu/lbm]\n",
"delta_s = s_final - s_initial #[Btu/(lbm*R)]\n",
"delta_h = Q #[Btu/lbm]\n",
"\n",
"# The work done on the atmosphere is given by\n",
"w = delta_h - delta_u #[Btulbm]\n",
"\n",
"# Results\n",
"print \"The increase in internal energy of the steam by adding the heat is %0.2f Btu/lbm\"%(delta_u)\n",
"print \"The increase in enthalpy of the steam by adding the heat is %0.2f Btu/lbm\"%(delta_h)\n",
"print \"The increase in entropy of the steam by adding the heat is %0.4f Btu/lbm\"%(delta_s)\n",
"print \"Work done by the piston expanding against the atmosphere is %0.2f Btu/lbm\"%(w)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The increase in internal energy of the steam by adding the heat is 38.10 Btu/lbm\n",
"The increase in enthalpy of the steam by adding the heat is 50.00 Btu/lbm\n",
"The increase in entropy of the steam by adding the heat is 0.0615 Btu/lbm\n",
"Work done by the piston expanding against the atmosphere is 11.90 Btu/lbm\n"
]
}
],
"prompt_number": 4
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 2.2 Page: 28\n"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
" \n",
"import math \n",
"\n",
"# Variables\n",
"T_in = 600. #[F] Input steam temperature\n",
"P_in = 200. #[psia] Input steam pressure\n",
"P_exit = 50. #[psia]\n",
"\n",
"# Because this is a steady-state, steady-flow process, we use \n",
"# (work per pound) = W/m = -( h_in - h_out )\n",
"\n",
"# From the steam table we can read the the inlet enthalpy and entropy as \n",
"h_in = 1322.1 #[Btu/lbm]\n",
"s_in = 1.6767 #[Btu/(lb*R)]\n",
"\n",
"# Now, we need the value of h_out\n",
"\n",
"# For a reversible adiabatic steady-state, steady-flow process, we have\n",
"# sum(s*m_in_minus_out) = ( s_in - s_out ) = 0\n",
"\n",
"# Which indicates that inlet and outlet entropies are same\n",
"# We can find the outlet temperature by finding the value of the temperature in the steam table\n",
"# For which the inlet entropy at 50 psia is the same as the inlet entropy, 1.6767 Btu/(lb*R). \n",
"# By linear interpolation in the table we find \n",
"T_in = 307.1 #[R]\n",
"\n",
"# and by the linear interpolation in the same table we find that\n",
"h_out = 1188.1 #[Btu/lb]\n",
"\n",
"# Calculations\n",
"# Thus, we find \n",
"W_per_pound = (h_in - h_out) #[Btu/lb]\n",
"\n",
"# Results\n",
"print \" The work output of the turbine of steam is %0.1f Btu/lb\"%(-W_per_pound)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
" The work output of the turbine of steam is -134.0 Btu/lb\n"
]
}
],
"prompt_number": 1
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 2.3 Page: 38"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
" \n",
"import math \n",
"\n",
"\n",
"# Variables\n",
"T = 500. #[F]\n",
"P = 680. #[psi]\n",
"\n",
"# Calculations\n",
"# It is reported in the book in the table A.1(page 417) that for water \n",
"# We know that T_r = T/T_c and P_r = P/P_c, so\n",
"T_c = 647.1*1.8 #[R]\n",
"P_c = 220.55*14.51 #[psia]\n",
"w = 0.345\n",
"T_r = (T+459.67)/T_c\n",
"P_r = P/P_c\n",
"z_0 = 1+P_r/T_r*(0.083-0.422/T_r**(1.6))\n",
"z_1 = P_r/T_r*(0.139-0.172/T_r**(4.2))\n",
"z = z_0+w*z_1\n",
"\n",
"# Results\n",
"print \"The compressibility factor of steam at the given state is %0.3f\"%(z)\n",
"# Based on the steam table (which may be considered as reliable as the experimental data, the value of z is 0.804.\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The compressibility factor of steam at the given state is 0.851\n"
]
}
],
"prompt_number": 6
},
{
"cell_type": "code",
"collapsed": false,
"input": [],
"language": "python",
"metadata": {},
"outputs": []
}
],
"metadata": {}
}
]
}
|
