{
"cells": [
{
"cell_type": "markdown",
"metadata": {},
"source": [
"# Chapter 10 - Refrigeration and Air Conditioning"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Example 1 - pg 10.42"
]
},
{
"cell_type": "code",
"execution_count": 1,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Power rating of the compressor-motor unit if the cop of the plant is 2.1 is (kW) = 40.4\n"
]
}
],
"source": [
"#pg 10.42\n",
"#calculate the Power rating of compressor-motor\n",
"#Input data\n",
"T1=273.;#The temperature of ice in K\n",
"T2=298.;#Temperature of water at room in K\n",
"COP=2.1;#Cop of the plant\n",
"ne=90.;#Overall electrochemical efficiency in percentage\n",
"w=15.;#Weight of ice produced per day in tonnes\n",
"cw=4.187;#Specific heat of water in kJ/kg degrees celcius\n",
"Li=335.;#Latent heat of ice in kJ/kg\n",
"mi=1.;#Mass of ice produced at 0 degrees celcius\n",
"\n",
"#Calculations\n",
"m=(w*1000.)/(24*60);#Mass of ice produced in kg/min\n",
"h=(mi*cw*(T2-T1))+Li;#Heat extracted from 1kg of water at 25 degrees celcius to produce 1kg of ice at 0 degrees celcius in kJ/kg\n",
"Q=m*h;#Total heat extracted in kJ\n",
"W=Q/COP;#Work done by the compressor in kJ/kg\n",
"P=W/(60.*(ne/100));#Power of compressor in kW\n",
"\n",
"#Output\n",
"print 'Power rating of the compressor-motor unit if the cop of the plant is 2.1 is (kW) = ',round(P,1)\n"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Example 2 - pg 10.43"
]
},
{
"cell_type": "code",
"execution_count": 2,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"The refrigeration capacity of the plant is (TR) = 0.541\n"
]
}
],
"source": [
"#pg 10.43\n",
"#calculate the refrigeration capacity\n",
"#Input data\n",
"m=400.;#Mass of fruits supplied to a cold storage in kg\n",
"T1=293.;#Temperature at which fruits are stored in K\n",
"T2=268.;#Temperature of cold storage in K\n",
"t=8.;#The time untill which fruits are cooled in hours\n",
"hfg=105.;#Latent heat of freezing in kJ/kg\n",
"Cf=1.25;#Specific heat of fruit\n",
"TR=210.;#One tonne refrigeration in kJ/min\n",
"\n",
"#Calculations\n",
"Q1=m*Cf*(T1-T2);#Sensible heat in kJ\n",
"Q2=m*hfg;#Latent heat of freezing in kJ\n",
"Q=Q1+Q2;#Heat removed from fruits in 8 hrs\n",
"Th=(Q1+Q2)/(t*60);#Total heat removed in one minute in kJ/kg\n",
"Rc=Th/TR;#Refrigerating capacity of the plant in TR\n",
"\n",
"#Output\n",
"print 'The refrigeration capacity of the plant is (TR) = ',round(Rc,3)\n"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Example 3 - pg 10.44"
]
},
{
"cell_type": "code",
"execution_count": 3,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"(a)COP of the machine when it is operated as a refrigerating machine is 5.0\n",
"(b)COP when it is operated as heat pump is 6.0\n",
"(c)COP or efficiency of the Heat engine is (percent) = 16.67\n"
]
}
],
"source": [
"#pg 10.44\n",
"#calculate the COP of machine in all cases\n",
"#Input data\n",
"T1=300.;#The maximum temperature at which carnot cycle operates in K\n",
"T2=250.;#The minimum temperature at which carnot cycle operates in K\n",
"\n",
"#Calculations\n",
"COPr=T2/(T1-T2);#COP of the refrigerating machine\n",
"COPh=T1/(T1-T2)#COP of heat pump\n",
"n=((T1-T2)/T1)*100;#COP or efficiency of the heat engine in percentage\n",
"\n",
"#Output data\n",
"print '(a)COP of the machine when it is operated as a refrigerating machine is ',COPr\n",
"print '(b)COP when it is operated as heat pump is ',COPh\n",
"print '(c)COP or efficiency of the Heat engine is (percent) = ',round(n,2)\n"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Example 4 - pg 10.45"
]
},
{
"cell_type": "code",
"execution_count": 4,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"(a)Capacity of the plant is (TR) = 48.31\n",
"(b)Time taken to achieve cooling is (hours) = 10.67\n"
]
}
],
"source": [
"#pg 10.45\n",
"#calculate the capacity of the plant and time taken\n",
"#Input data\n",
"m=20000.;#The storage capacity of fish in a storage plant in kg\n",
"T1=298.;#Supplied temperature of fish in K\n",
"T2=263.;#Temperature of cold storage in which fish are stored in K\n",
"T3=268.;#Freezing point of fish in K\n",
"Caf=2.95;#Specific heat of fish above freezing point in kJ/kg K\n",
"Cbf=1.25;#Specific heat of below freezing point in kJ/kg K\n",
"W=75.;#Work required by the plant in kW\n",
"TR=210.;#One tonne refrigeration in kJ/min\n",
"hfg=230.;#Latent heat of fish in kJ/kg\n",
"\n",
"#Calculations\n",
"COPr=T2/(T1-T2);#COP of reversed carnot cycle\n",
"COPa=0.3*COPr;#Given that actual COP is 0.3 times of reversed COP\n",
"Hr=(COPa*W)*60;#Heat removed by the plant in kJ/min\n",
"C=Hr/TR;#Capacity of the plant in TR\n",
"Q1=m*Caf*(T1-T3);#Heat removed from the fish above freezing point in kJ\n",
"Q2=m*Cbf*(T3-T2);#Heat removed from fish below freezing point in kJ\n",
"Q3=m*hfg;#Total latent heat of the fish in kJ\n",
"Q=Q1+Q2+Q3;#Total heat removed by the plant in kJ\n",
"T=(Q/Hr)/60;#Time taken to achieve cooling in hrs \n",
"\n",
"#Output data\n",
"print '(a)Capacity of the plant is (TR) = ',round(C,2)\n",
"print '(b)Time taken to achieve cooling is (hours) = ',round(T,2)\n"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Example 5 - pg 10.46"
]
},
{
"cell_type": "code",
"execution_count": 5,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Theoretical COP for a CO2 machine working at given temperatures = 4.39\n",
"The answer in textbook is wrong. Please check using a calculator\n"
]
}
],
"source": [
"#pg 10.46\n",
"#calculate the Theoretical COP\n",
"#Input data\n",
"T2=298.;#Maximum temperature at which CO2 machine works in K\n",
"T1=268.;#Minimum temperature at which CO2 machine works in K\n",
"sf1=-0.042;#Liquid entropy at 268 K in kJ/kg K\n",
"hfg1=245.3;#Latent heat of gas at 268 K in kJ/kg\n",
"sf2=0.251;#Liquid entropy in kJ/kg K\n",
"hfg2=121.4;#Latent heat of gas at 298 K in kJ/kg\n",
"hf1=-7.54;#Liquid enthalpy at 268 K in kJ/kg\n",
"hf2=81.3;#Liquid enthalpy at 298 K in kJ/kg\n",
"hf3=81.3;#Enthalpy at point 3 in graph in kJ/kg\n",
"\n",
"#Calculations\n",
"s2=sf2+(hfg2/T2);#Entropy at point 2 from the graph in kJ/kg K\n",
"x1=(s2-sf1)/(hfg1/T1);#Dryness fraction at point 1\n",
"h1=hf1+(x1*hfg1);#Enthalpy at point 1 in kJ/kg\n",
"h2=hf2+hfg2;#Enthalpy at point 2 in kJ/kg\n",
"COP=(h1-hf3)/(h2-h1);#Coefficient of performance for a CO2 machine working at given temperatures\n",
"\n",
"#Output data\n",
"print 'Theoretical COP for a CO2 machine working at given temperatures = ',round(COP,2)\n",
"print 'The answer in textbook is wrong. Please check using a calculator'"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Example 6 - pg 10.48"
]
},
{
"cell_type": "code",
"execution_count": 7,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"The capacity of refrigerator is (TR) = 24.0\n"
]
}
],
"source": [
"#pg 10.48\n",
"#calculate the capacity of refrigerator\n",
"#Input data\n",
"T2=298.;#Maximum temperature at which ammonia refrigerating system works in K\n",
"T1=263.;#Minimum temperature at which ammonia refrigerating system works in K\n",
"mf=5.;#Fluid flow rate in kg/min\n",
"sf1=0.5443;#Liquid entropy at 298 K in kJ/kg K\n",
"sf2=1.1242;#Liquid entropy at 263 K in kJ/kg K\n",
"hfg1=1297.68;#Latent heat at 298 K in kJ/kg\n",
"hfg2=1166.94;#Latent heat at 263 K in kJ/kg\n",
"hf1=135.37;#Liquid enthalpy at point 1 in graph in kJ/kg\n",
"hf2=298.9;#Liquid enthalpy at point 2 in graph in kJ/kg\n",
"TR=210.;#One tonne refrigeration in TR\n",
"\n",
"#Calculations\n",
"s2=sf2+(hfg2/T2);#Entropy at point 2 in kJ/kg\n",
"x1=(s2-sf1)/(hfg1/T1);#Dryness fraction at point 1\n",
"h1=hf1+(x1*hfg1);#Enthalpy at point 1 in kJ/kg\n",
"h=h1-hf2;#Heat extracted of refrigerating effect produced per kg of refrigerant in kJ/kg\n",
"ht=mf*h;#Total heat extracted at a fluid flow rate of 5 kg/min in kJ/min\n",
"C=ht/TR;#Capacity of refrigerating in TR\n",
"\n",
"#Output\n",
"print 'The capacity of refrigerator is (TR) = ',round(C,0)\n"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Example 7 - pg 10.49"
]
},
{
"cell_type": "code",
"execution_count": 8,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"The theoretical COP of a ammonia refrigerating machine working between given temperatures = 5.56\n"
]
}
],
"source": [
"#pg 10.49\n",
"#calculate the theoretical COP\n",
"#Input data\n",
"T1=263.;#Minimum temperature at which ammonia refrigerating machine works in K\n",
"T2=303.;#Maximum temperature at which ammonia refrigerating machine works in K\n",
"x1=0.6;#Dryness fraction of ammonia during suction stroke\n",
"sf1=0.5443;#Liquid entropy at 263 K in kJ/kg K\n",
"hfg1=1297.68;#Latent heat at 263 K in kJ/kg\n",
"sf2=1.2037;#Liquid entropy at 303 K in kJ/kg K\n",
"hfg2=1145.8;#Latent heat at 303 K in kJ/kg\n",
"hf1=135.37;#Liquid enthalpy at 263 K in kJ/kg\n",
"hf2=323.08;#Liquid enthalpy at 303 K in kJ/kg\n",
"\n",
"#Calculations\n",
"s1=sf1+((x1*hfg1)/T1);#Entropy at point 1 in kJ/kg K\n",
"x2=(s1-sf2)/(hfg2/T2);#Entropy at point 2 in kJ/kg K\n",
"h1=hf1+(x1*hfg1);#Enthalpy at point 1 in kJ/kg\n",
"h2=hf2+(x2*hfg2);#Enthalpy at point 2 in kJ/kg\n",
"COP=(h1-hf2)/(h2-h1);#Theoretical COP of ammonia refrigerating machine\n",
"\n",
"#Output\n",
"print 'The theoretical COP of a ammonia refrigerating machine working between given temperatures = ',round(COP,2)\n"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Example 8 - pg 10.51"
]
},
{
"cell_type": "code",
"execution_count": 9,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"The amount of ice produced is (kg/kW hr) = 11.44\n"
]
}
],
"source": [
"#pg 10.51\n",
"#calculate the amount of ice\n",
"#Input data\n",
"T1=263.;#Minimum temperature at which Vapour compression refrigerator using methyl chloride operates in K\n",
"T2=318.;#Maximum temperature at which Vapour compression refrigerator using methyl chloride operates in K\n",
"sf1=0.183;#Entropy of the liquid in kJ/kg K\n",
"hfg1=460.7;#Enthalpy of the liquid in kJ/kg\n",
"sf2=0.485;#Entropy of the liquid in kJ/kg K\n",
"hfg2=483.6;#Enthalpy of the liquid in kJ/kg\n",
"x2=0.95;#Dryness fraction at point 2\n",
"hf3=133.0;#Enthalpy of the liquid in kJ/kg\n",
"W=3600.;#Work to be spent corresponding to 1kW/hour\n",
"Cw=4.187;#Specific heat of water in kJ/kg degrees celcius\n",
"mi=1;#Mass of ice produced at 0 degrees celcius\n",
"Li=335.;#Latent heat of ice in kJ/kg\n",
"hf1=45.4;#Enthalpy of liquid at 263 K in kJ/kg\n",
"hf2=133.;#Enthalpy of liquid at 318 K in kJ/kg\n",
"\n",
"#Calculations\n",
"s2=sf2+((x2*(hfg2-hf2))/T2);#Enthalpy at point 2 in kJ/kg\n",
"x1=(s2-sf1)/((hfg1-hf1)/T1);#Dryness fraction at point 1\n",
"h1=hf1+(x1*hfg1);#Enthalpy at point 1 in kJ/kg\n",
"h2=hf2+(x2*hfg2);#Enthalpy at point 2 in kJ/kg\n",
"COP=(h1-hf3)/(h2-h1);#Theoretical COP\n",
"COPa=0.6*COP;#Actual COP which is 60 percent of theoretical COP\n",
"H=W*COPa;#Heat extracted or refrigeration effect produced per kW hour in kJ\n",
"Hw=(mi*Cw*10)+Li;#Heat extracted from water at 10 degrees celcius for the formation of 1 kg of ice at 0 degrees celcius\n",
"I=H/Hw;#Amount of ice produced in kg/kW hr\n",
"\n",
"#Output\n",
"print 'The amount of ice produced is (kg/kW hr) = ',round(I,2)\n"
]
}
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