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AnnyKZ [126]
2 years ago
5

A gas mixture containing 85.0 mole% N2 and the balance n-hexane flows through a pipe at a rate of 100.0 m3/h. The pressure is 2.

00 atm absolute and the temperature is 140.0°C.
a. What is the molar flow rate of the gas? kmol/h




b. To what temperature would the gas have to be cooled at constant pressure in order to begin condensing hexane? °C




c. To what temperature would the gas have to be cooled at constant pressure in order to condense 75.0% of the hexane? °C
Engineering
1 answer:
Nadusha1986 [10]2 years ago
7 0

Answer:

a.6531.53 mole/hr

b. 32.76 degC

c.  3.78 deg.c

Explanation:A gas mixture containing 85.0 mole% N2 and the balance n-hexane flows through a pipe at a rate of 100.0 m3/h. The pressure is 2.00 atm absolute and the temperature is 140.0°C.

a. What is the molar flow rate of the gas? kmol/h

b. To what temperature would the gas have to be cooled at constant pressure in order to begin condensing hexane? °C

c. To what temperature would the gas have to be cooled at constant pressure in order to condense 75.0% of the hexane? °C

Given V= Volume of gas mixture= 100m3/hr=10^5 Lt/hr P= 2.0 atm and T= 100 deg.c =100+273.15= 373.15K

n= moles of mixture= PV/RT , where  R=0.08206 L.atm/mole.K

n= 2.0*10^5/0.08206*373.15)=6531.53 mole/hr

b. To what temperature would the gas have to be cooled at constant pressure in order to begin condensing hexane? A gas mixture containing 85 mole% N2 and the°C

Condensation begins at a point at which the partial pressure of vapor =vapor pressure of liquid at the given temperature

Partial pressure of  hexane in the mixture= 0.15*2.0= 0.3 atm

so for saturation to begin, the vapor pressure shoudl correspond to 0.3 atm=30.39 Kpa

Antoine constant for Hexane

lnP  (Kpa)= 13.82- 2696/(T-48.833)  ( T is in K

ln(30.39 )= 13.8193- 2696/ (T-48.833)

, 3.414= 13,82-2696/(T-48.333)

2696/(T-48.833)= 13.82-3.414=10.405

T-46.833= 2696/10.414=259.08

259.08+46.833 K=305.91k

305.91k-273.15K

32.76C

c. To what temperature would the gas have to be cooled at constant pressure in order to condense 75% of the hexane? °C?

Vapor present in the gas mixture= 25%, Its partial pressure=0.15* 0.25*2.0 =0.075 atm= 7.59Kpa

from ln(7.59)= 13.82- 2696/ (T-48.333)

2.02 = 13.82- 2696/(T-48.333)

2696/(T-48.333)= 11.179

T-48.333= 2696/11.79=228.60

T= 228.6+48.333= 276.93 K= 3.78 deg.c

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Answer:

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Explanation:

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R = V^2/ P

R = 400^2/100000

Resistance in load = 1.6 Ohms

Current l = V / R

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V = 250 × 2.367

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B.) Voltage at the source side Vsource will be

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591.7/991.7 × 100 = 59.7%

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Answer:

<em>0.0386 hr</em>

<em></em>

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<em>There are 24 hrs in a day,</em> therefore rate in hrs will be

220/24 = 9.17 mole/m^2-hr

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molar mass of water = (1 x 2) + 16 = 18 kg/mole

therefore,

<em>mole of water = mass of water/molar mass of water</em>

mole of water = 0.4/18 = 0.02 mole

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A thermal energy storage unit consists of a large rectangular channel, which is well insulated on its outer surface and encloses
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Answer:

the temperature of the aluminum at this time is 456.25° C

Explanation:

Given that:

width w of the aluminium slab = 0.05 m

the initial temperature T_1 = 25° C

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h = 100 W/m²

The properties of Aluminium at temperature of 600° C by considering the conditions for which the storage unit is charged; we have ;

density ρ = 2702 kg/m³

thermal conductivity k = 231 W/m.K

Specific heat c = 1033 J/Kg.K

Let's first find the Biot Number Bi which can be expressed by the equation:

Bi = \dfrac{hL_c}{k} \\ \\ Bi = \dfrac{h \dfrac{w}{2}}{k}

Bi = \dfrac{hL_c}{k} \\ \\ Bi = \dfrac{100 \times \dfrac{0.05}{2}}{231}

Bi = \dfrac{2.5}{231}

Bi = 0.0108

The time constant value \tau_t is :

\tau_t = \dfrac{pL_cc}{h} \\ \\ \tau_t = \dfrac{p \dfrac{w}{2}c}{h}

\tau_t = \dfrac{2702* \dfrac{0.05}{2}*1033}{100}

\tau_t = \dfrac{2702* 0.025*1033}{100}

\tau_t = 697.79

Considering Lumped capacitance analysis since value for Bi is less than 1

Then;

Q= (pVc)\theta_1 [1-e^{\dfrac {-t}{ \tau_1}}]

where;

Q = -\Delta E _{st} which correlates with the change in the internal energy of the solid.

So;

Q= (pVc)\theta_1 [1-e^{\dfrac {-t}{ \tau_1}}]= -\Delta E _{st}

The maximum value for the change in the internal energy of the solid  is :

(pVc)\theta_1 = -\Delta E _{st}max

By equating the two previous equation together ; we have:

\dfrac{-\Delta E _{st}}{\Delta E _{st}{max}}= \dfrac{  (pVc)\theta_1 [1-e^{\dfrac {-t}{ \tau_1}}]} { (pVc)\theta_1}

Similarly; we need to understand that the ratio of the energy storage to the maximum possible energy storage = 0.75

Thus;

0.75=  [1-e^{\dfrac {-t}{ \tau_1}}]}

So;

0.75=  [1-e^{\dfrac {-t}{ 697.79}}]}

1-0.75=  [e^{\dfrac {-t}{ 697.79}}]}

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In(0.25) =  {\dfrac {-t}{ 697.79}}

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t = 1.386294361 × 697.79

t = 967.34 s

Finally; the temperature of Aluminium is determined as follows;

\dfrac{T - T _{\infty}}{T_1-T_{\infty}}= e ^ {\dfrac{-t}{\tau_t}}

\dfrac{T - 600}{25-600}= e ^ {\dfrac{-967.34}{697.79}

\dfrac{T - 600}{25-600}= 0.25

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T - 600 = -575 × 0.25

T - 600 = -143.75

T = -143.75 + 600

T = 456.25° C

Hence; the temperature of the aluminum at this time is 456.25° C

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