Chemical Engineering Question: The process below is part of a system that liquef

Question

Chemical Engineering Question:

The process below is part of a system that liquefies carbon dioxide (all streams are CO2). Stream 1 is at 20 degree C, 38 bar, passes through the heat exchanger where it becomes saturated liquid (stream 2) and is then throttled to 8 bar (stream 3) to produce a vapor-liquid mixture. Stream 3 is separated in a vapor/liquid separator, with all vapor exiting in stream 5 and all liquid as stream 4. Stream 6 is at 20 degree C, 38 bar and expands in a turbine. Stream 7, which contains 5% liquid, is mixed with stream 5. The resulting stream 8 is passed through the heat exchanger and exits at 20 degree C. Use the data given below and a basis of 1 mol/s in stream 1 to do the following:

a) Obtain the thermodynamic properties (T , P, H, S, phase) of all streams. If the state is vapor-liquid mixture, instead of phase report the liquid fraction.

b) Calculate all molar flow rates.

c) Calculate the work in the turbine.

d) Calculate the efficiency of the turbine.

e) Calculate the amount of heat that is exchanged between the two streams in the heat exchanger.

Summarize your results in the table below but you must show your work.

turbine exchanger VIL valve Data for CO2 T (PC) P (bar) H (kJ kg) S(kJ/kg K) Phase T (OC) P (bar) H (kJ/kg) S (kJ/kg K) Phase 0.0000 38 99.66 0.99753 liq -46.005 8 100.87 0.61414 sat liq 3.3047 38 208.19 1.0286 sat liq -46.005 8 433.86 2.0801 vapor 3.3047 38 428.76 1.8264. vapor 0.0000 8 476.57 2.2516 vapor 481.09 2.2679 vapor 485.60 2.2840 vapor 5.0000 38 432.10 1.8385 vapor 5.0000 8 10.000 38 441.02 1.8703 vapor 0.000 8 490.11 2.2998 vapor 15.000 38 449.00 1.8982 vapor 15.000 8 1.9235 vapor 20.000 8 456.37 494.62 2.3153 vapor 20.000 38 T (OC) P (bar) H (kJ/kg) S (kJ/kg K) phase kJ/s kJ/s eff

Explanation / Answer

Ans (a)

All the thrmodynamic properties (T,P,H, S ) of CO2 in the different streams can be calculated by

using the standard table:

For stream 1: Temperature T=200 C = 273 +20 K =293 K

Pressure P = 38 Bar

from standard table: average enthalpy h(av)= 9063 Kj/ mol and averahe entropy s(av)=212.660 KJ/mol/k

If the quality of this gas is x% then its entropy s1= (1-x) sf1 + xsg1

enthalpy h1= (1-x) hf1 + xhg1 ,

hf1 (or sf1) and hg1(or sg1) are the fluid and gas part of h(or s).

For stream 2:  The T & P values for this section can be found from given data table.

(Please Note: This table is not readable in the question-paper in its present form.)

From the standard table of chemicals, the h& s values can be calculated.

For saturated liquid, quality factor=0

Hence the enthalpy and entropy will be due to the fluid only.

Similarly we can calculate for all other states.

Ans(b) Molar flow rates:

Molar flow rate specifies how many moles of fluid passes therough a fixed point in the system

in a given amount of time.

Given in the stream 1 the gas enters at the rate 1 mol/s. Hence the molar flow rate of stream1 is 1mol/s.

Similarly one can calculate the other flow rates. The diagram will give the guideline for the numbers to

be plugged in.

Ans (c)

Work done in the turbine per unit mass flow = change in specific enthalpy from the entry point to the exit point of

the turbine.

As we are calculating all the h in section(a), using that data, work done can be calculated.

Ans (d)

Efficiency of the turbine is given by= work done/ input kinetic energy

Input kinetic energy can be calculated from the mass flow times the sqaure of velocities

(1/2 mv2).

The isentropic efficiency of turbine is given by actual output to ideal isentropic output.

Ans e

The amount of heat exchanged = Qin -Qout

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