Estimate the maximum expected thermal conductivity for a Cermet that contains 58 vol% titanium carbide (TiC) particles in a coba
1 answer:
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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.
Answer:
speed by mass attain is 55.86 m/s
Explanation:
given data
glucose = 10 g
mass = 100 kg
to find out
speed by mass attain
solution
we know glucose have 180 g molecular weight and
that 1 g glucose produce energy = 2816/180 × 10³ J
so here 10 g of glucose produce energy = 1.56 × J
so here energy release = 1/2 × mv²
1.56 × = 1/2 × (100)v²
v² = 3.12 × 10³
and v = 55.86 m/s
so speed by mass attain is 55.86 m/s
Answer:
Speed of aircraft ; (V_1) = 83.9 m/s
Explanation:
The height at which aircraft is flying = 3000 m
The differential pressure = 3200 N/m²
From the table i attached, the density of air at 3000 m altitude is; ρ = 0.909 kg/m3
Now, we will solve this question under the assumption that the air flow is steady, incompressible and irrotational with negligible frictional and wind effects.
Thus, let's apply the Bernoulli equation :
P1/ρg + (V_1)²/2g + z1 = P2/ρg + (V_2)²/2g + z2
Now, neglecting head difference due to high altitude i.e ( z1=z2 ) and V2 =0 at stagnation point.
We'll obtain ;
P1/ρg + (V_1)²/2g = P2/ρg
Let's make V_1 the subject;
(V_1)² = 2(P1 - P2)/ρ
(V_1) = √(2(P1 - P2)/ρ)
P1 - P2 is the differential pressure and has a value of 3200 N/m² from the question
Thus,
(V_1) = √(2 x 3200)/0.909)
(V_1) = 83.9 m/s
