path: root/Process_Heat_Transfer_by_D_Q_Kern/17-Cooling_Towers.ipynb

{
"cells": [
 {
		   "cell_type": "markdown",
	   "metadata": {},
	   "source": [
       "# Chapter 17: Cooling Towers"
	   ]
	},
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 17.1: Calculation_of_the_Enthalpy_of_Saturated_Air.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"printf('\t example 17.1 \n');\n",
"pw=0.4298; // psia, at 75F, table 7\n",
"pt=14.696; // psia\n",
"t=75;\n",
"Mw=18;\n",
"Ma=29;\n",
"X=(pw/(pt-pw))*(Mw/Ma);\n",
"printf('\t humidity is : %.4f lb water/lb air \n',X);\n",
"H=(X*t)+(1051.5*X)+(0.24*t); // eq 17.54\n",
"printf('\t enthalpy at 75F is : %.1f Btu/lb dry air \n',H);\n",
"// end"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 17.2: Calculation_of_the_Number_of_Diffusion_Units.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"printf('\t example 17.2 \n');\n",
"printf('\t approximate values are mentioned in the book \n');\n",
"printf('\t by numerical integration \n');\n",
"T1=85;\n",
"T2=120;\n",
"A=576; // ground area, from fig 17.12\n",
"L=1500*(500/576);\n",
"G=1400;\n",
"R=(L/G);\n",
"printf('\t R is : %.2f \n',R);\n",
"H1=39.1; // fig 17.12\n",
"H2=H1+(R*(T2-T1));\n",
"printf('\t H2 is : %.1f Btu \n',H2);\n",
"// The area between the saturation line and the operating line represents the potential for heat transfer\n",
"// at T=85F\n",
"Hs=50; // fig 17.12\n",
"d1=(Hs-H1);\n",
"printf('\t difference is : %.1f \n',d1);\n",
"//at t=90\n",
"Hs=56.7; // fig 17.12\n",
"H=43.7; // fig 17.12\n",
"d2=Hs-H;\n",
"printf('\t difference is : %.1f \n',d2);\n",
"d=(d1+d2)/(2);\n",
"printf('\t average of difference is : %.1f \n',d);\n",
"dT=5; // F\n",
"nd1=(dT/d);\n",
"printf('\t nd1 is : %.3f \n',nd1);\n",
"// similarly calculating nd at each temperature and adding them will give you total nd value\n",
"nd=1.70;\n",
"printf('\t number of diffusing units : %.2f \n',nd);\n",
"printf('\t log mean enthalpy difference \n');\n",
"dt=49.9; // diff. of enthalpies at top of the tower, from table in solution\n",
"db=10.9; // diff of enthalpies at bottom of the tower,from table in solution\n",
"LME=(dt-db)/(2.3*log10(dt/db));\n",
"printf('\t log mean of enthalpy : %.1f Btu/lb \n',LME);\n",
"nd=(T2-T1)/(LME);\n",
"printf('\t number of diffusing units are : %.2f \n',nd);\n",
"// The error is naturally larger the greater the range\n",
"//end"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 17.3: Calculation_of_the_Required_Height_of_Fill.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"printf('\t example 17.3 \n');\n",
"printf('\t approximate values are mentioned in the book \n');\n",
"// Since the loading is based on 1 ft2 of ground area\n",
"nd=1.7;\n",
"L=1302;\n",
"Kxa=115;\n",
"Z=(nd*L)/(Kxa);\n",
"printf('\t Z is : %.1f ft \n',Z);\n",
"HDU=(Z/nd);\n",
"printf('\t height of diffusion unit : %.1ff ft \n',HDU);\n",
"// end"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 17.4: Determination_of_a_Cooling_tower_Guarantee.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"printf('\t example 17.4 \n');\n",
"printf('\t approximate values are mentioned in the book \n');\n",
"// The area between the saturation line and the operating line represents the potential for heat transfer\n",
"// at T=79.3F\n",
"Hs=43.4; // fig 17.12\n",
"H=30.4; // fig 17.12\n",
"d1=(Hs-H);\n",
"printf('\t difference is : %.1f \n',d1);\n",
"//at t=85\n",
"Hs=50; // fig 17.12\n",
"H=35.7; // fig 17.12\n",
"d2=Hs-H;\n",
"printf('\t difference is : %.1f \n',d2);\n",
"d=(d1+d2)/(2);\n",
"printf('\t average of difference is : %.2f \n',d);\n",
"dT=(85-79.3); // F\n",
"nd1=(dT/d);\n",
"printf('\t nd1 is : %.3f \n',nd1);\n",
"// similarly calculating nd at each temperature and adding them will give you total nd value\n",
"nd=1.72;\n",
"printf('\t number of diffusing units : %.2f \n',nd);\n",
"// end"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 17.5: The_Recalculation_of_Cooling_tower_Performance.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"printf('\t example 17.5 \n');\n",
"printf('\t approximate values are mentioned in the book \n');\n",
"T1=85;\n",
"T2=120;\n",
"R=0.93; // R=(L/G), for 1500 gpm\n",
"printf('\t for 120percent of design \n');\n",
"R1=1.2*R;\n",
"printf('\t R is : %.3f \n',R1);\n",
"H1=39.1; // at 87.2F\n",
"H2=H1+(R1*(T2-T1)); \n",
"printf('\t H2 is : %.1f Btu \n',H2);\n",
"// The area between the saturation line and the operating line represents the potential for heat transfer\n",
"// at T=87.2F\n",
"Hs=53.1; // from table in the solution\n",
"d1=(Hs-H1);\n",
"printf('\t difference is : %.1f \n',d1);\n",
"//at t=90\n",
"Hs=56.7; // fig 17.12\n",
"H=42; // fig 17.12\n",
"d2=Hs-H;\n",
"printf('\t difference is : %.1f \n',d2);\n",
"d=(d1+d2)/(2);\n",
"printf('\t average of difference is : %.1f \n',d);\n",
"dT=(90-87.2); // F\n",
"nd1=(dT/d);\n",
"printf('\t nd1 is : %.3f \n',nd1);\n",
"// similarly calculating nd at each temperature and adding them will give you total nd value\n",
"nd=1.53;\n",
"printf('\t number of diffusing units : %.2f \n',nd);\n",
"printf('\t for 80 percent of design \n');\n",
"R2=0.8*R;\n",
"printf('\t R is : %.3f \n',R2);\n",
"H1=39.1; // at 87.2F\n",
"H2=H1+(R2*(T2-T1)); \n",
"printf('\t H2 is : %.0f Btu \n',H2);\n",
"// The area between the saturation line and the operating line represents the potential for heat transfer\n",
"// at T=82.5F\n",
"Hs=47.2; // from table in the solution\n",
"d1=(Hs-H1);\n",
"printf('\t difference is : %.1f \n',d1);\n",
"//at t=85\n",
"Hs=50; // fig 17.12\n",
"H=40.8; // fig 17.12\n",
"d2=Hs-H;\n",
"printf('\t difference is : %.1f \n',d2);\n",
"d=(d1+d2)/(2);\n",
"printf('\t average of difference is : %.1f \n',d);\n",
"dT=(85-82.5); // F\n",
"nd1=(dT/d);\n",
"printf('\t nd1 is : %.3f \n',nd1);\n",
"// similarly calculating nd at each temperature and adding them will give you total nd value\n",
"nd=1.92;\n",
"printf('\t number of diffusing units : %.2f \n',nd);\n",
"X=[1.115 0.93 0.74];\n",
"Y=[1.53 1.70 1.92];\n",
"plot2d(X,Y,style=3,rect=[0.7,1.4,1.3,2]);\n",
"xtitle('KxaV/L vs L/G','L/G','nd');\n",
"printf('\t trial 1 \n');\n",
"R3=1.1;\n",
"printf('\t R is : %.3f \n',R3);\n",
"H1=34.5; // at 87.2F\n",
"H2=H1+(R3*(T2-T1)); \n",
"printf('\t H2 is : %.0f Btu \n',H2);\n",
"// The area between the saturation line and the operating line represents the potential for heat transfer\n",
"// at T=85F\n",
"Hs=50; // from table in the solution\n",
"d1=(Hs-H1);\n",
"printf('\t difference is : %.1f \n',d1);\n",
"//at t=90\n",
"Hs=56.7; // fig 17.12\n",
"H=40; // fig 17.12\n",
"d2=Hs-H;\n",
"printf('\t difference is : %.1f \n',d2);\n",
"d=(d1+d2)/(2);\n",
"printf('\t average of difference is : %.1f \n',d);\n",
"dT=(90-85); // F\n",
"nd1=(dT/d);\n",
"printf('\t nd1 is : %.3f \n',nd1);\n",
"// similarly calculating nd at each temperature and adding them will give you total nd value\n",
"nd=1.48;\n",
"printf('\t number of diffusing units : %.2f \n',nd);\n",
"R3=1.19; // from fig 17.14\n",
"printf('\t L/G is : %.2f \n',R3);\n",
"printf('\t trial 2 \n');\n",
"R4=1.2;\n",
"printf('\t R4 is : %.3f \n',R4);\n",
"H1=34.5; // at 87.2F\n",
"H2=H1+(R4*(T2-T1)); \n",
"printf('\t H2 is : %.1f Btu \n',H2);\n",
"// The area between the saturation line and the operating line represents the potential for heat transfer\n",
"// at T=85F\n",
"Hs=50; // from table in the solution\n",
"d1=(Hs-H1);\n",
"printf('\t difference is : %.1f \n',d1);\n",
"//at t=90\n",
"Hs=56.7; // fig 17.12\n",
"H=40.5; // fig 17.12\n",
"d2=Hs-H;\n",
"printf('\t difference is : %.1f \n',d2);\n",
"d=(d1+d2)/(2);\n",
"printf('\t average of difference is : %.1f \n',d);\n",
"dT=(90-85); // F\n",
"nd1=(dT/d);\n",
"printf('\t nd1 is : %.3f \n',nd1);\n",
"// similarly calculating nd at each temperature and adding them will give you total nd value\n",
"nd=1.56;\n",
"printf('\t number of diffusing units : %.2f \n',nd);\n",
"R3=1.08; // from fig 17.14\n",
"printf('\t L/G is : %.2f \n',R3);\n",
"// end"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 17.6: Calculation_of_a_Direct_contact_Gas_Cooler.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
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	   },
	   "outputs": [],
"source": [
"printf('\t example 17.6 \n');\n",
"printf('\t approximate values are mentioned in the book \n');\n",
"// basis 1ft^2 ground area\n",
"//Assumption: 20 per cent of the initial vapor content of the gas enters the water body\n",
"X1=(1.69/(14.7-1.69))*(18/29);\n",
"printf('\t X1 : %.4f lb/lb \n',X1);\n",
"G=1500;\n",
"w1=G*X1;\n",
"printf('\t total water in inlet gas : %.2f lb/hr \n',w1);\n",
"// The inlet gas is at 300F and a 120F dew point. Use 0.25 Btu/(lb)(°F) for the specific heat of nitrogen\n",
"H1=(0.0807*120)+(0.0807*1025.8)+(0.45*0.0807*(300-120))+(0.25*300); // eq 17.55\n",
"printf('\t H1 : %.0f Btu/lb dry air \n',H1);\n",
"X2=(w1*(1-.2)/G);\n",
"printf('\t outlet gas humidity : %.5f lb/lb \n',X2);\n",
"pw=(X2*29*14.7/18)/(1+(X2*29/18));\n",
"printf('\t pw : %.3f psia \n',pw);\n",
"Tw=112.9; // F, from table 7 for above pw\n",
"// The outlet gas has a temperature of 200°F and a 112.9°F dew point\n",
"H2=(X2*Tw)+(X2*1029.8)+(X2*0.45*(200-Tw))+(0.25*200); // eq 17.55\n",
"printf('\t H2 : %.1f Btu/lb dry air \n',H2);\n",
"q=G*(H1-H2);\n",
"printf('\t total heat load : %.2e Btu/hr \n',q);\n",
"w2=q/(120-85);\n",
"printf('\t water loading : %.2e lb/hr \n',w2);\n",
"printf('\t interval 1 \n');\n",
"// (Kxa*delV/L)= 0 t0 0.05\n",
"nd=0.05; // nd=Kxa*V/L\n",
"Le=0.93; // fig 17.4 at 300F\n",
"C=(0.25)+(0.45*X1);\n",
"printf('\t C : %.3f Btu/(lb)*(F) \n',C);\n",
"haV=(nd*w2*Le*C);\n",
"printf('\t haV : %.1f Btu/(hr)*(F) \n',haV);\n",
"qc=(haV*(300-120));\n",
"printf('\t qc : %.2e Btu/hr \n',qc);\n",
"delT=(qc/(C*G));\n",
"printf('\t delT : %.1f F \n',delT);\n",
"T1=(300-delT);\n",
"printf('\t T(0.05) : %.1f F \n',T1);\n",
"delt=(qc/w2);\n",
"printf('\t delt : %.2f F \n',delt);\n",
"t1=(120-delt);\n",
"printf('\t t(0.05) : %.1f F \n',t1);\n",
"printf('\t interval 2 \n');\n",
"// (Kxa*delV/L)= 0.05 to 0.15\n",
"nd1=0.1;\n",
"haV1=(nd1*w2*Le*C);\n",
"printf('\t haV1 : %.1f Btu/(hr)*(F) \n',haV1);\n",
"qc1=(haV1*(T1-t1));\n",
"printf('\t qc1 : %.1e Btu/hr \n',qc1);\n",
"delT1=(qc1/(C*G));\n",
"printf('\t delT1 : %.1f F \n',delT1);\n",
"T2=(T1-delT1);\n",
"printf('\t T(0.15) : %.2f F \n',T2);\n",
"X3=0.0748; // at 117.6F\n",
"w3=(nd1*w2*(0.0807-X3));\n",
"printf('\t water diffused during interval : %.3f lb/hr \n',w3);\n",
"w4=(w1-w3);\n",
"printf('\t water remaining : %.2f lb/hr \n',w4);\n",
"l1=1027; // Btu/lb, l1= lamda at 117.6F\n",
"qd=(w3*l1);\n",
"printf('\t qd : %.0f Btu/hr \n',qd);\n",
"q1=(qd+qc1);\n",
"printf('\t q1 : %.0f Btu/hr \n',q1);\n",
"delt1=(q1/w2);\n",
"printf('\t delt1 : %.2f F \n',delt1);\n",
"t2=(t1-delt1);\n",
"printf('\t t(0.15) : %.1f F \n',t2);\n",
"X4=0.0640; // at 112.5\n",
"X5=(w4/G);\n",
"printf('\t X(112.5F) : %.4f lb/lb \n',X5);\n",
"printf('\t interval 3 \n');\n",
"// (Kxa*delV/L)= 0.15 to 0.25\n",
"nd1=0.1;\n",
"haV1=(nd1*w2*Le*C);\n",
"printf('\t haV1 : %.1f Btu/(hr)*(F) \n',haV1);\n",
"qc2=(haV1*(T2-t2));\n",
"printf('\t qc2 : %.2e Btu/hr \n',qc2);\n",
"delT2=(qc2/(C*G));\n",
"printf('\t delT2 : %.1f F \n',delT2);\n",
"T3=(T2-delT2);\n",
"printf('\t T(0.25) : %.1f F \n',T3);\n",
"w5=(nd1*w2*(X5-X4));\n",
"printf('\t water diffused during interval : %.3f lb/hr \n',w5);\n",
"w6=(w4-w5);\n",
"printf('\t water remaining : %.2f lb/hr \n',w6);\n",
"l2=1030; // Btu/lb, l1= lamda at 112.5F\n",
"qd1=(w5*l2);\n",
"printf('\t qd1 : %.2e Btu/hr \n',qd1);\n",
"q2=(qd1+qc2);\n",
"printf('\t q2 : %.3e Btu/hr \n',q2);\n",
"delt2=(q2/w2);\n",
"printf('\t delt2 : %.2f F \n',delt2);\n",
"t3=(t2-delt2);\n",
"printf('\t t(0.25) : %.1f F \n',t3);\n",
"X6=0.0533; // at 106.5\n",
"X7=(w6/G);\n",
"printf('\t X(106.5F) : %.4f lb/lb \n',X7);\n",
"// The calculations of the remaining intervals until a. gas temperature of 200°F is reached are shown in Fig. 17.17\n",
"w7=21.92; // total water diffused from table in solution\n",
"d=(w7/w1)*100;\n",
"printf('\t calculated diffusion : %.0f \n',d);\n",
"printf('\t Using some standard low-pressure-drop data \n');\n",
"// For G = 1500, extrapolate to L = 2040 on logarithmic coordinates. Kxa = 510.\n",
"ndt=.54; // from 1st table in solution\n",
"Kxa=510; // from 2nd table in solution\n",
"Z=(ndt*w2/Kxa);\n",
"printf('\t tower height : %.2f ft \n',Z);\n",
"A=(50000/G);\n",
"printf('\t cross section : %.1f ft^2 \n',A);\n",
"// end\n",
"\n",
"\n",
""
   ]
   }
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		   "source": [
			"## Example 17.7: Approximate_Calculation_of_a_Gas_Cooler.sce"
		   ]
		  },
  {
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"printf('\t example 17.7 \n');\n",
"printf('\t approximate values are mentioned in the book \n');\n",
"C=0.28; // assumption\n",
"w=50000; // lb/hr\n",
"G=1500;\n",
"Qs=(w*C*(500-200));\n",
"Qd=(w/G)*(22685); // qd=22685, from previous prblm\n",
"printf('\t sensible heat : %.1e Btu/hr \n',Qs);\n",
"printf('\t approximate diffusion : %.2e Btu/hr \n',Qd);\n",
"Q=(Qs+Qd);\n",
"printf('\t total heat : %.3e Btu/hr \n',Q);\n",
"// an allowance as high as 30 per cent of the sensible load can be made and the excess water compensated for by throttling when the tower is in operation\n",
"w1=(Q/(120-85));\n",
"printf('\t total water quantity : %.2e lb/hr \n',w1);\n",
"// If the maximum liquid loading is taken as 2040 lb/(hr)(ft'!), the required tower cross section\n",
"A=(w1/2040);\n",
"printf('\t tower cross section : %.1f ft^2 \n',A);\n",
"w3=(w/A);\n",
"printf('\t new gas rate : %.0f lb/(hr)(ft^2) \n',w3);\n",
"// The two terminal temperature differences are (200 - 85) and (500 - 120).\n",
"LMTD=((500-120)-(200-85))/(log((500-120)/(200-85)));\n",
"printf('\t LMTD : %.0f \n',LMTD);\n",
"dt=35;\n",
"N=(dt/LMTD); // eq 17.88\n",
"printf('\t haV/L : %.2f \n',N);\n",
"Le=0.93;\n",
"nd=(N/(C*Le));\n",
"printf('\t number diffusion units : %.2f \n',nd);\n",
"// By extrapolation for G = 718 and L = 2040,Kxa=215\n",
"L=2040;\n",
"Kxa=215;\n",
"Z=(nd*L/Kxa); // calculation mistake\n",
"printf('\t height of tower : %.1f ft \n',Z);\n",
"di=(A)^(1/2);\n",
"printf(' ground dimensions : %.1f ft \n',di);\n",
"// ground dimensions are 5.8*8.3*8.3 ft\n",
"// end"
   ]
   }
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