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diff --git a/A_TEXTBOOK_OF_ELECTRICAL_TECHNOLOGY_(VOL-III)_by_B.L.Thareja/chapter47.ipynb b/A_TEXTBOOK_OF_ELECTRICAL_TECHNOLOGY_(VOL-III)_by_B.L.Thareja/chapter47.ipynb new file mode 100644 index 00000000..c29e8c0a --- /dev/null +++ b/A_TEXTBOOK_OF_ELECTRICAL_TECHNOLOGY_(VOL-III)_by_B.L.Thareja/chapter47.ipynb @@ -0,0 +1,737 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# CHAPTER 47 : ELECTRIC HEATING\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## EXAMPLE 47.1 , PAGE NO :- 1841" + ] + }, + { + "cell_type": "code", + "execution_count": 17, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Length of wire = 16.07 m.\n", + "Diameter of wire = 2.72 mm.\n" + ] + } + ], + "source": [ + "'''A resistance oven employing nichrome wire is to be operated from 220 V single-phase supply and is to be rated at 16 kW.\n", + "If the temperature of the element is to be limited to 1,170°C and average temperature of the charge is 500°C, find the\n", + "diameter and length of the element wire.\n", + "Radiating efficiency = 0.57, Emmissivity=0.9, Specific resistance of nichrome=(109e–8)ohm-m.'''\n", + "\n", + "\n", + "P = 16000.0 #W (output power)\n", + "V = 220.0 #V (applied voltage)\n", + "rho = 109.0e-8 #ohm-m (resistivity)\n", + "e = 0.9 # (Emmisivity)\n", + "K = 0.57 # (Radiating efficiency)\n", + "T1 = 1170.0 + 273.0 #K (Temp of hot body)\n", + "T2 = 500.0 + 273.0 #K (Temp of cold body)\n", + "\n", + "#Now , l/d^2 = pi*V^2/4*rho*P = a .Therefore a is\n", + "a = 3.14*(V**2)/(4*rho*P) # (a = l/d^2) ------ 1\n", + " \n", + "#Using Stefan's law of radiation \n", + "H = 5.72*e*K*((T1/100)**4-(T2/100)**4) #W/m^2\n", + "\n", + "#Total heat dissipated = electrical power input\n", + "# (pi*d)*l*H = P . Therefore ,let b = l*d. So,\n", + "b = P/(H*3.14)\n", + "b2 = b**2 #(b2 = l^2*d^2)------------------------ 2\n", + "\n", + "#Multiplying 1 and 2.\n", + "l3 = a*b2 # ( = l^3)\n", + "l = l3**(1/3.0) #m (length)\n", + "d = b/l*1000 #mm (diameter)\n", + "\n", + "print \"Length of wire =\",round(l,2),\"m.\"\n", + "print \"Diameter of wire =\",round(d,2),\"mm.\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## EXAMPLE 47.2 , PAGE NO :- 1841" + ] + }, + { + "cell_type": "code", + "execution_count": 18, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Width of strip = 7.4 mm.\n" + ] + } + ], + "source": [ + "'''A 30-kW, 3-phase, 400-V resistance oven is to employ nickel-chrome strip 0.254 mm thick for the three star-connected heating\n", + "elements.If the wire temperature is to be 1,100°C and that of the charge to be 700°C, estimate a suitable width for the strip.\n", + "Assume emissivity = 0.9 and radiating efficiency to be 0.5 and resistivity of the strip material is 101.6e-8 ohm- m.What would be\n", + "the temperature of the wire if the charge were cold ?'''\n", + "\n", + "import math as m\n", + "\n", + "P = 30.0*1000 *(1/3.0) #W (Power/phase)\n", + "V = 400.0/m.sqrt(3) #V (Phase voltage)\n", + "rho = 101.6e-8 #ohm-m (resistivity)\n", + "e = 0.9 # (emmisivity)\n", + "K = 0.5 # (radiating efficiency)\n", + "t = 0.254e-3 #m (thickness of strip)\n", + "T1 = 1100.0 + 273 #K (Temp. of hot wire)\n", + "T2 = 700.0 + 273 #K (Temp. of charge) \n", + "R = V*V/P #ohm (Resistance => P = V^2/R)\n", + "\n", + "# R = rho*l/(w*t) l/w = R*t/rho = a.Therefore a is\n", + "a = R*t/rho #(=l/w)------------------------------------ 1\n", + "\n", + "#Using stefan's law\n", + "H = 5.72*e*K*((T1/100)**4-(T2/100)**4) #W/m^2\n", + "\n", + "#Surface area of strip = 2*w*l .\n", + "#Total Heat dissipated = electrical power => wl*2H = P . Let b = wl\n", + "b = P/(2*H) #(=wl)----------------------------------------- 2\n", + "\n", + "#Dividing 1 by 2\n", + "w2 = b/a #(=w*w)\n", + "w = m.sqrt(w2)*1000 #mm (width)\n", + "print \"Width of strip =\",round(w,2),\"mm.\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## EXAMPLE 47.3 , PAGE NO :- 1842" + ] + }, + { + "cell_type": "code", + "execution_count": 19, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Loading in kW = 13.51 kW.\n", + "Efficency of Tank = 87.41 %\n" + ] + } + ], + "source": [ + "'''A cubic water tank has surface area of 6.0 m^2 and is filled to 90% capacity six times daily. The water is heated from 20°C \n", + "to 65°C.The losses per square metre of tank surface per 1°C temperature difference are 6.3 W. Find the loading in kW and the\n", + "efficiency of the tank.Assume specific heat of water = 4,200 J/kg/°C and one kWh = 3.6 MJ.'''\n", + "\n", + "import math as m\n", + "sa = 6.0 #m^2 (surface area)\n", + "T2 = 65.0 #*C (final temp)\n", + "T1 = 20.0 #*C (initial temp)\n", + "loss = 6.3 #W/*C/m^2 (loss per square metre per 1*C)\n", + "s = 4200.0 #J/Kg/*C (Specific heat)\n", + "\n", + "#Now, sa = 6*l^2\n", + "l = m.sqrt(sa/6) #m\n", + "\n", + "#Volume = l^3\n", + "V = l**3 #m^3\n", + "\n", + "#Volume of water to be heated daily is\n", + "V2 = 6*V*0.9 #m^3\n", + "\n", + "#Since 1m^3 = 1000 kg => mass of water to be heated is\n", + "mass = V2*1000.0 #kg\n", + "\n", + "#Heat required to raise temp =\n", + "H = mass*s*(T2-T1) #MJ\n", + "H = H/(3.6*10e+5) #kWh\n", + "\n", + "#Daily loss from surface\n", + "L = 6*loss*(T2-T1)*(24.0/1000) #kWh\n", + "\n", + "#Total energy required\n", + "Tot = L + H #kWh \n", + "#(i) Loading in KW is\n", + "load = Tot/24 #kW\n", + "\n", + "#(ii)Efficiency of tank is\n", + "eff = (H/Tot)*100.0 # (% efficency)\n", + "print \"Loading in kW = \",round(load,2),\"kW.\"\n", + "print \"Efficency of Tank =\",round(eff,2),\"%\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## EXAMPLE 47.4 , PAGE NO :- 1844" + ] + }, + { + "cell_type": "code", + "execution_count": 20, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Power factor = 0.9487\n", + "Power drawn from supply = 900.0 kW.\n", + "Time required for melting steel = 40.0 minutes. and 46.0 seconds.\n" + ] + } + ], + "source": [ + "'''A 4-phase electric arc furnace has the following data :\n", + "Current drawn = 5000 A ; Arc voltage = 50 V\n", + "Resistance of transformer referred to secondary = 0.002 ohm\n", + "Resistance of transformer referred to secondary = 0.004 ohm\n", + "(i) Calculate the power factor and kW drawn from the supply.\n", + "(ii) If the overall efficiency of the furnace is 65%, find the time required to melt 2 tonnes of steel if\n", + "latent heat of steel = 8.89 kcal/kg, specific heat of steel = 0.12, melting point of steel = 1370°C and\n", + "initial temperature of steel = 20°C.'''\n", + "\n", + "import math as m\n", + "\n", + "I = 5000.0 #A (current drawn)\n", + "V = 50.0 #V (Arc Voltage) \n", + "Rs = 0.002 #ohm (transformer resistance on secondary)\n", + "Xs = 0.004 #ohm (transformer reactance on secondary)\n", + "T2 = 1370.0 #*C (final temp)\n", + "T1 = 20.0 #*C (initial temp)\n", + "\n", + "# Voltage drop due to resistance =\n", + "Vr = I*Rs #V\n", + "\n", + "# Voltage drop due to reactance =\n", + "Vx = I*Xs #V\n", + "\n", + "#Total Voltage is (Using vector sum)\n", + "V_tot = m.sqrt((V+Vr)**2 + Vx**2) #V\n", + "\n", + "#(i) Supply power factor is\n", + "pf = (V+Vr)/V_tot\n", + "\n", + "#Total Power drawn => P = 3*VI*(power factor)\n", + "P = 3*V_tot*I*pf/1000 #kW\n", + "\n", + "print \"Power factor =\",round(pf,4)\n", + "print \"Power drawn from supply =\",round(P,2),\"kW.\"\n", + "#Energy required to melt 2 tonnes of steel\n", + "m = 2000.0 #kg\n", + "s = 0.12 # (specific heat of steel)\n", + "L = 8.89 #kcal/kg (latent heat of steel)\n", + "\n", + "enrgy = m*s*(T2-T1) + m*L #kcal\n", + "enrgy = enrgy/860.0 #kWh\n", + "\n", + "#Utilised power\n", + "P = 0.65*P #kW\n", + "\n", + "#Time required for melting steel\n", + "Time = enrgy/P #hr\n", + "Time = Time*60 #min\n", + "Sec = (round(Time,2) - round(Time,-1) )*60 #sec\n", + "\n", + "print \"Time required for melting steel =\",round(Time,-1),\"minutes. and \",round(Sec),\"seconds.\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## EXAMPLE 47.5 , PAGE NO :- 1845" + ] + }, + { + "cell_type": "code", + "execution_count": 16, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "#Total KVA taken from supply line = 2145.63 KVA .\n" + ] + } + ], + "source": [ + "'''If a 3-phase arc furnace is to melt 10 tonne steel in 2 hours, estimate the average input to the furnace if overall\n", + "efficiency is 50%. If the current input is 9,000 A with the above kW input and the resistance and reactance of furnace leads\n", + "(including transformer) are 0.003 ohm and 0.005 ohm respectively, estimate the arc voltage and total kVA taken from the supply\n", + "Specific heat of steel = 444 J /kg/°C ,Latent heat of fusion of steel = 37.25 kJ/kg , Melting point of steel = 1,370 °C.'''\n", + "\n", + "from sympy import Symbol,solve,Eq,sqrt\n", + "import math as m\n", + "\n", + "mass = 10000.0 #kg (mass in kg)\n", + "t = 2.0 #hr (time taken to melt)\n", + "eff = 50.0 #% (overall efficiency)\n", + "I = 9000.0 #A (current input)\n", + "Rs = 0.003 #ohm (secondary resistance)\n", + "Xs = 0.005 #ohm (secondary reactance)\n", + "s = 444.0 #J/kg/*C (specific heat of steel)\n", + "L = 37250 #J/kg (latent heat of fusion)\n", + "T2 = 1370.0 #*C (final temp)\n", + "T1 = 20.0 #*C (initial temp)\n", + "\n", + "#Energy required to melt 10 tonnes of steel\n", + "\n", + "enrgy = mass*s*(T2-T1) + mass*L #J\n", + "enrgy = enrgy/(1000*3600) #kWh\n", + "\n", + "#Avg output power = energy/time\n", + "P = enrgy/t #kW\n", + "#Avg input power =\n", + "Pin = P/eff*100\n", + "\n", + "#Voltage drop due to resistance\n", + "Vr = I*Rs #V\n", + "#Voltage drop due to reactance\n", + "Vx = I*Xs #V\n", + "\n", + "#Now,Let Va is arc drop voltage\n", + "Va = Symbol('Va') #V\n", + "Vt = sqrt((Va+Vr)**2 + Vx**2) \n", + "pf = (Va+Vr)/Vt \n", + "#Total power input = 3*(Vt*It*pf)\n", + "\n", + "eq = Eq(Pin*1000,3*Vt*I*pf)\n", + "Va = solve(eq) #V\n", + "\n", + "Va1 = Va[0]\n", + "Vt = sqrt((Va1+Vr)**2 + Vx**2)\n", + "\n", + "#Total KVA taken from supply line =\n", + "power = 3*Vt*I/1000\n", + "\n", + "print \"#Total KVA taken from supply line =\",round(power,2),\"KVA .\"" + ] + }, + { + "cell_type": "markdown", + "metadata": { + "collapsed": true + }, + "source": [ + "## EXAMPLE 47.6 , PAGE NO :- 1850" + ] + }, + { + "cell_type": "code", + "execution_count": 27, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Efficiency of induction furnace = 59.35 %.\n" + ] + } + ], + "source": [ + "'''Determine the efficiency of a high-frequency induction furnace which takes 10 minutes to melt 2 kg of a aluminium initially \n", + "at a temperature of 20°C. The power drawn by the furnace is 5 kW, specific heat of aluminium = 0.212, melting point of \n", + "aluminium = 660° C and latent heat of fusion of aluminium. = 77 kcal/kg.'''\n", + "\n", + "m = 2.0 #kg (mass of alluminium)\n", + "L = 77.0 #kcal/kg (Latent heat of fusion)\n", + "T2 = 660.0 #*C (final temp)\n", + "T1 = 20.0 #*C (initial temp)\n", + "s = 0.212 # (specific heat of alluminium)\n", + "Pin = 5.0 #kW (input power)\n", + "#Heat required to melt alluminium\n", + "H1 = m*L #kcal\n", + "#Heat required to raise the temperature\n", + "H2 = m*s*(T2 - T1) #kcal\n", + "#Total heat\n", + "heat_tot = H1 + H2 #kcal\n", + "#Heat required per hour\n", + "enrgy = heat_tot/(10.0/60) #kcal\n", + "#Power delivered to alluminium\n", + "Power = enrgy/860 #kW\n", + "\n", + "eff = Power/Pin*100 #(% efficiency)\n", + "\n", + "print \"Efficiency of induction furnace = \",round(eff,2),\"%.\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## EXAMPLE 47.7 , PAGE NO :- 1850" + ] + }, + { + "cell_type": "code", + "execution_count": 21, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Power absorbed = 577.0 kW.\n", + "power factor = 0.83\n" + ] + } + ], + "source": [ + "'''A low-frequency induction furnace has a secondary voltage of 20V and takes 600 kW at 0.6 p.f. when the hearth is full. If the\n", + "secondary voltage is kept constant, determine the power absorbed and the p.f. when the hearth is half-full. Assume that the\n", + "resistance of the secondary circuit is doubled but the reactance remains the same.'''\n", + "\n", + "import math as m\n", + "\n", + "V = 20.0 #V (secondary voltage)\n", + "P = 600*1000 #W (Input Power)\n", + "pf = 0.6 # (power factor)\n", + "\n", + "#Inital secondary current using P = VI*pf\n", + "I = P/(V*pf) #A (secondary current)\n", + "\n", + "#Now, pf = cosQ , .'. sinQ = sqrt(1-pf^2)\n", + "Vr = V*pf #V (Voltage across resistance)\n", + "Vx = V*m.sqrt(1-pf**2) #V (Voltage across reactance)\n", + "\n", + "#As, Vr = I*R and Vx = I*X\n", + "\n", + "R = Vr/I #ohm\n", + "X = Vx/I #ohm\n", + "\n", + "#When hearth is half-full\n", + "R2 = 2*R\n", + "X2 = X\n", + "pf = R2/m.sqrt(R2**2 + X2**2) #(new power factor)\n", + "\n", + "Vr = V*pf #V (Voltage across resistance)\n", + "#As, Vr = I*R\n", + "I = Vr/R2 #A\n", + "\n", + "power = V*I*pf/1000 #kW\n", + "\n", + "print \"Power absorbed =\",round(power),\"kW.\"\n", + "print \"power factor =\",round(pf,2)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## EXAMPLE 47.8 , PAGE NO :- 1851" + ] + }, + { + "cell_type": "code", + "execution_count": 22, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Total furnace input = 119.37 kWh .\n" + ] + } + ], + "source": [ + "'''Estimate the energy required to melt 0.5 tonne of brass in a single-phase induction furnace. If the melt is to be carried out\n", + "in 0.5 hour, what must be the average power input to the furnace?\n", + "Specific heat of brass = 0.094\n", + "Latent heat of fusion of brass = 39 kilocal/kg\n", + "Melting point of brass = 920°C\n", + "Furnace efficiency = 60.2%\n", + "The temperature of the cold charge may be taken as 20°C.'''\n", + "\n", + "m = 0.5*1000 #kg (mass of brass)\n", + "s = 0.094 # (specific heat)\n", + "T2 = 920.0 #*C (final temp)\n", + "T1 = 20.0 #*C (initial temp)\n", + "L = 39.0 #kcal/kg(Latent heat of fusion)\n", + "eff = 60.2 # (% efficiency)\n", + "\n", + "#Total amount of heat req to melt 0.5 kg brass\n", + "H = m*L + m*s*(T2-T1) #kcal\n", + "H = H/860 #kWh\n", + "H_tot = H/(eff)*100 #kWh (input energy required)\n", + "\n", + "print \"Total furnace input =\",round(H_tot,2),\"kWh .\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## EXAMPLE 47.9 , PAGE NO :- 1851" + ] + }, + { + "cell_type": "code", + "execution_count": 23, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "height for maximum heat = 0.75 *H.\n" + ] + } + ], + "source": [ + "'''A low-frequency induction furnace whose secondary voltage is maintained constant at 10 V, takes 400 kW at 0.6 p.f. when the\n", + "hearth is full.Assuming the resistance of the secondary circuit to vary inversely as the height of the charge and reactance to \n", + "remain constant,find the height upto which the hearth should be filled to obtain maximum heat.'''\n", + "\n", + "\n", + "import math as m\n", + "V2 = 10.0 #V (secondary voltage)\n", + "P = 400.0*1000 #kW (power)\n", + "pf = 0.6 # (power factor)\n", + "\n", + "\n", + "#Secondary current is (Using P = VI*cosQ)\n", + "I = P/(V2*pf) #A\n", + "\n", + "#Impedance of secondary circuit is\n", + "Z = V2/I #ohm\n", + "#Now, R = Z*cosQ X = Z*sinQ\n", + "R = Z*pf #ohm (resistance)\n", + "X = Z*m.sqrt(1-pf**2) #ohm (reactance)\n", + "\n", + "#Let height of charge be 'x' times of the full hearth h = x*H\n", + "#Resistance varies inersely as height .Therefore,\n", + "# R' = R/x\n", + "\n", + "#Now ,for max heat resistance should be equal to reactance.Therefore,\n", + "x = R/X\n", + "\n", + "print \"height for maximum heat = \",round(x,2),\"*H.\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## EXAMPLE 47.10 , PAGE NO :- 1853" + ] + }, + { + "cell_type": "code", + "execution_count": 24, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Voltage = 798.04 V.\n" + ] + } + ], + "source": [ + "'''A slab of insulating material 150 cm^2 in area and 1 cm thick is to be heated by dielectric heating. The power required is \n", + "400 W at 30 MHz.Material has relative permittivity of 5 and p.f. of 0.05. Determine the necessary voltage. Absolute\n", + "permittivity = 8.854e - 12 F/m.'''\n", + "\n", + "import math as m\n", + "\n", + "P = 400.0 #W (power)\n", + "f = 30.0e+6 #Hz (frequency)\n", + "A = 150.0e-4 #m^2 (area)\n", + "d = 1.0e-2 #m (thickness)\n", + "er = 5.0 # (relative permitivity)\n", + "e0 = 8.89e-12#F/m (absolute permitivity)\n", + "pf = 0.05 # (power factor) \n", + "#Capacitance\n", + "C = (A/d)*(er*e0) #F\n", + "\n", + "#Now, P = (2*pi*f)*(C*V^2)*pf .Therefore V is\n", + "V = m.sqrt(P/(2*3.14*f*C*pf)) #V\n", + "print \"Voltage =\",round(V,2),\"V.\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## EXAMPLE 47.11 , PAGE NO :- 1853" + ] + }, + { + "cell_type": "code", + "execution_count": 25, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Voltage = 846.45 V.\n", + "Current through material = 9.45 A.\n", + "Frequency = 58.49 MHz.\n" + ] + } + ], + "source": [ + "'''An insulating material 2 cm thick and 200 cm^2 in area is to be heated by dielectric heating. The material has relative \n", + "permitivity of 5 and power factor of 0.05.Power required is 400 W and frequency of 40 MHz is to be used. Determine the necessary\n", + "voltage and the current that will flow through the material.If the voltage were to be limited to 700 V, what will be the \n", + "frequency to get the same loss? '''\n", + "\n", + "import math as m\n", + "\n", + "d = 2.0e-2 #m (Thickness)\n", + "A = 200e-4 #m^2 (Area)\n", + "er = 5 # (relative permitivity)\n", + "pf = 0.05 # (power factor)\n", + "f = 40.0e+6 #Hz (frequency)\n", + "P = 400.0 #W (power)\n", + "e0 = 8.89e-12 # (absolute permitivity)\n", + "\n", + "C = (A/d)*(e0*er) #F (Capacitance)\n", + "\n", + "#Now, P = 2*pi*f*C*V^2*pf\n", + "V = m.sqrt(P/(2*3.14*f*C*pf)) #V\n", + "\n", + "#Also, P = VI*cosQ .Therefore,current through material\n", + "I = P/(V*pf) #A\n", + "\n", + "#Heat produced is propotional to V^2*f. (V2/V1)^2 = (f1/f2)\n", + "f2 = f*(V/700)**2 #Hz\n", + "f2 = f2/(10e+5) #MHz \n", + "\n", + "print \"Voltage = \",round(V,2),\"V.\"\n", + "print \"Current through material =\",round(I,2),\"A.\"\n", + "print \"Frequency = \",round(f2,2),\"MHz.\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## EXAMPLE 47.12 , PAGE NO :- 1853" + ] + }, + { + "cell_type": "code", + "execution_count": 26, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Power input = 750.0 W .\n" + ] + } + ], + "source": [ + "'''A plywood board of 0.5*0.25*0.02 metre is to be heated from 25 to 125°C in 10 minutes by dielectric heating employing a\n", + "frequency of 30 MHz. Determine the power required in this heating process. Assume specific heat of wood 1500/J/kg/°C; \n", + "weight of wood 600 kg/m3 and efficiency of process 50%.'''\n", + "\n", + "\n", + "Vol = 0.5*0.25*0.02 #m^3 (Volume of plywood to be heated)\n", + "f = 30.0e+6 #Hz (frequency)\n", + "t = 10.0 #minutes (time)\n", + "T2 = 125.0 #*C (final temperature)\n", + "T1 = 25.0 #*C (initial temperature)\n", + "s = 1500.0 #J/kg/*C (specific heat of wood)\n", + "den = 600.0 #kg/m^3 (weight of wood)\n", + "eff = 50.0 #% (efficiency of process)\n", + "\n", + "wt = den*Vol #kg (weight of plywood)\n", + "#Heat required to raise the temp is\n", + "H = wt*s*(T2-T1) #J\n", + "H = H/3600 #Wh\n", + "\n", + "#As P = H/t .Therfore power required for heating\n", + "P = H/(10.0/60) #W\n", + "\n", + "#As efficiency is 50%\n", + "Pin = P/eff*100 #W\n", + "\n", + "print \"Power input =\",round(Pin,2),\"W .\"" + ] + }, + { + "cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], + "source": [] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.11" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} |