path: root/Heat_Transfer_In_SI_Units_by_J_P_Holman/10-Heat_Exchangers.ipynb

{
"cells": [
 {
		   "cell_type": "markdown",
	   "metadata": {},
	   "source": [
       "# Chapter 10: Heat Exchangers"
	   ]
	},
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 10.10: off_design_calculation_of_exchanger_in_example_10_4.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"clear;\n",
"clc;\n",
"printf('\t\t\tExample Number 10.10\n\n\n');\n",
"// off-design calculation of exchanger in example 10-4  \n",
"// Example 10.10 (page no.-544-546) \n",
"// solution\n",
"\n",
"m_dot_c = 68;// [kg/min] water flow rate\n",
"T1 = 35;// [degree celsius] initial temperature \n",
"T2 = 75;// [degree celsius] final temperature\n",
"Toe = 110;// [degree celsius] oil entering temperature \n",
"Tol = 75;// [degree celsius] oil leaving temperature\n",
"Cc = 4180;// [J/kg degree celsius] water specific heat capacity\n",
"Ch = 1900;// [J/kg degree celsius] heat capacity of oil\n",
"U = 320;// [W/square meter degree celsius] overall heat transfer coefficient\n",
"A = 15.814568;// [square meter] area of heat exchanger (from example 10-4)\n",
"// the flow rate of oil is calculated from the energy balance for the original problem:\n",
"m_dot_h = m_dot_c*Cc*(T2-T1)/(Ch*(Toe-Tol));// [kg/min]\n",
"// the capacity rates for the new conditions are calculated as \n",
"C_h = m_dot_h*Ch/60;// [W/degree celsius]\n",
"C_c = m_dot_c*Cc/60;// [W/degree celsius]\n",
"// so that the water (cold fluid) is the minimum fluid, and \n",
"C_min_by_C_max = C_c/C_h;\n",
"NTU_max = U*A/C_c;\n",
"// from figure 10-13(page no.-542) or table 10-3(page no.-543) the effectiveness is \n",
"E = 0.744;\n",
"// and because the cold fluid is the minimum, we can write \n",
"dT_cold = E*(Toe-T1);// [degree celsius]\n",
"// and the exit water temperature is \n",
"Tw_exit = T1+dT_cold;// [degree celsius]\n",
"// the total heat transfer under the new flow conditions is calculated  as \n",
"m_dot_c = 40;// [kg/min]\n",
"q = m_dot_c*Cc*dT_cold/60;// [W]\n",
"printf('exit water temperature is %f degree celcius',Tw_exit);\n",
"printf('\n\n the total heat transfer under the new flow conditions is %f kW',q/1000);\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
""
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 10.11: cross_flow_exchanger_with_both_fluid_unmixed.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"clear;\n",
"clc;\n",
"printf('\t\t\tExample Number 10.11\n\n\n');\n",
"// cross-flow exchanger with both fluid unmixed  \n",
"// Example 10.11 (page no.-547-549) \n",
"// solution\n",
"\n",
"pa = 101325;// [Pa] pressure of air\n",
"Ti = 15.55;// [degree celsius] initial temperature of air\n",
"Tf = 29.44;// [degree celsius] final temperature of air\n",
"Thw = 82.22;// [degree celsius] hot water temperature\n",
"U = 227;// [W/square meter degree celsius] overall heat transfer coefficient\n",
"S = 9.29;// [square meter] total surface area of heat exchanger\n",
"R = 287;// [] universal gas constant\n",
"Cc = 1006;// [J/kg degree celsius] specific heat of air \n",
"Ch = 4180;// [J/kg degree celsius] specific heat of water\n",
"// the heat transfer is calculated from the energy balance on the air. first, the inlet air density is \n",
"rho = pa/(R*(Ti+273.15));// [kg/cubic meter]\n",
"// so the mass flow of air (the cold fluid) is \n",
"mdot_c = 2.36*rho;// [kg/s]\n",
"// the heat transfer is then \n",
"q = mdot_c*Cc*(Tf-Ti);// [W]\n",
"// from the statement of the problem we do not know whether the air or water is the minimum fluid. a trial and error procedur must be used with figure  10-15(page no.-545) or table 10-3(page no.-543).\n",
"// we assume that the air is the minimum fluid and then check out our assumption. then\n",
"Cmin = mdot_c*Cc;// [W/degree celsius]\n",
"NTU_max = U*S/Cmin;\n",
"// and the effectiveness based on the air as the minimum fluid is \n",
"E = (Tf-Ti)/(Thw-Ti);\n",
"// entering figure 10-15, we are unable to match these quantities with the curves. this require that the hot fluid be the minimum. we must therefore assume values for the water flow rate until we are able to match the performance as given by figure 10-15 or table 10-3. we first note that\n",
"Cmax = mdot_c*Cc;// [W/degree celsius]           (a)\n",
"// NTU_max = U*S/Cmin;                            (b)\n",
"// E = dT_h/(Thw-Ti)                             (c)\n",
"// dT_h = q/Cmin                                 (d)\n",
"\n",
"// now we assume different values for Cmin abd calculate different-different values for NTU_max, dT_h, and E\n",
"\n",
"// for \n",
"Cmin_by_Cmax1 = 0.5;\n",
"Cmin1 = Cmin_by_Cmax1*Cmax;// [W/degree celsius]\n",
"NTU_max1 = U*S/Cmin1;\n",
"dT_h1 = q/Cmin1;// [degree celsius]\n",
"E1_c1 = dT_h1/(Thw-Ti);// calculated\n",
"E1_t1 = 0.65;// from table \n",
"\n",
"// for \n",
"Cmin_by_Cmax2 = 0.25;\n",
"Cmin2 = Cmin_by_Cmax2*Cmax;// [W/degree celsius]\n",
"NTU_max2 = U*S/Cmin2;\n",
"dT_h2 = q/Cmin2;// [degree celsius]\n",
"E1_c2 = dT_h2/(Thw-Ti);// calculated\n",
"E1_t2 = 0.89;// from table \n",
"\n",
"// for \n",
"Cmin_by_Cmax3 = 0.22;\n",
"Cmin3 = Cmin_by_Cmax3*Cmax;// [W/degree celsius]\n",
"NTU_max3 = U*S/Cmin3;\n",
"dT_h3 = q/Cmin3;// [degree celsius]\n",
"E1_c3 = dT_h3/(Thw-Ti);// calculated\n",
"E1_t3 = 0.92;// from table \n",
"\n",
"// we estimate the water-flow rate as about\n",
"Cmin = 660;// [W/degree celsius]\n",
"mdot_h = Cmin/Ch;// [kg/s]\n",
"// the exit water temperature is accordingly\n",
"Tw_exit = Thw-q/Cmin;// [degree celsius]\n",
"printf('the exit water temperature is %f degree celsius',Tw_exit);\n",
"printf('\n\n the heat transfer is %f kW',q/1000);"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 10.12: comparison_of_single_or_two_exchanger_options.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"clear;\n",
"clc;\n",
"printf('\t\t\tExample Number 10.12\n\n\n');\n",
"// comparison of single- or two-exchanger options  \n",
"// Example 10.12 (page no.-549-551) \n",
"// solution\n",
"\n",
"mdot_c = 1.25;// [kg/s] water flow rate\n",
"Ti = 35;// [degree celsius] initial temperature of water\n",
"Tf = 80;// [degree celsius] final temperature of water\n",
"Toi = 150;// [degree celsius] initial temperature of oil\n",
"Tof = 85;// [degree celsius] final temperature of oil\n",
"U = 850;// [W/square meter degree celsius] overall heat transfer coefficient\n",
"Cp_water = 4180;// [] specific heat of water\n",
"Cp_oil = 2000;// [J/kg degree celsius]  \n",
"// we calculate the surface area required for both alternatives and then compare costs. for the one large exchanger \n",
"q = mdot_c*Cp_water*(Tf-Ti);// [W]\n",
"mdot_c_into_Cp_water = mdot_c*Cp_water;// [W/degree celsius]\n",
"mdot_h_into_Cp_oil = q/(Toi-Tof);// [W/degree celsius]\n",
"Cmin = mdot_h_into_Cp_oil;// [W/degree celsius]\n",
"Cmax = mdot_c_into_Cp_water;// [W/degree celsius]\n",
"// so that oil is the minimum fluid:\n",
"Eh = (Toi-Tof)/(Toi-Ti);\n",
"Cmin_by_Cmax = Cmin/Cmax;\n",
"// from figure 10-13(page no.-542), \n",
"NTU_max = 1.09;\n",
"A = NTU_max*Cmin/U;// [square meter]\n",
"// we now wish to calculate the surface-area requirement for the two small exchanger because U*A and Cmin are the same for each exchanger. \n",
"// this requires that the effectiveness be the same for each exchanger. thus,\n",
"// E1 = (Toi-Toe_1)/(Toi-Ti) = E2 = (Toi-Toe_2)/(Toi-Tw2)                                            (a)\n",
"// where the nomenclature for the temperatures is indicated in the sketch. because the oil flow is the same in each exchanger and the average exit oil temperature must be 85 degree celsius, we may write\n",
"// (Toe_1+Toe_2)/2 = 85                                                                              (b)\n",
"// an energy balance on the second heat exchanger gives\n",
"// mdot_c_into_Cp_water*(Tf-Tw2) = mdot_h_into_Cp_oil*(Toi-Toe_2)/2                                  (c)\n",
"// we now have three equations (a),(b), and (c) which may be solved for the three unknowns Toe_1, Toe_2, and Tw2. \n",
"// eliminating Tw2, and Toe_1 from equation (a) by the help of equation (b) and (c)\n",
"deff('[y] = H(Toe_2)','y = (Toi-(170-Toe_2))/(Toi-Ti) - (Toi-Toe_2)/(Toi-(Tf-(mdot_h_into_Cp_oil*(Toi-Toe_2)/(mdot_c_into_Cp_water*2))))');\n",
"Toe_2 = fsolve(1,H);// [degree celsius]\n",
"Toe_1 = (170-Toe_2);// [degree celsius]\n",
"Tw2 = (Tf-(mdot_h_into_Cp_oil*(Toi-Toe_2)/(mdot_c_into_Cp_water*2)));// [degree celsius]\n",
"// the effectiveness can then be calculated as \n",
"E1 = (Toi-Toe_1)/(Toi-Ti);\n",
"E2 = E1;\n",
"// from figure 10-13(page no.-542), we obtain \n",
"NTU_max = 1.16;\n",
"// so that \n",
"A1 = NTU_max*Cmin/(U*2);// [square meter]\n",
"printf('we have find that %f square meter of area is required for each of small exchangers, or a total of %f square meter',A1,2*A1);\n",
"printf('\n\n the area required in the one larger exchanger is %f square meter',A);\n",
"printf('\n\n the cost per unit area is greater so that the most economical choice would be the single larger exchanger ');"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 10.13: shell_and_tube_exchangeras_air_heater.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"clear;\n",
"clc;\n",
"printf('\t\t\tExample Number 10.13\n\n\n');\n",
"// shell and tube exchangeras air heater  \n",
"// Example 10.13 (page no.-551-552) \n",
"// solution\n",
"\n",
"To = 100;// [degree celsius] temperature of hot oil\n",
"m_dot_a = 2;// [kg/s] flow rate of air\n",
"T1 = 20;// [degree celsius] initial temperature of air \n",
"T2 = 80;// [degree celsius] final temperature of air\n",
"Cp_o = 2100;// [J/kg degree celsius] specific heat of the oil\n",
"Cp_a = 1009;// [J/kg degree celsius] specific heat of the air\n",
"m_dot_o = 3;// [kg/s] flow rate of oil\n",
"U = 200;// [W/square meter] overall heat transfer coefficient\n",
"// the basic energy balance is m_dot_o*Cp_o*(To-Toe) = m_dot_a*Cp_a*(T2-T1)\n",
"Toe = To-m_dot_a*Cp_a*(T2-T1)/(m_dot_o*Cp_o);// [degree celsius]\n",
"// we have\n",
"m_dot_h_into_Ch = m_dot_o*Cp_o;// [W/degree celsius]\n",
"m_dot_c_into_Cc = m_dot_a*Cp_a;// [W/degree celsius]\n",
"// so the air is minimum fluid\n",
"C = m_dot_c_into_Cc/m_dot_h_into_Ch;\n",
"// the effectiveness is \n",
"E = (T2-T1)/(To-T1);\n",
"// now we may use either figure 10-16(page no.-546) or the analytical relation from table 10-4(page no.-543) to obtain NTU. \n",
"// for this problem we choose to use the table \n",
"NTU = -(1+C^(2))^(-1/2)*log((2/E-1-C-(1+C^2)^(1/2))/(2/E-1-C+(1+C^2)^(1/2)));\n",
"// now, we calcuate the area as \n",
"A = NTU*m_dot_c_into_Cc/U;// [square meter]\n",
"printf('area required for the heat exchanger is %f square meter',A);\n",
"\n",
"\n",
"\n",
"\n",
""
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 10.14: ammonia_condenser.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"clear;\n",
"clc;\n",
"printf('\t\t\tExample Number 10.14\n\n\n');\n",
"// ammonia condenser  \n",
"// Example 10.14 (page no.-552-553) \n",
"// solution\n",
"\n",
"Ta = 50;// [degree celsius] temperature of entering ammonia vapour\n",
"Tw1 = 20;// [degree celsius] temperature of entering water\n",
"q = 200;// [kW] total heat transfer required\n",
"U = 1;// [kW/square meter degree celsius] overall heat transfer coefficient\n",
"Tw2 = 40;// [degree celsius] temperature of exiting water\n",
"Cw = 4.18;// [kJ/kg degree celsius] specific heat of water\n",
"// the mass flow can be calculated from the heat transfer with\n",
"m_dot_w = q/(Cw*(Tw2-Tw1));// [kg/s]\n",
"// because this is the condenser the water is the minimum fluid and \n",
"C_min = m_dot_w*Cw;// [kW/degree celsius]\n",
"// the value of NTU is obtained from the last entry of table 10-4(page no.-543), with\n",
"E = 0.6;// effectiveness\n",
"NTU = -log(1-E);\n",
"// so that area is calculated as \n",
"A = C_min*NTU/U;// [square meter]\n",
"// when the flow rate is reduced in half the new value of NTU is \n",
"NTU1 = U*A/(C_min/2);\n",
"// and the effectiveness is computed from the last entry of table 10-3(page no.-543):\n",
"E1 = 1-exp(-NTU1);\n",
"// the new water temperature difference is computed as \n",
"dT_w = E1*(Ta-Tw1);// [degree celsius]\n",
"// so that the heat transfer is \n",
"q1 = C_min*dT_w/2;// [kW]\n",
"printf('the area to achieve a heat exchanger effectiveness of 60%% with an exit water temperature of 40 degree celsius is %f square meter',A);\n",
"printf('\n\n by reducing the flow rate we have lowered the heat transfer by %d percent',(q-q1)*100/q);\n",
"\n",
"\n",
""
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 10.15: crossflow_exchanger_as_energy_conservation_device.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"clear;\n",
"clc;\n",
"printf('\t\t\tExample Number 10.15\n\n\n');\n",
"// crossflow exchanger as energy conservation device  \n",
"// Example 10.15 (page no.-553-555) \n",
"// solution\n",
"\n",
"q = 210000;// [W] heat to be removed from atmospheric air\n",
"m_dot_h = 1200/60;// [kg/s] hot air flow rate\n",
"m_dot_c = m_dot_h;// [kg/s] cold air flow rate\n",
"Ta1 = 25;// [degree celsius] atmospheric air temperature \n",
"Ta2 = 0;// [degree celsius] temperature of air entering from out-door conditions \n",
"U = 30;// [W/m degree celsius] overall heat transfer coefficient\n",
"Cp = 1005;// [J/kg degree celsius] specific heat of air\n",
"\n",
"//*************calculation 1. the design value for the area of the heat exchanger **************//\n",
"\n",
"// the hot and cold fluids have the same flow rate \n",
"// and \n",
"Ch = m_dot_h*Cp;// [W/degree celsius]\n",
"Cc = m_dot_c*Cp;// [W/degree cslsius]\n",
"Cmin_by_Cmax = 1;// for use in table 10-3(page no.-543)\n",
"// the energy balance gives q = Ch*dT_h = Cc*dT_c\n",
"// and \n",
"dT_h = q/Ch;// [degree celsius]\n",
"dT_c = q/Cc;// [degree celsius]\n",
"// the heat exchanger effectiveness is \n",
"E = dT_h/(Ta1-Ta2);\n",
"// consulting table 10-3(page no.-543) for a cross flow exchanger with both fluids unmixed, and inserting the value \n",
"C = 1;\n",
"// we have \n",
"deff('[y] = f(N)','y = E-1+exp(N^(0.22)*(exp(-N^(0.78))-1))');\n",
"N = fsolve(1,f);\n",
"// solving above to get the value of NTU\n",
"// area is \n",
"A = N*Ch/U;// [square meter]\n",
"printf('the design value for the area of heat exchanger is %f square meter',A);\n",
"\n",
"//*************calculation 2. the percent reduction in heat transfer rate if the flow rate is reduced by 50% while keeping the inlet temperatures and   value of U constant ******************//\n",
"\n",
"// we now examine the effect of reducing the flow rate by half, while keeping the inlet temperatures and value of U the same. \n",
"// note that the flow rate of both fluids is reduced because they are physically the same fluid. this means that the value of Cmin_by_Cmax will remain the same at a value of 1.0.\n",
"// the new value of Cmin is \n",
"Cmin = Cc/2;// [W/degree celsius] \n",
"// so that NTU is \n",
"N = U*A/Cmin;\n",
"// equation (b) may be used for the calculation of effectiveness \n",
"E = 1-exp(N^(0.22)*(exp(-N^(0.78))-1));\n",
"// the temperature difference for each fluid is then \n",
"dT = E*(Ta1-Ta2);// [degree celsius]\n",
"// the resulting heat transfer is then \n",
"q_dot = m_dot_c*Cp*dT/2;// [W]\n",
"printf('\n\nthe percent reduction in heat transfer rate if the flow rate is reduced by 50%% is %f ',(q-q_dot)*100/q);\n",
"\n",
"//*************calculation 3. the percent reduction in heat transfer rate if the flow rate is reduced by 50% and the value of U varies as mass flow to   the 0.8 power, with the same inlet temperature conditions\n",
"\n",
"// finally, we examine the effect of reducing the flow rate by 50 percent coupled with reduction in overall heat-transfer coefficient under the assumption that U varies as m_dot^(0.8) or, correspondingly, as Cmin^(0.8)\n",
"// still keeping the area constant, we would find that NTU varies as N = U*A/Cmin ~ C^(0.8)*C^(-1) = C^(-0.2)\n",
"// our new value of N under these conditions would be \n",
"N1 = 0.8*(Cmin/Cc)^(-0.2);\n",
"// inserting this value in equation (b) above for the effectiveness \n",
"E1 = 1-exp(N1^(0.22)*(exp(-N1^(0.78))-1));\n",
"// the corresponding temperature difference in each fluid is \n",
"dT = E1*(Ta1-Ta2);// [degree celsius]\n",
"// the heat transfer is calculated as \n",
"q1 = Cmin*dT;// [W]\n",
"printf('\n\n the percent reduction in heat transfer is %f ',(q-q1)*100/q);"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 10.16: heat_transfer_coefficient_in_compact_exchanger.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"clear;\n",
"clc;\n",
"printf('\t\t\tExample Number 10.16\n\n\n');\n",
"// heat-transfer coefficient in compact exchanger   \n",
"// Example 10.16 (page no.-556-557) \n",
"// solution\n",
"\n",
"p = 101325;// [Pa] pressure of air\n",
"T = 300;// [K] temperature of entering air\n",
"u = 15;// [m/s] velocity of air\n",
"// we obtain the air properties from table A-5(page no.-607) \n",
"rho = 1.1774;// [kg/cubic meter] density of air\n",
"Cp = 1005.7;// [J/kg degree celsius] specific heat of air\n",
"mu = 1.983*10^(-5);// [kg/m s] viscosity of air\n",
"Pr = 0.708;// prandtl number\n",
"// from figure 10-19(page no.-557) we have\n",
"Ac_by_A = 0.697;\n",
"sigma = Ac_by_A;\n",
"Dh = 3.597*10^(-3);// [m] \n",
"// the mass velocity is thus \n",
"G = rho*u/sigma;// [kg/square meter s]\n",
"// and the reynolds number is \n",
"Re = Dh*G/mu;\n",
"// from figure 10-19(page no.-557) we can read\n",
"St_into_Pr_exp_2_by_3 = 0.0036;\n",
"// and the heat transfer coefficient is \n",
"h = St_into_Pr_exp_2_by_3*G*Cp*(Pr)^(-2/3);// [W/square meter degree celsius]\n",
"printf('heat-transfer coefficient is %f W/square meter degree celsius',h);\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
""
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 10.17: transient_response_of_thermal_energy_storage_system.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"clear;\n",
"clc;\n",
"printf('\t\t\tExample Number 10.17\n\n\n');\n",
"// transient response of thermal-energy storage system\n",
"// Example 10.17 (page no.-559-562)\n",
"// solution\n",
"\n",
"Rinf = 0.176;// [degree celsius square meter/W] overall R value of material\n",
"A = 2.25;// [square meter] inlet flow area\n",
"l = 3;// [m] rock bed length\n",
"// properties of the rock are:\n",
"rho_r = 1281.4;// [kg/cubic meter]\n",
"Cr = 0.880;// [kJ/kg degree celsius]\n",
"kr = 0.87;// [W/m degree celsius]\n",
"Ti = 5;// [degree celsius] initial temperature of rock bed\n",
"Ta = 40;// [degree celsius] air temperature \n",
"Tinf = Ta;// [degree celsius]\n",
"p = 101.325;// [kPa] pressure of air\n",
"Ts = 5;// [degree celsius] surrounding temperature\n",
"v1 = 0.3;// [m/s] inlet velocity 1\n",
"v2 = 0.9;// [m/s] inlet velocity 2\n",
"Cpa = 1.004;// [kJ/kg degree celsius]\n",
"R = 0.287;// [kJ/kg K] universal gas constant\n",
"// it can be seen that the axial energy conduction is small compared to the mass energy transport.\n",
"// for a 35 degree celsius temperature difference over a 0.6 length \n",
"dx = l/5;// [m]\n",
"q_cond = kr*A*(Ta-Ti)/dx;// [W]                                                (a)\n",
"// the density of air at 40 degree celsius\n",
"rho_a = p/(R*(Ta+273));// [kg/cubic meter]                                     (b)\n",
"// and the mass flow rate at 0.3 m/s is \n",
"mdot_a = rho_a*A*v1;// [kg/s]                                                    (c)\n",
"// the corresponding energy transport for a temperature difference of 35 degree celsius is \n",
"q = mdot_a*Cpa*(Ta-Ti);// [kW]                                                   (d)\n",
"// and this is much larger than the value in equation (a).\n",
"// we now write an energy balance for one of the axial nodes as \n",
"// energy transported in - energy transported out - energy lost to surroundings = rate of energy accumulation of node\n",
"// or mdot_a*Cpa*(Tm_o^(t)-Tm^(t)) - (Tm^(t)-Tinf)*P*dx/Rinf = rho_r*Cr*dVr*(Tm^(t+1)-Tm^(t))/dt                      (e)\n",
"// where the exit temperature from node m is assumed to be the rock temperatre of that node(Tm^(t)). equation (e) may be solved to give \n",
"// Tm^(t+1) = F*mdot_a*Cpa*Tm_o^(t) + [1-F*(mdot_a*Cpa-P*dx/Rinf)]*Tm^(t) + F*P*dx*Tinf/Rinf                      (f)\n",
"// where\n",
"//          F = dt/(rho_r*Cr*dVr)\n",
"// here P is perimeter and dx is the increment.\n",
"P = 4*1.5;// [m]\n",
"// the stability requirement is such that the coefficient on the Tm^(t) terms cannot be negative. using dx = 0.6m, we find that the maximum value of \n",
"dx = 0.6;// [m]\n",
"Fmax = 6.4495*10^(-4);\n",
"// which yields a maximum time increment of \n",
"tmax = 0.54176;// [h]\n",
"// with a velocity of 0.9 m/s the maximum time increment for stability is\n",
"tmax_v2 = 0.1922;// [h]\n",
"// for the calculations we select the following values of dt with the resultant values of F:\n",
"\n",
"// for v1\n",
"dt1 = 0.2;// [h]\n",
"F1 = 2.38095*10^(-4);\n",
"// for v2\n",
"dt2 = 0.1;// [h]\n",
"F2 = 1.190476*10^(-4);\n",
"\n",
"// with the appropriate properties and these values inserted into equation(f) there results\n",
"// for v1\n",
"// Tm^(t+1) = F1*mdot_a*Cpa*Tm_o^(t) + [1-F1*(mdot_a*Cpa+P*dx/Rinf)]*Tm^(t) + F1*P*dx*Tinf/Rinf                    (g)\n",
"// for v2\n",
"// Tm^(t+1) = F2*mdot_a*Cpa*Tm_o^(t) + [1-F2*(mdot_a*Cpa+P*dx/Rinf)]*Tm^(t) + F2*P*dx*Tinf/Rinf                    (h)\n",
"\n",
"// the energy storage relative to 5 degree celsius can then be calculated from \n",
"E_t = 0;\n",
"i = 1;\n",
"T1 = 40;\n",
"T2 = 5;\n",
"T3 = 5;\n",
"T4 = 5;\n",
"T5 = 5;\n",
"    for i = 1:100\n",
"    T2 = (F2*mdot_a*Cpa*1000*T1 + [1-F2*(mdot_a*Cpa*1000-P*dx/Rinf)]*T2 + F2*P*dx*Tinf/Rinf);\n",
"    T3 = (F2*mdot_a*Cpa*1000*T2 + [1-F2*(mdot_a*Cpa*1000-P*dx/Rinf)]*T3 + F2*P*dx*Tinf/Rinf);\n",
"    T4 = (F2*mdot_a*Cpa*1000*T3 + [1-F2*(mdot_a*Cpa*1000-P*dx/Rinf)]*T4 + F2*P*dx*Tinf/Rinf);\n",
"    T5 = (F2*mdot_a*Cpa*1000*T4 + [1-F2*(mdot_a*Cpa*1000-P*dx/Rinf)]*T5 + F2*P*dx*Tinf/Rinf);\n",
"    Temp(i,:) = [T1 T2 T3 T4 T5];\n",
"    E_t = (dt1/F1)*[(T1-5)+(T2-5)+(T3-5)+(T4-5)+(T5-5)];\n",
"    val(i) = i;\n",
"    val1(i) = E_t;\n",
"    end\n",
"\n",
"E_t = 0;\n",
"i = 1;\n",
"T1 = 40;\n",
"T2 = 5;\n",
"T3 = 5;\n",
"T4 = 5;\n",
"T5 = 5;\n",
"    for i = 1:100\n",
"    T2 = (F1*mdot_a*Cpa*1000*T1 + [1-F1*(mdot_a*Cpa*1000-P*dx/Rinf)]*T2 + F1*P*dx*Tinf/Rinf);\n",
"    T3 = (F1*mdot_a*Cpa*1000*T2 + [1-F1*(mdot_a*Cpa*1000-P*dx/Rinf)]*T3 + F1*P*dx*Tinf/Rinf);\n",
"    T4 = (F1*mdot_a*Cpa*1000*T3 + [1-F1*(mdot_a*Cpa*1000-P*dx/Rinf)]*T4 + F1*P*dx*Tinf/Rinf);\n",
"    T5 = (F1*mdot_a*Cpa*1000*T4 + [1-F1*(mdot_a*Cpa*1000-P*dx/Rinf)]*T5 + F1*P*dx*Tinf/Rinf);\n",
"    Temp(i,:) = [T1 T2 T3 T4 T5];\n",
"    E_t = (dt1/F1)*[(T1-5)+(T2-5)+(T3-5)+(T4-5)+(T5-5)];\n",
"    val2(i) = i;\n",
"    val3(i) = E_t;\n",
"    end\n",
"plot(val,val1,val2,val3);\n",
"legend('v = 0.3m/s','v = 0.9m/s');\n",
"xlabel('time(h)');\n",
"ylabel('E(t) kJ ');\n",
"printf('the result of the calculations are shown in the accompanying figure');\n",
""
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 10.18: variable_properties_analysis_of_a_duct_heater.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"clear;\n",
"clc;\n",
"printf('\t\t\tExample Number 10.18\n\n\n');\n",
"// variable-properties analysis of a duct heater\n",
"// Example 10.18 (page no.-562-564)\n",
"// solution\n",
"\n",
"d = 0.3;// [m] diameter of duct\n",
"Tma = 700;// [K] temperature of hot air\n",
"E = 0.6;// emissivity of outside duct surface\n",
"Tinf = 20+273;// [K] room temperature\n",
"// air properties at 700 K\n",
"rho = 0.5030;// [kg/cubic meter] density of air\n",
"mu = 3.332*10^(-5);// [kg/m s] viscosity of air\n",
"k = 0.05230;// [W/m degree celsius] heat transfer coefficient\n",
"Pr = 0.684;// prandtl no. of air\n",
"A = %pi*d^(2)/4;// [square meter] area of duct\n",
"sigma = 5.669*10^(-8);// [W/square meter K^(4)]\n",
"P = %pi*d;// [m]\n",
"Cp = 1083.5;// [J/kg degree celsius]\n",
"// this is a problem where a numerical solution must be employed.\n",
"// we choose a typical section of the duct with length dx and perimeter P as shown inn figure example 10-18A(page no.-562) and make the energy balances.\n",
"// we assume that resistance of the duct wall is negligible. \n",
"// inside the duct the energy balance is \n",
"// mdot_a*Cp*Tma = hi*P*dx*(Tma-Tmw)+mdot_a*Cp*Tm_po_a               (a)\n",
"// where hi is the convection heat transfer coefficient on the inside which may be calculated from(the flow is turbulent)\n",
"// Nu = hi*d/k = 0.023*Re_d^(0.8)*Pr^(0.3)                            (b)\n",
"// with the properties evaluated at the bulk temperature of air(Tma). the energy balance for the heat flow through the wall is\n",
"// qconv_i = qconv_o+qrad_o\n",
"// or, by using convection coefficients and radiation terms per unit area,\n",
"// hi*(Tma-Tmw) = hc*(Tmw-Tinf)+sigma*E*(Tmw^(4)-Tinf^(4))             (c)\n",
"// where the outside convection coefficient can be calculated from the free convection relation \n",
"// hc = 0.27*((Tmw-Tinf)/d)^(1/4)                                      (d)\n",
"// inserting this relation in equation (c) gives\n",
"// hi*(Tma-Tmw) = 0.27*(Tmw-Tinf)^(5/4)/d^(1/4)+sigma*E*(Tmw^(4)-Tinf^(4))              (e)\n",
"// equation (a) may be solved for Tm_po_a to give\n",
"// Tm_po_a = (1-hi*P*dx/(mdot_a*Cp))_m*Tma + (hi*P*dx/(mdot_a*Cp))_m*Tmw                  (f)\n",
"\n",
"// for \n",
"x=180;\n",
"mdot_a = [0.14 0.45 0.68];// [kg/s]\n",
"for i = 1:3\n",
"\n",
"v = mdot_a(i)/(A*rho);// [m/s]\n",
"Re_d = d*v*rho/mu;\n",
"hi = k*0.023*Re_d^(0.8)*Pr^(0.3)/d;// [W/square meter degree celsius]\n",
"\n",
"\n",
"for dx = 1:1:179\n",
"    for Tmw = 295:1:715\n",
"        Z = (hi/dx)*(Tma-Tmw)-0.27*(Tmw-Tinf)^(5/4)/d^(1/4)-sigma*E*(Tmw^(4)-Tinf^(4));\n",
"        if (Z>0 & Z<40) then\n",
"            Tmw_new = Tmw;\n",
"        end\n",
"    end\n",
"    for Tm_po_a = 275:1:715\n",
"        X = Tm_po_a-(1-(hi/dx)*P*dx/(mdot_a(i)*Cp))*Tmw_new + ((hi/dx)*P*dx/(mdot_a(i)*Cp))*Tmw_new;\n",
"        if (X>0 & X<5) then\n",
"            Tm_po_a_new = Tm_po_a;\n",
"        end\n",
"    end\n",
"    q_by_A = (hi/dx)*(Tma-Tmw_new);// [W/square meter]\n",
"    val1(dx,i) = q_by_A;\n",
"    val(dx) = dx;\n",
"    val2(dx,i) = Tmw_new;\n",
"    val3(dx,i) = Tm_po_a_new;\n",
"end\n",
"end\n",
"scf(1);\n",
"plot(val,val1(:,1),val,val1(:,2),val,val1(:,3));\n",
"legend('mdot_a=0.14','mdot_a=0.45','mdot_a=0.68');\n",
"xlabel('Duct Length x,m');\n",
"ylabel('Local Heat Flux q / A,W / m^2');\n",
"xgrid();\n",
"title('Heat Flux');\n",
"\n",
"scf(2);\n",
"plot(val,val2(:,1),val,val2(:,2),val,val2(:,3));\n",
"legend('Tw=0.14','Tw=0.45','Tw=0.68');\n",
"xlabel('Duct Length x,m');\n",
"ylabel('Local Wall Temperature Tw K');\n",
"xgrid();\n",
"title('Temperature Profile');\n",
"\n",
"scf(3);\n",
"plot(val,val3(:,1),val,val3(:,2),val,val3(:,3));\n",
"legend('Ta=0.14','Ta=0.45','Ta=0.68');\n",
"xlabel('Duct Length x,m');\n",
"ylabel('Local Air Temperature Ta K');\n",
"xgrid();\n",
"title('Temperature Profile');\n",
"printf('plots are shown as :');"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 10.1: overall_heat_transfer_coefficient_for_pipe_in_air.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"clear;\n",
"clc;\n",
"printf('\t\t\tExample Number 10.1\n\n\n');\n",
"// overall heat transfer coefficient for pipe in air   \n",
"// Example 10.1 (page no.-520-522) \n",
"// solution\n",
"\n",
"Tw = 98;// [degree celsius] temperature of hot water \n",
"k_p = 54;// [W/m degree celsius] heat transfer coefficient of pipe\n",
"Ta = 20;// [degree celsius] atmospheric air temperature\n",
"u = 0.25;// [m/s] water velocity\n",
"// from appendix A the dimensions of 2-in schedule 40 pipe are \n",
"ID = 0.0525;// [m]\n",
"OD = 0.06033;// [m]\n",
"// the properties of water at 98 degree celsius are \n",
"rho = 960;// [kg/cubic meter] \n",
"mu = 2.82*10^(-4);// [kg/m s]\n",
"k_w = 0.68;// [W/m degree celsius]\n",
"Pr = 1.76;// prandtl number\n",
"// the reynolds number is \n",
"Re = rho*u*ID/mu;\n",
"// and since turbulent flow is encountered, we may use equation(6-4):\n",
"Nu = 0.023*Re^(0.8)*Pr^(0.4);\n",
"hi = Nu*k_w/ID;// [W/square meter degree celsius]\n",
"// for unit length of pipe the thermal resistance of the steel is \n",
"Rs = log(OD/ID)/(2*%pi*k_p);\n",
"// again, on a unit length basis the thermal resistance on the inside is \n",
"Ai = %pi*ID;// [square meter]\n",
"Ri = 1/(hi*Ai);\n",
"Ao = %pi*OD;// [square meter]\n",
"// the thermal resistance for outer surface is as yet unknown but is written, for unit lengths, is  Ro = 1/(ho*Ao)              (a)\n",
"// from table 7-2(page no.-339), for laminar flow, the simplified relation for ho is \n",
"// ho = 1.32*(dT/d)^(1/4) = 1.32*((To-Ta)/OD)^(1/4)                                                                              (b)\n",
"// where To is the unknown outside pipe surface temperature. we designate the inner pipe surface as Ti and the water temperature as Tw; then the energy balance requires \n",
"// (Tw-Ti)/Ri = (Ti-To)/Rs = (To-Ta)/Ro                                                                                          (c)\n",
"// combining equations (a) and (b) gives \n",
"// (To-Ta)/Ro = %pi*OD*1.32*(To-Ta)^(5/4)/OD^(1/4)\n",
"// this relation may be introduced into equation (c) to yield two equations with the two unknowns Ti and To:\n",
"\n",
"// (Tw-Ti)/Ri = (Ti-To)/Rs              (1)\n",
"// (Ti-To)/Rs = %pi*OD*1.32*(To-Ta)^(5/4)/OD^(1/4)             (2)\n",
"// this is a non-linear equation which can be solved as \n",
"for Ti = 50:0.001:100\n",
"    Q = ((Ti-(Ti-(Tw-Ti)*(Rs/Ri)))/Rs)-(%pi*OD*1.32*((Ti-(Tw-Ti)*(Rs/Ri))-Ta)^(5/4)/OD^(1/4));\n",
"    if Q>0 & Q<6 then\n",
"        Tinew = Ti;\n",
"    else\n",
"        Ti = Ti;\n",
"    end\n",
"end\n",
"Ti = Tinew;// [degree celsius]\n",
"To = (Ti-(Tw-Ti)*(Rs/Ri));// [Degree celsius]\n",
"// as a result, the outside heat transfer coefficient and thermal resistance are\n",
"ho = 1.32*((To-Ta)/OD)^(1/4);// [W/square meter degree celsius]\n",
"Ro = 1/(OD*7.91*%pi);// \n",
"// the overall heat transfer coefficient based on the outer area is written in terms of these resistances as \n",
"Uo = 1/(Ao*(Ri+Ro+Rs));// [W/area degree celsius]\n",
"// in this calculation we used the outside area for 1.0 m length as  Ao\n",
"// so \n",
"Uo = Uo;// [W/square meter degree celsius]\n",
"printf('overall heat transfer coefficient is %f W/square meter degree celsius',Uo);\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
""
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 10.2: overall_heat_transfer_coefficient_for_pipe_exposed_to_steam.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"clear;\n",
"clc;\n",
"printf('\t\t\tExample Number 10.2\n\n\n');\n",
"// overall heat transfer coefficient for pipe exposed to steam\n",
"// Example 10.2 (page no.-523-524) \n",
"// solution\n",
"\n",
"p = 101325;// [Pa] pressure of steam\n",
"Tg = 100;// [degree celsius] temperature of steam\n",
"// we have already determined the inside convection heat-transfer coefficient in example(10.1) as \n",
"hi = 1961;// [W/square meter]\n",
"// the water film properties are \n",
"rho = 960;// [kg/cubic meter] density\n",
"mu_f = 2.82*10^(-4);// [kg/m s]\n",
"kf = 0.68;// [W/m degree celsius]\n",
"hfg = 2255*10^(3);// [J/kg]\n",
"g = 9.8;// [m/s^(2)] acceleration due to gravity\n",
"d = 0.06033;// [m] diameter of the pipe\n",
"// the convection coefficient for condensation on the outside of the pipe is obtained by using equation(9-12)\n",
"// h_o = 0.725*[(rho^(2)*g*hfg*kf^(3))/(mu_f*d*(Tg-To))]^(1/4)                            (a)\n",
"Ao = %pi*d;// [square meter] outside area\n",
"// outside thermal resistance per unit length is \n",
"// R_o = 1/(h_o*A_o)                                                                         (b)\n",
"// the energy balance requires \n",
"// [Tg-To]/R_o = [To-Ti]/R_s = [Ti-Tw]/R_i                                                  (c)\n",
"// from example 10.1 we have\n",
"Ri = 3.092*10^(-3);\n",
"Rs = 4.097*10^(-4);\n",
"Tw = 98;// [degree celsius]\n",
"// equation (b) and (c) may be combined to give \n",
"// (Tg-To)^(3/4)/3403 = (To-Ti)/Rs              (1)\n",
"// (To-Ti)/Rs = (Ti-Tw)/Ri             (2)\n",
"// this is a non-linear equation which can be solved as\n",
"for Ti = 98.1:0.01:99.75\n",
"    P = ((Tg-(Ti+Rs*(Ti-Tw)/Ri))^(3/4))*3403-(((Ti+Rs*(Ti-Tw)/Ri)-Ti)/Rs);\n",
"    if P>(-10) & P<0 then\n",
"        Tinew = Ti;\n",
"    else\n",
"        Ti = Ti;\n",
"    end\n",
"    \n",
"end\n",
"Ti = Tinew;// [degree celsius]\n",
"To = (Ti+Rs*(Ti-Tw)/Ri);// [degree celsius]\n",
"// the exterior heat-transfer coefficient and thermal resistance then become\n",
"ho = 0.725*[(rho^(2)*g*hfg*kf^(3))/(mu_f*d*(Tg-To))]^(1/4);// [W/square meter degree celsius]\n",
"Ro = 1/(ho*Ao);\n",
"// based on unit length of pipe, the overall heat transfer coefficient is \n",
"Uo = 1/(Ao*(Ri+Ro+Rs));// [W/area degree celsius]\n",
"// since Ao and the R's were per unit length\n",
"// so \n",
"Uo = Uo;// [W/square meter degree celsius]\n",
"printf('overall heat transfer coefficient is %f W/square meter degree celsius',Uo);\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
""
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 10.3: influence_of_fouling_factor.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"clear;\n",
"clc;\n",
"printf('\t\t\tExample Number 10.3\n\n\n');\n",
"// influence of fouling factor\n",
"// Example 10.2 (page no.-524-525) \n",
"// solution\n",
"\n",
"// the fouling factor influences the heat transfer coefficient on the inside of the pipe. we have\n",
"Rf = 0.0002;\n",
"// using \n",
"h_clean = 1961;// [W/square meter degree celsius]\n",
"// we obtain \n",
"hi = 1/[Rf+(1/h_clean)];// [W/square meter degree celsius]\n",
"printf('the percent reduction because of fouling factor is %f ',(h_clean-hi)*100/h_clean);"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 10.4: calculation_of_heat_exchanger_size_from_known_temperatures.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"clear;\n",
"clc;\n",
"printf('\t\t\tExample Number 10.4\n\n\n');\n",
"// calculation of heat exchanger size from known temperatures\n",
"// Example 10.4 (page no.-532-533) \n",
"// solution\n",
"\n",
"m_dot = 68;// [kg/min] water flow rate \n",
"U = 320;// [W/square meter degree celsius] overall heat transfer coefficient\n",
"T1 = 35;// [degree celsius] initial temperature \n",
"T2 = 75;// [degree celsius] final temperature\n",
"Toe = 110;// [degree celsius] oil entering temperature \n",
"Tol = 75;// [degree celsius] oil leaving temperature\n",
"Cw = 4180;// [J/kg degree celsius] water specific heat capacity\n",
"// the total heat transfer is determined from the energy absorbed by the water:\n",
"q = m_dot*Cw*(T2-T1);// [J/min]\n",
"q = q/60;// [W]\n",
"// since all the fluid temperatures are known, the LMTD can be calculated by using the temperature scheme in figure 10-7b(page no.-530)\n",
"dT_m = ((Toe-Tol)-(T2-T1))/log((Toe-Tol)/(T2-T1));// [degree celsius]\n",
"// then, since  q = U*A*dT_m\n",
"A = q/(U*dT_m);// [square meter] area of heat-exchanger\n",
"printf('area of heat-exchanger is %f square meter ',A);"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 10.5: shell_and_tube_heat_exchanger.sce"
		   ]
		  },
  {
"cell_type": "code",
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"clear;\n",
"clc;\n",
"printf('\t\t\tExample Number 10.5\n\n\n');\n",
"// shell-and-tube heat exchanger\n",
"// Example 10.5 (page no.-533-534) \n",
"// solution\n",
"\n",
"// to solve this problem, we determine a correction factor from figure 10-8 to be used with the LMTD calculated on the basis of counterflow exchanger.\n",
"// the parameters according to the nomenclature of figure 10-8(page no.-532) are \n",
"T1 = 35;// [degree celsius]\n",
"T2 = 75;// [degree celsius]\n",
"t1 = 110;// [degree celsius]\n",
"t2 = 75;// [degree celsius]\n",
"P = (t2-t1)/(T1-t1);\n",
"R = (T1-T2)/(t2-t1);\n",
"// so the correction factor is \n",
"F = 0.81;// from figure 10-10(page no.-534)\n",
"// and the heat transfer is q = U*A*F*dT_m\n",
"// so that. from example 10-4 we have \n",
"U = 320;// [W/square meter degree celsius] overall heat transfer coefficient\n",
"q = 189493.33;// [W]\n",
"dT_m = 37.44;// [degree celsius]\n",
"A = q/(U*F*dT_m);// [square meter]\n",
"printf('area required for this exchanger is %f square meter',A)\n",
""
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 10.6: design_of_shell_and_tube_heat_exchanger.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"clear;\n",
"clc;\n",
"printf('\t\t\tExample Number 10.6\n\n\n');\n",
"// design of shell-and-tube heat exchanger\n",
"// Example 10.5 (page no.-534-536) \n",
"// solution\n",
"\n",
"m_dot_c = 3.8;// [kg/s] water flow rate\n",
"Ti = 38;// [degree celsius] initial temperature of water\n",
"Tf = 55;// [degree celsius] final temperature of water\n",
"m_dot_h = 1.9;// [kg/s] water flow rate entering the exchanger\n",
"Te = 93;// [degree celsius] entering water temperature\n",
"U = 1419;// [W/square meter degree celsius] overall heat transfer coefficient\n",
"d = 0.019;// [m] diameter of tube\n",
"v_avg = 0.366;// [m/s] average water velocity in exchanger\n",
"Cc = 4180;// [] specific heat of water\n",
"Ch = Cc;// [] specific heat \n",
"rho = 1000;// [kg/cubic meter] density of water\n",
"// we first assume one tube pass and check to see if it satisfies the conditions of this problem. the exit temperature of the hot water is calculated from\n",
"dTh = m_dot_c*Cc*(Tf-Ti)/(m_dot_h*Ch);// [degree celsius]\n",
"Th_exit = Te-dTh;// [degree celsius]\n",
"// the total required heat transfer is obtained for the cold fluid is \n",
"q = m_dot_c*Cc*(Tf-Ti);// [W]\n",
"// for a counterflow exchanger, with the required temperature \n",
"LMTD = ((Te-Tf)-(Th_exit-Ti))/log((Te-Tf)/(Th_exit-Ti));// [degree celsius]\n",
"dTm = LMTD;// [degree celsius]\n",
"A = q/(U*dTm);// [square meter]\n",
"// using the average water velocity in the tubes and the flow rate, we calculate the total area with\n",
"A1 = m_dot_c/(rho*v_avg);// [square meter]\n",
"// this area is the product of number of tubes and the flow area per tube:\n",
"n = A1*4/(%pi*d^(2));// no. of tubes\n",
"n = ceil(n);// rounding of value of n because no. of pipe is an integer value\n",
"// the surface area per tube per meter of length is \n",
"S = %pi*d;// [square meter/tube meter]\n",
"// we recall that the total surface area required for a one tube pass exchanger was calculated above .\n",
"// we may thus compute the length of tube for this type of exchanger from \n",
"L = A/(S*n);// [m]\n",
"// this length is greater than the allowable 2.438 m, so we must use more than one tube pass. when we increase the number of passes, we correspondingly increase the total surface area required because of the reduction in LMTD caused by the correction factor F.\n",
"// we next try two tube passes. from figure 10-8(page no.-532) \n",
"F = 0.88;\n",
"A_total = q/(U*F*dTm);// [square meter]\n",
"// the number of tubes per pass is still 37 because of the velocity requirement. for the two pass exchanger the total surface area is now related to the length by\n",
"L1 = A_total/(2*S*n);// [m]\n",
"// this length is within the 2.438 m requirement, so the final design choice is \n",
"printf('number of tubes per pass = %f',n);\n",
"printf('\n\n number of passes = 2');\n",
"printf('\n\n length of tube per pass = %f m',L1);\n",
""
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 10.7: cross_flow_exchanger_with_one_fluid_mixed.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"clear;\n",
"clc;\n",
"printf('\t\t\tExample Number 10.7\n\n\n');\n",
"// cross flow exchanger with one fluid mixed \n",
"// Example 10.7 (page no.-537) \n",
"// solution\n",
"\n",
"m_dot = 5.2;// [kg/s] mass flow rate\n",
"T1 = 130;// [degree celsius] temperature of entering steam\n",
"T2 = 110;// [degree celsius] temperature of leaving steam\n",
"t1 = 15;// [degree celsius] temperature of entering oil\n",
"t2 = 85;// [degree celsius] temperature of leaving oil\n",
"c_oil = 1900;// [J/kg degree celsius] heat capacity of oil\n",
"c_steam = 1860;// [J/kg degree celsius] heat capacity of steam\n",
"U = 275;// [W/square meter degree celsius] overall heat transfer coefficient\n",
"//the total heat transfer may be obtained from an energy balance on the steam \n",
"q = m_dot*c_steam*(T1-T2);// [W]\n",
"// we can solve for the area from equation (10-13). the value of dT_m is calculated as if the exchanger were counterflow double pipe,thus\n",
"dT_m = ((T1-t2)-(T2-t1))/log((T1-t2)/(T2-t1));// [degree celsius]\n",
"// t1,t2 is representing the unmixed fluid(oil) and T1,T2 is representing the mixed fluid(steam) so that:\n",
"// we calculate \n",
"R = (T1-T2)/(t2-t1);\n",
"P = (t2-t1)/(T1-t1);\n",
"// consulting figure 10-11(page no.-534) we find \n",
"F = 0.97;\n",
"// so the area is calculated from \n",
"A = q/(U*F*dT_m);// [square meter]\n",
"printf('surface area of heat exchanger is %f square meter',A);"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 10.8: effects_of_off_design_flow_rates_for_exchanger_in_previous_example.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
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"source": [
"clear;\n",
"clc;\n",
"printf('\t\t\tExample Number 10.8\n\n\n');\n",
"// effects of off-design flow rates for exchanger in example 10-7 \n",
"// Example 10.8 (page no.-537-538) \n",
"// solution\n",
"\n",
"// we did not calculate the oil flow in example 10-7 but can do so now from \n",
"q = 193;// [kW]\n",
"c_oil = 1.9;// [J/kg degree celsius] heat capacity of oil\n",
"t1 = 15;// [degree celsius] temperature of entering oil\n",
"t2 = 85;// [degree celsius] temperature of leaving oil\n",
"m_dot_o = q/(c_oil*(t2-t1));// [kg/s]\n",
"// the new flow rate will be half this value \n",
"m_dot_o = m_dot_o/2;// [kg/s]\n",
"// we are assuming the inlet temperatures remain the same at 130 degree celsius for the steam and 15 degree celsius for the oil.\n",
"// the new relation for the heat transfer is q = m_dot_o*c_oil*(Teo-15) = m_dot_s*cp*(130-Tes)                       (a)\n",
"// but the exit temperatures, Teo and Tes are unknown. furthermore, dT_m is unknown without these temperatures, as are the values of R and P from figure  10-11(page no.-535). this means we must use an iterative procedure to solve for the exit temperatures using equation (a) and   q = U*A*F*dT_m          (b)\n",
"// the general procedure is to assume values of the exit temperatures until the q's agree between equations(a) and (b).\n",
"printf('the objective of this example is to show that an iterative procedure is required when the inlet and outlet temperatures are not known or easily calculated');\n",
"printf('\n\n there is no need to go through this iteration because it can be avoided by using the techniques described in section 10-6');\n",
""
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 10.9: off_design_calculation_using_E_NTU_method.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
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"clear;\n",
"clc;\n",
"printf('\t\t\tExample Number 10.9\n\n\n');\n",
"// off-design calculation using E-NTU method  \n",
"// Example 10.9 (page no.-542-544) \n",
"// solution\n",
"\n",
"m_dot_o = 0.725;// [kg/s] oil flow rate\n",
"m_dot_s = 5.2;// [kg/s] steam flow rate\n",
"t1 = 15;// [degree celsius] temperature of entering oil\n",
"T1 = 130;// [degree celsius] temperature of entering steam\n",
"c_oil = 1900;// [J/kg degree celsius] heat capacity of oil\n",
"c_steam = 1860;// [J/kg degree celsius] heat capacity of steam\n",
"// for the steam \n",
"Cs = m_dot_s*c_steam;// [W/degree celsius]\n",
"// for the oil\n",
"Co = m_dot_o*c_oil;// [W/degree celsius]\n",
"// so the oil is minium fluid. we thus have\n",
"C_min_by_C_max = Co/Cs;\n",
"U = 275;// [W/square meter degree celsius] overall heat transfer coefficient\n",
"A = 10.83;// [square meter] surface area of heat exchanger\n",
"NTU = U*A/Co;\n",
"// we choose to use the table and note that Co(minimum) is unmixed and Cs(maximum) is mixed so that the first relation in the table 10-3(page no.-543)  applies.\n",
"// we therfore calculate E(effectiveness) as \n",
"E = (1/C_min_by_C_max)*{1-exp(-C_min_by_C_max*(1-exp(-NTU)))};\n",
"// if we were using figure 10-14(page no.-544) we would have to evaluate \n",
"C_mixed_by_C_unmixed = Cs/Co;\n",
"// and would still determine \n",
"E = 0.8;// approximately\n",
"// now, using the effectiveness we can determine the temperature difference of the minimum fluid(oil as)\n",
"dT_o = E*(T1-t1);// [degree celsius]\n",
"// so that heat transfer is \n",
"q = m_dot_o*c_oil*(dT_o);// [W]\n",
"q_initial = 193440;// [W] heat transfer when oil flow rate is 100 %\n",
"printf('we find a reduction in the oil flow rate of 50 %% causes a reduction in heat transfer of only %f %%',(q_initial-q)*100/q_initial);\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
""
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