{
"cells": [
 {
		   "cell_type": "markdown",
	   "metadata": {},
	   "source": [
       "# Chapter 8: Thermodynamic Cycles"
	   ]
	},
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 8.10: Determination_of_temperature_of_air.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"clear;\n",
"clc;\n",
"funcprot(0);\n",
"\n",
"//Example - 8.10\n",
"//Page number - 307\n",
"printf('Example - 8.10 and Page number - 307\n\n');\n",
"\n",
"//Given\n",
"P_high = 40;//[bar]\n",
"P_low = 5;//[bar]\n",
"m_1 = 0.5;//[kg/s] - Rate of mass moving through the expander\n",
"m_2 = 0.1;//[kg/s] - Rate of mass of vapour mixing with air\n",
"e = 0.7;//Efficiency\n",
"\n",
"//At state 3,(40 bar and 200 K),enthalpy and entropy is given by\n",
"H_3 = 179.7;//[kJ/kg]\n",
"S_3 = 5.330;//[kJ/kg-K]\n",
"\n",
"//If isentropic conditions exits in the turbine then state 11 is at 5 bar\n",
"S_11 = 5.330;//[kJ/kg-K]\n",
"//From given compressed air tables at 5 bar and 120 K,\n",
"H_11_1 = 113.6;//[kJ/kg]\n",
"S_11_1 = 5.455;//[kJ/kg-K]\n",
"//At 5 bar and 100 K \n",
"H_11_2 = 90.6;//[kJ/kg]\n",
"S_11_2 = 5.246;//[kJ/kg-K]\n",
"//The enthalpy has to be determined when S = S_3\n",
"//Solving by interpolation we get\n",
"H_11_s = ((H_11_1 - H_11_2)*(S_3 - S_11_2))/(S_11_1 - S_11_2) + H_11_2;//[kJ/kg]\n",
"\n",
"//The adiabatic efficiency of tyrbine is given by\n",
"//(H_3 - H_11_a)/(H_3 - H_11_s) = e\n",
"H_11_a = H_3 - e*(H_3 - H_11_s);//[kJ/kg] - Actual enthalpy\n",
"\n",
" //At 5 bar,the saturated enthalpy is given to be\n",
"H_8 = 88.7;//[kJ/kg]\n",
"//From enthalpy balance during mixing we get,\n",
"//0.1*H_8 + 0.5*H_11_a = 0.6*H_9\n",
"H_9 = (m_2*H_8 + m_1*H_11_a)/(m_1 + m_2);//[kJ/kg]\n",
"\n",
"//From given compressed air tables at 5 bar and 140 K,\n",
"H_9_1 = 135.3;//[kJ/kg]\n",
"//At 5 bar and 120 K \n",
"H_9_2 = 113.6;//[kJ/kg]\n",
"//By interpolation we get\n",
"T_9 = ((H_9 - H_11_1)*(140 - 120))/(H_9_1 - H_11_1) + 120;//[K]\n",
"\n",
"printf(' The temperature of air entering the second heat exchanger is %f K\n\n',T_9);"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 8.1: Calculation_of_work_done.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"clear;\n",
"clc;\n",
"//Example - 8.1\n",
"//Page number - 287\n",
"printf('Example - 8.1 and Page number - 287\n\n');\n",
"//Given\n",
"P_1 = 30;//[bar]\n",
"P_2 = 0.04;//[bar]\n",
"//(1).Carnot cycle\n",
"//It has been reported in the book that at 30 bar pressure (saturated) :\n",
"H_liq_1 = 1008.42;//[kJ/kg]\n",
"H_vap_1 = 2804.2;//[kJ/kg]\n",
"S_liq_1 = 2.6457;//[kJ/kg-K]\n",
"S_vap_1 = 6.1869;//[kJ/kh-K]\n",
"//Therefore, H_1 = H_liq_1, H_2 = H_vap_1, S_1 = S_liq_1 and S_2 = S_vap_1\n",
"H_1 = H_liq_1;\n",
"H_2 = H_vap_1;\n",
"S_1 = S_liq_1;\n",
"S_2 = S_vap_1;\n",
"//At 0.04 bar pressure (saturated) :\n",
"H_liq_2 = 121.46;//[kJ/kg]\n",
"H_vap_2 = 2554.4;//[kJ/kg]\n",
"S_liq_2 = 0.4226;//[kJ/kg-K]\n",
"S_vap_2 = 8.4746;//[kJ/kh-K]\n",
"//Dryness fraction at state 3 can be found the fact that S_3 = S_2 \n",
"x_3 = (S_2 - S_liq_2)/(S_vap_2 - S_liq_2);\n",
"H_3 = H_liq_2*(1 - x_3) + x_3*H_vap_2;//[kJ/kg]\n",
"//Dryness fraction at state 4 can be found the fact that S_4 = S_1\n",
"x_4 = (S_1 - S_liq_2)/(S_vap_2 - S_liq_2);\n",
"H_4 = H_liq_2*(1 - x_4) + x_4*H_vap_2;//[kJ/kg]\n",
"//Work done by turbine W_tur = -delta_H = -(H_3 - H_2)\n",
"W_tur = H_2 - H_3;//[kJ/kg]\n",
"//Work supplied by boiler,\n",
"q_H = H_2 - H_1;//[kJ/kg]\n",
"//Work transfer in compressor is given by\n",
"W_com = -(H_1 - H_4);//[kJ/kg]\n",
"//Efficiency can now be calculated as\n",
"//n = (Net work done/Work supplied by boiler)\n",
"n_carnot = (W_tur + W_com)/q_H;\n",
"//Efficiency of the Carnot cycle can also be determined from the formula\n",
"// n = 1 - (T_L/T_H), Where T_L is saturated temperature at 0.04 bar and T_H is saturated temperature at 30 bar\n",
"printf('(1).Carnot cycle\n\n');\n",
"printf('The work done by the turbine is %f kJ/kg\n\n',W_tur);\n",
"printf('The heat transfer in the boiler is %f kJ/kg\n\n',q_H);\n",
"printf('The cycle efficiency is %f\n\n\n',n_carnot);\n",
"//(2).Rankine cycle\n",
"//The enthalpies at state 2 and 3 remain as in the Carnot cycle\n",
"//Saturated liquid enthalpy at 0.04 bar is \n",
"H_4_prime = H_liq_2;\n",
"//Saturated liquid volume at 0.04 bar as reported in the book is\n",
"V_liq = 0.001004;//[m^(3)/kg]\n",
"//Work transfer in pump can be calculated as\n",
"W_pump = -V_liq*(P_1 - P_2)*100;//[kJ/kg]\n",
"//Work transfer around pump gives, W_pump = -delta_H =  -(H_1_prime - H_4_prime);\n",
"H_1_prime = H_4_prime - W_pump;//[kJ/kg]\n",
"//Heat supplied to boiler is\n",
"q_H_prime = H_2 - H_1_prime;//[kJ/kg]\n",
"//Work done by turbine is\n",
"W_tur_prime = H_2 - H_3;//[kJ/kg]\n",
"//Efficiency can now be calculated as\n",
"//n = (Net work done/Heat input)\n",
"n_rankine = (W_tur_prime + W_pump)/q_H_prime;//\n",
"printf('(2).Rankine cycle\n\n');\n",
"printf('The work done by the turbine is %f kJ/kg\n\n',W_tur_prime);\n",
"printf('The heat transfer in the boiler is %f kJ/kg\n\n',q_H_prime);\n",
"printf('The cycle efficiency is %f',n_rankine);"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 8.2: Calculation_of_efficiency_of_Rankine_cycle.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"clear;\n",
"clc;\n",
"\n",
"//Example - 8.2\n",
"//Page number - 288\n",
"printf('Example - 8.2 and Page number - 288\n\n');\n",
"\n",
"//Given\n",
"T_max = 700+273.15;//[K] - Maximum temperature.\n",
"P_boiler = 10*10^(6);//[Pa] - Constant pressure in the boiler\n",
"P_condenser = 10*10^(3);//[Pa] - Constant pressure in the condenser\n",
"\n",
"//At state 2 i.e, at 700 C and 10 MPa,it has been reported in the book that from steam table\n",
"S_2 = 7.1687;//[kJ/kg-K] - Entropy\n",
"H_2 = 3870.5;//[kJ/kg] - Enthalpy\n",
"\n",
"//At state 3 i.e, at 700 C and 10 KPa,\n",
"S_3 = S_2;//[kJ/kg-K]- Entropy \n",
"\n",
"//For sturated steam at 10 kPa, it has been reported in the book that from steam table\n",
"S_liq = 0.6493;//[kJ/kg-K]- Entropy of saturated liquid\n",
"S_vap = 8.1502;//[kJ/kg-K] - Enthalpy of saturated liquid\n",
"//Therefore steam is saturated and its dryness factor can be calculated as\n",
"x = (S_2 - S_liq)/(S_vap - S_liq);\n",
"\n",
"//The enthalpy at state 3 is now calculated. For steam at 10 kPa,it has been reported in the book that from steam table\n",
"H_liq = 191.83;//[kJ/kg]\n",
"H_vap = 2584.7;//[kJ/kg]\n",
"//Therefore enthalpy at state 3 is\n",
"H_3 = H_liq*(1-x) + H_vap*x;//[kJ/kg]\n",
"\n",
"//Work done by the turbine \n",
"W_tur = -(H_3 - H_2);//[kJ/kg]\n",
"\n",
"//Now we have to calculate work input to the pump\n",
"//State 4:Saturated liquid at 10 kPa\n",
"//State 4:Compressed liquid at 10 MPa\n",
"//Since volume of liquid does not get affected by pressure we take volume of saturated liquid at 10 kPa,\n",
"V_liq = 0.001010;//[m^(3)/kg]\n",
"\n",
"//Work transfer in the pump is\n",
"W_pump = -V_liq*(P_boiler - P_condenser)*10^(-3);//[kJ/kg]\n",
"\n",
"//Energy balance around pump gives, W_pump = -delta_H =  -(H_1 - H_4)\n",
"H_4 = H_liq;// Enthalpy at state 4 (saturated liquid at 10 kPa)\n",
"H_1 = H_4 - W_pump;//[kJ/kg]\n",
" \n",
"//Heat supplied to boiler is\n",
"q_H = H_2 - H_1;//[kJ/kg]\n",
"\n",
"//Efficiency can now be calculated as\n",
"//n = (Net work done/Heat input)\n",
"n_rankine = (W_tur + W_pump)/q_H;\n",
"\n",
"printf('The efficiency of the Rankine cycle is found to be %f',n_rankine);\n",
"\n",
"//Now let us determine the efficiency of Carnot cycle. The maximun temperature is 700 C and minimum temperature is that of saturated steam at 10 kPa,\n",
"T_min = 45.81 + 273.15;//[K] - From steam table as reported in the book\n",
"n_carnot = 1-(T_min/T_max);\n",
"//Note that the efficiency of Rankine cycle is less than that of carnot cycle.\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
""
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 8.3: Calculatrion_of_COP_of_carnot_refrigerator_and_heat_rejected.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"clear;\n",
"clc;\n",
"\n",
"//Example - 8.3\n",
"//Page number - 291\n",
"printf('Example - 8.3 and Page number - 291\n\n');\n",
"\n",
"//Given\n",
"W = 1.1;//[kW] - Work done per ton of refrigeration \n",
"//1 ton refrigeration = 3.517 kW, therefore\n",
"H = 3.517;//[kW] - Heat absorbed\n",
"T_low = -30 + 273.15;//[K] - Low temperature maintained\n",
"\n",
"//COP can be calculated as\n",
"//COP = (Heat absorbed/Work done)\n",
"COP = H/W;\n",
"\n",
"//For reversed carnot cycle, COP = T_low/(T_high - T_low). Solving this we get\n",
"T_high = (T_low/COP) + T_low;//[K] - Higher temperature\n",
"\n",
"//Heat rejected is\n",
"H_rej = W + H;//[kW];\n",
"\n",
"printf('The COP is %f\n\n',COP);\n",
"printf('The higher temperature of the cycle is %f K\n\n',T_high);\n",
"printf('The heat rejected per ton of refrigeration is %f kW\n\n',H_rej);"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 8.4: Calculation_of_minimum_power_required.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"clear;\n",
"clc;\n",
"\n",
"//Example - 8.4\n",
"//Page number - 292\n",
"printf('Example - 8.4 and Page number - 292\n\n');\n",
"\n",
"//Given\n",
"T_high = 20 + 273.15;//[K] - High temperature\n",
"T_low = 0 + 273.15;//[K] - Low temperature\n",
"Q_H = 10;//[kW] - Heat supplied\n",
"\n",
"//If 'Q_H' is the rate at which heat is taken from surrounding and 'W' is the rate at which work is done,then\n",
"// Q_H = W + Q_L\n",
"//(Q_H/Q_L) = (T_high/T_low)\n",
"//Also for a reversible cycle, (Q_H/Q_L) = 1 + (W/Q_L). Solving we get,\n",
"Q_L = (T_low/T_high)*Q_H;//[kW]\n",
"W = (Q_H - Q_L) ;//[kW]\n",
" \n",
"printf('The minimum power required is %f kW',W);"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 8.5: Determination_of_COP_and_power_required.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"clear;\n",
"clc;\n",
"\n",
"//Example - 8.5\n",
"//Page number - 292\n",
"printf('Example - 8.5 and Page number - 292\n\n');\n",
"\n",
"//Given\n",
"T_high = 40 + 273.15;//[K] - High temperature\n",
"T_low = -20 + 273.15;//[K] - Low temperature\n",
"C = 10;//[tons of refrigeration] - Capacity\n",
"//1 ton refrigeration = 3.517 kW, therefore\n",
"H = C*3.517;//[kW] - Heat absorbed\n",
"\n",
"//For reversed carnot cycle, COP = T_low/(T_high - T_low)\n",
"COP = T_low/(T_high - T_low);\n",
"\n",
"// COP = (Refrigerating effect)/(Work input), therefore power required is given by\n",
"P = (H/COP);//[kW]\n",
"\n",
"printf('The COP is %f\n\n',COP);\n",
"printf('The power required is %f kW',P);"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 8.6: Determination_of_maximum_refrigeration_effect.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"clear;\n",
"clc;\n",
"\n",
"//Example - 8.6\n",
"//Page number - 292\n",
"printf('Example - 8.6 and Page number - 292\n\n');\n",
"\n",
"//Given\n",
"COP = 4;//Coefficient of performance\n",
"P = 10;//[kW] - Work done on the cycle\n",
"\n",
"//For reversed carnot cycle, COP = T_low/(T_high - T_low)\n",
"//ratio = (T_high/T_low),therefore\n",
"ratio = -1/(COP + 1);\n",
"\n",
"// Refrigerating effect = (COP)*Work input, therefore refrigeration is given by\n",
"H = COP*P;//[kW]\n",
"\n",
"//Maximum refrigearation in tons is given by\n",
"H_max = (H/3.517);\n",
"\n",
"printf('The maximum refrigeration value is %f ton',H_max);\n",
""
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 8.7: Determination_of_refrigeration_effect_power_consumed_and_COP_of_refrigerator.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"clear;\n",
"clc;\n",
"\n",
"//Example - 8.7\n",
"//Page number - 292\n",
"printf('Example - 8.7 and Page number - 292\n\n');\n",
"\n",
"//Given\n",
"m = 0.6;//[kg/s] - mass flow rate\n",
"T_low = -20+273.15;//[K] - Temperature at which vapour enters the compressor\n",
"T_high = 30+273.15;//[K] - Temperature at which vapour leaves the condenser\n",
"\n",
"//From saturated refrigeration-12 tables we get,at -20 C\n",
"H_1 = 178.74;//[kJ/kg] - (H_1 = H_vap)\n",
"P_1 = 0.15093;//[MPa] - (P_1 = P_sat)\n",
"P_4 = P_1;\n",
"S_1 = 0.7087;//[kJ/kg-K] - (S_1 = S_vap)\n",
"S_2 = S_1;\n",
"\n",
"//At 30 C\n",
"P_2 = 0.7449;//[MPa] - (P_2 = P_sat)\n",
"P_3 = P_2;\n",
"H_3 = 64.59;//[kJ/kg] - (H_3 = H_liq)\n",
"H_4 = H_3;\n",
"S_3 = 0.24;//[kJ/kg-K] - (S_3 = S_liq)\n",
"\n",
"//It is assumed that presssure drop in the evaporator and condenser are negligible. The heat transfer rate in the evaporator is\n",
"Q_L = m*(H_1 - H_4);\n",
"\n",
"printf('The heat transfer rate in the evaporator is %f kW\n\n',Q_L);\n",
"\n",
"//At state 2 (P = 0.7449 MPa and S = 0.7087 kJ/kg-K) and looking in the superheated tables we have to calculate the enthalpy at state 2\n",
"\n",
"//At P = 0.7 MPa and S = 0.6917 kJ/kg-K,\n",
"H_11 = 200.46;//[kJ/kg]\n",
"\n",
"//At P = 0.7 MPa and S = 0.7153 kJ/kg-K,\n",
"H_12 = 207.73;//[kJ/kg]\n",
"\n",
"//Thus at P = 0.7 MPa and S = 0.7087 kJ/kg-K, enthalpy is given by\n",
"H_13 = ((S_2 -0.6917)/(0.7153 - 0.6917))*(H_12 - H_11) + H_11;//[kJ/kg]\n",
"\n",
"//At P = 0.8 MPa and S = 0.7021 kJ/kg-K,\n",
"H_21 = 206.07;//[kJ/kg]\n",
"\n",
"//At P = 0.8 MPa and S = 0.7253 kJ/kg-K,\n",
"H_22 = 213.45;//[kJ/kg]\n",
"\n",
"//Thus at P = 0.8 MPa and S = 0.7087 kJ/kg-K, enthalpy is given by\n",
"H_23 = ((S_2 -0.7021)/(0.7253 - 0.7021))*(H_22 - H_21) + H_21;//[kJ/kg]\n",
"\n",
"//At P = 0.7449 MPa, S = 0.7087 kJ/kg-K, the enthalpy is\n",
"H_2 = ((0.7449 - 0.7)/(0.8 - 0.7))*(H_23 - H_13) + H_13;//[kJ/kg]\n",
"\n",
"//Power consumed by the compressor is\n",
"W_comp = m*(H_2 - H_1);//[kW]\n",
"\n",
"printf('The power consumed by the compressor is %f kW\n\n',W_comp);\n",
"\n",
"//Heat removed in evaporator/work done on compressor\n",
"COP_R = Q_L/W_comp;\n",
"\n",
"printf('The COP the refrigerator is %f kW\n\n',COP_R);\n",
"\n",
"\n",
"//At -20 C,saturated conditions \n",
"H_liq = 17.82;//[kJ/kg]\n",
"H_vap = 178.74;//[kJ/kg]\n",
"x_4 = (H_4 - H_liq)/(H_vap - H_liq);\n",
"\n",
"printf('The dryness factor of refrigerant after the expansion valve is %f\n\n',x_4);\n",
"\n",
"//The heat transfer rate in the condenser is\n",
"Q_H = m*(H_3 - H_2);//[kW]\n",
"\n",
"printf('The heat transfer rate in the condenser is %f kW\n\n',Q_H);\n",
"\n",
"//If the cycle would have worked as a pump then,\n",
"//COP_HP = (Heat supplied from condenser/Work done on compressor)\n",
"COP_HP = (-Q_H)/W_comp;\n",
"\n",
"printf('The COP if cycle would work as a heat pump is %f kW\n\n',COP_HP);\n",
"\n",
"//If the cycle would have been a reversed Carnot cycle then\n",
"COP_C = T_low/(T_high - T_low);\n",
"\n",
"printf('The COP if cycle would run as reversed Carnot cycle is %f kW\n\n',COP_C);"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 8.8: Calculation_of_amount_of_air.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"clear;\n",
"clc;\n",
"\n",
"//Example - 8.8\n",
"//Page number - 300\n",
"printf('Example - 8.8 and Page number - 300\n\n');\n",
"\n",
"//Given\n",
"//From compressor to the expansion valve the pressure is 200 bar and from expansion valve to the inlet of compressor the pressure is 1 bar.\n",
"//Point 5 is saturated liquid at 1 bar and point 6 is saturated vapour at 1 bar\n",
"\n",
"//It has been reported in the book that at state 1 (310 K, 1 bar)\n",
"H_1 = 310.38;//[kJ/kg]\n",
"//At state 2 (310 K, 200 bar)\n",
"H_2 = 277.7;//[kJ/kg]\n",
"//At state 5 (1 bar,saturated liquid)\n",
"H_5 = -122.6;//[kJ/kg]\n",
"//At state 6 (1 bar,saturated vapur)\n",
"H_6 = 77.8;//[kJ/kg]\n",
"\n",
"//The enthalpy at point 3 is same at point 4 as the expansion is isenthalpic\n",
"\n",
"//The mass condensed is 1 kg and therefore m_1 = m+6 + 1\n",
"\n",
"//Enthalpy balance around heat exchanger\n",
"//m_2*H_2 + m_2*H_6 = m_3*H_3 + m_7*H_7\n",
"\n",
"//Enthalpy balance around separator\n",
"//m_4*H_4 = m_5*H_5 + m_6*H_6\n",
"//It can be seen that m_1 = m_2 = m_3 = m_4\n",
"//and m_6 = m_7 = m_1 - 1\n",
"\n",
"//Substituting the values for enthalpy balance around heat exchanger we get,\n",
"//m_1*H_2 + (m_1 - 1)*(H_6) = m_1*H_3 + (m_1 - 1)*H_1\n",
"//and substituting the values for enthalpy balance around seperator we get\n",
"//m_1*H_3 = (1)*(-122.6) + (m_1 - 1)*77.8\n",
"//H_3 = ((1)*(-122.6) + (m_1 - 1)*77.8)/m_1\n",
"//Substituting the expression for 'H_3' in the above equation and then solving for m_1, we get\n",
"deff('[y]=f(m_1)','y=m_1*H_2+(m_1-1)*(H_6)-m_1*(((1)*(-122.6) + (m_1 - 1)*77.8)/m_1)-(m_1-1)*H_1');\n",
"m_1 = fsolve(4,f);//[kg]\n",
"//Thus to liquify 1 kg of air compression of m_1 kg of air is carried out.\n",
"\n",
"//Now substituting this value of m_1 to get the value of H_3,\n",
"H_3 = ((1)*(-122.6) + (m_1 - 1)*77.8)/m_1;//[kJ/kg]\n",
"\n",
"//From given compressed air table we see at 200 bar and 160 K,\n",
"H_3_1 = 40.2;//[kJ/kg]\n",
"\n",
"//At 200 bar and 180 K,\n",
"H_3_2 = 79.8;//[kJ/kg]\n",
"//By interpolation we get,\n",
"T_3 = ((H_3 - H_3_1)*(180 - 160))/(H_3_2 - H_3_1) + 160;//[K]\n",
"\n",
"printf('Temperature before throttling is %f',T_3);"
   ]
   }
,
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		   "source": [
			"## Example 8.9: Calculation_of_amount_of_air_and_temperature.sce"
		   ]
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"clear;\n",
"clc;\n",
"funcprot(0);\n",
"\n",
"//Example - 8.9\n",
"//Page number - 304\n",
"printf('Example - 8.9 and Page number - 304\n\n');\n",
"\n",
"//Given\n",
"//At 1 bar, 310 K \n",
"H_1 = 310.38;//[kJ/kg]\n",
"//At 200 bar, 310 K\n",
"H_2 = 277.7;//[kJ/kg]\n",
"//At 1 bar, Saturated liquid\n",
"H_7 = -122.6;//[kJ/kg]\n",
"//At 1 bar, Saturated vapour\n",
"H_8 = 77.8;//[kJ/kg]\n",
"//At 200 bar, 200 K\n",
"H_3 = 117.6;//[kJ/kg]\n",
"//At 1 bar, 100 K\n",
"H_11 = 98.3;//[kJ/kg]\n",
"\n",
"//(1)\n",
"//For 1 kg of liquid air obtained,the overall enthalpy balance is\n",
"//m_2*H_2 = W - 122.6 + (m_2 - 1)*H_1\n",
"//W = - 0.8*m_2*(H_11 - H_3)\n",
"//Overall enthalpy balance equation becomes\n",
"//H_2*m_2 = 15.44*m_2 - H_7 + (m_2 - 1)*H_1, solving\n",
"m_2_prime = (H_7 - H_1)/(H_2 - 15.44 - H_1);\n",
"\n",
"printf('The number of kilograms of air compressed per kg of liquid air produced is %f kg\n\n',m_2_prime);\n",
"\n",
"//(2)\n",
"//Enthalpy balance around separator is \n",
"//0.2*m_2*H_5 = -H_7 + (0.2*m_2 - 1)*H_8, solving\n",
"m_2 = m_2_prime;\n",
"H_5_prime = ((0.2*m_2-1)*H_8 - H_7)/(0.2*m_2);\n",
"\n",
"//At point 5, P = 200 bar and enthalpy is\n",
"H_5_1 = -33.53;//[kJ/kg]\n",
"//From compressed air tables at 200 bar and 140 K,\n",
"H_5_2 = 0.2;//[kJ/kg]\n",
"//At 200 bar and 120 K,\n",
"H_5_3 = -38.0;//[kJ/kg]\n",
"//Solving by interpolation we get\n",
"T_5 = ((H_5_1 - H_5_3)*(140 - 120))/(H_5_2 - H_5_3) + 120;//[K]\n",
"\n",
"printf('The temperature of air before throttling is %f K\n\n',T_5);\n",
"\n",
"//(3)\n",
"//During mixing of streams 8 and 11 to produce stream 9, the enthalpy balance is\n",
"// (0.2*m_2 - 1)*H_8 + 0.8*m_2*H_11 = (m_2 - 1)*H_9,Solving for H_9\n",
"\n",
"H_9_prime = ((0.2*m_2-1)*H_8+0.8*m_2*H_11)/(m_2 - 1);\n",
"\n",
"//From given compressed air tables at 1 bar and 100 K,\n",
"H_9_1 = H_11;\n",
"//At 1 bar and 90 K \n",
"H_9_2 = 87.9;//[kJ/kg]\n",
"//Solving by interpolation we get\n",
"T_9 = ((H_9_prime - H_9_2)*(100 - 90))/(H_9_1 - H_9_2) + 90;//[K]\n",
"\n",
"printf('The temperature of stream entering second heat exchanger is %f K\n\n',T_9);\n",
"\n",
"//(4)\n",
"//Enthalpy balance around first heat exchanger is\n",
"//H_2*m_2 + (m_2 - 1)*H_10 = H_3*m-2 + (m-2 - 1)*H_1, solving for H_10\n",
"\n",
"H_10_prime = ((m_2 - 1)*H_1 + H_3*m_2 - H_2*m_2)/(m_2 - 1);\n",
"\n",
"//From given compressed air tables at 1 bar and 140 K,\n",
"H_10_1 = 139.1;//[kJ/kg]\n",
"//At 1 bar and 120 K \n",
"H_10_2 = 118.8;//[kJ/kg]\n",
"//Solving by interpolation we get\n",
"T_10 = ((H_10_prime - H_10_2)*(140 - 120))/(H_10_1 - H_10_2) + 120;//[K]\n",
"\n",
"printf('The temperature of stream exiting second heat exchanger is %f K\n\n',T_10);"
   ]
   }
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