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Chapter 7: Problem 134

Engine oil at \(105^{\circ} \mathrm{C}\) flows over the surface of a flat plate whose temperature is \(15^{\circ} \mathrm{C}\) with a velocity of \(1.5 \mathrm{~m} / \mathrm{s}\). The local drag force per unit surface area \(0.8 \mathrm{~m}\) from the leading edge of the plate is (a) \(21.8 \mathrm{~N} / \mathrm{m}^{2}\) (b) \(14.3 \mathrm{~N} / \mathrm{m}^{2}\) (c) \(10.9 \mathrm{~N} / \mathrm{m}^{2}\) (d) \(8.5 \mathrm{~N} / \mathrm{m}^{2}\) (e) \(5.5 \mathrm{~N} / \mathrm{m}^{2}\) (For oil, use \(\nu=8.565 \times 10^{-5} \mathrm{~m}^{2} / \mathrm{s}, \rho=864 \mathrm{~kg} / \mathrm{m}^{3}\) )

Short Answer

Expert verified
Answer: The local drag force per unit surface area is approximately 10.9 N/m².

Step by step solution

01

Calculate the Reynolds number at the given location

To calculate the Reynolds number at \(0.8\mathrm{~m}\) from the leading edge of the plate, we use the formula: \(Re=\frac{Vx}{\nu}\), where \(V=1.5\mathrm{~m} / \mathrm{s}\) is the velocity of the oil, \(x=0.8\mathrm{~m}\) is the distance from the leading edge, and \(\nu=8.565\times10^{-5}\mathrm{~m}^{2}/\mathrm{s}\) is the kinematic viscosity of the oil. $$Re=\frac{(1.5\mathrm{~m/s})(0.8\mathrm{~m})}{8.565\times10^{-5}\mathrm{~m}^{2}/\mathrm{s}}\approx 14,067$$
02

Use the Blasius equation to find the local drag coefficient \(C_f\)

The Blasius equation for the local drag coefficient of a flat plate in a laminar flow is: \(C_f=\frac{0.664}{Re^{1/2}}\). Using the calculated Reynolds number from Step 1: $$C_f=\frac{0.664}{\sqrt{14,067}}\approx0.005618$$
03

Calculate the local drag force per unit surface area

To find the local drag force per unit area, we use the drag force formula: \(F =\frac{1}{2}\rho V^2 C_f\). Here, \(\rho=864\mathrm{~kg} / \mathrm{m}^{3}\) is the density of the oil. Substituting the values, we get: $$F =\frac{1}{2}(864\mathrm{~kg/m}^{3})(1.5\mathrm{~m/s})^2(0.005618) \approx 10.89\mathrm{~N/m}^{2}$$ Thus, the local drag force per unit surface area at the given location is approximately \(10.9\mathrm{~N/m}^{2}\) which corresponds to option (c).

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Most popular questions from this chapter

Steam at \(250^{\circ} \mathrm{C}\) flows in a stainless steel pipe \((k=15 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) whose inner and outer diameters are \(4 \mathrm{~cm}\) and \(4.6 \mathrm{~cm}\), respectively. The pipe is covered with \(3.5-\mathrm{cm}-\) thick glass wool insulation \((k=0.038 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) whose outer surface has an emissivity of \(0.3\). Heat is lost to the surrounding air and surfaces at \(3^{\circ} \mathrm{C}\) by convection and radiation. Taking the heat transfer coefficient inside the pipe to be \(80 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), determine the rate of heat loss from the steam per unit length of the pipe when air is flowing across the pipe at \(4 \mathrm{~m} / \mathrm{s}\). Evaluate the air properties at a film temperature of \(10^{\circ} \mathrm{C}\) and \(1 \mathrm{~atm}\).

Liquid mercury at \(250^{\circ} \mathrm{C}\) is flowing in parallel over a flat plate at a velocity of \(0.3 \mathrm{~m} / \mathrm{s}\). Surface temperature of the \(0.1-\mathrm{m}\)-long flat plate is constant at \(50^{\circ} \mathrm{C}\). Determine \((a)\) the local convection heat transfer coefficient at \(5 \mathrm{~cm}\) from the leading edge and \((b)\) the average convection heat transfer coefficient over the entire plate.

To defrost ice accumulated on the outer surface of an automobile windshield, warm air is blown over the inner surface of the windshield. Consider an automobile windshield \(\left(k_{w}=1.4 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\right)\) with an overall height of \(0.5 \mathrm{~m}\) and thickness of \(5 \mathrm{~mm}\). The outside air ( \(1 \mathrm{~atm}\) ) ambient temperature is \(-20^{\circ} \mathrm{C}\) and the average airflow velocity over the outer windshield surface is \(80 \mathrm{~km} / \mathrm{h}\), while the ambient temperature inside the automobile is \(25^{\circ} \mathrm{C}\). Determine the value of the convection heat transfer coefficient, for the warm air blowing over the inner surface of the windshield, necessary to cause the accumulated ice to begin melting. Assume the windshield surface can be treated as a flat plate surface.

Wind at \(30^{\circ} \mathrm{C}\) flows over a \(0.5\)-m-diameter spherical tank containing iced water at \(0^{\circ} \mathrm{C}\) with a velocity of \(25 \mathrm{~km} / \mathrm{h}\). If the tank is thin-shelled with a high thermal conductivity material, the rate at which ice melts is (a) \(4.78 \mathrm{~kg} / \mathrm{h} \quad\) (b) \(6.15 \mathrm{~kg} / \mathrm{h}\) (c) \(7.45 \mathrm{~kg} / \mathrm{h}\) (d) \(11.8 \mathrm{~kg} / \mathrm{h}\) (e) \(16.0 \mathrm{~kg} / \mathrm{h}\) (Take \(h_{i f}=333.7 \mathrm{~kJ} / \mathrm{kg}\), and use the following for air: \(k=\) \(0.02588 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, \operatorname{Pr}=0.7282, v=1.608 \times 10^{-5} \mathrm{~m}^{2} / \mathrm{s}, \mu_{\infty}=\) \(\left.1.872 \times 10^{-5} \mathrm{~kg} / \mathrm{m} \cdot \mathrm{s}, \mu_{\mathrm{s}}=1.729 \times 10^{-5} \mathrm{~kg} / \mathrm{m} \cdot \mathrm{s}\right)\)

In flow across tube banks, how does the heat transfer coefficient vary with the row number in the flow direction? How does it vary with in the transverse direction for a given row number?
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