Heat Transfer - Engineering

Q1:

The Sieder-Tate correlation for heat transfer in turbulent flow in pipe gives Nu α Re0.8, where, Nu is the Nusselt number and Re is the Reynolds number for the flow. Assuming that this relation is valid, the heat transfer co-efficient varies with the pipe diameter (D) as

A D)-1.8

B (D))-0.2

C (D)0.2

D (D)1.8

ANS:B - (D))-0.2

Given that the Sieder-Tate correlation for heat transfer in turbulent flow in a pipe gives 0.8Nu∝Re0.8, where Nu is the Nusselt number and Re is the Reynolds number, we can express this relation as: 0.8Nu=C⋅Re0.8 where C is a constant. The Nusselt number (Nu) is defined as hD/k, where ℎh is the convective heat transfer coefficient, D is the pipe diameter, and k is the thermal conductivity of the fluid. Substituting Nu=hD/k and rearranging, we get: ℎ=h=DNu⋅k​ Substituting 0.8Nu=C⋅Re0.8 into the equation for ℎh, we get: ℎ=h=DC⋅Re0.8⋅k​ Now, Re=μρ⋅V⋅D​, where ρ is the fluid density, V is the fluid velocity, and μ is the dynamic viscosity of the fluid. Substituting Re=μρ⋅V⋅D​ into the equation for ℎh, we get: ℎ=)0.8⋅h=DC⋅(μρ⋅V⋅D​)0.8⋅k​ Now, let's analyze how ℎh varies with D. We have D in the denominator and D raised to the power of 0.80.8 in the numerator. So, the variation of ℎh with respect to D is: ℎ∝0.8−1=−0.2h∝D0.8−1=D−0.2 Therefore, the heat transfer coefficient ℎh varies with the pipe diameter D as −0.2D−0.2. So, the correct choice is −0.2(D)−0.2.