Map A continuous reaction requires a steady flow rate of ethylene. Ethylene is s

Question

Map A continuous reaction requires a steady flow rate of ethylene. Ethylene is supplied by a 55.7-L gas cylind initially at 2200.0 psig and 25.0°C. Once leaving the cylinder, a regulator reduces the pressure to 15.0 psig, and a valve allows 0.900 mL/s of ethylene at 25.0°C to flow into the reactor At this flow rate, how long would it take before the gas cylinder needs to be replaced (before the cylinder pressure reaches 15.0 psig)? Assume that the cylinder doesn't change temperature. Use the law of corresponding states and compressibility charts to describe ethylene within the cylinder. Atmospheric pressure is 1 atm. First, label the process flow diagram initial cylinder conditions pressure regulator flow valve to the reactor V= 55.7 mL/s V= 0.900 ninitial = unknown mol C2H4 Pi2200.0 unknown molC2Has unknown mol C2Ha/s nitial psig P215.0 psig initial25.0 gas cylinder T2 25.0 final cylinder conditions 0.900 2200.0 150 0 V= unknown nfinal-unknown mol C2H4 Pinal-unknown psig final = unknown unknown 55.7 25.0 What is the reduced temperature of ethylene at 25.0°C? Number

Explanation / Answer

For real gases you have to use the comprensibility factor to ee the real behavior of gas in a media. The correspondent states law comes from Van der waals. And they are several of eos for different cases and topics. And all of them obbey this fundamental law.

First, calculate the reduced temperature of ethylene at 25°C. The initial temperature is:

T = 25°C = 536.67 R (Ranking)

Critical temperature : 9.2 °C = 508.23

P = 2200 psig = 154.67 atm

Critical pressure : 50.42 bar = 49.76 atm

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1. Tr = T/Tc = 536.67/508.23 = 1.0556

2. Pr = P/Pc = 154.67 atm/49.76 atm = 3.1083

3. From the Standing and Katz, Factor Z chart:

Z = 0.42

4. The initial moles C2H4:

PV = ZnRT

P, R & T are constants!

R = 10.73 psi.ft3/(lbmol.oR)

V = 55.7 L = 0.0557 m3 = 1.96 ft3

n = 2200 psig x 1.96 ft3 / 0.42 x 10.73 psi.ft3/(lbmol.oR) x 536.67 R

ni = 1.7828 lb/moles C2H4

5. The final moles C2H4:

n = (15 psig x 1.96ft3) / (0.42 x 10.73 x 536.67)

nf = 0.012 lb/moles C2H4

6. The dn C2H4:

dn = ni - nf = 1.7828 - 0.012 = 1.7706 lb/moles

7. Time required to flow at 0.900 ml/s = 0.9x10-3 L/s

t = V/Q = 55.7 L / (0.9x10-3 L/s)

t = 61.8x103 seg = 0.71 days!

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