1 Problem 20.6 (WWWR) For a horizontal cylinder 2 27 8 16 9 6 1 1 .0 559 / .0 387 60.
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SCHEMATIC: ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional (radial) conduction, (3) No internal heat generation, (4) Insulation has uniform properties independent of temperature and position. 2.1, for this one-dimensional (cylindrical) radial system has the form This relation requires that the product of the radial temperature gradient, d T/dr, and the radius, r, remains constant throughout the insulation.
For our situation, the temperature distribution must appear as shown in the sketch.
D The cross section of a storm window is shown in the sketch.
How much heat will be lost through a window measuring 1.83m by 3.66 m on a cold day when the inside and outside air temperatures are, respectively, 295K and 250K?FIND: (a) Equivalent thermal circuit, (b) Expression for heater temperature, (c) Ratio of outer and inner heat flows and conditions for which ratio is minimized.ASSUMPTIONS: (1) One-dimensional, steady-state conduction, (2) Constant properties, (3) Isothermal heater, (4) Negligible contact resistance(s).Don't show me this again This is one of over 2,200 courses on OCW. Find materials for this course in the pages linked along the left. Use OCW to guide your own life-long learning, or to teach others. This is because more energy must be lost from the substance of greater heat capacity to drop the temperature. Heat capacities and heat transfer coefficients are themselves temperature dependent quantities, so my answers are only valid for modest $\Delta T$.For example, the heat capacity of hydrogen gas changes in very interesting ways as you get to really low temperatures, due to the equilibrium ratio of ortho/para hydrogen (both hydrogen, but with different nuclear spin states).St 2.1× 10 − 3 5.55× 10 − 7 1.45× 10 − 6 1.22× 10 − 6 7.21× 10 − 7 Problem 19.8 (WWWR) 2 sin 900 2500sin 1.in / , in qx x ab AL q Wm x m A ⎛⎞ππ⎛ ⎞ = ⎜⎟= ⎜ ⎟ ⎝⎠ ⎝ ⎠ 77.4 83.5 1 cos 1.What will be the average temperature of the air between the glass panes? Ts =° 100 c Tcsurrounding= 25 ° Air space 0.8cm wide Window glass--0.32 cm thick Solution: q R Ri − ==× − ==× 3 0 − ==× q W ∑R Air space 0.8cm wide 250k hi 295k h 0 Window glass 0.32 cm thick Ti To Ri RGL RAS RGL Ro KNOWN: Hot water pipe covered with thick layer of insulation.FIND: Sketch temperature distribution and give brief explanation to justify shape.