22-Agric-B8 Food Process Engineering (Part 1) · May 2017
Question 3 of 10: Sizing a counter-current double-pipe heat exchanger
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
04-Agric-B8, Food Process Engineering (Part 1) — National Exams, May 2017. 3 hours duration, closed book (one aid sheet permitted). Ten questions are printed, grouped into four sections (I–IV); a complete exam paper requires six. All ten are solved below as a complete study set.
Reference texts: Toledo, R.T., Fundamentals of Food Process Engineering (this exam's own cited source for its Stumbo g-table and steam-table appendix); Geankoplis, C.J., Transport Processes and Separation Process Principles (evaporator design and steam economy); Incropera, F.P. & DeWitt, D.P., Fundamentals of Heat and Mass Transfer (unsteady-state Heisler-chart conduction, composite cylindrical walls); Singh, R.P. & Heldman, D.R., Introduction to Food Engineering, and Cleland, A.C., Food Refrigeration Processes (freezing-time prediction, modified Plank equation).
Question 3: Sizing a counter-current double-pipe heat exchanger (15 marks)
Given. A double-pipe counter-current exchanger heats a liquid food in the inner pipe using hot water in the annulus.
Given data
Stream
ṁ (kg/s)
C₀ (kJ/kg·K)
Tᵢⁿ (°C)
T₀™ₜ (°C)
Liquid food (inner pipe)
0.5
4.00
20
60 (target)
Hot water (annulus, counter-current)
1.0
4.18
90
? (find)
Additional data: U = 2000 W/(m²·K), inner pipe diameter D = 0.05 m.
Find. (a) The water outlet temperature; (b) the exchanger length required.
Counter-current temperature profile: hot water (annulus) enters where the food leaves, and vice versa.
Approach. An overall energy balance (duty fixed by the food side) gives the water outlet temperature; the log-mean temperature difference for counter-current flow then sizes the required area and length.
Duty from the food side. $$Q = \dot m_f C_{pf}(T_{out}-T_{in}) = (0.5)(4000)(60-20) = \boxed{80.0\ \text{kW}}$$
Water outlet temperature from an energy balance. $$Q=\dot m_w C_{pw}(T_{w,in}-T_{w,out}) \;\Rightarrow\; T_{w,out}=T_{w,in}-\frac{Q}{\dot m_w C_{pw}}=90-\frac{80\,000}{(1.0)(4180)}=90-19.14=\boxed{70.9\ ^\circ\text{C}}$$ The water only needs to give up 19.1°C of its own temperature range to supply the food's 40°C rise, because its mass-flow×specific-heat (thermal capacity rate) is more than double the food's.
Log-mean temperature difference (counter-current). Hot water enters where food leaves, so ΔT₁ is taken at the food-exit end and ΔT₂ at the food-inlet end:
$$\Delta T_1=T_{w,in}-T_{f,out}=90-60=30\ ^\circ\text{C},\qquad \Delta T_2=T_{w,out}-T_{f,in}=70.9-20=50.9\ ^\circ\text{C}$$
$$LMTD=\frac{\Delta T_1-\Delta T_2}{\ln(\Delta T_1/\Delta T_2)}=\frac{30-50.9}{\ln(30/50.9)}=\boxed{39.5\ ^\circ\text{C}}$$
Required area and length. $$A=\frac{Q}{U\cdot LMTD}=\frac{80\,000}{(2000)(39.5)}=1.012\ \text{m}^2$$
$$L=\frac{A}{\pi D}=\frac{1.012}{\pi(0.05)}=\boxed{6.44\ \text{m}}$$
