Question

An adiabatic heat exchanger is one for which no heat is exchanged with the surroundings. All of the heat lost by the hot stream is transferred to the cold stream in this adiabatic process. In a proposed process, propane gas enters a continuous adiabatic heat exchanger at 45 °C and 265 kPa and exits at 255 °C. Superheated steam at 300 °C and 7.0 bar enters the exchanger flowing countercurrently to the propane and exits as a saturated liquid at the same pressure.

Taking as a basis 100 mol of propane fed to the exchanger, draw and label a process flowchart. Include in your labeling the volume of propane fed (m³), the mass of steam fed (kg), and the volume of steam fed (m³).

Calculate values of the labeled specific enthalpies in the following inlet-outlet table for this process.

References: H2O (l, 0.01 °C), C3H8 (g, 45 °C)
Species	nin		nout
C3H8	100 mol	(kJ/mol)	100 mol	(kJ/mol)
H2O	mW (kg)	(kJ/kg)	mW (kg)	(kJ/kg)

Use an energy balance to calculate the required mass feed rate of the steam. Then calculate the volumetric feed ratio of the two streams (m³ steam fed/m³ propane fed). Assume ideal gas behavior for the propane but not the steam and recall that the exchanger is adiabatic.

Calculate the heat transferred from the water to the propane (kJ/m³ propane fed). (Hint: Do an energy balance on either the water or the propane rather than on the entire heat exchanger.)

Adiabatic heat exchanger

Answer 1

From the steam table, the temperature of saturated water at 7 bar = 164.96 degree c or 438.11 k m_progane C_propane (Delta T)_p = -m_steam C_p steam(Delta T)_s (C_p)_propne = 115.58 J/mol k[for Temperature range of (350 - 550k)] (C_p)_H_2O = -203.6 + 1523.29T - 3196.4T^2 + 2474.45T^3 m_propane(115.58)(528.15 - 318.15) uparrow (100mol) = -m_s integral^438.11_573.15 (C_p)_s dT Putting all the values provided we can find out M_s \r\n Heat exchanged = m_p(C_p)_propane Delta T = 100 times 115.58 times 210 = 2427.18 kJ