Answer:
Actual COP = 5.368
Maximum theoretical COP = 6.368
Explanation:
Given - An ideal vapor compression refrigeration cycle operates with a condenser pressure of 900 kPa. The temperature at the inlet to the compressor is -5oC.
To find - If this device operates using R134a as the working fluid. Calculate the actual COP of this device as well as the maximum theoretical COP.
Proof -
Given that,
An ideal vapor compression refrigeration cycle operates with a condenser pressure of 900 kPa.
From Refrigerant 134-a Table
At T1 = -5°C
h1 = 247.505 KJ/kg
S1 = 0.93434 KJ/kg
At P2 = 900 KPa
S1 = S2
h2 = 274.679 Kj/Kg
h3 = h4 = 101.61 KJ/g
So,
Compressor work (Wc) = h2 - h1
= 274.679 - 247.505
= 27.174
⇒Compressor work (Wc) = 27.174 KJ/kg
Now,
Heat out (Qout) = h2 - h3
= 274.679 - 101.61
= 173.069
⇒Heat out (Qout) = 173.069 KJ/kg
Now,
Heat input (Qin) = h1 - h4
= 274.505 - 101.61
= 145.895
⇒Heat input (Qin) = 145.895 KJ/kg
So,
Actual COP at the refrigerator is -
(COP)R = (Qin)/(Wc)
= (145.895)/ (27.174)
= 5.368
⇒Actual COP = 5.368
Now,
Maximum theoretical COP is -
(COP) = (Qout)/(Wc)
= (173.069)/ (27.174)
= 6.368
⇒Maximum theoretical COP = 6.368
Answer:
a) Check explanation for this
b)Rate law is ![Rate = \frac{k_{1}k_{4} }{k_{3}+ 2k_{4} } [H_{2} ]](https://tex.z-dn.net/?f=Rate%20%3D%20%5Cfrac%7Bk_%7B1%7Dk_%7B4%7D%20%20%7D%7Bk_%7B3%7D%2B%202k_%7B4%7D%20%20%7D%20%5BH_%7B2%7D%20%5D)
c) The rate does not depend on the concentration of CO₂
Explanation:
a) Elementary steps for the RWGS reaction:
- Dissociative adsorption of the H₂ Molecule
(Fast process)
- Reversible Reaction between CO₂ and H
(Fast Process)
- Slow dissociation of COOH into gaseous CO and absorbed OH
(Slow process)
- Fast hydrogenation of the OH to form H₂O
(Fast process)
b) Derivation of the rate law
We need to determine the rate law for H, OH and COOH because these are the intermediates for this reaction.
The steady state approximation is applied to a consecutive reaction with a slow first step and a fast second step (k1≪k2). If the first step is very slow in comparison to the second step, there is no accumulation of intermediate product.
Rate of consumption = Rate of production
For COOH:
Using steady state approximation
![\frac{d[COOH]}{dt} = 0](https://tex.z-dn.net/?f=%5Cfrac%7Bd%5BCOOH%5D%7D%7Bdt%7D%20%3D%200)
![k_{2} [CO_{2} ][H] = k_{3} [COOH] k_{4} [COOH]\\](https://tex.z-dn.net/?f=k_%7B2%7D%20%5BCO_%7B2%7D%20%5D%5BH%5D%20%3D%20k_%7B3%7D%20%5BCOOH%5D%20k_%7B4%7D%20%5BCOOH%5D%5C%5C)
![[COOH] = \frac{k_{2} [CO_{2} ][H]}{k_{3}k_{4} } \\](https://tex.z-dn.net/?f=%5BCOOH%5D%20%3D%20%5Cfrac%7Bk_%7B2%7D%20%5BCO_%7B2%7D%20%5D%5BH%5D%7D%7Bk_%7B3%7Dk_%7B4%7D%20%20%7D%20%5C%5C)
For H:
![\frac{d[H]}{dt} = 0](https://tex.z-dn.net/?f=%5Cfrac%7Bd%5BH%5D%7D%7Bdt%7D%20%3D%200)
![k_{1}[H_{2}] = k_{2}[CO_{2} [H]+k_{5} [ OH][H]](https://tex.z-dn.net/?f=k_%7B1%7D%5BH_%7B2%7D%5D%20%3D%20k_%7B2%7D%5BCO_%7B2%7D%20%5BH%5D%2Bk_%7B5%7D%20%5B%20OH%5D%5BH%5D)
![[H]= \frac{k_{1}[H_{2}] }{k_{5}[OH] +k_{2}[CO_{2}]}\\](https://tex.z-dn.net/?f=%5BH%5D%3D%20%5Cfrac%7Bk_%7B1%7D%5BH_%7B2%7D%5D%20%20%7D%7Bk_%7B5%7D%5BOH%5D%20%2Bk_%7B2%7D%5BCO_%7B2%7D%5D%7D%5C%5C)
For OH:
![\frac{d[OH]}{dt} = 0](https://tex.z-dn.net/?f=%5Cfrac%7Bd%5BOH%5D%7D%7Bdt%7D%20%3D%200)
![k_{4} [COOH] = k_{5} [OH][H]\\\k[OH] = \frac{k_{4} [COOH]}{k_{5} H}\\](https://tex.z-dn.net/?f=k_%7B4%7D%20%5BCOOH%5D%20%3D%20k_%7B5%7D%20%5BOH%5D%5BH%5D%5C%5C%5Ck%5BOH%5D%20%3D%20%5Cfrac%7Bk_%7B4%7D%20%5BCOOH%5D%7D%7Bk_%7B5%7D%20H%7D%5C%5C)
The rate of the overall reaction is determined by the slowest step of the reaction. The slowest process is the dissociation of COOH
Therefore the overall rate of reaction is:
![Rate = k_{4} [COOH]\\](https://tex.z-dn.net/?f=Rate%20%3D%20k_%7B4%7D%20%5BCOOH%5D%5C%5C)
![Rate = k_{4} \frac{k_{2} [CO_{2} ][H]}{k_{3}k_{4} }\\Rate = k_{4} \frac{k_{2}[CO_{2}]\frac{k_{1}[H_{2}] }{k_{5}[OH] +k_{2}[CO_{2}]} }{k_{3}k_{4}}\\Rate = k_{4} \frac{k_{2}[CO_{2}]\frac{k_{1}[H_{2}] }{k_{5}\frac{k_{4}COOH }{k_{5}H } +k_{2}[CO_{2}]} }{k_{3}k_{4}}](https://tex.z-dn.net/?f=Rate%20%3D%20k_%7B4%7D%20%20%5Cfrac%7Bk_%7B2%7D%20%5BCO_%7B2%7D%20%5D%5BH%5D%7D%7Bk_%7B3%7Dk_%7B4%7D%20%20%7D%5C%5CRate%20%3D%20k_%7B4%7D%20%20%5Cfrac%7Bk_%7B2%7D%5BCO_%7B2%7D%5D%5Cfrac%7Bk_%7B1%7D%5BH_%7B2%7D%5D%20%20%7D%7Bk_%7B5%7D%5BOH%5D%20%2Bk_%7B2%7D%5BCO_%7B2%7D%5D%7D%20%20%7D%7Bk_%7B3%7Dk_%7B4%7D%7D%5C%5CRate%20%3D%20k_%7B4%7D%20%20%5Cfrac%7Bk_%7B2%7D%5BCO_%7B2%7D%5D%5Cfrac%7Bk_%7B1%7D%5BH_%7B2%7D%5D%20%20%7D%7Bk_%7B5%7D%5Cfrac%7Bk_%7B4%7DCOOH%20%7D%7Bk_%7B5%7DH%20%7D%20%20%2Bk_%7B2%7D%5BCO_%7B2%7D%5D%7D%20%20%7D%7Bk_%7B3%7Dk_%7B4%7D%7D)
Simplifying the equation above, the rate law becomes
![Rate = \frac{k_{1}k_{4} }{k_{3}+ 2k_{4} } [H_{2} ]](https://tex.z-dn.net/?f=Rate%20%3D%20%5Cfrac%7Bk_%7B1%7Dk_%7B4%7D%20%20%7D%7Bk_%7B3%7D%2B%202k_%7B4%7D%20%20%7D%20%5BH_%7B2%7D%20%5D)
c) It is obvious from the rate law written above that the rate of the RWBG reaction does not depend on the concentration of CO₂
Answer:
The answer is true
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Explanation:
Answer:
nothing much what class r u in
Answer:
a) 0.487
b) refrigeration load = 5.46w
c) cop = 2.24
d)ref load max = 12.43kw
Explanation: