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Ejercicio con funciones bessel

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JasminSeufert

This document contains the step-by-step work and calculations to solve a heat transfer problem involving conduction between two surfaces at different temperatures. The key steps are: 1) An equation for conductive heat transfer is used to relate the temperature difference to the heat transfer coefficient, area, and thermal conductivity. 2) The temperature of the second surface, T2, is calculated to be 197.4°C. 3) Various parameters such as thickness, thermal conductivity, and temperatures are provided for the two surfaces.

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a) qcond = qconv . -kA(T2-T1)/dx = h(Ts-Talr)A Despejando para la temperatura T2 - T1 = hdx(T2-Talr)/-k T2 - 200 = (180)(0.004)(T2-8)/(-52) T2 = 197,4 °C Para R1 x = 0,9303 m = 18,6052 h = 180 I0(x) = 1,2283 R1 = 0,0500 k = 52 I1(x) = 0,5173 R2 = 0,1000 t = 0,02 K0(x) = 0,4656 T0 = 200 K1(x) = 0,6791 mR1 = 0,9303 Tamb = 8 2/piK0(x) = 0,2964 mR2 = 1,8605 2/piK1(x) = 0,4323 Para R2 x = 1,8605 I0(x) = 2,0716 I1(x) = 1,3952 b) Q = 1172,54 K0(x) = 0,1353 n = 0,79 K1(x) = 0,1683 2/piK0(x) = 0,0861 2/piK1(x) = 0,1072 Seufert García Ofelia Jasmín, Mecanismos de Transferencia 2014.

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Ejercicio con funciones bessel

  • 1. a) qcond = qconv . -kA(T2-T1)/dx = h(Ts-Talr)A Despejando para la temperatura T2 - T1 = hdx(T2-Talr)/-k T2 - 200 = (180)(0.004)(T2-8)/(-52) T2 = 197,4 °C Para R1 x = 0,9303 m = 18,6052 h = 180 I0(x) = 1,2283 R1 = 0,0500 k = 52 I1(x) = 0,5173 R2 = 0,1000 t = 0,02 K0(x) = 0,4656 T0 = 200 K1(x) = 0,6791 mR1 = 0,9303 Tamb = 8 2/piK0(x) = 0,2964 mR2 = 1,8605 2/piK1(x) = 0,4323 Para R2 x = 1,8605 I0(x) = 2,0716 I1(x) = 1,3952 b) Q = 1172,54 K0(x) = 0,1353 n = 0,79 K1(x) = 0,1683 2/piK0(x) = 0,0861 2/piK1(x) = 0,1072 Seufert García Ofelia Jasmín, Mecanismos de Transferencia 2014.