Answer:
The heat loss rate through one of the windows made of polycarbonate is 252W. If the window is made of aerogel, the heat loss rate is 16.8W. If the window is made of soda-lime glass, the heat loss rate is 1190.4W.
The cost associated with the heat loss through the windows for an 8-hour flight is:
For aerogel windows: $17.472 (most efficient)
For polycarbonate windows: $262.08
For soda-lime glass windows: $1,238.016 (least efficient)
Explanation:
To calculate the heat loss rate through the window, we can use a model of heat transmission by conduction throw flat wall. Using unidimensional Fourier law:
In this case:
If we replace the data provided by the problem we get the heat loss rate through one of the windows of each material (we only have to change the thermal conductivities).
To obtain the thermal conductivity of the soda-lime glass we use the graphic attached to this answer (In this case for soda-lime glass k₃₀₀=0.992w/m·K).
To calculate the cost associated with the heat loss through the windows for an 8-hour flight we use this formula (using the heat loss rate calculated in each case):
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Answer:
B. The thickness of the heated region near the plate is increasing.
Explanation:
First we know that, a boundary layer is the layer of fluid in the immediate vicinity of a bounding surface where the effects of viscosity are significant. The fluid is often slower due to the effects of viscosity. Advection i.e the transfer of heat by the flow of liquid becomes less since the flow is slower, thereby the local heat transfer coefficient decreases.
From law of conduction, we observe that heat transfer rate will decrease based on a smaller rate of temperature, the thickness therefore increases while the local heat transfer coefficient decreases with distance.
Answer:
79 kW.
Explanation:
The equation for enthalpy is:
H2 = H1 + Q - L
Enthalpy is defined as:
H = G*(Cv*T + p*v)
This is specific volume.
The gas state equation is:
p*v = R*T (with specific volume)
The specific gas constant for air is:
287 K/(kg*K)
Then:
T1 = 60 + 273 = 333 K
T2 = 200 + 273 = 473 K
p1*v1 = 287 * 333 = 95.6 kJ/kg
p2*v2 = 287 * 473 = 135.7 kJ/kg
The Cv for air is:
Cv = 720 J/(kg*K)
So the enthalpies are:
H1 = 0.8*(0.72 * 333 + 95.6) = 268 kW
H2 = 0.8*(0.72 * 473 + 135.7) = 381 kW
Ang the heat is:
Q = 34 kW
Then:
H2 = H1 + Q - L
381 = 268 + 34 - L
L = 268 + 34 - 381 = -79 kW
This is the work from the point of view of the air, that's why it is negative.
From the point of view of the machine it is positive.
