A cylinder fitted with a frictionless piston contains 2 kg of R-134a at 3.5 bar and 100 C. The cylinder is now cooled so that th
1 answer:
Answer:
The answer to the question is
The heat transferred in the process is -274.645 kJ
Explanation:
To solve the question, we list out the variables thus
R-134a = Tetrafluoroethane
Intitial Temperaturte t₁ = 100 °C
Initial pressure = 3.5 bar = 350 kPa
For closed system we have m₁ = m₂ = m
ΔU = m×(u₂ - u₁) = ₁Q₂ -₁W₂
For constant pressure process we have
Work done = W = = P×ΔV = P × (V₂ - V₁) = P×m×(v₂ - v₁)
From the tables we have
State 1 we have h₁ = (490.48 +489.52)/2 = 490 kJ/kg
State 2 gives h₂ = 206.75 + 0.75 × 194.57= 352.6775 kJ/kg
Therefore Q₁₂ = m×(u₂ - u₁) + W₁₂ = m × (u₂ - u₁) + P×m×(v₂ - v₁)
= m×(h₂ - h₁) = 2.0 kg × (352.6775 kJ/kg - 490 kJ/kg) =-274.645 kJ
You might be interested in
Answer:
Explanation:
From the question we are told that:
Dimension
Thickness
Normal tensile force on top side
Normal tensile force on right side
Elastic modulus,
Generally the equation for Normal Strain X is mathematically given by
Therefore
For Top
Where
For Right side
Where
Area=L*B*T
Generally the equation for elongation is mathematically given by
For top
For Right
Answer:
The displacement from t = 0 to t = 10 s, is -880 m
Distance is 912 m
Explanation:
. . . . . . . . . . A
integrate above equation we get
from information given in the question we have
t = 1 s, s = -10 m
so distance s will be
-10 = 12 - 1 + C,
C = -21
we know that acceleration is given as
[FROM EQUATION A]
Acceleration at t = 4 s, a(4) = -24 m/s^2
for the displacement from t = 0 to t = 10 s,
the distance the particle travels during this time period:
let v = 0,
t = 2 s
Distance