Question

Q1. A hydroelectric power plant uses the reaction of ethylene (?2?4) with oxygen to generate vapour. The products of this reaction are carbon dioxide and water. Additionally, the reaction produces −1323.16?? per mol of ethylene consumed. a) How much ethylene is required by the power plant if 8500 kg of steam at 120 °C is produced in an hour? (assume 100% yield) (12.5 marks) b) Recalculate your answer for a 75 % yield. (12.5 marks) Assumptions. The heat capacities of water are C?? = 4.19 kJ⁄kg°C, ??? = 1.996 ??⁄??°C, the heat of ?? vaporization for water is ∆h? = 2256.9 ⁄??, and the power plant takes water from a source at 15 °C.

Answer 1

a) Given: Ethylene reacts with oxygen to give CO₂ and H₂O. The energy released during the reaction is -1323.16 KJ/mol

8500 Kg of steam is to be produced in an hour at 120 C from the water available at 15 C.

Cpl = 4.19 KJ/KgC ; Cpv = 1.996 KJ/KgC ; Hv = 2256.9 KJ/Kg

First lets calculate the energy required for converting water at 15 C to steam at 120 C.  

Basis: 1 hour of operation

Heat required for the conversion of water to steam = Heat required to increase the temperature of water from 15 C to 100 C + Heat of vapourization + Heat required to increase the temperature of steam from 100 to 120 C

Heat required to increase the temperature of water from 15 C to 100 C = m * Cpl * T

m = 8500 Kg ; Cpl = 4.19 KJ/KgC ; T = 100 - 15 = 85 C;

Heat required to increase the temperature of water from 15 C to 100 C = 8500 * 4.19 * 85

= 3027275 KJ

Heat of vapourization = m* Hv

m = 8500 Kg ; Hv = 2256.9 KJ/Kg

Heat of vapourization = 8500*2256.9 = 19183650 KJ

Heat required to increase the teperature of steam from 100 to 120 C =  m * Cpv * T

m = 8500 Kg ; Cpl = 1.996 KJ/KgC ; T = 120 -100 = 20 C;

Heat required to increase the teperature of steam from 100 to 120 C = 8500*1.996*20

= 339320 KJ

Heat required for the conversion of water to steam = 3027275 KJ + 19183650 KJ + 339320 KJ

Heat required for the conversion of water to steam = 22550245 KJ

Heat from ethylene reaction = Heat required for the conversion of water to steam = 22550245 KJ

Energy released per mole ethylene converted = -1323.16 KJ

Moles of to ethylene to be reacted to give out 22550245 KJ = 22550245 KJ / energy releases per mole converted

= 22550245/1323.16 = 17042.72 mole

Ethylene required for 22550245 KJ of energy (100 % conversion) = 17042.72 mol = 17.043 Kmol

in Kg, 1 kmol = 28.05 Kg

Ethylene required for 22550245 KJ of energy (100 % conversion) = 17.043 Kmol * 28.05 Kg/1 Kmol = 478.056 Kg

b) 75% conversion.

The energy required is released only by 75% of the ethylene supplied and from previous calculations it amounts to 17.043 Kmol or 478.056 Kg. Using direct variation,

i.e 17.043 Kmol = 75%

x Kmol = 100%

gives, x = 17.043*100/75 = 22.724 Kmol

in Kg,

22.274 kmol = 22.274 * 28.05 Kg/ 1 Kmol = 637.408 Kg

Ethylene required for 75% conversion = 22.274 Kmol or  637.408 Kg

All the above quantities are for one hour basis, so in a continuos process, only the units will change to Kmol/hr or Kg/hr.