Benzene as a saturated liquid at 5,000 mmHg is fed to a vaporizer. The product f

Question

Benzene as a saturated liquid at 5,000 mmHg is fed to a vaporizer. The product from the vaporizer is a superheat vapor at 165°C and 4,550 mmHg.

For a flow rate of 2,500 moles/min of benzene, how much is needed (kW)?

If steam is available at 500°C and 20 bar and condenses to a saturated liquid at 20 bars, how much steam (kg/hr) is needed to vaporize the benzene?

Know: Hv = 30.765 kJ/mol, Tb = 80.10°C, SG = 0.879, Cp (l) = 126.5 * 10^-3 + 23.4 * 10^-5 T (°C), Cp (v) = 74.06 * 10^-3 + 32.95 * 10^-5T - 25.20 * 10^-8 T^2 + 77.57*10^-12 T^3

Explanation / Answer

Saturated liquid Benzene conditions

Pressure Pl = 5000 mmHg

Temperature can be predicted from Antoine Equation log10Pl = A - B/(Tl + C) where A = 6.88, B=1196.76 and C = 219.161, Tl in K, Pl in mmHg

Substitute these values and calculate Tl at Pl=5000 mmHg

Tl = 157.06 °C

Saturated Vapor Benzene conditions

Pressure Pv = 4550 mmHg

Temperature can be predicted from Antoine Equation log10Pl = A - B/(Tl + C) where A = 6.88, B=1196.76 and C = 219.161, Tv in K, Pv in mmHg

Substitute these values and calculate Tv at Pv=4550 mmHg

Tv = 152.27 °C

Benzene flow rate m = 2500 moles/min

= 2500/60

= 41.667 moles/sec

kW needed to vaporise benzene = latent heat of vaporization + sensible heat required to superheat the saturated vapor

Note : Since liquid is already saturated sensible heat of liquid is will not be considered for the calculation

Cp(v) = 74.06 * 10^-3 + 32.95 * 10^-5 T - 25.2 * 10^-8 T^2 + 77.57 * 10^-12 T^3

For T in above relation substitute Tsv = 165 ° C

Cp(v) = 0.1219 kJ/mol °C

deltaHv heat of vaporization = 30.765 kJ/mol

kW needed = (m * deltaHv) + (m * Cp(v) * (Tsv - Tv))

= (41.667 * 30.765) +(41.667 * 0.1219 * (165 - 152.27))

= 1346.54 kW

Steam Calculation

From steam table

Steam saturation temperature at 20 bar is Ts = 212.37 °C.

Temperature Tss = 500 °C

Specific Heat Cps = 3.0248 kJ/kg °C

lateny heat of vaporization/condensation hc = 1888.65 kJ/kg

Q = msCps(Tss - Ts) +mshs

Q=1346.54 kW

= 1346.54 * 3600

= 4847544 kJ/hr

sustitute all values and solve for ms

Steam required ms = 1757.20 kg/hr

Related Questions

Navigate

• Browse (All)
• Browse B
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.