U of U CHEN 3453 - Lecture 11 - Transient Conduction - Lumped Capacitance

Unformatted text preview:

Transient Conduction:Lumped CapacitanceCH EN 3453 – Heat Transfer Reminders…• Homework #5 due Friday next week• Midterm #1 coming up Wed. October 1– Covers chapters 1, 2, 3, 4, 5• Engineering career fair Sept 23– Professionalism!!• Bethany not available for office hours todaySome Thought Exercises• Imagine an apple pie coming out of the oven at 200°C and being set on a window sill at 20°C. How long will it take to cool?• Imagine a 5-cm steel ball coming out of a furnace at 530°C and being placed in a room at 30°C. How long will it take to cool?The Biot Number• If Bi < 0.1 then the lumped capacitance approach can be used– Eq. 5.5 to find time to reach a given T– Eq. 5.6 to find T after a given time– Eq. 5.8a to find total heat gain (loss) for given time• L depends on geometry– General approach is L = V/As• L/2 for wall with both sides exposed• ro/2 for long cylinder• ro/3 for sphere– Conservative approach is to use the maximum length• L/2 for wall exposed on both sides• ro for cylinder or sphere (preferred to use this) Bi =hLkTime to Reach Temperaturet =ρVchAslnθiθEq. 5.5θi = Ti – T∞ = difference between initial object temperature and bulk fluid temperatureθ = T – T∞ = difference between object temperature and bulk fluid temperature at time tLumped Capacitance Equations• Time to reach specifiedtemperature (5.5):• Temperature after specified time (5.6):• Thermal time constant (5.7):• Heat transferred during heating (5.8a):θθi=T − T∞Ti− T∞= exp −hAsρVc⎛⎝⎜⎞⎠⎟t⎡⎣⎢⎤⎦⎥τt=1hAs⎛⎝⎜⎞⎠⎟ρVc( )= RtCtQ =ρVc( )θi1 − exp −tτt⎛⎝⎜⎞⎠⎟⎡⎣⎢⎢⎤⎦⎥⎥t =ρVchAslnθiθExample - Steel Ball•How long does it take for the 5-cm steel ball to cool to 30°C?D = 0.05 m ρ = 7854 kg/m3cp = 434 J/kg·Kh = 25 W/m2·K•What is the temperature of the ball after 30 minutes?•What was the total energy loss (Joules) from the ball?Other relations• Temperature after a given time• Thermal time constant• Heat transferred during heating or coolingθθi=T − T∞Ti− T∞= exp −hAsρVc⎛⎝⎜⎞⎠⎟t⎡⎣⎢⎤⎦⎥τt=1hAs⎛⎝⎜⎞⎠⎟ρVc( )= RtCtQ =ρVc( )θi1 − exp −tτt⎛⎝⎜⎞⎠⎟⎡⎣⎢⎢⎤⎦⎥⎥About that Pie…(Can we use a lumped analysis approach?)930 cm3 cmρ = 1000 kg/m3 c = 4184 J/kg·Kk = 0.68 W/m·KEffect of Biot


View Full Document

U of U CHEN 3453 - Lecture 11 - Transient Conduction - Lumped Capacitance

Download Lecture 11 - Transient Conduction - Lumped Capacitance
Our administrator received your request to download this document. We will send you the file to your email shortly.
Loading Unlocking...
Login

Join to view Lecture 11 - Transient Conduction - Lumped Capacitance and access 3M+ class-specific study document.

or
We will never post anything without your permission.
Don't have an account?
Sign Up

Join to view Lecture 11 - Transient Conduction - Lumped Capacitance 2 2 and access 3M+ class-specific study document.

or

By creating an account you agree to our Privacy Policy and Terms Of Use

Already a member?