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Chapter 7: Problem 139

Air at \(25^{\circ} \mathrm{C}\) flows over a 5 -cm-diameter, 1.7-m-long smooth pipe with a velocity of \(4 \mathrm{~m} / \mathrm{s}\). A refrigerant at \(-15^{\circ} \mathrm{C}\) flows inside the pipe and the surface temperature of the pipe is essentially the same as the refrigerant temperature inside. The drag force exerted on the pipe by the air is (a) \(0.4 \mathrm{~N}\) (b) \(1.1 \mathrm{~N}\) (c) \(8.5 \mathrm{~N}\) (d) \(13 \mathrm{~N}\) (e) \(18 \mathrm{~N}\) (For air, use \(\nu=1.382 \times 10^{-5} \mathrm{~m}^{2} / \mathrm{s}, \rho=1.269 \mathrm{~kg} / \mathrm{m}^{3}\) )

Short Answer

Expert verified
Answer: To determine the drag force acting on the pipe, follow these steps: 1. Calculate the Reynolds number using the given values: \(Re = \dfrac{\rho v d}{\nu}\). 2. Determine the flow regime (laminar, transitional, or turbulent) based on the calculated Reynolds number. 3. Identify the drag coefficient (\(C_D\)) based on the flow regime. 4. Calculate the drag force using the formula: \(F_D = \dfrac{1}{2} \rho v^2 A C_D\). 5. Compare the calculated drag force with the given options to determine the correct answer.

Step by step solution

01

Calculate the Reynolds Number

First, we will compute the Reynolds number using the formula: \(Re = \dfrac{\rho v d}{\nu}\) Where \(Re\) is the Reynolds number, \(\rho\) is the density of air, \(v\) is the flow velocity, \(d\) is the diameter of the pipe, and \(\nu\) is the kinematic viscosity of air. Given values are \(\rho= 1.269 \mathrm{~kg/m}^{3}\), \(v = 4 \mathrm{~m/s}\), \(d = 0.05 \mathrm{~m}\), and \(\nu = 1.382 \times 10^{-5} \mathrm{~m}^{2}/\mathrm{s}\).
02

Determine the Flow Regime

We will now use the Reynolds number calculated in Step 1 to determine the flow regime. As a rule of thumb: - If \(Re < 2000,\) the flow is laminar. - If \(2000 < Re < 4000,\) the flow is transitional. - If \(Re > 4000,\) the flow is turbulent.
03

Determine the Drag Coefficient

Once we know the flow regime, we have to determine the drag coefficient (\(C_D\)) using the appropriate correlation: - For laminar flow (\(Re < 2000\)), \(C_D = \dfrac{24}{Re}\) - For turbulent flow (\(Re > 4000\)), Blasius correlation is \(C_D = 0.079 \cdot Re^{-0.25}\)
04

Calculate the Drag Force

Now that we have the drag coefficient, we can calculate the drag force using the formula: \(F_D = \dfrac{1}{2} \rho v^2 A C_D\) Where \(F_D\) is the drag force, \(A\) is the area of the pipe (cross-sectional), and \(C_D\) is the drag coefficient found in step 3.
05

Choose the Correct Answer

Using the calculated drag force from step 4, we can now compare the value to the given options to choose the correct answer.

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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 \(1 \mathrm{~atm}\).

Solar radiation is incident on the glass cover of a solar collector at a rate of \(700 \mathrm{~W} / \mathrm{m}^{2}\). The glass transmits 88 percent of the incident radiation and has an emissivity of \(0.90\). The entire hot water needs of a family in summer can be met by two collectors \(1.2 \mathrm{~m}\) high and \(1 \mathrm{~m}\) wide. The two collectors are attached to each other on one side so that they appear like a single collector \(1.2 \mathrm{~m} \times 2 \mathrm{~m}\) in size. The temperature of the glass cover is measured to be \(35^{\circ} \mathrm{C}\) on a day when the surrounding air temperature is \(25^{\circ} \mathrm{C}\) and the wind is blowing at \(30 \mathrm{~km} / \mathrm{h}\). The effective sky temperature for radiation exchange between the glass cover and the open sky is \(-40^{\circ} \mathrm{C}\). Water enters the tubes attached to the absorber plate at a rate of \(1 \mathrm{~kg} / \mathrm{min}\). Assuming the back surface of the absorber plate to be heavily insulated and the only heat loss to occur through the glass cover, determine \((a)\) the total rate of heat loss from the collector, \((b)\) the collector efficiency, which is the ratio of the amount of heat transferred to the water to the solar energy incident on the collector, and \((c)\) the temperature rise of water as it flows through the collector.

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 the effect of surface roughness on the friction drag coefficient in laminar and turbulent flows?

Air \((\operatorname{Pr}=0.7, k=0.026 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) at \(200^{\circ} \mathrm{C}\) flows across 2-cm-diameter tubes whose surface temperature is \(50^{\circ} \mathrm{C}\) with a Reynolds number of 8000 . The Churchill and Bernstein convective heat transfer correlation for the average Nusselt number in this situation is $$ \mathrm{Nu}=0.3+\frac{0.62 \mathrm{Re}^{0.5} \mathrm{Pr}^{0.33}}{\left[1+(0.4 / \mathrm{Pr})^{0.67}\right]^{0.25}} $$ (a) \(8.5 \mathrm{~kW} / \mathrm{m}^{2}\) (b) \(9.7 \mathrm{~kW} / \mathrm{m}^{2}\) (c) \(10.5 \mathrm{~kW} / \mathrm{m}^{2}\) (d) \(12.2 \mathrm{~kW} / \mathrm{m}^{2}\) (e) \(13.9 \mathrm{~kW} / \mathrm{m}^{2}\)
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