A cylindrical tank is required to contain a gage pressure 670 kPakPa . The tank is to be made of A516 grade 60 steel with a maxi
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:
Total wight =640.7927 KN
Explanation:
Given that
do= 61 cm
L =120
t= 0.9 cm
That is why inner diameter of the pipe
di= 61 - 2 x 0.9 cm
di=59.2 cm
Water density ,ρ = 1 kg/L = 1000 kg/m³
Weight of the pipe ,wt = 2500 N/m
wt = 2500 x 120 N = 300,000 N
The wight of the water
wt ' = ρ V g
wt'=340792.47 N
That is why total wight
Total wight = wt + wt'
Total wight =300,000+ 340792.47 N
Total wight =640,792.47 N
Total wight =640.7927 KN
Answer:
Explanation:
The question is: <em>Find the magnitude of the centripetal acceleration of the star with mass m₂</em>
The <em>centripetal acceleration</em> is the quotient of the centripetal force and the mass.
Thus, you can write the equations for each star:
As per Newton's third law, the centripetal forces are equal in magnitude. Then:
Now you can clear :