When charging an R-410A system that uses a water-cooled condenser, you must first charge with vapor to a pressure of at least __
1 answer:
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Answer:
a. 81 kj/kg
b. 420.625K
c. 101.24kj/kg
Explanation:
t1 = 360
p1 = 0.4mpa
p2 = 1.20
y = 1.13
substitute these values into the equation
when we cross multiply
t2 = 360 * 1.1347
= 408.5
a. the work required in the firs compressor
w=c(t2-t1)
c=1.67x10³
t1 = 360
t2 = 408.5
w = 1670(408.5-360)
= 1670*48.5
= 80995 J
= 81KJ/kg
b.
n = 80%
t2 = 408.5
t1 = 360
0.80 = 408.5-360 ÷ t'2-360
cross multiply to get the value of t'2
0.80(t'2-360) = 48.5
0.80t'2 - 288 = 48.5
0.8t'2 = 48.5+288
0.8t'2 = 336.5
t'2 = 336.5/0.8
= 420.625
this is the temperature at the exit of the first compressor
c. cooling requirement
w = c(t2-t1)
= 1.67x10³(420.625-360)
= 1670*60.625
= 101243.75
= 101.24kj/kg
Answer:
Volume of face centered cubic cell=
Explanation:
Consider the face centered cubic cell:
1 atom at each corner of cube.
1 atom at center of each face.
Consider the one face (ABCD) as shown in attachment for calculation:
Length of the all sides of face centered cubic cell is L.
Volume of face centered cubic cell= L^3
Now Consider the figure shown in attachment:
According to Pythagoras theorem on ΔADC.
(a is the atomic radius)
(Put in the formula of Volume)
Volume of face centered cubic cell= L^3
Volume of face centered cubic cell=
Volume of face centered cubic cell=
Volume of face centered cubic cell=
