Answer:
Explanation:
You can utilize barbed clusters to store inadequate grids. On the off chance that there are a great many lines yet each line has just 4 or 5 associations with different segments, at that point as opposed to utilizing a 1000x1000 cluster you can utilize a 1000 line rough exhibit while you simply store the components that the present section has association with another segment. Other utilization can be done on account of query tables. Query tables will be tables which have different qualities concerning a solitary key where the quantity of qualities isn't fixed. Aside from this, barbed clusters have an exceptionally set number of utilization cases. Multidimensional exhibits then again have plenty of utilizations. It is utilized to store a great deal of information reliably on the grounds that the greater part of the information is put away is steady concerning which section compares to what information. Aside from that it very well may be utilized to make thick diagrams or sparse(not effective), plotting information. Another utilization case would be used as an impermanent stockpiling for the figurings that need to tail them and utilize the past information like in powerful programming.
Explanation:
Thermodynamics system :
Thermodynamics system is a region or space in which study of matters can be done.The system is separated from surroundings by a boundary this boundary maybe flexible or fixed it depends on situations.The out side the system is called surroundings.
Generally thermodynamics systems are of three types
1.Closed system(control mass system)
Only energy transfer take place ,no mass transfer take place.
2.Open system(control volume system)
Both mass as well as energy transfer take place.
3.Isolated system
Neither mass or nor energy transfer take place.
At steady state ,property is did not changes with respect to time.
Answer:
COP = 3.828
W' = 39.18 Kw
Explanation:
From the table A-11 i attached, we can find the entropy for the state 1 at -20°C.
h1 = 238.43 KJ/Kg
s1 = 0.94575 KJ/Kg.K
From table A-12 attached we can do the same for states 3 and 4 but just enthalpy at 800 KPa.
h3 = h4 = hf = 95.47 KJ/Kg
For state 2, we can calculate the enthalpy from table A-13 attached using interpolation at 800 KPa and the condition s2 = s1. We have;
h2 = 275.75 KJ/Kg
The power would be determined from the energy balance in state 1-2 where the mass flow rate will be expressed through the energy balance in state 4-1.
W' = m'(h2 - h1)
W' = Q'_L((h2 - h1)/(h1 - h4))
Where Q'_L = 150 kW
Plugging in the relevant values, we have;
W' = 150((275.75 - 238.43)/(238.43 - 95.47))
W' = 39.18 Kw
Formula foe COP is;
COP = Q'_L/W'
COP = 150/39.18
COP = 3.828
Answer:
a. Heat removal rate will increase
b. Heat removal rate will decrease
Explanation:
Given that
One end of rod is connected to the furnace and rod is long.So this rod can be treated as infinite long fin.
We know that heat transfer in fin given as follows
We know that area
Now when diameter will triples then :
So the new heat transfer will increase by 3 times.
Now when copper rod will replace by aluminium rod :
As we know that thermal conductivity(K) of Aluminium is low as compare to Copper .It means that heat transfer will decreases.
