Chapter 2.8, Problem 100P

To determine

The plotting of rate of heat transfer against the convection heat transfer coefficient for the surface emissivities of 0.1, 0.5, 0.8, and 1. Also, discuss the results.

Answer to Problem 100P

The plotting of rate of heat transfer against the convection heat transfer coefficient for the surface emissivities of 0.1, 0.5, 0.8, and 1 are shown in Figure (1) and results are discussed as below.

Explanation of Solution

Calculate the rate of heat transfer by convection.

$\begin{array}{c}{\dot{Q}}_{\text{conv}}=hA\Delta T\\ =h\left(\pi D^2\right)\left(T_s-T_o\right)\end{array}$ (I)

Here, change in the temperature is $\Delta T$, surface temperature is $T_S$, final temperature is $T_o$, the diameter of a spherical ball is D, the convection heat transfer coefficient is h, and area of a spherical ball is A.

Calculate the rate of heat transfer by radiation.

$\begin{array}{c}{\dot{Q}}_{\text{rad}}=\epsilon \sigma A\left(T_s^4-T_o^4\right)\\ =\epsilon \sigma \left(\pi D^2\right)\left(T_s^4-T_o^4\right)\end{array}$ (II)

Here, surface temperature is $T_S$, exit temperature is $T_o$, emissivity of the ball is $\epsilon$, and Stefan-Bolzmann constant is $\sigma$.

Calculate the total rate of heat transfer from the ball.

${\dot{Q}}_{\text{total}}={\dot{Q}}_{\text{conv}}+{\dot{Q}}_{\text{rad}}$ (III)

Conclusion:

Let us solve for $\epsilon =0.8$.

Substitute $15{\text{W/m}}^{\text{2}}\cdot \text{°C}$ for h, 9 cm for D, $110°\text{C}$ for $T_S$, and $20°\text{C}$ for $T_o$ in Equation (I).

$\begin{array}{c}{\dot{Q}}_{\text{conv}}=15{\text{W/m}}^{\text{2}}\cdot \text{°C}\left(\pi \times {\left(9\text{cm}\right)}^2\right)\left(110°\text{C}-20°\text{C}\right)\\ =15{\text{W/m}}^{\text{2}}\left(\pi \times 81{\text{cm}}^2\times {\left(0.01\frac{\text{m}}{\text{cm}}\right)}^2\right)\left(90\right)\\ =34.35\text{W}\end{array}$

Substitute $5.67\times {10}^{-8}{\text{W/m}}^{\text{2}}\cdot {\text{K}}^{\text{4}}$ for $\sigma$, 9 cm for D, $110°\text{C}$ for $T_S$, and $20°\text{C}$ for $T_o$, and 0.8 for $\epsilon$ in Equation (II).

$\begin{array}{c}{\dot{Q}}_{\text{rad}}=0.8\left(5.67\times {10}^{-8}{\text{W/m}}^{\text{2}}\cdot {\text{K}}^{\text{4}}\right)\left(\pi \times {\left(\text{9}\text{cm}\right)}^2\right)\left({\left(110°\text{C}\right)}^4-{\left(20°\text{C}\right)}^4\right)\\ =\left[\begin{array}{c}0.8\times 5.67\times {10}^{-8}\frac{\mathrm{W}}{{\text{m}}^{\text{2}}\cdot {\text{K}}^{\text{4}}}\times \pi \times {\left(9\text{cm}\times 0.01\frac{\text{m}}{1\text{cm}}\right)}^2\\ \left[{\left\{\left(110+273\right)\text{K}\right\}}^4-{\left\{\left(20+273\right)\text{K}\right\}}^4\right]\end{array}\right]\\ =16.33\text{W}\end{array}$

Substitute 34.35 W for ${\dot{Q}}_{\text{conv}}$ and 16.33 W for ${\dot{Q}}_{\text{rad}}$ in Equation (III).

$\begin{array}{c}{\dot{Q}}_{\text{total}}=34.35\text{W}+16.33\text{W}\\ =50.7\text{W}\end{array}$

Follow the above process to calculate the rate of heat transfer against the convection heat transfer coefficient for the surface emissivities of 0.1 and 0.5 using spreadsheet including equations (I), (II), and (III) as in table (1).

$h,\left({\text{W/m}}^{\text{2}}\cdot {\text{K}}^{\text{4}}\right)$	${\dot{Q}}_{\text{conv}}$	${\dot{Q}}_{\text{rad}}@\epsilon =0.1$	${\dot{Q}}_{\text{total}}$	${\dot{Q}}_{\text{rad}}@\epsilon =0.5$	${\dot{Q}}_{\text{total}}$
5	11.4511	2.04127	13.4924	10.2064	21.6575
7.5	17.1767	2.04127	19.2179	10.2064	27.383
10	22.9022	2.04127	24.9435	10.2064	33.1086
12.5	28.6278	2.04127	30.669	10.2064	38.8341
15	34.3533	2.04127	36.3946	10.2064	44.5597
17.5	40.0789	2.04127	42.1201	10.2064	50.2852
20	45.8044	2.04127	47.8457	10.2064	56.0108
22.5	51.53	2.04127	53.5712	10.2064	61.7363
25	57.2555	2.04127	59.2968	10.2064	67.4619
27.5	62.9811	2.04127	65.0224	10.2064	73.1874
30	68.7066	2.04127	70.7479	10.2064	78.913

Continue table (1) for $\epsilon =0.8\text{and}1.0$ as in Table (2).

${\dot{Q}}_{\text{rad}}@\epsilon =0.8$	${\dot{Q}}_{\text{total}}$	${\dot{Q}}_{\text{rad}}@\epsilon =1$	${\dot{Q}}_{\text{total}}$
16.3302	27.7813	20.4127	31.8638
16.3302	33.5068	20.4127	37.5894
16.3302	39.2324	20.4127	43.3149
16.3302	44.9579	20.4127	49.0405
16.3302	50.6835	20.4127	54.766
16.3302	56.4091	20.4127	60.4916
16.3302	62.1346	20.4127	66.2172
16.3302	67.8602	20.4127	71.9427
16.3302	73.5857	20.4127	77.6683
16.3302	79.3113	20.4127	83.3938
16.3302	85.0368	20.4127	89.1194

Show the plotting of rate of heat transfer against the convection heat transfer coefficient for the surface emissivities of 0.1, 0.5, 0.8, and 1.0 using Table (1) and (2) as in Figure (1).