A fluid \(\left(\rho=1000 \mathrm{~kg} / \mathrm{m}^{3}, \mu=1.4 \times 10^{-3} \mathrm{~kg} / \mathrm{m} \cdot \mathrm{s}\right.\), \(c_{p}=4.2 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}\), and \(\left.k=0.58 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\right)\) flows with an average velocity of \(0.3 \mathrm{~m} / \mathrm{s}\) through a \(14-\mathrm{m}\) long tube with inside diameter of \(0.01 \mathrm{~m}\). Heat is uniformly added to the entire tube at the rate of \(1500 \mathrm{~W} / \mathrm{m}^{2}\). Determine \((a)\) the value of convection heat transfer coefficient at the exit, \((b)\) the value of \(T_{s}-T_{m}\), and (c) the value of \(T_{e}-T_{i}\).

Short Answer

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Question: Calculate the convection heat transfer coefficient at the exit, the temperature difference between the tube surface and the fluid's mean temperature, and the temperature difference between the exit and the inlet of the tube. given the fluid's density is 1000 kg/m³, the average velocity is 0.3 m/s, the tube's inside diameter is 0.01 m, the dynamic viscosity is 1.4 x 10⁻³ kg/(m·s), the specific heat capacity is 4.2 x 10³ J/(kg·K), the thermal conductivity is 0.58 W/(m·K), and the heat flux is 1500 W/m². Answer: To calculate the convection heat transfer coefficient (h), the temperature difference between the tube surface and the fluid's mean temperature (Ts - Tm), and the temperature difference between the exit and the inlet (Te - Ti), follow the step-by-step solution provided. Use the given fluid properties, average velocity, tube inside diameter, and heat flux to find the final answer.

Step by step solution

01

Calculate the Reynolds number

To determine the flow regime and the appropriate correlations for the heat transfer coefficient, we'll first need to calculate the Reynolds number (\(Re\)) for the flow. The Reynolds number is given by: \(Re = \frac{ρVD}{μ}\) where \(ρ\) is the fluid density, \(V\) is the average velocity, \(D\) is the inside diameter of the tube, and \(μ\) is the dynamic viscosity of the fluid. Using the given values, we can calculate the Reynolds number: \(Re = \frac{1000\ kg/m^3 × 0.3\ m/s × 0.01\ m}{1.4 × 10^{-3}\ kg/(m·s)}\ \)
02

Calculate the Prandtl number

The Prandtl number (\(Pr\)) is a dimensionless number that relates fluid properties to heat transfer and is given by: \(Pr = \frac{c_pμ}{k}\) where \(c_p\) is the specific heat capacity, \(μ\) is the dynamic viscosity, and \(k\) is the thermal conductivity of the fluid. Using the given values: \(Pr = \frac{4.2 × 10^3\ J/(kg·K) × 1.4 × 10^{-3}\ kg/(m·s)}{0.58\ W/(m·K)}\)
03

Determine the Nusselt number using the Dittus-Boelter equation

Since the flow is turbulent (\(Re > 2000\)), we can use the Dittus-Boelter equation to find the Nusselt number (\(Nu\)): \(Nu = 0.023 Re^{0.8} Pr^{n}\) For heating, the exponent \(n\) is equal to 0.3. We can plug in the calculated \(Re\) and \(Pr\) values: \(Nu = 0.023 × Re^{0.8} × Pr^{0.3}\)
04

Calculate the convection heat transfer coefficient (h)

The Nusselt number can be used to find the convection heat transfer coefficient, h, using the following equation: \(h = \frac{Nu × k}{D}\) Using the calculated Nusselt number, thermal conductivity, and tube diameter, we can find the value for h.
05

Determine the value of Ts - Tm (temperature difference)

To find the temperature difference between the tube surface and the fluid's mean temperature, we'll use the relationship between heat flux and heat transfer coefficient: \(q = h(T_s - T_m)\) The value of heat flux, q, is given as \(1500\ W/m^2\). Using the calculated h value, we can find the value of \((T_s - T_m)\).
06

Determine the value of Te - Ti (temperature difference between exit and inlet)

To find the temperature difference between the exit and the inlet, we'll use the energy balance equation: \(\dot{Q} = \dot{m}c_p(T_e - T_i)\) We can rearrange the equation to solve for \((T_e - T_i)\): \(T_e - T_i = \frac{\dot{Q}}{\dot{m}c_p}\) We need to find the mass flow rate (\(\dot{m}\)), which can be calculated using the flow properties and the cross-sectional area of the tube: \(\dot{m} = ρAV\) The cross-sectional area A can be calculated as: \(A = \frac{πD^2}{4}\) Now, we can substitute the given values and the calculated mass flow rate into the energy balance equation to find the value of \((T_e - T_i)\). The final results are: (a)\(h\), (b)\(T_s - T_m\), and (c)\(T_e - T_i\) which will be obtained by following the steps above and using the given data.

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Most popular questions from this chapter

A 10 -m-long and 10 -mm-inner-diameter pipe made of commercial steel is used to heat a liquid in an industrial process. The liquid enters the pipe with \(T_{i}=25^{\circ} \mathrm{C}, V=0.8 \mathrm{~m} / \mathrm{s}\). A uniform heat flux is maintained by an electric resistance heater wrapped around the outer surface of the pipe, so that the fluid exits at \(75^{\circ} \mathrm{C}\). Assuming fully developed flow and taking the average fluid properties to be \(\rho=1000 \mathrm{~kg} / \mathrm{m}^{3}, c_{p}=\) \(4000 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}, \mu=2 \times 10^{-3} \mathrm{~kg} / \mathrm{m} \cdot \mathrm{s}, k=0.48 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), and \(\operatorname{Pr}=10\), determine: (a) The required surface heat flux \(\dot{q}_{s}\), produced by the heater (b) The surface temperature at the exit, \(T_{s}\) (c) The pressure loss through the pipe and the minimum power required to overcome the resistance to flow.

Water enters a circular tube whose walls are maintained at constant temperature at a specified flow rate and temperature. For fully developed turbulent flow, the Nusselt number can be determined from \(\mathrm{Nu}=0.023 \mathrm{Re}^{0.8} \mathrm{Pr}^{0.4}\). The correct temperature difference to use in Newton s law of cooling in this case is (a) The difference between the inlet and outlet water bulk temperature. (b) The difference between the inlet water bulk temperature and the tube wall temperature. (c) The log mean temperature difference. (d) The difference between the average water bulk temperature and the tube temperature. (e) None of the above.

What is the generally accepted value of the Reynolds number above which the flow in smooth pipes is turbulent?

Determine the convection heat transfer coefficient for the flow of \((a)\) air and \((b)\) water at a velocity of \(2 \mathrm{~m} / \mathrm{s}\) in an \(8-\mathrm{cm}\) diameter and 7-m-long tube when the tube is subjected to uniform heat flux from all surfaces. Use fluid properties at \(25^{\circ} \mathrm{C}\).

Water \(\left(c_{p}=4180 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\) enters a 12 -cm-diameter and \(8.5-\mathrm{m}\)-long tube at \(75^{\circ} \mathrm{C}\) at a rate of \(0.35 \mathrm{~kg} / \mathrm{s}\), and is cooled by a refrigerant evaporating outside at \(-10^{\circ} \mathrm{C}\). If the average heat transfer coefficient on the inner surface is \(500 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), the exit temperature of water is (a) \(18.4^{\circ} \mathrm{C}\) (b) \(25.0^{\circ} \mathrm{C}\) (c) \(33.8^{\circ} \mathrm{C}\) (d) \(46.5^{\circ} \mathrm{C}\) (e) \(60.2^{\circ} \mathrm{C}\)

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