Question

Question 7 of 19 Question 7 18 points Save Answa 3 15.1 Wm"C. CP-40 JA "C) is heated coefficient inside the oven was in the air is 25 WmC. Answer the folowing questions based on the information give when placed (out) n to e oven The convectve heat tansler ceoeficent nside the oven was the a25Wner te in the oven. The oven has a unfom inside temperahure ol 700 C The bal has a unform temperature of 2 ーーーーコCalculate the characterstic Ienghofte sphercal number. Bi._(A00058 LC"-m(A00210 B.00106 C00035 Dooosn 00129 С 0.0048 D 0.0068) Can we se his spherical ball s a smped systee(AYs No Determine Pe time reqared for temperature of the bal to rise (increase) to 300 СПО .-. (A 156 B 280 С 348 D 96, Determne to time repiredfor temperatureofthe bal to rise encrease, to 95% o, "w naal temperature dterence to% "-MA 976 B4M Cica D 2465) Coose the leter corresponingtothe corect answer and type inthe blank space. For example, dyour answeraB''ype B

Answer 1

Given -:

Diameter of spherical stainless steel ball, D = 2.1 cm

Density = 8085 Kg/m³

Thermal conductivity, K = 15.1 W/m^oC

C_p = 480 J/Kg.K

T_oven = 700 ^oC = 973 K, T_ball = 25 ^oC = 298 K

Heat transfer coefficient, H = 25 Wm²/^oC.

Find -:  L_c , B_i , Time constant

Analysis -:

1.) Characteristic Length of the Spherical balls (Lc).

(Volume)ball Le = (Surface Area)ball

4T7-2

$L_c = \frac{D}{6}$

20.35cm 6

Lc 0.0035772

2.) Biot number (B_i)

$B_i = \frac{HL_c}{K}$

$B_i = \frac{25*0.0035}{15.1}$

$\Rightarrow B_i = \frac{25*0.0035}{15.1} = 0.005794$

3.) If the B_i less than 0.1, then the system is considered as Lumped system.

As we can see that if the B_i < 0.1, we can use below formula in order to calculate for the Temperature of the ball at any instant of time.

$\frac{T_{required \,\, Temperature } - T_{surrounding }}{T_{initial\,\, temperature} - T_{surrounding }} = \exp\left( - \frac{HA_s}{\rho C_p V}t \right )$

4.) Time constant ,T

$T = \frac{1}{\tau}$

HAs

$\Rightarrow \tau = \frac{\rho L_c C_p}{H}$

$\Rightarrow \tau = \frac{8085 *0.0035* 480}{25} = 543.312$

$\Rightarrow T = \frac{1}{543.312} = 0.00184 ts^{-1}$

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