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1 answer:
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Answer:
please find attached.
Explanation:
Answer:
18 kJ
Explanation:
Given:
Initial volume of air = 0.05 m³
Initial pressure = 60 kPa
Final volume = 0.2 m³
Final pressure = 180 kPa
Now,
the Work done by air will be calculated as:
Work Done = Average pressure × Change in volume
thus,
Average pressure = = 120 kPa
and,
Change in volume = Final volume - Initial Volume = 0.2 - 0.05 = 0.15 m³
Therefore,
the work done = 120 × 0.15 = 18 kJ
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