Answer:
See explaination
Explanation:
for a reverse carnot cycle T-S diagram is a rectangle which i have shown
net work for a complete cycle must be equal to net heat interaction.
Kindly check attachment for the step by step solution of the given problem.
Answer:The Urban heat island temperature will be REDUCED.
Two Impacts of Rooftop gardens
1) provision of shade against Sunlight.
2) It helps to purify the air around the building.
Explanation: Rooftop gardens are gardens made on top of the roofs of buildings, it is a Green initiative aimed at helping to improve the overall Environment.
Rooftop gardens have several significant benefits which includes
Reduction of the surrounding temperatures and the Urban heat Island temperatures.
Rooftop gardens helps to shade the roof from the direct impacts of harsh weather conditions.
Generally, plants are known as air purifiers as they remove the excess Carbondioxide around the environment through photosynthesis, and they also help to release water vapor which will help to improve the humidity of the environment.
Answer:
(a) Surface energy is greater than grain boundary energy due to the fact that the bonds of the atoms on the surface are lower than those of the atoms at the grain boundary. The energy is also directly proportional to the number of bonds created.
(b) The energy of a high-angle grain boundary is higher than that of a small-angle grain boundary because the high-angle grain boundary has a higher misalignment and smaller number of bonds than a small-angle grain boundary.
Explanation:
(a) Surface energy is greater than grain boundary energy due to the fact that the bonds of the atoms on the surface are lower than those of the atoms at the grain boundary. The energy is also directly proportional to the number of bonds created.
(b) The energy of a high-angle grain boundary is higher than that of a small-angle grain boundary because the high-angle grain boundary has a higher misalignment and smaller number of bonds than a small-angle grain boundary.
Answer:
14.506°C
Explanation:
Given data :
flow rate of water been cooled = 0.011 m^3/s
inlet temp = 30°C + 273 = 303 k
cooling medium temperature = 6°C + 273 = 279 k
flow rate of cooling medium = 0.02 m^3/s
Determine the outlet temperature
we can determine the outlet temperature by applying the relation below
Heat gained by cooling medium = Heat lost by water
= ( Mcp ( To - 6 ) = Mcp ( 30 - To )
since the properties of water and the cooling medium ( water ) is the same
= 0.02 ( To - 6 ) = 0.011 ( 30 - To )
= 1.82 ( To - 6 ) = 30 - To
hence To ( outlet temperature ) = 14.506°C
Answer:
The pressure drop is 269.7N/m^2
Explanation:
∆P = ∆h × rho × g
∆h = 3.2cm = 3.2/100 = 0.032m, rho = 860kg/m^3, g = 9.8m/s^2
∆P = 0.032×860×9.8 = 269.7N/m^2
