A package of aluminum foil contains 50. Ft2 of foil, which weighs approximately 7.0 oz . Aluminum has a density of 2.70 g/cm3. W
1 answer:
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Answer:
a) the work (W) done during the process is -2043.25 kJ
b) the work (W) done during the process is -2418.96 kJ
Explanation:
Given the data in the question;
mass of water vapor m = 10 kg
initial pressure P₁ = 550 kPa
Initial temperature T₁ = 340 °C
steam cooled at constant pressure until 60 percent of it, by mass, condenses; x = 100% - 60% = 40% = 0.4
from superheated steam table
specific volume v₁ = 0.5092 m³/kg
so the properties of steam at p₂ = 550 kPa, and dryness fraction
x = 0.4
specific volume v₂ = v + xv
v₂ = 0.001097 + 0.4( 0.34261 - 0.001097 )
v₂ = 0.1377 m³/kg
Now, work done during the process;
W = mP₁( v₂ - v₁ )
W = 10 × 550( 0.1377 - 0.5092 )
W = 5500 × -0.3715
W = -2043.25 kJ
Therefore, the work (W) done during the process is -2043.25 kJ
( The negative, indicates work is done on the system )
b)
What would the work done be if steam were cooled at constant pressure until 80 percent of it, by mass, condenses
x₂ = 100% - 80% = 20% = 0.2
specific volume v₂ = v + x₂v
v₂ = 0.001097 + 0.2( 0.34261 - 0.001097 )
v₂ = 0.06939 m³/kg
Now, work done during the process will be;
W = mP₁( v₂ - v₁ )
W = 10 × 550( 0.06939 - 0.5092 )
W = 5500 × -0.43981
W = -2418.96 kJ
Therefore, the work (W) done during the process is -2418.96 kJ
Answer:
Range of arrow = 225.09 meter
Final horizontal velocity = 34.47 m/s
Explanation:
We have equation of motion , where u is the initial velocity, t is the time taken, a is the acceleration and s is the displacement.
Considering the vertical motion of arrow ( up direction as positive)
We have u = 45 sin40 = 28.93 m/s, s = -20 m, a = acceleration due to gravity = .
t = 6.53 seconds or t = -0.63 seconds
So time = 6.53 seconds.
Considering the horizontal motion of arrow
u = 45 cos 40 = 34.47 m/s, t = 6.53 s, a =
So range of arrow = 225.09 meter
Horizontal velocity will not change , final horizontal velocity = 34.47 m/s.