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{
"metadata": {
"name": ""
},
"nbformat": 3,
"nbformat_minor": 0,
"worksheets": [
{
"cells": [
{
"cell_type": "heading",
"level": 1,
"metadata": {},
"source": [
"Chapter 22 : Introduction to Energy Balances for Process without Reaction"
]
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
" Example 22.1 page no. 651\n"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"# Application of the energy balance to a closed system\n",
"\n",
"# Variables\n",
"#Assume that properties of water can be used to substitute properties of solution\n",
"# Given\n",
"V = 1.673 ;\t\t\t# Volume of closed vessel-[cubic metre]\n",
"m = 1. ;\t\t\t# mass of saturated liquid vaporized-[kg]\n",
"Pi = 1. ;\t\t\t# Initial pressure -[atm]\n",
"Ti = 10. ;\t\t\t# Initial temperature -[degree C]\n",
"Pf = 1. ;\t\t\t# final pressure -[atm]\n",
"Tf = 100. ;\t\t\t# final temperature -[degree C]\n",
"\n",
"# Use steam table to obtain additional information at given condition\n",
"Ui = 35. ;\t\t\t# Initial enthalpy-[kJ/kg]\n",
"Uf = 2506.0 ;\t\t\t# Final enthalpy -[kJ/kg]\n",
"\n",
"# Calculations\n",
"# Use eqn. 22.2 after modifiying it using given conditions(W = 0,del_KE = 0 and del_PE = 0 )\n",
"Q = m*(Uf - Ui) ;\t\t\t# Heat transferred to the vessel - [kJ]\n",
"\n",
"# Results\n",
"print 'Heat transferred to the vessel is %.1f kJ . '%Q\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Heat transferred to the vessel is 2471.0 kJ . \n"
]
}
],
"prompt_number": 1
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 22.2 page no. 652\n"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"# Calculation of U using American engineering units\n",
"\n",
"# Variables\n",
"# Given\n",
"T1 = 80. ;\t\t\t# Initial temperature -[degree F]\n",
"T1 = 40. ;\t\t\t# final temperature -[degree F]\n",
"\n",
"# Additional data obtained from steam table at given temperatures and corresponding vapour pressures\n",
"p1 = 0.5067 ;\t\t\t# Initial saturation pressure-[psia]\n",
"p2 = 0.1217 ;\t\t\t# Final saturation pressure-[psia]\n",
"V1 = 0.01607 ;\t\t\t# Initial specific volume - [cubic feet/lb]\n",
"V2 = 0.01602 ;\t\t\t# Final specific volume - [cubic feet/lb]\n",
"H1 = 48.02 ;\t\t\t#Initial specific enthalpy -[Btu/lb]\n",
"H2 = 8.05 ;\t\t\t# Final specific enthalpy -[Btu/lb]\n",
"\n",
"# Calculations\n",
"del_P = p2 - p1 ;\t\t\t# Change in pressure -[psia]\n",
"del_V = V2 - V1 ;\t\t\t# Change in specific volume -[cubic feet/lb]\n",
"del_H = H2 - H1 ;\t\t\t# Change in specific enthalpy -[Btu/lb]\n",
"del_pV = p2*144*V2/778. - p1*144*V1/778. ;\t\t\t# Change in pv-[Btu]\n",
"del_U = del_H - del_pV ;\t\t\t# Change in specific internal energy - [Btu/lb]\n",
"del_E = del_U ;\t\t\t# Change in specific total energy(since KE=0,PE=0 and W=0) -[Btu/lb]\n",
"\n",
"# Results\n",
"print 'Change in pressure is %.3f psia . '%del_P\n",
"print 'Change in specific volume is %.5f cubic feet/lb (negligible value) . '%del_V\n",
"print 'Change in specific enthalpy is %.2f Btu/lb . '%del_H\n",
"print 'Change in specific internal energy is %.2f Btu/lb . '%del_U\n",
"print 'Change in specific total energy is %.2f Btu/lb . '%del_E\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Change in pressure is -0.385 psia . \n",
"Change in specific volume is -0.00005 cubic feet/lb (negligible value) . \n",
"Change in specific enthalpy is -39.97 Btu/lb . \n",
"Change in specific internal energy is -39.97 Btu/lb . \n",
"Change in specific total energy is -39.97 Btu/lb . \n"
]
}
],
"prompt_number": 2
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
" Example 22.3 page no. 662\n"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"# use of the general energy balance to analyze an open , unsteady-state system\n",
"\n",
"# Variables\n",
"#Lets take tank to be system\n",
"# Given\n",
"T = 600. ; \t\t\t# Temperature of steam -[K]\n",
"P = 1000. ;\t\t\t# Pressure of steam -[kPa]\n",
"\n",
"# Calculations\n",
"# Additional data for steam obtained from CD database at T and P\n",
"U = 2837.73 ;\t\t\t# Specific internal energy-[kJ/kg]\n",
"H = 3109.44 ;\t\t\t# Specific enthalpy -[kJ/kg]\n",
"V = 0.271 ;\t\t\t# Specific volume -[cubic metre/kg]\n",
"# By the reduced equation \n",
"Ut2 = H ;\t\t\t# Internal energy at final temperature-[kJ/kg]\n",
"\n",
"# Results\n",
"print 'The specific internal energy at final temperature is %.2f kJ/kg. Now use two properties\\\n",
"of the steam (P = %i kPa and Ut2 = %.2f kJ/kg) to find final temperature (T) from steam table. \\\n",
"From steam table we get T = 764 K.'%(Ut2,P,Ut2)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The specific internal energy at final temperature is 3109.44 kJ/kg. Now use two propertiesof the steam (P = 1000 kPa and Ut2 = 3109.44 kJ/kg) to find final temperature (T) from steam table. From steam table we get T = 764 K.\n"
]
}
],
"prompt_number": 3
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
" Example 22.4 page no. 669\n"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"# application of the energy balance to a open, steady-state system, a heat exchanger\n",
"\n",
"# Variables\n",
"# Take milk plus water in tank to be system\n",
"# Given\n",
"T1_water = 70. ;\t\t\t# Temperature of entering water -[degree C]\n",
"T2_water = 35. ;\t\t\t# Temperature of exiting water -[degree C]\n",
"T1_milk = 15. ;\t\t\t#Temperature of entering milk -[degree C]\n",
"T2_milk = 25. ;\t\t\t#Temperature of exiting milk -[degree C]\n",
"\n",
"# Get additional data from steam table for water and milk,assuming milk to have same properties as that of water.\n",
"H_15 = 62.01 ;\t\t\t#Change in specific internal energy-[kJ/kg]\n",
"H_25 = 103.86 ;\t\t\t#Change in specific internal energy-[kJ/kg]\n",
"H_35 = 146.69 ;\t\t\t#Change in specific internal energy-[kJ/kg]\n",
"H_70 = 293.10 ;\t\t\t#Change in specific internal energy-[kJ/kg]\n",
"\n",
"# Assumptions to simplify Equation 22.8 are:\n",
"print 'Assumptions to simplify Equation 22.8 are:'\n",
"print '1. Change in KE and PE of system = 0.'\n",
"print '2. Q = 0 ,because of way we picked the system,it is is well insulated.'\n",
"print '3. W = 0,work done by or on the system.'\n",
"\n",
"# Calculations\n",
"#Basis m_milk = 1 kg/min , to directly get the answer .\n",
"m_milk = 1 ;\t\t\t# Mass flow rate of milk-[kg/min]\n",
"# By applying above assumtions eqn. 22.8 reduces to del_H = 0 .Using it get m_water-\n",
"m_water = (m_milk*(H_15 - H_25))/(H_35 - H_70) ; \t\t\t# Mass flow rate of water-[kg/min]\n",
"m_ratio = m_water/m_milk ;\t\t\t# Mass flow rate of water per kg/min of milk-[kg/min]\n",
"\n",
"# Results\n",
"print 'Mass flow rate of water per kg/min of milk is %.2f (kg water/min )/(kg milk/min).'%m_ratio\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Assumptions to simplify Equation 22.8 are:\n",
"1. Change in KE and PE of system = 0.\n",
"2. Q = 0 ,because of way we picked the system,it is is well insulated.\n",
"3. W = 0,work done by or on the system.\n",
"Mass flow rate of water per kg/min of milk is 0.29 (kg water/min )/(kg milk/min).\n"
]
}
],
"prompt_number": 4
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
" Example 22.5 page no. 670\n"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"# Calculation of power needed to pump water in an open, steady state system\n",
"\n",
"# Variables\n",
"# Take pipe between initial and final level of water\n",
"# Given\n",
"h_in = -20. ;\t\t\t# Depth of water below ground-[ft]\n",
"h_out = 5. ;\t\t\t# Height of water level above ground-[ft]\n",
"h = h_out - h_in ;\t\t\t# Total height to which water is pumped-[ft]\n",
"V = 0.50 ;\t\t\t# Volume flow rate of water - [cubic feet/s]\n",
"ef = 100.; \t\t\t# Efficiency of pump - [%] \n",
"g = 32.2; \t\t\t# Acceleration due to gravity -[ft/square second] \n",
"gc = 32.2 ;\t\t\t#[(ft*lbm)/(second square*lbf)]\n",
"\n",
"M = V * 62.4 ;\t\t\t# mass flow rate - [lbm/s]\n",
"PE_in = 0 ;\t\t\t# Treating initial water level to be reference level\n",
"\n",
"# Calculations\n",
"PE_out = (M*g*h*1.055)/(gc*778.2) ;\t\t\t# PE of discharged water -[lbm*(square feet/square second)]\n",
"W = PE_out - PE_in ;\t\t\t#Work done on system = power delivered by pump, (since we are using mass flow rate and pump efficiency is 100 % , so W = Power) -[kW]\n",
"\n",
"# Results\n",
"print 'The electric power required by the pump is %.2f kW. '%W\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The electric power required by the pump is 1.06 kW. \n"
]
}
],
"prompt_number": 5
},
{
"cell_type": "code",
"collapsed": true,
"input": [],
"language": "python",
"metadata": {},
"outputs": []
}
],
"metadata": {}
}
]
}
|
