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kozerog [31]
3 years ago
6

Convection ovens operate on the principle of inducing forced convection inside the oven chamber with a fan. A small cake is to b

e baked in an oven when the con- vection feature is disabled. For this situation, the free convection coefficient associated with the cake and its Steel sheet Concrete pad Ts, s Tb Ts Tsur Vs Pelec t s Lo ε Air T[infinity], h , 56 Chapter 1 ! Introduction CH001.qxd 2/24/11 12:03 PM Page 56 pan is hfr ! 3 W/m2 ! K. The oven air and wall are at temperatures T! ! Tsur ! 180)C. Determine the heat flux delivered to the cake pan and cake batter when they are initially inserted into the oven and are at a tem- perature of Ti ! 24)C. If the convection feature is acti- vated, the forced convection heat transfer coefficient is hfo ! 27 W/m2 ! K. What is the heat flux at the batter or pan surface when the oven is operated in the convection mode? Assume a value of 0.97 for the emissivity of the cake batter and pan.

Engineering
1 answer:
svetlana [45]3 years ago
5 0

complete question:

attached

Answer:

2356.11 W/m^2

6100.11 W/m^2

Explanation:

Assumptions:

1. Steady-state conditions.

2. The cake is placed in a large surrounding.

3. Heat flux delivered to the cake is due to convection and radiation.  

Case 1

Since convection feature is disabled the mode of heat transfer associated with this situation is through free convection and radiation.  

q''(free) = [q''(free convection+q''(radiation) ]W/m^2

            = h_free(T_infinty - T_i) + εσ(T_air^4 - T_i^4)

            = 3 W/m^2K(180°C - 24°C) + 0.97*5.67*10^-8*[(180+273K)^4 -  

               (24+273K)^4 ]

            = 468 +1881.11

            = 2356.11 W/m^2

Case 2

Since convection feature is enabled or activated the mode of heat transfer associated with this situation is through forced convection and radiation.  

q''(free) = [q''(forced convection+q''(radiation) ]W/m^2

            = h_forced(T_infinty - T_i) + εσ(T_air^4 - T_i^4)

            = 27 W/m^2K(180°C - 24°C) + 0.97*5.67*10^-8*[(180+273K)^4 -  

               (24+273K)^4 ]

            = 4212 +1881.11

            = 6100.11 W/m^2

1. The total heat flux is is 2.58 times higher when the convection feature is activated. Therefore the cake will bake faster during this condition.  

2. The contribution of convection heat flux under natural(free) convection is very low as compared to the contribution during forced convection.  

3. The heat transfer due to radiation is same in both the cases.  

4. Only 19.9 % of the total heat flux is contributed by free convection in the first case.  

5. In the second case 69 % of the total heat flux is contributed by forced convection.  

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P₁ - P₂ + ρgh₁ -  ρgh₂ = 1/2ρv₂² - 1/2ρv₁²

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where ΔP = pressure difference = 12.35 kPa = 12350 Pa

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From the flow rate equation Q = A₁v₁ = A₂v₂ and v₁ = A₂v₂/A₁ = d₂²v₂/d₁² where v₁ and v₂ represent the inlet and exit velocities from the pipe and d₁ = 0.95 m and d₂ = 0.44 m represent the diameters at the top end and lower end of the pipe respectively.

Substituting v₁ into (1), we have

ΔP + ρgΔh = 1/2ρ(v₂² - (d₂²v₂/d₁² )²)

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So, the inlet velocity v₁ = 6.66 m/s, the exit velocity v₂ = 7.32 m/s and the volume flow rate Q = 4.72 m³/s

7 0
3 years ago
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