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2 years ago

Chlorine is one of the important commodity chemicals for the global economy. Before the advent of large scale

Engineering

1 answer:

artcher [175]2 years ago

7 0

The composition of gas in the feed, the percentage conversion and the

theoretical yield are combined to give the product stream composition.

Response:

The composition of gas in the product stream are;

HCl: 0.4 kmol/h, Cl₂: 1.6 kmol/h, H₂O: 1.6 kmol/h, O₂: 0.5 kmol/h

<h3>How can percentage conversion give the contents of the product stream?</h3>

The amount of oxygen used = 30% exceeding the theoretical amount

Number of moles of hydrochloric acid = 4 kmol/h

Percentage conversion = 80%

Required:

The composition of the gas in the product feed.

Solution;

The given reaction is; 4HCl + O₂ $\longrightarrow$ 2Cl₂ + 2H₂O

$Percentage \ conversion = \mathbf{ \dfrac{Moles \ of \ limiting \ reactant \ reacted}{Moles \ of \ limiting \ reactant \ supplied \ in \ the \, feed}}$

Which gives;

$80 \% = \mathbf{ \dfrac{Moles \ of \ limiting \ reactant \ reacted}{4 \, kmol/h}}$

Moles of limiting reactant reacted = 4 kmol/h × 0.80 = 3.6 kmol/h

Which gives;

Number of moles of HCl in the stream = 4 kmol/h - 3.6 kmol/h = 0.4 kmol/h

Number of moles of Cl₂ produced = 2 kmol/h × 0.8 = 1.6 kmol/h

Similarly;

Number of moles of H₂O produced = 2 kmol/h × 0.8 = 1.6 kmol/h

Number of moles of O₂ in the product stream = 30% × 1 kmol/h + 20% × 1 kmol/h = 0.5 kmol/h

The composition of the production stream is therefore;

HCl: 0.4 kmol/h

Cl₂: 1.6 kmol/h

H₂O: 1.6 kmol/h

O₂: 0.5 kmol/h

Learn more about theoretical and actual yield here:

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Assume you are an observer standing at a point along a three-lane roadway. All vehicles in lane 1 are traveling at 30 mi/h, all

garik1379 [7]

Answer:

Explanation:

Given data;

In line 1, v1 = 30mi/hr

in line 2, v2 = 45mi/hr

in line 3, v3 = 60mi/hr

therefore time mean speed = v1 + v2 + v3 /n

= VT = 45mi/hr

space mean speed ; Vs

harmonic mean = 1/V = 1/v1 + 1/v2 + 1/v3

V = 13.85mi/hr

Hence Vs = V x n = 3 x 13.85 = 41.55mi/hr

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2 years ago

Describe the testing process you might use for a new tablet operating system.

rusak2 [61]

Answer:

You may execute some or all Test Cases based on the requirement in all versions of Mobile that is 2g, 3g, and 4g. Tablets and smartphones have special operating-system needs.

Explanation:

3 0

2 years ago

Taking the convection heat transfer coefficient on both sides of the plate to be 860 W/m2 ·K, deter- mine the temperature of the

Aleks [24]

Answer:

Hello your question is incomplete attached below is the complete question

answer :

a) 95.80°C

b) 8.23 MW

Explanation:

Convection heat transfer coefficient = 860 W/m^2 . k

a) Calculate for the temp of sheet metal when it leaves the oil bath

first step : find the Biot number 

Bi = hLc / K ------- ( 1 )

where : h = 860 W/m^2 , Lc = 0.0025 m , K = 60.5 W/m°C

Input values into equation 1 above

Bi = 0.036 which is < 1 ( hence lumped parameter analysis can be applied )

next : find the time constant 

t ( time constant ) = h / P*Cp *Lc --------- ( 2 )

where : p = 7854 kg/m^3 , Lc = 0.0025 m , h = 860 W/m^2, Cp = 434 J/kg°C

Input values into equation 2 above

t ( time constant ) = 0.10092 s^-1

Determine the elapsed time

T = L / V = 9/20 = 0.45 min

∴ temp of sheet metal when it leaves the oil bath

= (T(t) - 45 ) / (820 - 45) = e^-(0.10092 * 27 )

T∞ = 45°C

Ti = 820°C

hence : T(t) = 95.80°C

b) Calculate the required rate of heat removal form the oil

Q = mCp ( Ti - T(t) ) ------------ ( 3 )

m = ( 7854 *2 * 0.005 * 20 ) = 26.173 kg/s

Cp = 434 J/kg°C

Ti = 820°C

T(t) = 95.80°C

Input values into equation 3 above

Q = 8.23 MW

6 0

2 years ago

A gas turbine plant operates on the regenerative Brayton cycle with two stages of reheating and two stages of intercooling betwe

Colt1911 [192]

Answer:

Explanation:

Given

T₁ = 300 K,

T₃ = 350 K

P₁, P₃ = 100 kPa

P₂, P₄ = 1200 kPa

T₆ = 1400 K

T₈ = 1300 K

$\eta_{Compressor}$ , $\eta_{Turbine}$ = 80 %

Effectiveness of regenerator = 75 %

Step by step solution is given as

Point 1

T₁= 300 K

Point 2

$\eta_{c} = \frac{h_{2s}- h_{1} }{h_{2} -h_{1} } = \frac{c_{p}( T_{2s}- T_{1} }{c_{p}(T_{2}- T_{1})}$ from which T₂ = T₁ + $\frac{( T_{2s}- T_{1} )}{\eta_{c} }$

From the above therefore to find T₂, we look for the value of $T_{2s}$ as follows

$\frac{T_{2s} }{T_{1} } = (\frac{P_{2} }{P_{1} })^{\frac{k-1}{k} }$ from which $T_{2s} = 300 K(\frac{1200 kPa}{100kPa} )^{\frac{0.4}{1.4} } =$ 610.18 K

Therefore T₂ = T₁ + $\frac{( T_{2s}- T_{1} )}{\eta_{c} }$ = 300 K + $\frac{610.18-300}{0.8}$ = 687.726 K

Point 3

T₃ = 350 K, inlet condition of 2nd compressor

Point 4

The value of T₄ is calculated as follows

$\eta_{c} = \frac{h_{4s}- h_{3} }{h_{4} -h_{3} } = \frac{c_{p}( T_{4s}- T_{3} }{c_{p}(T_{4}- T_{3})}$ from which T₄ = T₃ + $\frac{( T_{4s}- T_{3} )}{\eta_{c} }$

also since

$\frac{T_{4s} }{T_{3} } = (\frac{P_{4} }{P_{3} })^{\frac{k-1}{k} }$ we have $T_{4s} = 350 K(\frac{1200 kPa}{100kPa} )^{\frac{0.4}{1.4} } =$ 711.87 K

T4 = 802.34 K

Point 5

$\epsilon_{regen} =\frac{q_{regen,act} }{q_{regen,max}} = \frac{h_{5}- h_{4} }{h_{9} -h_{4} } = \frac{c_{p}( T_{5}- T_{4} }{c_{p}(T_{9}- T_{4})}$ from which T₅ =T₄ + $\epsilon_{regen}$ ×(T₉ - T₄)

Therefore we look for T₉ first before determining T₅

Point 6

T₆ is given as = 1400 K

Point 7

Here we also have

$\eta_{t} = \frac{h_{6}- h_{7} }{h_{6} -h_{7s} } = \frac{c_{p}( T_{6}- T_{7} }{c_{p}(T_{6}- T_{7s})}$ from which T₇ = T₆ - $\eta_{t}{( T_{6}- T_{7s} )}$

also $\frac{T_{7s} }{T_{6} } = (\frac{P_{7} }{P_{6} })^{\frac{k-1}{k} }$ which gives $T_{7s} = 1400 K(\frac{P_{7} }{P_{6} } )^{\frac{0.4}{1.4} }$ here we assume efficient cycle with pressure ratio = √12 thus we have

$T_{7s} = 1400 K(\frac{1}{12 } } )^{\frac{0.4}{1.4} }$ = 688.32 K

and T₇ = T₆ - $\eta_{t}{( T_{6}- T_{7s} )}$ = 1400 - 0.8(1400-688.32) = 830.656 K

Point 8

T₈ = 1400 K

Point 9

Here we have

$\eta_{t} = \frac{h_{8}- h_{9} }{h_{8} -h_{9s} } = \frac{c_{p}( T_{8}- T_{9} }{c_{p}(T_{8}- T_{9s})}$ from which T₉ = T₈ - $\eta_{t}{( T_{8}- T_{9s} )}$

and $\frac{T_{9s} }{T_{8} } = (\frac{P_{9} }{P_{8} })^{\frac{k-1}{k} }$ → $T_{9s} = 1400 K(\frac{1}{12 } } )^{\frac{0.4}{1.4}}$ = 688.32 K

Therefore T₉ = 830.656 K

Therefore for point 5 we have T₅ =T₄ + $\epsilon_{regen}$ ×(T₉ - T₄) = 711.87 K + 0.75×(830.656 K-711.87 K) = 914.21 K

T₅ = 800.9595 K

(a) Back work ratio is given as T₁/T₉ = 300K/981.66 K = 0.3056

(b) Thermal efficiency under cold air standard is given as

$\eta_{th,regen}$ = $1-(\frac{T_{1} }{T_{6} })(r_{p} )^{\frac{k-1}{k} }$ = $1-\frac{300}{1300}\frac{1200}{100}^{\frac{0.4}{1.4} }$ =0.5606 or 53.06 %

$\eta _{turbine} = \frac{W_{real} }{W_{rev} }$ = Wrev = 1 - TL/TH = 1 -300/1400 = 0.786

Therefore we have

$\eta_{II} =$ 0.53/0.786 =0.675

(d) The exergy is given by $\epsilon = \frac{E_{out}-E_{in} }{W }$ for the compressor

and for the turbine we have

$\epsilon =\frac{W_{net} +W_{c} }{E_{in} - E_{out} }$

6 0

2 years ago

jeka57 [31]

Answer:

False

Explanation:

When we open the gate of refrigerator then the temperature of kitchen get increase.

But we think how it is possible that temperature of kitchen get increases.Actually the heat rejecting coil(condenser) is also present inside the room that is why it reject more heat as compare to heat absorbed by evaporator coil.This is the reason to increases the temperature of kitchen.

If heat rejecting coil put outside the kitchen then it will produces cooling effect even door of refrigerator open.This concept use in air conditioning.

8 0

2 years ago