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
Answer:
I. 3.316 kW
II. 1.218 kW
III. 2.72
Explanation:
At state 1, the enthalpy and entropy are determined using the given data from A-13.
At P1 = 200kpa and T1 = 0,
h1 = 253.07 kJ/kg
s1 = 0.9699 kJ/kgK
At state 2, the isentropic enthalpy is determined at P2 = 1400kpa and s1 = s2 by interpolation. Thus
h2(s) = 295.95 kJ/kg
The actual enthalpy is then gotten by
h2 = h1 + [h2(s) - h1]/n
h2 = 253.07 + [295.95 - 253.07]/0.88
h2 = 253.07 + 48.73
h2 = 301.8 kJ/kg
h3 = h4 = 120.43 kJ/kg
Heating load is determined from energy balance, thus,
Q'l = m'(h1 - h4)
Q'l = 0.025(253.07 - 120.43)
Q'l = 0.025 * 132.64
Q'l = 3.316 kW
Power is determined by using
W' = m'(h2 - h1)
W'= 0.025(301.8 - 253.07)
W'= 0.025 * 48.73
W'= 1.218 kW
The Coefficient Of Performance is Q'l / W'
COP = 3.316/1.218
COP = 2.72
Answer:
V=1601gal
Explanation:
Hello! This problem is solved as follows,
First we must raise the equation that defines the pressure at the bottom of the tank with the purpose of finding the height that olive oil reaches.
This is given as the sum due to the atmospheric pressure (1atm = 101.325kPa), and the pressure due to the weight of the olive oil, taking into account the above, the following equation is inferred.
P=Poil+Patm
P=total pressure or absolute pressure=26psi=179213.28Pa
Patm= the atmospheric pressure =101325Pa
Poil=pressure due to the weight of olive oil=0.86αgh
α=density of water=1000kg/m^3
g=gravity=9.81m/s^2
h= height that olive oil reaches
solving
P=Poil+Patm
P=Patm+0.86αgh
[/tex]
Now we can use the equation that defines the volume of a cylinder.
V=
D=3ft=0.9144m
h=9.23m
solving
finally we use conversion factors to find the volume in gallons
Answer:
Data structures are important as it allows the user to insert, update, arrange, rearrange, delete, and retrieve data in an efficient manner, from the database. And to accomplish the said tasks algorithms are used. It is used to manipulate the stored data within the database in the required manner. …
Explanation:
hope you like my ans
ANSWER:
Aerospace Engineering. ...
Chemical Engineering. ...
Biomedical Engineering.
EXPLANATION:
This is all i know but ... I hope this helps~