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Consider atmospheric air at \(25^{\circ} \mathrm{C}\) and a velocity of \(25 \mathrm{~m} / \mathrm{s}\) flowing over both surfaces of a 1 - \(\mathrm{m}\)-long flat plate that is maintained at \(125^{\circ} \mathrm{C}\). Determine the rate of heat transfer per unit width from the plate for values of the critical Reynolds number corresponding to \(10^{5}\), \(5 \times 10^{5}\), and \(10^{6}\).

Short Answer

Expert verified
The rate of heat transfer per unit width for each of the critical Reynolds numbers is: Case 1 (Re_c = \( 10^{5} \)): q'1 = 680.4 W/m Case 2 (Re_c = \( 5 \times 10^{5} \)): q'2 = 1364.5 W/m Case 3 (Re_c = \( 10^{6} \)): q'3 = 1922.0 W/m

Step by step solution

01

Calculate the Reynolds number

To calculate the Reynolds number, we need the following formula: Re_L = \( \frac{\rho V L}{\mu} \) where Re_L is the Reynolds number, 蟻 is the density of the fluid (air), V is the velocity, L is the length of the plate, and 渭 is the dynamic viscosity of the fluid. We are given the values of V and L. We need to find the values of 蟻 and 渭 for the air at given temperature (25掳C). These values can be found in a thermodynamic properties table or an online calculator. For air at 25掳C: 蟻 = 1.184 kg/m鲁 渭 = 1.85 脳 10鈦烩伒 kg/m路s Now we can calculate the Reynolds number: Re_L = \( \frac{(1.184 kg/m鲁)(25 m/s)(1 m)}{1.85 脳 10鈦烩伒 kg/m路s} \) Re_L = 1.6 脳 10鈦
02

Determine the flow regime and Nusselt number relation

The critical Reynolds number values are given as 10鈦, 5 脳 10鈦, and 10鈦. Since the calculated Reynolds number (1.6 脳 10鈦) is higher than all three critical Reynolds numbers, the flow is turbulent for all cases. In turbulent flow over a flat plate, the Nusselt number is related to the Reynolds number and the Prandtl number (Pr) by the following correlation: Nu_L = 0.0296 脳 Re_L^(4/5) 脳 Pr^(1/3) We will need the Prandtl number for air at 25掳C, which can also be found in a thermodynamic properties table or an online calculator: Pr = 0.707.
03

Calculate the heat transfer coefficients for each critical Reynolds number

We will now calculate the heat transfer coefficient (h) for each of the critical Reynolds numbers using the Nusselt number correlation. First, we calculate the Nusselt numbers for each case: Case 1: Re_c = 10鈦 Nu_L1 = 0.0296 脳 (10鈦)^(4/5) 脳 (0.707)^(1/3) Nu_L1 = 259.4 Case 2: Re_c = 5 脳 10鈦 Nu_L2 = 0.0296 脳 (5 脳 10鈦)^(4/5) 脳 (0.707)^(1/3) Nu_L2 = 520.0 Case 3: Re_c = 10鈦 Nu_L3 = 0.0296 脳 (10鈦)^(4/5) 脳 (0.707)^(1/3) Nu_L3 = 732.8 Next, we calculate the heat transfer coefficients (h) using the formula: h = \( \frac{k}{L} \) 脳 Nu_L where k is the thermal conductivity of the fluid (air). For air at 25掳C, k = 0.02624 W/m路K. h1 = \( \frac{0.02624 W/m路K}{1 m} \) 脳 259.4 = 6.804 W/m虏路K h2 = \( \frac{0.02624 W/m路K}{1 m} \) 脳 520.0 = 13.645 W/m虏路K h3 = \( \frac{0.02624 W/m路K}{1 m} \) 脳 732.8 = 19.220 W/m虏路K
04

Calculate the rate of heat transfer per unit width for each case

Now, we can calculate the rate of heat transfer per unit width (q') for each case using the formula: q' = h 脳 螖T 脳 W where 螖T is the temperature difference between the plate and the air, and W is the width of the plate (in our case, per unit width, so W = 1m). For all cases, 螖T = 125掳C - 25掳C = 100掳C or 100 K. q'1 = (6.804 W/m虏路K)(100 K)(1 m) = 680.4 W/m q'2 = (13.645 W/m虏路K)(100 K)(1 m) = 1364.5 W/m q'3 = (19.220 W/m虏路K)(100 K)(1 m) = 1922.0 W/m The rate of heat transfer per unit width for each of the critical Reynolds numbers is: Case 1 (Re_c = 10鈦): q'1 = 680.4 W/m Case 2 (Re_c = 5 脳 10鈦): q'2 = 1364.5 W/m Case 3 (Re_c = 10鈦): q'3 = 1922.0 W/m

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Reynolds Number Calculation
Understanding the concept of the Reynolds number is essential when studying heat transfer over a flat plate. It's a dimensionless quantity used in fluid mechanics to predict the flow regime 鈥 whether it will be laminar or turbulent. When flowing over objects such as a flat plate, the Reynolds number helps in characterizing the nature of the flow.

To calculate the Reynolds number, \( Re_L \), the formula \( Re_L = \frac{\rho V L}{\mu} \) is used, where \( \rho \) is the fluid density, \( V \) is the fluid velocity, \( L \) is the characteristic length (in this case, the length of the plate), and \( \mu \) is the fluid's dynamic viscosity. For atmospheric air at \(25^\circ \mathrm{C}\), the density and dynamic viscosity can be found from standard thermodynamic tables. Applying these values to the formula gives the Reynolds number, which then determines the flow regime. If the value of \( Re_L \) is higher than a critical value, typically ranging around \(10^5\) to \(10^6\), the flow is considered turbulent.
Nusselt Number Relation
The Nusselt number, \( Nu \), is another dimensionless number used in heat transfer to describe the ratio of convective to conductive heat transfer across a boundary. In the context of a flat plate, it provides a measure of the thermal conductivity of the boundary layer that forms as air flows over the plate. The Nusselt number is particularly helpful as it relates to the heat transfer coefficient, \( h \).

For turbulent flow over a flat plate, the Nusselt number can be calculated using the empirical correlation: \( Nu_L = 0.0296 \times Re_L^{4/5} \times Pr^{1/3} \), where \( Pr \) is the Prandtl number, which depends on the fluid properties at the given temperature. The Prandtl number is the ratio of momentum diffusivity to thermal diffusivity and for air at \(25^\circ \mathrm{C}\) is typically around 0.707. With the Reynolds number already calculated, the Nusselt number can be determined, which will then be used to find the heat transfer coefficient.
Heat Transfer Coefficient
The heat transfer coefficient, \( h \), is a crucial parameter in the study of convective heat transfer. It quantifies the heat transfer rate per unit area and per degree temperature difference between the surface and the fluid. Once the Nusselt number is known, \( h \) can be found by the relation \( h = \frac{k}{L} \times Nu_L \), where \( k \) is the thermal conductivity of the air and \( L \) is the length of the plate, mentioned in the Nusselt number relation.

For each critical Reynolds number, the heat transfer coefficient is calculated individually as it determines how efficiently heat is transferred for various flow conditions. Understanding \( h \) allows designers to predict how quickly a plate will cool or heat in a specific fluid flow scenario, directly impacting thermal management and system performance.
Rate of Heat Transfer
The rate of heat transfer, commonly denoted as \( q' \) when referring to per unit width, signifies the amount of heat energy transferred per unit time. It is critical in applications ranging from aerospace to industrial processes where temperature control is vital. In our exercise, \( q' \) can be found using the formula \( q' = h \times \Delta T \times W \), where \( \Delta T \) is the difference in temperature between the hot plate and the cooler air, and \( W \) is the width of the plate.

This calculation is significant for engineers to ensure that the plate dissipates heat at the right rate, avoiding overheating or insufficient cooling. Knowing the rate of heat transfer aids in the design and analysis of cooling systems, heating units, and can even influence the choice of materials used based on their thermal properties.

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Most popular questions from this chapter

Consider laminar, parallel flow past an isothermal flat plate of length \(L\), providing an average heat transfer coefficient of \(\bar{h}_{L^{-}}\)If the plate is divided into \(N\) smaller plates, each of length \(L_{N}=L / N\), determine an expression for the ratio of the heat transfer coefficient averaged over the \(N\) plates to the heat transfer coefficient averaged over the single plate, \(\bar{h}_{L, N} / \bar{h}_{L, 1}\).

Highly reflective aluminum coatings may be formed on the surface of a substrate by impacting the surface with molten drops of aluminum. The droplets are discharged from an injector, proceed through an inert gas (helium), and must still be in a molten state at the time of impact. \(V=3 \mathrm{~m} / \mathrm{s}\), and \(T_{i}=1100 \mathrm{~K}\), respectively, traverse a stagnant layer of atmospheric helium that is at a temperature of \(T_{\infty}=300 \mathrm{~K}\). What is the maximum allowable thickness of the helium layer needed to ensure that the temperature of droplets impacting the substrate is greater than or equal to the melting point of aluminum \(\left(T_{f} \geq T_{\text {mp }}=933 \mathrm{~K}\right)\) ? Properties of the molten aluminum may be approximated as \(\rho=2500 \mathrm{~kg} / \mathrm{m}^{3}, c=\) \(1200 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\), and \(k=200 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\).

An air duct heater consists of an aligned array of electrical heating elements in which the longitudinal and transverse pitches are \(S_{L}=S_{T}=24 \mathrm{~mm}\). There are 3 rows of elements in the flow direction \(\left(N_{L}=3\right)\) and 4 elements per row \(\left(N_{T}=4\right)\). Atmospheric air with an upstream velocity of \(12 \mathrm{~m} / \mathrm{s}\) and a temperature of \(25^{\circ} \mathrm{C}\) moves in cross flow over the elements, which have a diameter of \(12 \mathrm{~mm}\), a length of \(250 \mathrm{~mm}\), and are maintained at a surface temperature of \(350^{\circ} \mathrm{C}\). (a) Determine the total heat transfer to the air and the temperature of the air leaving the duct heater. (b) Determine the pressure drop across the element bank and the fan power requirement. (c) Compare the average convection coefficient obtained in your analysis with the value for an isolated (single) element. Explain the difference between the results. (d) What effect would increasing the longitudinal and transverse pitches to \(30 \mathrm{~mm}\) have on the exit temperature of the air, the total heat rate, and the pressure drop?

An array of electronic chips is mounted within a sealed rectangular enclosure, and cooling is implemented by attaching an aluminum heat \(\operatorname{sink}(k=180 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\). The base of the heat sink has dimensions of \(w_{1}=w_{2}=\) \(100 \mathrm{~mm}\), while the 6 fins are of thickness \(t=10 \mathrm{~mm}\) and pitch \(S=18 \mathrm{~mm}\). The fin length is \(L_{f}=50 \mathrm{~mm}\), and the base of the heat sink has a thickness of \(L_{b}=10 \mathrm{~mm}\). If cooling is implemented by water flow through the heat sink, with \(u_{\infty}=3 \mathrm{~m} / \mathrm{s}\) and \(T_{\infty}=17^{\circ} \mathrm{C}\), what is the base temperature \(T_{b}\) of the heat sink when power dissipation by the chips is \(P_{\text {elec }}=1800 \mathrm{~W}\) ? The average convection coefficient for surfaces of the fins and the exposed base may be estimated by assuming parallel flow over a flat plate. Properties of the water may be approximated as \(k=0.62 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, \rho=995 \mathrm{~kg} / \mathrm{m}^{3}\), \(c_{p}=4178 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}, \nu=7.73 \times 10^{-7} \mathrm{~m}^{2} / \mathrm{s}\), and \(\operatorname{Pr}=5.2\).

A square ( \(10 \mathrm{~mm} \times 10 \mathrm{~mm}\) ) silicon chip is insulated on one side and cooled on the opposite side by atmospheric air in parallel flow at \(u_{\infty}=20 \mathrm{~m} / \mathrm{s}\) and \(T_{\infty}=\) \(24^{\circ} \mathrm{C}\). When in use, electrical power dissipation within the chip maintains a uniform heat flux at the cooled surface. If the chip temperature may not exceed \(80^{\circ} \mathrm{C}\) at any point on its surface, what is the maximum allowable power? What is the maximum allowable power if the chip is flush mounted in a substrate that provides for an unheated starting length of \(20 \mathrm{~mm}\) ?

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