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;
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:
hello some parts of your question is missing attached below is the missing part ( the required fig and table )
answer : The solar collector surface area = 7133 m^2
Explanation:
Given data :
Rate of energy input to the collectors from solar radiation = 0.3 kW/m^2
percentage of solar power absorbed by refrigerant = 60%
Determine the solar collector surface area
The solar collector surface area = 7133 m^2
attached below is a detailed solution of the problem
Answer:
D Reflection
Explanation:
It reflects the light from the sun. Then when the earth gets in the way, it casts a shadow causing crescent moons.
Answer:
(N-1) × (L/2R) = (N-1)/2
Explanation:
let L is length of packet
R is rate
N is number of packets
then
first packet arrived with 0 delay
Second packet arrived at = L/R
Third packet arrived at = 2L/R
Nth packet arrived at = (n-1)L/R
Total queuing delay = L/R + 2L/R + ... + (n - 1)L/R = L(n - 1)/2R
Now
L / R = (1000) / (10^6 ) s = 1 ms
L/2R = 0.5 ms
average queuing delay for N packets = (N-1) * (L/2R) = (N-1)/2
the average queuing delay of a packet = 0 ( put N=1)