Answers
Answer:
The Answer is C!
Explanation:
Related Questions
Un mol de gas ideal realiza un trabajo de 3000 J sobre su entorno, cuando se expande de manera isotermica a una temperatura de 58°C, cuando su volumen inicial es de 25 L. Determinar el volumen final
Answers
Answer:
74,4 litros
Explanation:
Dado que
W = nRT ln (Vf / Vi)
W = 3000J
R = 8,314 JK-1mol-1
T = 58 + 273 = 331 K
Vf = desconocido
Vi = 25 L
W / nRT = ln (Vf / Vi)
W / nRT = 2.303 log (Vf / Vi)
W / nRT * 1 / 2.303 = log (Vf / Vi)
Vf / Vi = Antilog (W / nRT * 1 / 2.303)
Vf = Antilog (W / nRT * 1 / 2.303) * Vi
Vf = Antilog (3000/1 * 8,314 * 331 * 1 / 2,303) * 25
Vf = 74,4 litros
A gas stream contains 18.0 mole% hexane and the remainder nitrogen. The stream flows to a condenser, where its temperature is reduced and some of the hexane is liquefied. The hexane mole fraction in the gas stream leaving the condenser is 0.0500. Liquid hexane condensate is recovered at a rate of 1.50 L/min.
(a) What is the flow rate of the gas stream leaving the condenser in mol/min? (Hint : First calculate the molar flow rate of the condensate and note that the rates at which C6H14 and N2 enter the unit must equal the total rates at which they leave in the two exit streams.)
(b) What percentage of the hexane entering the condenser is recovered as a liquid?
Answers
Answer:
A. 72.34mol/min
B. 76.0%
Explanation:
A.
We start by converting to molar flow rate. Using density and molecular weight of hexane
= 1.59L/min x 0.659g/cm³ x 1000cm³/L x 1/86.17
= 988.5/86.17
= 11.47mol/min
n1 = n2+n3
n1 = n2 + 11.47mol/min
We have a balance on hexane
n1y1C6H14 = n2y2C6H14 + n3y3C6H14
n1(0.18) = n2(0.05) + 11.47(1.00)
To get n2
(n2+11.47mol/min)0.18 = n2(0.05) + 11.47mol/min(1.00)
0.18n2 + 2.0646 = 0.05n2 + 11.47mol/min
0.18n2-0.05n2 = 11.47-2.0646
= 0.13n2 = 9.4054
n2 = 9.4054/0.13
n2 = 72.34 mol/min
This value is the flow rate of gas that is leaving the system.
B.
n1 = n2 + 11.47mol/min
72.34mol/min + 11.47mol/min
= 83.81 mol/min
Amount of hexane entering condenser
0.18(83.81)
= 15.1 mol/min
Then the percentage condensed =
11.47/15.1
= 7.59
~7.6
7.6x100
= 76.0%
Therefore the answers are a.) 72.34mol/min b.) 76.0%
Please refer to the attachment .
g A smooth pipeline with a diameter of 5 cm carries glycerin at 20 degrees Celsius. The flow rate in the pipe is 0.01 m3/s. What is the friction factor
Answers
Answer:
The friction factor is 0.303.
Explanation:
The flow velocity ([tex]v[/tex]), measured in meters per second, is determined by the following expression:
[tex]v = \frac{4\cdot \dot V}{\pi \cdot D^{2}}[/tex] (1)
Where:
[tex]\dot V[/tex] - Flow rate, measured in cubic meters per second.
[tex]D[/tex] - Diameter, measured in meters.
If we know that [tex]\dot V = 0.01\,\frac{m^{3}}{s}[/tex] and [tex]D = 0.05\,m[/tex], then the flow velocity is:
[tex]v = \frac{4\cdot \left(0.01\,\frac{m^{3}}{s} \right)}{\pi\cdot (0.05\,m)^{2}}[/tex]
[tex]v \approx 5.093\,\frac{m}{s}[/tex]
The density and dinamic viscosity of the glycerin at 20 ºC are [tex]\rho = 1260\,\frac{kg}{m^{3}}[/tex] and [tex]\mu = 1.5\,\frac{kg}{m\cdot s}[/tex], then the Reynolds number ([tex]Re[/tex]), dimensionless, which is used to define the flow regime of the fluid, is used:
[tex]Re = \frac{\rho\cdot v \cdot D}{\mu}[/tex] (2)
If we know that [tex]\rho = 1260\,\frac{kg}{m^{3}}[/tex], [tex]\mu = 1.519\,\frac{kg}{m\cdot s}[/tex], [tex]v \approx 5.093\,\frac{m}{s}[/tex] and [tex]D = 0.05\,m[/tex], then the Reynolds number is:
[tex]Re = \frac{\left(1260\,\frac{kg}{m^{3}} \right)\cdot \left(5.093\,\frac{m}{s} \right)\cdot (0.05\,m)}{1.519 \frac{kg}{m\cdot s} }[/tex]
[tex]Re = 211.230[/tex]
A pipeline is in turbulent flow when [tex]Re > 4000[/tex], otherwise it is in laminar flow. Given that flow has a laminar regime, the friction factor ([tex]f[/tex]), dimensionless, is determined by the following expression:
[tex]f = \frac{64}{Re}[/tex]
If we get that [tex]Re = 211.230[/tex], then the friction factor is:
[tex]f = \frac{64}{211.230}[/tex]
[tex]f = 0.303[/tex]
The friction factor is 0.303.
An unknown impedance Z is connected across a 380 V, 60 Hz source. This causes a current of 5A to flow and 1500 W is consumed. Determine the following: a. Real Power (kW) b. Reactive Power (kvar) c. Apparent Power (kVA) d. Power Factor e. The impedance Z in polar and rectangular form
Answers
Answer:
a) Real Power (kW) = 1.5 kW
b) Reactive Power (kvar) is 1.1663 KVAR
c) Apparent Power (kVA) is 1.9 KVA
d) the Power Factor cos∅ is 0.7894
e) the impedance Z in polar and rectangular form is 76 ∠ 37.87° Ω
Explanation:
Given that;
V = 380v
i = 5A
P = 1500 W
determine;
a) Real Power (kW)
P = 1500W = 1.5 kW
therefore Real Power (kW) = 1.5 kW
b) Reactive Power (kvar)
p = V×i×cos∅
cos∅ = p / Vi
cos∅ = 1500 / ( 380 × 5 ) = 0.7894
∅ = cos⁻¹ (0.7894)
∅ = 37.87°
Q = VIsin∅
Q = 380 × 5 × sin( 37.87° )
Q = 1.1663 KVAR
Therefore Reactive Power (kvar) is 1.1663 KVAR
c) Apparent Power (kVA)
S = P + jQ
= ( 1500 + J 1166.3 ) VA
S = 1900 ∠ 37.87° VA
S = 1.9 KVA
Therefore Apparent Power (kVA) is 1.9 KVA
d) Power Factor
p = V×i×cos∅
cos∅ = p / Vi
cos∅ = 1500 / ( 380 × 5 ) = 0.7894
Therefore the Power Factor cos∅ is 0.7894
e) The impedance Z in polar and rectangular form
Z = 380 / ( S∠-37.87) = V/I
Z = ( 60 + j 46.647) Ω
Z = 76 ∠ 37.87° Ω
Therefore the impedance Z in polar and rectangular form is 76 ∠ 37.87° Ω