Answer:
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Explanation:
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Tech A says that in some cases, the electronic brake control module can be programmed with a new tire size to restore proper ele
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Answer:
- the mass flow rate of air is 7.53 kg/s
- the exit area is 0.108 m²
Explanation:
Given the data in the question;
lets take a look at the steady state energy equation;
m" = W" / [ (h₁ - h₂ ) -
]
Now at;
T₁ = 900K, h₁ = 932.93 k³/kg
T₂ = 500 K, h₂ = 503.02 k³/kg
so we substitute, in our given values
m" = [ 3200 kW × ] / [ (932.93 - 503.02 )k³/kg -
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m" = 7.53 kg/s
Therefore, the mass flow rate of air is 7.53 kg/s
now, Exit area A₂ = v₂m" / V₂
we know that; pv = RT
so
A₂ = RT₂m" / P₂V₂
so we substitute
A₂ = {[ ×
×
] / [(1 bar)(100 m/s )]} |
||
A₂ = 0.108 m²
Therefore, the exit area is 0.108 m²
Answer:
T = 167 ° C
Explanation:
To solve the question we have the following known variables
Type of surface = plane wall ,
Thermal conductivity k = 25.0 W/m·K,
Thickness L = 0.1 m,
Heat generation rate q' = 0.300 MW/m³,
Heat transfer coefficient hc = 400 W/m² ·K,
Ambient temperature T∞ = 32.0 °C
We are to determine the maximum temperature in the wall
Assumptions for the calculation are as follows
- Negligible heat loss through the insulation
- Steady state system
- One dimensional conduction across the wall
Therefore by the one dimensional conduction equation we have
During steady state
= 0 which gives
From which we have
Considering the boundary condition at x =0 where there is no heat loss
= 0 also at the other end of the plane wall we have
hc (T - T∞) at point x = L
Integrating the equation we have
from which C₁ is evaluated from the first boundary condition thus
0 = from which C₁ = 0
From the second integration we have
From which we can solve for C₂ by substituting the T and the first derivative into the second boundary condition s follows
→ C₂ =
T(x) = and T(x) = T∞ +
∴ Tmax → when x = 0 = T∞ +
Substituting the values we get
T = 167 ° C