A 1m Diameter steel ball ( with thermocouple leads attached to it) with properties: \rho=7800...


A 1m Diameter steel ball ( with thermocouple leads attached to it) with properties:

{eq}\rho=7800 kg/m^{3}, k=35 W/m.^{\circ}C, c_{p}=460 J/kg.^{\circ}C {/eq} and initially at a uniform temperature {eq}T_{0}=450^{\circ}C {/eq} is immediately placed in a furnace that maintains its temperature at {eq}T_{0}=100^{\circ}C {/eq}. If the convection heat transfer coefficient is {eq}10 W/m^{2}.^{\circ}C {/eq}. Calculate:

a) The time require for the ball to reach {eq}T_{0}=150^{\circ}C {/eq}

b) Since you have a thermocouple attached to the ball and want to check the temperature one hour after the ball was placed in the furnace, what would be temperature in the thermocouple display?.

c) What would be dissipated energy of the ball one hour and 15 minutes after the ball was placed in the furnace?.

Lumped Capacitance analysis

This analysis is used to calculate the heat transfer, temperature, time of the bodies participating, but it is only valid for the substance in which there is no internal variation of the temperature within the body.

Answer and Explanation:

Given Data

  • The diameter of the steel ball is {eq}D = 1\;{\rm{m}}. {/eq}
  • The density of the steel ball is {eq}\rho = 7800\;{\rm{kg/}}{{\rm{m}}^{\rm{3}}}. {/eq}
  • The thermal conductivity of the steel ball is {eq}k = 35\;{\rm{W/m}}. {/eq}
  • The specific heat capacity of the steel ball is {eq}{c_p} = 460\;{\rm{J/kg}} \cdot ^\circ {\rm{C}}. {/eq}
  • The initial uniform temperature is {eq}{T_0} = 450^\circ {\rm{C}}. {/eq}
  • The temperature of the furnace is {eq}{T_\infty } = 100^\circ {\rm{C}} {/eq}
  • The convective heat transfer coefficient is {eq}h = 10\;{\rm{W/}}{{\rm{m}}^{\rm{2}}}{\rm{^\circ C}}. {/eq}

(a) According to the lumped capacitance analysis calculating the biot number as,

{eq}Bi = \dfrac{{h{L_c}}}{k} {/eq}

We know that the characteristic length for the sphere is calculated as,

{eq}\begin{align*} {L_c} &= \dfrac{V}{{{A_s}}}\\ &= \dfrac{{\dfrac{4}{3}\pi {R^3}}}{{4\pi {R^2}}}\\ &= \dfrac{R}{3} \end{align*} {/eq}

Therefore, the biot number is calculated as,

{eq}\begin{align*} Bi &= \dfrac{{h{L_c}}}{k}\\ &= \dfrac{{10 \times \dfrac{{0.5}}{3}}}{{35}}\\ &= 0.0476 \end{align*} {/eq}

Which is less than 0.1 therefore lumped system can be applicable.

The time can be calculated as,

{eq}\begin{align*} \dfrac{{T - {T_\infty }}}{{{T_0} - {T_\infty }}} &= {e^{ - \left( {\dfrac{{hAt}}{{\rho V{c_p}}}} \right)}}\\ \dfrac{{150 - 100}}{{450 - 100}} &= {e^{ - \left( {\dfrac{{10 \times 3 \times t}}{{7800 \times 0.5 \times 460}}} \right)}}\\ 0.14 &= {e^{ - 1.6722 \times {{10}^{ - 5}} \times t}}\\ t &= 117576.4177\;\sec \\ &= 32.66\;{\rm{hr}} \end{align*} {/eq}

(b) The temperature of the thermocouple after one hour,

{eq}\begin{align*} \dfrac{{T - {T_\infty }}}{{{T_0} - {T_\infty }}} &= {e^{ - \left( {\dfrac{{hAt}}{{\rho V{c_p}}}} \right)}}\\ \dfrac{{T - 100}}{{450 - 100}} &= {e^{ - \left( {\dfrac{{10 \times 3 \times 3600}}{{7800 \times 0.5 \times 460}}} \right)}}\\ T - 100 &= 350 \times {e^{ - 1.6722 \times {{10}^{ - 5}} \times 3600}}\\ T &= 429.55^\circ {\rm{C}} \end{align*} {/eq}

(c) The dissipated energy of the ball is calculated as,

{eq}\begin{align*} Q &= \rho V{c_p}\left( {{T_i} - {T_\infty }} \right)\left[ {{e^{ - \left( {\dfrac{{hAt}}{{\rho V{c_p}}}} \right)}} - 1} \right]\\ &= 7800 \times \dfrac{{4\pi }}{3} \times {0.5^3} \times 460 \times \left( {450 - 100} \right)\left[ {{e^{ - \left( {\dfrac{{10 \times 3 \times 75 \times 60}}{{7800 \times 0.5 \times 460}}} \right)}} - 1} \right]\\ &= - 47.66\;{\rm{MW}} \end{align*} {/eq}

This, negative sign indicates heat loss over balls.

Learn more about this topic:

Thermal Expansion & Heat Transfer

from High School Physics: Help and Review

Chapter 17 / Lesson 12

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