Question

A beaker contains a fluid of density \(\rho\ \mathrm{kg/m^3}\), specific heat \(S\ \mathrm{J/kg^\circ C}\) and viscosity \(\eta\). The beaker is filled up to height h. To estimate the rate of heat transfer per unit area \((\dot Q/A)\) by convection when beaker is put on a hot plate, a student proposes that it should depend on \(\eta\), \(\left(\frac{S\Delta\theta}{h}\right)\) and \(\left(\frac{1}{\rho g}\right)\) when \(\Delta\theta\) (in \(^\circ\)C) is the difference in the temperature between the bottom and top of the fluid. In that situation the correct option for \((\dot Q/A)\) is :

• A. \(\eta\,\frac{S\Delta\theta}{h}\)
• B. \(\eta\left(\frac{S\Delta\theta}{h}\right)\left(\frac{1}{\rho g}\right)\)
• C. \(\frac{S\Delta\theta}{\eta h}\)
• D. \(\left(\frac{S\Delta\theta}{\eta h}\right)\left(\frac{1}{\rho g}\right)\)

Hint:

Write dimensions of Q/A (M T^-3) and of the three proposed quantities, then match exponents.

Answer 1

The Correct Option is A

Dimensions: \([\dot Q/A]=\frac{\mathrm{W}}{\mathrm{m^2}}=M T^{-3}\).
\([\eta]=M L^{-1}T^{-1}\); \(\left[\frac{S\Delta\theta}{h}\right]=\frac{L^2T^{-2}}{L}=LT^{-2}\); \(\left[\frac{1}{\rho g}\right]=\frac{1}{ML^{-3}\cdot LT^{-2}}=M^{-1}L^{2}T^{2}\).
Let \(\dot Q/A=\eta^a\left(\frac{S\Delta\theta}{h}\right)^b\left(\frac{1}{\rho g}\right)^c\). Then M: \(a-c=1\); L: \(-a+b+2c=0\); T: \(-a-2b+2c=-3\).
Solving: \(a=1+c,\ b=1-c\), and the T equation gives \(3c-3=-3\Rightarrow c=0\), so \(a=1,\ b=1\).
\(\dot Q/A=\eta\frac{S\Delta\theta}{h}\).
Answer: (1) \(\eta\,\frac{S\Delta\theta}{h}\)