Air at \(15^{\circ} \mathrm{C}\) and \(1 \mathrm{~atm}\) flows over a \(0.3\)-m-wide plate at \(65^{\circ} \mathrm{C}\) at a velocity of \(3.0 \mathrm{~m} / \mathrm{s}\). Compute the following quantities at \(x=0.3 \mathrm{~m}\) : (a) Hydrodynamic boundary layer thickness, \(\mathrm{m}\) (b) Local friction coefficient (c) Average friction coefficient (d) Total drag force due to friction, \(\mathrm{N}\) (e) Local convection heat transfer coefficient, W/m² \(\mathbf{K}\) (f) Average convection heat transfer coefficient, W/m² \(\mathrm{K}\) (g) Rate of convective heat transfer, W

Short Answer

Expert verified
Question: Determine the values of the following quantities at a distance of 0.3 m from the leading edge of the heated flat plate: (a) hydrodynamic boundary layer thickness, (b) local friction coefficient, (c) average friction coefficient, (d) total drag force due to friction, (e) local convection heat transfer coefficient, (f) average convection heat transfer coefficient, and (g) rate of convective heat transfer. Answer: The required quantities are as follows: (a) Hydrodynamic boundary layer thickness: \(\delta = 0.0021\ \mathrm{m}\) (b) Local friction coefficient: \(C_{f} = 0.00293\) (c) Average friction coefficient: \(C_{f_{avg}} = 0.00584\) (d) Total drag force due to friction: \(F_{d} = 0.0474\ \mathrm{N}\) (e) Local convection heat transfer coefficient: \(h_{x} = 93.7\ \mathrm{W/ m^2 K}\) (f) Average convection heat transfer coefficient: \(h_{avg} = 189.4\ \mathrm{W/ m^2 K}\) (g) Rate of convective heat transfer: \(q_{conv} = 849.4\ \mathrm{W}\)

Step by step solution

01

Determine the properties of the air

First, let's find the film temperature by averaging the temperatures of the air and the heated plate: $$T_{film} = \frac{T_{air} + T_{plate}}{2}$$ Using the provided values, calculate the film temperature: $$T_{film} = \frac{15 + 65}{2} = 40^{\circ} \mathrm{C}$$ Now, look up the properties of air at the film temperature \(40^{\circ} \mathrm{C}\). Here are the necessary values: $$\rho = 1.127 \, \mathrm{kg/ m^3}$$ $$\mu = 1.97 × 10^{-5} \, \mathrm{kg/ m\cdot s}$$ $$k= 0.028 \, \mathrm{W/ m\cdot K}$$ $$C_{p}= 1007 \, \mathrm{J/ kg\cdot K}$$
02

Calculate the Reynolds number

Now, we need to find the Reynolds number at \(x = 0.3\, \mathrm{m}\). The Reynolds number can be calculated as: $$Re_{x} = \frac{\rho U x}{\mu}$$ Substituting the known values, we get: $$Re_{x} = \frac{1.127\, (3)(0.3)}{1.97 × 10^{-5}} = 5.14 \times 10^4$$ Since the Reynolds number is less than \(5 \times 10^5\), the flow is laminar.
03

Calculate part (a) - Hydrodynamic boundary layer thickness

To find the hydrodynamic boundary layer thickness, apply the laminar boundary layer correlation: $$\delta = \frac{5x}{\sqrt{Re_{x}}}$$ Substitute the values to get: $$\delta = \frac{5\, (0.3)}{\sqrt{5.14 × 10^4}} = 0.0021 \, \mathrm{m}$$
04

Calculate part (b) - Local friction coefficient

To find the local friction coefficient, apply the following formula: $$C_{f} = \frac{0.664}{\sqrt{Re_{x}}}$$ Using the calculated Reynolds number, we get: $$C_{f} = \frac{0.664}{\sqrt{5.14 \times 10^4}} = 0.00293$$
05

Calculate part (c) - Average friction coefficient

Calculate the average friction coefficient using the following formula: $$C_{f_{avg}} = \frac{1.328}{\sqrt{Re_{x}}}$$ Using the calculated Reynolds number, we get: $$C_{f_{avg}} = \frac{1.328}{\sqrt{5.14 \times 10^4}} = 0.00584$$
06

Calculate part (d) - Total drag force due to friction

To find the total drag force due to friction, apply this formula: $$F_{d} = C_{f_{avg}} \frac{1}{2} \rho U^{2} A$$ Here, \(A = x \, w\), since the plate is \(0.3\, \mathrm{m}\) wide. Substituting the known values: $$F_{d} = 0.00584 \frac{1}{2} (1.127) (3)^{2} (0.3) (0.3) = 0.0474 \, \mathrm{N}$$
07

Calculate part (e) - Local convection heat transfer coefficient

To find the local convection heat transfer coefficient, apply the following correlation for laminar flow: $$h_{x} = C_{hf}\left(\frac{k^3 \rho C_{p} Ux}{\mu}\right)^{1/2}$$ Here, \(C_{hf} = 0.664\). Substituting the values, we get: $$h_{x} = 0.664\left(\frac{0.028^3 \cdot 1.127 \cdot 1007 \cdot 3 \cdot 0.3}{1.97 \times 10^{-5}}\right)^{1/2} = 93.7 \, \mathrm{W/ m^2 K}$$
08

Calculate part (f) - Average convection heat transfer coefficient

To find the average convection heat transfer coefficient, apply the following correlation for laminar flow: $$h_{avg} = C_{h_{avg}}\left(\frac{k^3 \rho C_{p} Ux}{\mu }\right)^{1/2}$$ Here, \(C_{h_{avg}} = 0.664\). Substituting the values, we get: $$h_{avg} = 1.128\left(\frac{0.028^3 \cdot 1.127 \cdot 1007 \cdot 3 \cdot 0.3}{1.97 \times 10^{-5}}\right)^{1/2} = 189.4 \, \mathrm{W/ m^2 K}$$
09

Calculate part (g) - Rate of convective heat transfer

Finally, to find the rate of convective heat transfer, use this formula: $$q_{conv} = h_{avg} A \Delta T$$ Calculate the heat transfer rate using the known values: $$q_{conv} = (189.4) (0.3) (0.3) (65 - 15) = 849.4 \, \mathrm{W}$$ In summary, the required quantities are as follows: (a) \(\delta = 0.0021\ \mathrm{m}\) (b) \(C_{f} = 0.00293\) (c) \(C_{f_{avg}} = 0.00584\) (d) \(F_{d} = 0.0474\ \mathrm{N}\) (e) \(h_{x} = 93.7\ \mathrm{W/ m^2 K}\) (f) \(h_{avg} = 189.4\ \mathrm{W/ m^2 K}\) (g) \(q_{conv} = 849.4\ \mathrm{W}\)

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Most popular questions from this chapter

Air at \(15^{\circ} \mathrm{C}\) and \(1 \mathrm{~atm}\) flows over a \(0.3\)-m-wide plate at \(65^{\circ} \mathrm{C}\) at a velocity of \(3.0 \mathrm{~m} / \mathrm{s}\). Compute the following quantities at \(x=x_{\mathrm{cr}}\) : (a) Hydrodynamic boundary layer thickness, \(\mathrm{m}\) (b) Local friction coefficient (c) Average friction coefficient (d) Total drag force due to friction, \(\mathrm{N}\) (e) Local convection heat transfer coefficient, \(\mathrm{W} / \mathrm{m}^{2} \cdot \mathrm{K}\) (f) Average convection heat transfer coefficient, W/m² \(\cdot \mathrm{K}\) (g) Rate of convective heat transfer, W

A heated long cylindrical rod is placed in a cross flow of air at \(20^{\circ} \mathrm{C}(1 \mathrm{~atm})\) with velocity of \(10 \mathrm{~m} / \mathrm{s}\). The rod has a diameter of \(5 \mathrm{~mm}\) and its surface has an emissivity of \(0.95\). If the surrounding temperature is \(20^{\circ} \mathrm{C}\) and the heat flux dissipated from the rod is \(16000 \mathrm{~W} / \mathrm{m}^{2}\), determine the surface temperature of the rod. Evaluate the air properties at \(70^{\circ} \mathrm{C}\).

What is flow separation? What causes it? What is the effect of flow separation on the drag coefficient?

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Hydrogen gas at \(1 \mathrm{~atm}\) is flowing in parallel over the upper and lower surfaces of a 3-m-long flat plate at a velocity of \(2.5 \mathrm{~m} / \mathrm{s}\). The gas temperature is \(120^{\circ} \mathrm{C}\) and the surface temperature of the plate is maintained at \(30^{\circ} \mathrm{C}\). Using the EES (or other) software, investigate the local convection heat transfer coefficient and the local total convection heat flux along the plate. By varying the location along the plate for \(0.2 \leq x \leq 3 \mathrm{~m}\), plot the local convection heat transfer coefficient and the local total convection heat flux as functions of \(x\). Assume flow is laminar but make sure to verify this assumption. 7-31 Carbon dioxide and hydrogen as ideal gases at \(1 \mathrm{~atm}\) and \(-20^{\circ} \mathrm{C}\) flow in parallel over a flat plate. The flow velocity of each gas is \(1 \mathrm{~m} / \mathrm{s}\) and the surface temperature of the 3 -m-long plate is maintained at \(20^{\circ} \mathrm{C}\). Using the EES (or other) software, evaluate the local Reynolds number, the local Nusselt number, and the local convection heat transfer coefficient along the plate for each gas. By varying the location along the plate for \(0.2 \leq x \leq 3 \mathrm{~m}\), plot the local Reynolds number, the local Nusselt number, and the local convection heat transfer coefficient for each gas as functions of \(x\). Discuss which gas has higher local Nusselt number and which gas has higher convection heat transfer coefficient along the plate. Assume flow is laminar but make sure to verify this assumption.

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