Countercurrent Heat Exchanger

Question

A countercurrent heat exchanger is being used to heat a liquid food from 15 to 70°C. The heat exchanger has a 23-mm internal and 10 km length with an overall heat-transfer coefficient, based on the inside area, of 2000W/(m2 K). Water, the heating medium, enters the heat exchanger at 95°C and leaves at 85°C. Determine the flow rates for the product and water that will provide the conditions described. Used specific heats of 3.7 kJ/(kg K) for product and 4.18 kJ/(Kg K) for water.

Answer 1

We need to first determine the heat rate, using the usual formula

 

H = Q/t = h A ΔT

 

We are given h=2000 W/m²/K, but need to calculate the area A and the temp gradient ΔT. The area is

 

A =2*Pi*r*L

 

I'm assuming the 23 mm is the internal diameter, not the radius, so r=0.023/2 = 0.0115 m, with L = 10 km = 10,000 m (which sounds like a lot to me). Then

 

A =2*Pi*(0.0115)(10000) = 722 m².

 

For ΔT, the formula for a countercurrent heat exchanger is

 

ΔT = (ΔT₁ - ΔT₂)/ln (ΔT₁/ΔT₂), with

ΔT₁ = T_hot-in - T_cold-in and ΔT₂ = T_hot-out - T_cold-out

In this case,

ΔT₁ = 95-15 = 80 °C and ΔT₂= 85-70 = 15 °C

so that

ΔT =(80-15)/ln(80/15) = 38.8 °C.

 

Therefore,

H = 2000 (722) (38.8) = 5.6*10⁷ W

 

On the food side,

H =Q_food/t = (m_f/t) c_f ΔT

so that

5.6*10⁷ W = (m_f/t) 3700*(70-15)

so the flow rate is

m_f/t = 275 kg/s.

 

On the water side,

5.6*10⁷ W = (m_w/t) 4180*(95-85)

m_w/t = 1340 kg/s.

 

Sounds like a lot of food and water to me -- are you sure this thing is 10 km long??