Answer:
The tube surface temperature immediately after installation is 120.4°C and after prolonged service is 110.8°C
Explanation:
The properties of water at 100°C and 1 atm are:
pL = 957.9 kg/m³
pV = 0.596 kg/m³
ΔHL = 2257 kJ/kg
CpL = 4.217 kJ/kg K
uL = 279x10⁻⁶Ns/m²
KL = 0.68 W/m K
σ = 58.9x10³N/m
When the water boils on the surface its heat flux is:
For copper-water, the properties are:
Cfg = 0.0128
The heat flux is:
qn = 0.9 * 18703.42 = 16833.078 W/m²
The tube surface temperature immediately after installation is:
Tinst = 100 + 20.4 = 120.4°C
For rough surfaces, Cfg = 0.0068. Using the same equation:
ΔT = 10.8°C
The tube surface temperature after prolonged service is:
Tprolo = 100 + 10.8 = 110.8°C
Answer:
The correct option is;
c. the exergy of the tank can be anything between zero to P₀·V
Explanation:
The given parameters are;
The volume of the tank = V
The pressure in the tank = 0 Pascal
The pressure of the surrounding = P₀
The temperature of the surrounding = T₀
Exergy is a measure of the amount of a given energy which a system posses that is extractable to provide useful work. It is possible work that brings about equilibrium. It is the potential the system has to bring about change
The exergy balance equation is given as follows;
![X_2 - X_1 = \int\limits^2_1 {} \, \delta Q \left (1 - \dfrac{T_0}{T} \right ) - [W - P_0 \cdot (V_2 - V_1)]- X_{destroyed}](https://tex.z-dn.net/?f=X_2%20-%20X_1%20%3D%20%5Cint%5Climits%5E2_1%20%7B%7D%20%5C%2C%20%5Cdelta%20Q%20%5Cleft%20%281%20-%20%5Cdfrac%7BT_0%7D%7BT%7D%20%5Cright%20%29%20-%20%5BW%20-%20P_0%20%5Ccdot%20%28V_2%20-%20V_1%29%5D-%20X_%7Bdestroyed%7D)
Where;
X₂ - X₁ is the difference between the two exergies
Therefore, the exergy of the system with regards to the environment is the work received from the environment which at is equal to done on the system by the surrounding which by equilibrium for an empty tank with 0 pressure is equal to the product of the pressure of the surrounding and the volume of the empty tank or P₀ × V less the work, exergy destroyed, while taking into consideration the change in heat of the system
Therefore, the exergy of the tank can be anything between zero to P₀·V.
Answer:
The break force that must be applied to hold the plane stationary is 12597.4 N
Explanation:
p₁ = p₂, T₁ = T₂
The heat supplied = 
× Heating value of jet fuel
The heat supplied = 0.5 kg/s × 42,700 kJ/kg = 21,350 kJ/s
The heat supplied = 
· 
= 20 kg/s
The heat supplied = 20* = 21,350 kJ/s
= 1.15 kJ/kg
T₃ = 21,350/(1.15*20) + 485.03 = 1413.3 K
p₂ = p₁ × p₂/p₁ = 95×9 = 855 kPa
p₃ = p₂ = 855 kPa
T₃ - T₄ = T₂ - T₁ = 485.03 - 280.15 = 204.88 K
T₄ = 1413.3 - 204.88 = 1208.42 K
T₅ = 1208.42*(2/2.333) = 1035.94 K
= √(1.333*287.3*1035.94) = 629.87 m/s
The total thrust = × 
= 20*629.87 = 12597.4 N
Therefore;
The break force that must be applied to hold the plane stationary = 12597.4 N.
Answer:
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Explanation:
