ANS:A -

The equivalent diameter (DeD_eDe​) for pressure drop calculations in a fluid flowing through a rectangular cross-section channel with sides xxx and yyy can be determined using the hydraulic diameter concept. For a rectangular duct, the hydraulic diameter DhD_hDh​ is given by: Dh=4×AreaWetted perimeterD_h = \frac{4 \times \text{Area}}{\text{Wetted perimeter}}Dh​=Wetted perimeter4×Area​ For a rectangular duct with sides xxx and yyy:

• Area (AAA) = x×yx \times yx×y
• Wetted perimeter (PPP) = 2×(x+y)2 \times (x + y)2×(x+y)

Therefore, the hydraulic diameter DhD_hDh​ is: Dh=4×x×y2×(x+y)=2xyx+yD_h = \frac{4 \times x \times y}{2 \times (x + y)} = \frac{2xy}{x + y}Dh​=2×(x+y)4×x×y​=x+y2xy​ This hydraulic diameter DhD_hDh​ is often used as the equivalent diameter DeD_eDe​ for pressure drop calculations in rectangular ducts. Hence, for a rectangular cross-section channel with sides xxx and yyy, the equivalent diameter DeD_eDe​ is: De=2xyx+yD_e = \frac{2xy}{x + y}De​=x+y2xy​ This equivalent diameter is used because it represents the diameter of a circular cross-section that has the same pressure drop and flow characteristics as the given rectangular duct.