Question

A long 2-in-diameter rod with surface temperature of \(200^{\circ} \mathrm{F}\) is submerged in a bath of fluid. Determine the Grashof and Rayleigh numbers if the fluid is (a) liquid water at \(40^{\circ} \mathrm{F}\), (b) liquid ammonia at \(40^{\circ} \mathrm{F}\), (c) engine oil at \(50^{\circ} \mathrm{F}\), and \((d)\) air at \(40^{\circ} \mathrm{F}(1 \mathrm{~atm})\).

Short answer

Question: Determine the Grashof and Rayleigh numbers for each fluid. Answer: (a) Liquid water at \(40^{\circ} \mathrm{F}\): Gr = \(4.65\times 10^9\), Ra = \(6.09\times 10^{10}\) (b) Liquid ammonia at \(40^{\circ} \mathrm{F}\): Gr = \(6.60\times 10^9\), Ra = \(5.85\times 10^9\) (c) Engine oil at \(50^{\circ} \mathrm{F}\): Gr = \(1.965\times 10^5\), Ra = \(1.70\times 10^8\) (d) Air at \(40^{\circ} \mathrm{F}(1 \mathrm{~atm})\): Gr = \(9.78\times 10^{7}\), Ra = \(6.96\times 10^{7}\)

Step-by-step solution

Convert temperatures to Celsius

First, convert fluid temperature \(40^{\circ} \mathrm{F}\) to Celsius: \(T_f = (40 - 32)\times 5/9 = 4.44^{\circ} \mathrm{C}\)

Find properties at the given temperature

At \(4.44^{\circ} \mathrm{C}\), the properties of water are as follows: \(\beta = 0.000214\, /{}^{\circ}\text{C}\) \(\nu = 1.307 \times 10^{-6} \, m^{2}/s\) \(Pr = 13.1\)

Calculate Grashof number (Gr)

Using the given properties and formula for Grashof number: \(Gr = \frac{9.81\times 0.000214\times (200-4.44) \times 0.0508^3}{(1.307\times 10^{-6})^2} = 4.65\times 10^9\)

Calculate Rayleigh number (Ra)

Using Grashof and Prandtl numbers, we can calculate the Rayleigh number: \(Ra = Gr \times Pr = 4.65\times 10^9 \times 13.1 = 6.09\times 10^{10}\) (b) Liquid ammonia at \(40^{\circ} \mathrm{F}\):

Convert temperatures to Celsius

\(T_f = 4.44^{\circ} \mathrm{C}\) (already calculated)

Find properties at the given temperature

At \(4.44^{\circ} \mathrm{C}\), the properties of ammonia are as follows: \(\beta = 0.000961\, /{}^{\circ}\text{C}\) \(\nu = 2.83 \times 10^{-5} \, m^{2}/s\) \(Pr = 0.887\)

Calculate Grashof number (Gr)

\(Gr = \frac{9.81\times 0.000961\times (200-4.44) \times 0.0508^3}{(2.83\times 10^{-5})^2} = 6.60\times 10^9\)

Calculate Rayleigh number (Ra)

\(Ra = Gr \times Pr = 6.60\times 10^9 \times 0.887 = 5.85\times 10^9\) (c) Engine oil at \(50^{\circ} \mathrm{F}\):

Convert temperatures to Celsius

First, convert fluid temperature \(50^{\circ} \mathrm{F}\) to Celsius: \(T_f = (50 - 32)\times 5/9 = 10.0^{\circ} \mathrm{C}\)

Find properties at the given temperature

At \(10.0^{\circ} \mathrm{C}\), the properties of engine oil are as follows: \(\beta = 0.000695\, /{}^{\circ}\text{C}\) \(\nu = 1.20 \times 10^{-4} \, m^{2}/s\) \(Pr = 867\)

Calculate Grashof number (Gr)

\(Gr = \frac{9.81\times 0.000695\times (200-10) \times 0.0508^3}{(1.20\times 10^{-4})^2} = 1.965\times 10^5\)

Calculate Rayleigh number (Ra)

\(Ra = Gr \times Pr = 1.965\times 10^5 \times 867 = 1.70\times 10^8\) (d) Air at \(40^{\circ} \mathrm{F}(1 \mathrm{~atm})\):

Convert temperatures to Celsius

\(T_f = 4.44^{\circ} \mathrm{C}\) (already calculated)

Find properties at the given temperature

At \(4.44^{\circ} \mathrm{C}\) and 1 atm pressure, the properties of air are as follows: \(\beta = 3.42\times 10^{-3}\, /{}^{\circ}\text{C}\) \(\nu = 1.49 \times 10^{-5} \, m^{2}/s\) \(Pr = 0.712\)

Calculate Grashof number (Gr)

\(Gr = \frac{9.81\times 3.42\times 10^{-3}\times (200-4.44) \times 0.0508^3}{(1.49\times 10^{-5})^2} = 9.78\times 10^{7}\)

Calculate Rayleigh number (Ra)

\(Ra = Gr \times Pr = 9.78\times 10^{7} \times 0.712 = 6.96\times 10^{7}\) In conclusion, we have determined the Grashof and Rayleigh numbers for each fluid as follows: (a) Liquid water at \(40^{\circ} \mathrm{F}\): Gr = \(4.65\times 10^9\), Ra = \(6.09\times 10^{10}\) (b) Liquid ammonia at \(40^{\circ} \mathrm{F}\): Gr = \(6.60\times 10^9\), Ra = \(5.85\times 10^9\) (c) Engine oil at \(50^{\circ} \mathrm{F}\): Gr = \(1.965\times 10^5\), Ra = \(1.70\times 10^8\) (d) Air at \(40^{\circ} \mathrm{F}(1 \mathrm{~atm})\): Gr = \(9.78\times 10^{7}\), Ra = \(6.96\times 10^{7}\)