Answer:
(a) Final temperature is 151.2 K
(b) Change in the specific internal energy is -30.798 kJ/kg
Explanation:
(a) P1 = P2 = 200 kPa
V1 = 12.322 m^3
V2 = 1/2 × V1 = 1/2 × 12.322 = 6.161 m^3
mass of refrigerant-134a = 100 kg
MW of refrigerant-134a (C2H2F4) = (12×2) + (2×1) + (19×4) = 24 + 2 +76 = 102 g/mol
number of moles of refrigerant-134a (n) = mass/MW = 100/102 = 0.98 kgmol
R = 8.314 kJ/kgmol.K
Final temperature (T2) = P2V2/nR = 200×6.16/0.98×8.314 = 151.2 K
(b) Change in specific internal energy = change in internal energy (∆U)/mass (m)
∆U = Cv(T2 - T1)
Cv = 20.785 kJ/kgmol.K
T1 = P1V1/nR = 200×12.322/0.98×8.314 = 302.4 K
∆U = 20.785(151.2 - 302.4) = 20.785 × -151.2 = -3142.7 kJ/kgmol × 0.98 kgmol = -3079.8 kJ
Change in specific internal energy = -3079.8 kJ/100 kg = -30.798 kJ/kg
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Answer:
Modulus of elasticity is dependent on the elemental constitution of 1080 steel and various heat treatment don't affect this elemental composition but only affects the mechanical properties(like strength , hardness, ductility, malleability) of the 1080 steel that are affected by plastic deformation of the 1080 steel
Explanation:
Generally the underlying physical reason why when we conduct various heat treatments on 1018 steel we expect the modulus of elasticity to stay about the same for every heat treatment is
Modulus of elasticity is dependent on the elemental constitution of 1080 steel and various heat treatment don't affect this elemental composition but only affects the mechanical properties(like strength , hardness, ductility, malleability) of the 1080 steel that are affected by plastic deformation of the 1080 steel
Answer:
$$\begin{align*}
P(Y-X=m | Y > X) &= \sum_{k} P(Y-X=m, X=k | Y > X) \\ &= \sum_{k} P(Y-X=m | X=k, Y > X) P(X=k | Y > X) \\ &= \sum_{k} P(Y-k=m | Y > k) P(X=k | Y > X).\end{split}$$
Explanation:
\eqalign{
P(Y-X=m\mid Y\gt X)
&=\sum_kP(Y-X=m,X=k\mid Y\gt X)\cr
&=\sum_kP(Y-X=m\mid X=k,Y\gt X)\,P(X=k\mid Y>X)\cr
&=\sum_kP(Y-k=m\mid Y\gt k)\,P(X=k\mid Y\gt X)\cr
}
P(Y-X=m | Y > X) &= \sum_{k} P(Y-X=m, X=k | Y > X) \\ &= \sum_{k} P(Y-X=m | X=k, Y > X) P(X=k | Y > X) \\ &= \sum_{k} P(Y-k=m | Y > k) P(X=k | Y > X).\end{split}$$
Answer:
Letter D
Explanation:
AutoCAD provides eleven different ways to create arcs. The different options are used based on the geometry conditions of the design. To create an arc, you can specify various combinations of center, endpoint, start point, radius, angle, chord length, and direction values.