Question

Heat transfer need help

A cylinder, 6 mm in diameter, has a length of 24 mm. One end of the cylinder is open while all other surfaces have external insulation. The inside of the cylinder is diffuse and gray having an emissivity of 0.8 and a uniform temperature of 1000 K. (a) Draw a sketch and state your assumptions (b) Determine the radiant power leaving the open end of the cylinder. () Calculate the effective emissivity of the cylinder (see note below). (d) The length of the cylinder is decreased to 6mm, what is the effective emissivity of the cylinder? (e) If the length of the cylinder is decreased further, do you expect an increase or decrease in the effective emissivity? NOTE: The effective emissivity of any cavity is defined as the ratio of the radiant power leaving the cavity to that from a blackbody having the area of the opening and a temperature of the inner surfaces of the cavity.

Answer 1

Answer:

Step by step solution with proper assumptions is given below. In case of any issue or query, kindly message in comment box.

(a)

Az Ahower: (a) L=24mml Alg 8,2.8 T -100 ok D=6 mm Ebium Ebz new J2 mewww 92 q

Assumptions:

• The hypothetical surface A2 is a blackbody which hhave temperature 0 K
• Cavity surface is isothermal, opaqua and diffuse gray opaque and diffuse-gray

Analysis

Noe by aproximating the A2 as the blackbody at 0 K implies that all the radiation incident on A2 from the cavity results from emmison by the walls and escapes to the surrondings. It implies that A2, $\varepsilon _{2}=1$ , and J2=Eb2=0

(b) Now from the thermal circuit as shown above, the heat radiated through the hole A2 is given as

$q_{1}=\left ( \frac{E_{b1}-E_{b2}}{\left ( \frac{1-\varepsilon _{1}}{\varepsilon 1A_{1}} \right )+\frac{1}{A_{1}F_{12}}+\frac{1-\varepsilon _{2}}{\varepsilon 2A_{2}}} \right )..................1$

Again as we can see from the figure that F21 = 1

and also we know that A1 F12 = A2 F21

Now calculating the area as

$A_{1}=\frac{\Pi D^{2}}{4}+ \Pi DL$

$A_{1}=\Pi D\left ( \frac{D}{4}+L \right )$

$A_{1}=\Pi *.006\left ( \frac{.006}{4}+.024\right )$

$A_{2}=\frac{\Pi *D^{2}}{4}$

$A_{2}=\frac{\Pi *.006^{2}}{4}$

Now substituting all the values in equation 1.

Where in equation 1

$E_{b1}=\sigma T^{4}$

$q_{1}=\left ( \frac{5.67*10^{-8}(1000^{4}-0)}{\left ( \frac{1-.8}{.8*4.807*10^{-4}} \right )+\frac{1}{2.827*10^{-5}}+0} \right )$

As we know that A1F12=A2F21

As F21=1

So A1F12=A2

Solving the equation

Answer

(c) The effective emissivity, $\varepsilon _{e}$ , of the cavity is defined as the ratio of the radiant power leaving the cavity to that from a blackbody having the same area of the cavity opening and at the temperature of the inner surfaces of the cavity. For the cavity above,

$\varepsilon _{e}=\frac{q_{1}}{A_{2}\sigma T^{4}}$

Putting all the values

$\varepsilon _{e}=\frac{1.580}{5.67*10^{-8}*2.827*10^{-5}*1000^{4}}$

$\varepsilon _{e}=.986$ Answer

(d) & (e)

As in case of the depth of the hole increases, Aterm which is shown as (1 - ε1)/ε1 A1 goes to zero such that the remaining term in the denominator 1/A1 F12 = 1/A2 F21. That is, as L increases, q1 implies that A2 F21 Eb1. This implies that $\varepsilon _{e}=1$ as L increases. In case of the ratio of L/D = 10, One can expect the value of $\varepsilon _{e}=.999$ or better also and when the length decrease it acts as vice a versa

If the length is decreased furthur from the 6 mm, the effective emmisivity will decrease.