Question

In a thermal system, water enters a \(25-\mathrm{mm}\)-diameter and \(23-\mathrm{m}\)-long circular tube with a mass flow rate of $0.1 \mathrm{~kg} / \mathrm{s}\( at \)25^{\circ} \mathrm{C}$. The heat transfer from the tube surface to the water can be expressed in terms of heat flux as \(\dot{q}_{s}(x)=a x\). The coefficient \(a\) is $400 \mathrm{~W} / \mathrm{m}^{3}\(, and the axial distance from the tube inlet is \)x$ measured in meters. Determine \((a)\) an expression for the mean temperature \(T_{m}(x)\) of the water, \((b)\) the outlet mean temperature of the water, and (c) the value of a uniform heat flux \(\dot{q}_{s}\) on the tube surface that would result in the same outlet mean temperature calculated in part (b). Evaluate water properties at \(35^{\circ} \mathrm{C}\).

Short answer

Answer: The mean temperature expression is: \(T_{m}(x)= 25 + \dfrac{1102.13}{418.1}\)

Step-by-step solution

Find the given parameters and notations

The given parameters are the diameter (D) of the tube, the length (L) of the tube, the mass flow rate (m_dot), the inlet temperature (T_inlet), the heat flux equation (q_dot_s), and the coefficient (a). Let's list these parameters for easy reference: - Diameter (D) = 25 mm - Length (L) = 23 m - Mass flow rate (m_dot) = 0.1 kg/s - Inlet temperature (T_inlet) = 25°C - Heat flux equation (q_dot_s) = ax - Coefficient (a) = 400 W/m³

Calculate the area of the tube cross-section

First, we need to determine the area (A) of the tube cross-section. The area of a circle is given by \(A = \pi r^2\), where r is the radius of the circle. Since the diameter is given, we can find the radius (r) by dividing the diameter by 2. r = D/2 r = 25 mm / 2 = 12.5 mm = 0.0125 m Now, calculate the area (A) of the tube cross-section: A = \(\pi (0.0125)^{2}\) A = 4.909 × 10⁻⁴ m²

Calculate the total heat transfer rate

We can find the total heat transfer rate (Q_dot_total) by integrating the heat flux equation along the length of the tube: \(\dot{Q}_{total}(x) = \int_{0}^{L} a x A dx\) Substitute the given values of a and A: \(\dot{Q}_{total}(x)= \int_{0}^{23} 400 x (4.909 \times 10^{-4}) dx\) Evaluate the integral: \(\dot{Q}_{total}(x)= (4.909 \times 10^{-4}) \left[\dfrac{400}{2}x^{2} \right]_0^{23} = 1102.13 \thinspace \mathrm{W}\)

Determine the mean temperature expression

The mean temperature can be determined using the energy balance equation: \(T_{m}(x)= T_{inlet} + \dfrac{\dot{Q}_{total}(x)}{\dot{m} C_{p}}\) where \(C_p\) is the specific heat of water, which should be evaluated at 35°C. The specific heat of water at 35°C is approximately \(C_p\) = 4181 J/kg·K. Now, plug in the values for \(\dot{Q}_{total}(x)\), \(\dot{m}\), and \(C_p\): \(T_{m}(x)= 25 + \dfrac{1102.13}{0.1 \times 4181} = 25 + \dfrac{1102.13}{418.1}\) This gives us the mean temperature expression: \(T_{m}(x)= 25 + \dfrac{1102.13}{418.1}\)

Calculate the outlet mean temperature

To calculate the outlet mean temperature, we will use the given length (L) of the tube, which is 23 meters. \(T_{m}(23) = 25 + \dfrac{1102.13}{418.1} = 25 + \dfrac{263.23}{418.1}\) The outlet mean temperature of the water is therefore: \(T_{m}(23) = 25 + 2.63 = 27.63\thinspace^{\circ}\thinspace \mathrm{C}\)

Determine the uniform heat flux

To find the value of the uniform heat flux (q_dot_s_uniform), we will use the energy balance equation: \(\dot{q}_{s_{uniform}}= \dfrac{\dot{Q}_{total}(23)}{A L}\) Substitute the values of \(\dot{Q}_{total}(23)\), A, and L into the equation: \(\dot{q}_{s_{uniform}} = \dfrac{1102.13}{4.909 \times 10^{-4} \times 23}\) Calculate the uniform heat flux: \(\dot{q}_{s_{uniform}} = 10416.8 \thinspace \mathrm{W/m^2}\) #Summary# a) The mean temperature expression is: \(T_{m}(x)= 25 + \dfrac{1102.13}{418.1}\) b) The outlet mean temperature of the water is 27.63°C. c) The value of the uniform heat flux is approximately 10416.8 W/m².