{ "cells": [ { "cell_type": "markdown", "metadata": {}, "source": [ "# Cogeneració" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "**PAU 2007 S3 4B**\n", "![](http://www.jorts.net/PAU/dbase/img/Screenshots_2013-05-01-18-26-30.png)" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "Primer farem un diagrama del flux d'energia, ubicant les dades i les incògnites:\n", "\n", "![](img/diagramaCogeneracio.png)\n", "\n", "Introduim les dades (S.I.):\n", "\n" ] }, { "cell_type": "code", "execution_count": 2, "metadata": {}, "outputs": [], "source": [ "Eelec = 65*10**6*3600\n", "mr = 75*10**3\n", "p = 11.8*10**6\n", "ma = 3*10**6\n", "deltat = 40\n", "ce = 4.18*10**3" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "**a)** Calculem la calor Qc generada cada dia:\n", "\n" ] }, { "cell_type": "code", "execution_count": 3, "metadata": {}, "outputs": [ { "data": { "text/plain": [ "885000000000.0" ] }, "execution_count": 3, "metadata": {}, "output_type": "execute_result" } ], "source": [ "Qc = mr*p\n", "Qc" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "Ara podem calcular el rendiment mitja de la planta $\\eta_{elèc}$ com el quocient de l'energia generada i l'energia consumida cada dia:\n", "\n" ] }, { "cell_type": "code", "execution_count": 5, "metadata": {}, "outputs": [ { "data": { "text/plain": [ "0.26440677966101694" ] }, "execution_count": 5, "metadata": {}, "output_type": "execute_result" } ], "source": [ "etaelec = Eelec/Qc\n", "etaelec" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "Arrodonint, $\\eta_{elèc}$ = 0,2644" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "**b)** Calculem la calor Q no aprofitada per la planta\n", "\n", "$Q = (1 - \\eta_{elèc}) \\cdot Q_{c}$" ] }, { "cell_type": "code", "execution_count": 6, "metadata": {}, "outputs": [ { "data": { "text/plain": [ "651000000000.0" ] }, "execution_count": 6, "metadata": {}, "output_type": "execute_result" } ], "source": [ "Q = (1-etaelec)*Qc\n", "Q" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "La calor util Qu que escalfa l'aigua el podem calcular com\n", "\n", "$Q_{u} = m_{a} \\cdot c_{e} \\cdot \\Delta t$\n" ] }, { "cell_type": "code", "execution_count": 7, "metadata": {}, "outputs": [ { "data": { "text/plain": [ "501600000000.0" ] }, "execution_count": 7, "metadata": {}, "output_type": "execute_result" } ], "source": [ "Qu = ma*ce*deltat\n", "Qu" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "El rendiment tèrmic mitja $\\eta_{tèrmic}$ serà el quocient entre la calor útil (Qu) i la que no s'ha aprofitat a la transformació elèctrica (Q)" ] }, { "cell_type": "code", "execution_count": 8, "metadata": {}, "outputs": [ { "data": { "text/plain": [ "0.7705069124423963" ] }, "execution_count": 8, "metadata": {}, "output_type": "execute_result" } ], "source": [ "etatermic = Qu/Q\n", "etatermic" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "Arrodonint, $\\eta_{tèrmic} = 0,7705$" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "**c)** Com un dia té 86400 s :" ] }, { "cell_type": "code", "execution_count": 9, "metadata": {}, "outputs": [ { "data": { "text/plain": [ "2708333.3333333335" ] }, "execution_count": 9, "metadata": {}, "output_type": "execute_result" } ], "source": [ "Pelec = Eelec/86400\n", "Pelec" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "Arrodonint $P_{elèc} = 2,708 \\hspace{1mm} MW$" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "En un dia el volum d'aigua escalfada sera la massa d'aigua dividida per la densitat:" ] }, { "cell_type": "code", "execution_count": 10, "metadata": {}, "outputs": [ { "data": { "text/plain": [ "3000.0" ] }, "execution_count": 10, "metadata": {}, "output_type": "execute_result" } ], "source": [ "Va = ma / 1000\n", "Va" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "Per tant, el cabal -que és aquest volum dividit pels 86400 s d'un dia- serà, en m³/s:" ] }, { "cell_type": "code", "execution_count": 11, "metadata": {}, "outputs": [ { "data": { "text/plain": [ "0.034722222222222224" ] }, "execution_count": 11, "metadata": {}, "output_type": "execute_result" } ], "source": [ "q = Va/86400\n", "q" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "Arrodonint, $q = 34,72 \\hspace{1mm} l/s$" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## Rendiment total de la planta de cogeneració" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "Si ens preguntem quin és el rendiment total, primer podríem calcular l'energia total aprofitada, tant en producció elèctrica com en l'escalfament de l'aigua\n", "\n", "$E_{util} = E_{elèc} + Q_{u} = Q{c} \\cdot \\eta_{elèc} + Q{c} \\cdot (1 - \\eta_{elèc}) \\cdot \\eta_{tèrmic}$\n", "\n", "dividint per la calor consumida obtenim el rendiment total:\n", "\n", "$\\eta_{} = \\eta_{elèc} + (1 - \\eta_{elèc}) \\cdot \\eta_{tèrmic}$\n", "\n", "en el nostre cas" ] }, { "cell_type": "code", "execution_count": 14, "metadata": {}, "outputs": [ { "data": { "text/plain": [ "0.8311864406779661" ] }, "execution_count": 14, "metadata": {}, "output_type": "execute_result" } ], "source": [ "eta = etaelec+(1-etaelec)*etatermic\n", "eta" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "Arrodonint, tenim un rendiment total $\\eta = 83,12 \\%$" ] }, { "cell_type": "code", "execution_count": null, "metadata": {}, "outputs": [], "source": [] } ], "metadata": { "kernelspec": { "display_name": "Python 3", "language": "python", "name": "python3" }, "language_info": { "codemirror_mode": { "name": "ipython", "version": 3 }, "file_extension": ".py", "mimetype": "text/x-python", "name": "python", "nbconvert_exporter": "python", "pygments_lexer": "ipython3", "version": "3.7.3" } }, "nbformat": 4, "nbformat_minor": 4 }