Question

13) A heat exchanger is shown below. You may assume fully developed laminar conditions for the hot and cold fluids. The triangular shape housing the hot fluid is an equilateral triangle. The bottom shape housing the cold fluid is a perfect square. One side of the square is length “s," and the square lines up perfectly with the bottom of the triangle. All outer surfaces of both the triangular and square duct can be assumed to be perfectly adiabatic. The drawing is not to scale. Using the diagram and properties below, what is the log mean temperature difference if there is 2000W of total heat transfer along the entire length of the heat exchanger. (30 Points) S kf.h kfc L 25.4 mm 1 W/(m·K) 5 W/(mK) 100 m The ci T hi T. C,0

Answer 1

Solt Griven Data :- 1 S = 25.4mm > kfin= 1 w/mk kefica 5W/mK, Lo 100m For Duct :- Triangular S=25.4mm Area √3 gr 4 = √3x (25-45² > Aca 279.36 mm 2 Perimeter , p = 35 = 3x 25.4 76.2 mm Hydraulic Diameter, Ina YAC P = 4 X 279.36 76.2 14.664 mm

of Nusselt Number triangular cross-section with 2 Uniform Temperature is Nu = 2.49 hnx Dn 2.ug kgin h, K( 0.0 14 66 4 ) 2.ug 1 Ihn 169.8 W/m²K ar Square Cross-Section Duct S = 25.4 mm Area , sz (25.42 645. 16 mm² Ac= Perimeter , 45a 4x 25.4 p= p= 101.6 mm

Hydraulic Diameter, ona LAC = 16:K ( Sبا - w 101 6 Dn= ۳۲ ا.aS Nusselt Number of square cross-section with Uniform Temperature is Nu = a98 را h۸ - 98 . ۸۸ (506as ) a98 5 hea 5860 61 ww- Now, Overall Heat Transfer Coofficient is given by: ۴ 0 مح he دا ده 169 08 58 60 61 0- 131. 683 ۱۱۰K

Now, Area of SL Heat Transfer, A- A- (0.0254)* 100 ܠܨ A- 2.54 m² Now, Q UA (LMTO) 2000 - 131.683 X 2.54 x ( LMTD) LM TD= 5.98°C the :. Log Mean Temperature Difference of given heat exchanger is 5.98°C Completed