Question

A jacketed-agitated vessel, fitted with a turbine agitator, is used for heating a water stream from \(10^{\circ} \mathrm{C}\) to $54^{\circ} \mathrm{C}$. The average heat transfer coefficient for water at the vessel's inner wall can be estimated from \(\mathrm{Nu}=\) $0.76 \mathrm{Re}^{2 / 3} \mathrm{Pr}^{1 / 3}\(. Saturated steam at \)100^{\circ} \mathrm{C}$ condenses in the jacket, for which the average heat transfer coefficient in $\mathrm{kW} / \mathrm{m}^{2} \cdot \mathrm{K}$ is: \(h_{o}=13.1\left(T_{g}-T_{w}\right)^{-0.25}\). The vessel dimensions are: \(D_{t}=0.6 \mathrm{~m}, H=0.6 \mathrm{~m}\), and \(D_{a}=0.2 \mathrm{~m}\). The agitator speed is \(60 \mathrm{rpm}\). Calculate the mass rate of water that can be heated in this agitated vessel steadily.

Short answer

In a jacketed-agitated vessel used for heating a water stream from \(10^{\circ} \mathrm{C}\) to \(54^{\circ} \mathrm{C}\), determine the mass flow rate of water that can be heated in this vessel steadily. The mass flow rate of water that can be heated in this agitated vessel steadily is approximately \(0.380 \mathrm{~kg/s}\).

Step-by-step solution

Determine Reynolds number and Prandtl number

First, we need to calculate the Reynolds number (Re) and the Prandtl number (Pr) for the water. Using the given vessel dimensions and agitator speed, we can determine the kinematic viscosity and thermal diffusivity of water and use them to calculate Re and Pr, respectively. The kinematic viscosity of water at approximately \(20^{\circ}C\) is \(\nu = 1 \times 10^{-6} \mathrm{~m}^2/\mathrm{s}\), and its thermal diffusivity is \(\alpha = 1.45 \times 10^{-7} \mathrm{~m}^2/\mathrm{s}\). Reynolds number: \(\mathrm{Re} = \frac{D_t N}{\nu}\) Prandtl number: \(\mathrm{Pr} = \frac{c_p \mu}{k}\), where \(c_p\) is the heat capacity at constant pressure, \(\mu\) is the dynamic viscosity, and \(k\) is the thermal conductivity. Since we are not given these values for water, we will instead use typical values for water: \(c_p = 4.2 \mathrm{~kJ / kg \cdot K}\), \(\mu = 1 \times 10^{-3} \mathrm{~N \cdot s / m^2}\), and \(k = 0.6 \mathrm{~W / m \cdot K}\). Calculate Re and Pr: Re = (0.6 m)(60 rpm)(1/0.0167 Hz)(1/1*10^{-6} m^2/s) = 21600 Pr = (4.2 kJ/kgK)(1*10^{-3} Ns/m^2)/(0.6 W/mK) = 7 Remember these results for the forthcoming steps.

Calculate heat transfer coefficient for water inside the vessel

Next, we calculate the heat transfer coefficient for water inside the vessel. Meanwhile, we'll make use of the given Nusselt number correlation: \(\mathrm{Nu} = 0.76 \mathrm{Re}^{2/3} \mathrm{Pr}^{1/3}\). The Nusselt number is the ratio of convective to conductive heat transfer across a boundary, and it can be expressed as \(\mathrm{Nu} = \frac{h_i D_t}{k}\), where \(h_i\) is the heat transfer coefficient inside the vessel and \(D_t\) is the vessel diameter. Calculate the Nusselt number, and then use it to determine \(h_i\): Nu = 0.76 * (21600)^(2/3) * (7)^(1/3) = 160.3 h_i = (160.3 * 0.6 W/mK) / 0.6 m = 160.3 W/m^2K

Calculate overall heat transfer coefficient

Now we need to calculate the overall heat transfer coefficient \(U\) using the given function for the average heat transfer coefficient in the jacket, \(h_o = 13.1(T_g - T_w)^{-0.25}\). The inner wall temperature \(T_w\) can be assumed to be \(54^{\circ}C\), and the jacket steam temperature is given as \(T_g = 100^{\circ}C\). Calculate the jacket heat transfer coefficient \(h_o\): h_o = 13.1(100-54)^{-0.25} = 5.43 kW/m^2K To find the overall heat transfer coefficient \(U\), we take the reciprocal of the sum of the reciprocals of \(h_i\) and \(h_{\mathrm{o}}\): \(\frac{1}{U} = \frac{1}{h_{\mathrm{i}}} + \frac{1}{h_{\mathrm{o}}}\) \(U = \frac{1}{\frac{1}{160.3} + \frac{1}{5430}} = 154.4 \mathrm{~W/m^2K}\)

Calculate heat transfer area

The heat transfer area (\(A\)) can be calculated using the given vessel dimensions, assuming that heat transfer occurs through the entire inner surface of the cylindrical vessel: \(A = 2 \pi (D_t/2)(H)\) \(A = 2 \pi (0.6\mathrm{m}/2)(0.6\mathrm{m}) = 1.131 \mathrm{~m}^2\)

Calculate energy transfer rate

To find the energy transfer rate \(Q\), we will use the following equation: \(Q = U A (T_g - T_w)\) \(Q = 154.4 \mathrm{~W/m^2K} * 1.131 \mathrm{~m}^2 * (100^{\circ} \mathrm{C} - 54^{\circ} \mathrm{C}) = 7.792 \times 10^3 \mathrm{~W}\)

Calculate mass flow rate of water

Finally, we can calculate the mass flow rate of water using the energy balance equation: \(\dot{m} = \frac{Q}{c_p (T_{out} - T_{in})}\) \(\dot{m} = \frac{7.792 \times 10^3 \mathrm{~W}}{4.2 \mathrm{~kJ/kgK} * (54^{\circ} \mathrm{C} - 10^{\circ} \mathrm{C})} = 0.380 \mathrm{~kg/s}\) The mass flow rate of water that can be heated in this agitated vessel steadily is approximately \(0.380 \mathrm{~kg/s}\).