Mini-project 4: Heat Exchangers

MECH3820 Heat Transfer

Question:

You’ve been tasked with analyzing a new heat exchanger system for small-scale dairy farms which pre-cools the raw milk coming from the cows before entering the larger chilled storage tank. The heat exchanger is of a shell-and-tube-type with 4 shells, and within each shell there are 22 tubes each making 2 passes. The tubes are 2 m long, with an inner diameter of 8 mm, and an outer diameter of 12 mm. You may ignore the length of the turning sections at the end, except that each straight length of pipe acts like a new pipe entrance for Nusselt number calculations.

The flow passage in the shell has an effective hydraulic diameter of 6.01 mm, and an effective flow cross-sectional area of 845.7 mm2. The whole heat exchanger is made from AISI 304 Stainless Steel (cf. the property table, Table 1, below for relevant heat exchanger material properties; you may assume the properties are constant unless otherwise specified). The milk enters the tubes at 38.6°C, while groundwater at 10°C is pumped into the shell. During milking, each cow produces approximately 1.74 kg of milk per minute. Assume the properties of the milk and groundwater are constant, as given in Table 1, and that the system is operating at steady state conditions.

1.If you milk 20 cows simultaneously and have a groundwater mass flow rate of 2 kg/s, what are the outlet temperatures of the milk and water?

2.2.Create a contour plot of the milk outlet temperature as a function of the number of cows being milked, from 5 to 50, and groundwater flow rate, from 0.05 to 3.1 kg/s.

• You may not use any rainbow-type color map for your contour plot. The default Matlab color map (parula) is acceptable. Please include all intermediate calculations as well as all codes used to answer these questions.

You may find the relations on the next page helpful as you set up your solutions:

Shell-and-Tube Heat Exchanger Relations

Tube Nusselt Number relations

-For laminar flow (Re < 2300), from Sieder and Tate (1936):

Gz = Di L ReD Pr NuD = max ( 1.86Gz1/3 ( μ μs )0.14 , 3.66

-For turbulent and transitional flow (Re > 2300), from Gnielinski (1976):

NuD = ( f 8 ) (Re − 1000) Pr 1 + 12.7 ( f 8 )1/2 ( Pr2/3 − 1).

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