Question

I need the energy balance around the heat exchanger given T1=40.8c (temperature of entering stream), T2= 150c(Temperature of leaving stream).

cp of methanol = 2.55 KJ/Kg.k

cp of water = 4.18 KJ/Kg.k

flow rate of methanol = 139.92 Kmol/hr

the flow rate of water = 21.634 Kmol/hr

Answer 1

Given data

T1 (methanol stream inlet) = 40.8 °C

T2 (methanol stream exit) = 150 °C

Cpm of methanol = 2.55 KJ/Kg.k

Cpw of water = 4.18 KJ/Kg.k

Molar flow rate of methanol = 139.92 Kmol/hr

Mass flow rate of methanol = molar flow x molecular weight

= 139.92 Kmol/hr x 32.04 kg/kmol

m1 = 4483.04 kg/hr

Molar flow rate of water = 21.634 Kmol/hr

Mass flow rate of water = molar flow x molecular weight

= 21.634 Kmol/hr x 18 kg/kmol

m2 = 389.412 kg/hr

Assumption - Methanol is being heated from water

Let water inlet temperature = T3

Water outlet temperature = T4

T3 > 100 °C > T4

Boiling point of methanol = 65 °C

energy balance around the heat exchanger

Energy absorbed by liquid methanol + latent heat of vaporization + energy absorbed by methanol vapor = Energy released by water vapor + latent heat of condensation + Energy released by liquid water

m1 x Cpm x (65-T1) + m1 x Hvap + m1 x Cpm x (T2-T1) = m2 x Cpw x (T4 - T3) +

4483.04 kg/hr x 2.55 KJ/Kg.K x (65 - 40.8)K + 1098.6 kJ/kg x 4483.04 kg/hr + 4483.04 kg/hr x 1.60 KJ/Kg.K x (150 - 65)K = 389.412 kg/hr x 2.1 KJ/Kg.k x (T4 - 100)K + 2260 kJ/kg x 389.412 kg/hr + 389.412 kg/hr x 4.18 KJ/Kg.k x (100-T3) K

5811409.5824 kJ/hr = 817.7652(T4-100) + 880071.12 + 1627.74216(100 - T3)

If we know either T4 or T3, then we can calculate the other.