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{
"metadata": {
"name": ""
},
"nbformat": 3,
"nbformat_minor": 0,
"worksheets": [
{
"cells": [
{
"cell_type": "heading",
"level": 1,
"metadata": {},
"source": [
"Chapter 27 : Ideal Processes Efficiency and the Mechanical Energy Balance"
]
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 27.1 page no. 838\n"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"# Calculation of the Work done during Evaporation of a Liquid\n",
"# Solution E27.1\n",
"\n",
"# Variables\n",
"V_w = 1. ;\t\t\t# Volume of given water -[L]\n",
"P_atm = 100. ;\t\t\t# Atmospheric pressure - [kPa]\n",
"\n",
"#W = -p*del_V\n",
"V_H2O = 0.001043 ;\t\t\t# Specific volume of water from steam table according to book- [cubic metre] \n",
"V_vap = 1.694 ;\t\t\t# Specific volume of vapour from steam table according to book- [cubic metre] \n",
"V1 = 0 ;\t\t\t# Initial volume of H2O in bag-[cubic metre]\n",
"\n",
"# Calculations\n",
"V2 = (V_w*V_vap)/(1000*V_H2O) ;\t\t\t# Final volume of water vapour -[cubic metre] \n",
"W = -P_atm*(V2 -V1)* 1000 ;\t\t\t# Work done by saturated liquid water -[J]\n",
" \n",
"# Results \n",
"print ' Work done by saturated liquid water is %.3e J.'%W\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
" Work done by saturated liquid water is -1.624e+05 J.\n"
]
}
],
"prompt_number": 5
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
" Example 27.2 page no. 840\n"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"# Calculation of Work in a Batch Process\n",
"# Solution E27.2\n",
"\n",
"# Variables\n",
"m_N2 = 1. ;\t\t\t# Moles of N2 taken -[kg mol]\n",
"p = 1000.;\t\t\t# Pressure of cylinder-[kPa]\n",
"T = 20. + 273. ;\t\t\t# Temperature of cylinder -[K]\n",
"a_pis = 6. ;\t\t\t# Area of piston - [square centimetre]\n",
"m_pis = 2. ;\t\t\t# Mass of pston - [kg]\n",
"R = 8.31 ;\t\t\t# Ideal gas constant - [(kPa*cubic metre)/(K * kgmol)]\n",
"\n",
"# Calculations\n",
"V = (R*T)/p ;\t\t\t# Specific volue of gas at initial stage -[cubic metre/kg mol]\n",
"V1 = V * m_N2 ;\t\t\t# Initial volume of gas - [cubic metre]\n",
"V2 = 2.*V1 ;\t\t\t# Final volume of gas according to given condition -[cubic metre]\n",
"\n",
"# Assumed surrounding pressure constant = 1 atm\n",
"p_atm = 101.3 ;\t\t\t# Atmospheric pressure-[kPa]\n",
"del_Vsys = V2 -V1 ;\t\t\t# Change in volume of system -[cubic metre]\n",
"del_Vsurr = - del_Vsys ;\t\t\t# Change in volume of surrounding -[cubic metre]\n",
"W_surr = -p_atm*del_Vsurr ;\t\t\t# Work done by surrounding - [kJ]\n",
"W_sys = -W_surr ;\t\t\t# Work done by system - [kJ]\n",
"\n",
"# Results\n",
"print ' Work done by gas(actually gas + piston system) is %.0f kJ.'%W_sys\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
" Work done by gas(actually gas + piston system) is -247 kJ.\n"
]
}
],
"prompt_number": 6
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
" Example 27.3 page no. 845\n"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"# Efficiency of Power Generation by a Hydroelectric Plant\n",
"# Solution \n",
"\n",
"# Variables\n",
"p_plant = 20. ;\t\t\t# Power generated by plant-[MW]\n",
"h = 25. ;\t\t\t# Height of water level - [m]\n",
"V = 100. ;\t\t\t# Flow rate of water -[cubic metre/s]\n",
"d_water = 1000. ;\t\t\t# Density of water - [ 1000 kg / cubic metre]\n",
"g = 9.807 ;\t\t\t# Acceleration due to gravity-[m/square second]\n",
"\n",
"# Calculations\n",
"M_flow = V*d_water ;\t\t\t# Mass flow rate of water -[kg/s]\n",
"del_PE = M_flow*g*h ;\t\t\t# Potential energy change of water per second -[W]\n",
"eff = (p_plant*10**6) /(del_PE) ;\t\t\t# Efficiency of plant \n",
"\n",
"# Results\n",
"print ' Efficiency of plant is %.2f .'%eff\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
" Efficiency of plant is 0.82 .\n"
]
}
],
"prompt_number": 7
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
" Example 27.4 page no. 845\n"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"# Calculation of Plant Efficiency\n",
"# Solution Fig.E27.4\n",
"\n",
"# Variables\n",
"LHV = 36654. ;\t\t\t# LHV value of fuel - [kJ/ cubic metre]\n",
"Q1 = 16. ;\t\t\t#- [kJ/ cubic metre]\n",
"Q2 = 0 ;\t\t\t#- [kJ/ cubic metre]\n",
"Q3 = 2432. ;\t\t\t#- [kJ/ cubic metre]\n",
"Q4 = 32114. ;\t\t\t#- [kJ/ cubic metre]\n",
"Q41 = 6988. ;\t\t\t#- [kJ/ cubic metre]\n",
"Q8 = 1948. ;\t\t\t#- [kJ/ cubic metre]\n",
"Q9 = 2643. ;\t\t\t#- [kJ/ cubic metre]\n",
"Q81 = 2352. - Q8 ;\t\t\t# - [kJ/ cubic metre]\n",
"Q567 = 9092. ;\t\t\t# Sum of Q5, Q6 and Q7- [kJ/ cubic metre]\n",
"\n",
"# Calculations and Results\n",
"#(a)\n",
"G_ef = (LHV+ Q1 +Q2 + Q3 - Q9)/(LHV) ;\t\t\t# Gross efficiency\n",
"print '(a) Gross efficiency is %.3f .'%G_ef\n",
"\n",
"#(b)\n",
"T_ef = (Q567+Q8)/(LHV+ Q1 +Q2 + Q3) ;\t\t\t#Thermal efficiency \n",
"print ' (b) Thermal efficiency is %.3f .'%T_ef\n",
"\n",
"#(c)\n",
"C_ef = Q4/(Q4 + Q41) ;\t\t\t# Combustion efficiency\n",
"print ' (c) Combustion efficiency is %.3f .'%C_ef\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"(a) Gross efficiency is 0.995 .\n",
" (b) Thermal efficiency is 0.282 .\n",
" (c) Combustion efficiency is 0.821 .\n"
]
}
],
"prompt_number": 8
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 27.5 page no. 850"
]
},
{
"cell_type": "code",
"collapsed": true,
"input": [
"# Comparison of the reversible work for a batch process with that of a flow process operating under the same conditions.\n",
"\n",
"from scipy.integrate import quad\n",
"\n",
"# Variables\n",
"V1 = 5. ;\t\t\t# Volume of gas initially - [cubic feet]\n",
"P1 = 1. ;\t\t\t# Initial pressure - [atm]\n",
"P2 = 10. ;\t\t\t# Final pressure - [atm]\n",
"T1 = 100. + 460 ;\t\t\t# initial temperature - [degree Rankine]\n",
"R = 0.7302 ;\t\t\t# Ideal gas constant -[(cubic feet*atm)/(lb mol)*(R)]\n",
"\t\t\t#Equation of state pV**1.4 = constant\n",
"\n",
"\n",
"# Calculations and Results\n",
"V2 = V1*(P1/P2)**(1/1.4) ;\t\t\t# Final volume - [cubic feet] \n",
"\n",
"def f(V):\n",
" return -(P1)*(V1/V)**(1.4)\n",
" \n",
"W1_rev = quad(f,V1,V2)[0] ;\t\t\t# Reversible work done in compresion in a horizontal cylinder with piston -[cubic feet *atm]\n",
"W1 = W1_rev *1.987/.7302 ;\t\t\t# Conversion to Btu -[Btu]\n",
"print '(a)Reversible work done in compression in a horizontal cylinder with piston is %.1f Btu . '%W1\n",
"\n",
"#(b)\n",
"n1 = (P1*V1)/(R*T1) ;\t\t\t# Number of moles of gas\n",
"\n",
"def f1(P):\n",
" return (V1)*(P1/P)**(1/1.4)\n",
"W2_rev = quad(f1,P1,P2)[0]\t\t# Reversible work done in compresion in a rotary compressor -[cubic feet *atm]\n",
"W2 = W2_rev *1.987/.7302 ;\t\t\t# Conversion to Btu -[Btu]\n",
"\n",
"print '(b)Reversible work done in a rotary compressor is %.1f Btu . '%W2\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"(a)Reversible work done in compression in a horizontal cylinder with piston is 31.7 Btu . \n",
"(b)Reversible work done in a rotary compressor is 44.3 Btu . \n"
]
}
],
"prompt_number": 1
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 27.6 page no. 853"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"# Application of the mechanical energy balance to the pumping water\n",
"\n",
"from scipy.integrate import quad\n",
"# Variables\n",
"m_water = 1. ;\t\t\t# Mass flow rate of water -[lb/min]\n",
"P1 = 100. ;\t\t\t# Initial pressure - [psia]\n",
"P2 = 1000. ;\t\t\t# Final pressure - [psia]\n",
"T1 = 80. + 460 ;\t\t\t# initial temperature - [degree Rankine]\n",
"T2 = 100. + 460 ;\t\t\t# final temperature - [degree Rankine]\n",
"h = 10. ;\t\t\t# Difference in water level between entry and exit of stream-[ft]\n",
"g = 32.2 ;\t\t\t# Accleration due to gravity - [ft/ square second]\n",
"gc = 32.2 ;\t\t\t#[(ft*lbm)/(lbf*square second)]\n",
"\n",
"v1 = .01607 ;\t\t\t# specific volume of liquid water at 80 degree F -[cubic feet/lbm]\n",
"v2 = .01613 ;\t\t\t# specific volume of liquid water at 100 degree F -[cubic feet/lbm] \n",
"v= 0.0161 ;\t\t\t# -[cubic feet/lbm]\n",
"\n",
"# Calculations\n",
"del_PE = (h*g)/(gc*778) ;\t\t\t# Change in potential energy - [Btu/lbm]\n",
"\n",
"def f(P):\n",
" return (v)*(12**2/778.)\n",
" \n",
"PV_work = quad(f,P1,P2)[0]\t\t\t# PV work done -[Btu/lbm]\n",
"#From eqn. (A)\n",
"W = PV_work + del_PE ;\t\t\t# Work per minute required to pump 1 lb water per minute - [Btu/lbm]\n",
"\n",
"# Results\n",
"print ' Work per minute required to pump 1 lb water per minute is %.2f Btu/lbm . '%W\n",
"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
" Work per minute required to pump 1 lb water per minute is 2.69 Btu/lbm . \n"
]
}
],
"prompt_number": 1
},
{
"cell_type": "code",
"collapsed": false,
"input": [],
"language": "python",
"metadata": {},
"outputs": [],
"prompt_number": 2
}
],
"metadata": {}
}
]
}
|
