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Ugo [173]
3 years ago
8

A long rod of 60-mm diameter and thermophysical properties rho=8000 kg/m^3, c=500J/kgK, and k=50 W/mK is initally at a uniform t

emperature in a forced convection furnace maintained at 750K. The convection coeffiecent is estimated to be 1000 W/m^2K.
What is the centerline temperature of the rod when the surface temperature is 550K?
Engineering
1 answer:
Monica [59]3 years ago
3 0

Answer:

Tc = 424.85 K

Explanation:

Given that,

D = 60 mm = 0.06 m

\rho = 8000 kg/m^3

k = 50 w/m . kc = 500 j/kg.k

h_{\infty} = 1000 w/m^2t_{\infity} = 750 kt_w = 500 K

surface area = As = \pi dL \\\frac{As}{L} = \pi D = \pi \timeS 0.06

HEAT FLOW Q  is

Q = h_{\infty} As (T_[\infty} - Tw)  = 1000 \pi\times 0.06 (750-500)

 = 47123.88 w per unit length of rod

volumetric heat rate

q = \frac{Q}{LAs}

= \frac{47123.88}{\frac{\pi}{4} D^2 \times 1}

q = 1.66\times 10^{7} w/m^3

Tc = \frac{- qR^2}{4K} + Tw

= \frac{ - 1.67\times 10^7 \times (\frac{0.06}{2})^2}{4\times 50} +  500

  = 424.85 K

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and

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<h3>The  Algorithm and decimal equivalent on the seven-segment displays HEX3-0</h3>

Generally,  the program will be written using a  cpulator simulator in order to attain best result.

We are to

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This will be the Algorithm of the Nios II assembly program .

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