Question

A heat exchanger is used to transfer heat from a steam stream to a chlorine gas stream. 100 kg of steam at 1 bar and 250°C enters the heat exchanger and leaves the exchanger as saturated steam at 1 bar. Chlorine gas enters the exchanger at 1 bar and 33.0°C and leaves the exchanger at 132.0°C. Assume the heat exchanger is adiabatic.

A)What is the specific enthalpy of the inlet steam stream? (Use liquid water at 0.01°C as a reference state.)

B)What is the specific enthalpy of the outlet steam stream? (Use the same reference state.)

C) What is the specific enthalpy of the inlet chlorine gas stream? (Use chlorine gas at 1 bar and 33.0°C as a reference state.)

D) What is the specific enthalpy of the outlet chlorine gas stream? (Use chlorine gas at 1 bar and 33.0°C as a reference state.)

E)How many moles of chlorine gas enter the heat exchanger?

F)What is the ratio of the volume of inlet steam to the volume of inlet chlorine gas? Use the ideal gas law to calculate the volume of chlorine entering the process.

G)How much heat is transferred from the steam stream per cubic meter of chlorine gas?

Answer 1

Part a

specific enthalpy of the inlet steam stream at 1 bar and 250°C

H si = 2974.01 kJ/kg

Part b

specific enthalpy of the outlet steam stream at 1 bar as saturated steam

H so = 2675.27 kJ/kg

Part C

specific enthalpy of the inlet chlorine gas stream at 1 bar and 33.0°C with the same reference

H Ci= 0 kJ/mol

Part d

specific enthalpy of the outlet chlorine gas stream at 132.0°C

dHco = Cp dT

Hco = ( 33.05060 + 12.22940*T -12.06510*T^2 + 4.385330*T^3 - 0.159494/T^2) dT

T is in T(K)/1000

= 33.05060 (0.405-0.306)+ (12.22940/2)*(0.405²-0.306²) -(12.06510/3)* (0.405³-0.306³) + (4.385330/4)* (0.405⁴-0.306⁴) + 0.159494*(1/0.405 - 1/0.306)

Hco = 3.4429 kJ/mol

Part e

From the adiabatic overall energy balance

H = 0 = ms x (Hso - Hsi) + nc x (Hco - Hci)

ms x (Hso - Hsi) = - nc x (Hci - Hci)

100 kg x (2675.27 - 2974.01)kJ/kg = - nc x ( 3.4429 - 0) kJ/mol

29874 = 3.4429 nc

nc = 8676.987 mol

Part f

Volume of inlet steam at 1 bar & 250°C

Specific volume = 2.4064 m3/kg

Volume of inlet steam = mass x Specific volume

= 100 kg x 2.4064 m3/kg

= 240.64 m3

Volume of inlet chlorine gas = nRT/P

= 8676.987 mol x 8.314 J/mol-K x (273+33)K / 10^5 Pa

= 220.75 m3

volume of inlet steam/volume of inlet chlorine gas

= 240.64 m3/220.75 m3

= 1.09

Part g

Heat transfer from the steam = ms x (Hso - Hsi)

= 100 kg x (2675.27 - 2974.01)kJ/kg

= 29874 kJ

Heat transfer from the steam/volume of inlet chlorine gas

= 29874 kJ/220.75 m3

= 135.33 kJ/m3