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Act like a helpful tutor and exlain me step-by-step solution with explanation and final answer:1. The design of a shell and tube heat exchanger necessitates a rigorous evaluation of the overall heat transfer coefficient, which is a composite parameter accounting for all thermal resistances in the system. The diagram below illustrates the cross-section of a single tube, showing the inner fluid film, the inner fouling layer, cuter fouling layer are tube wall, and outer fluid film, with their respective thermal resistances and the temperature profile across them. «For a counter-flow shell and tube heat exchanger operating under steady-state conditions, with constant fluid properties and a uniform overall heat transfer coefficient U, derive the expression for U based on the summation of the individual thermal resistances offered by the inner and outer fluid film, the tube wall material, and the fouling layers on both the tube-side and shell-side surfaces, considering the ozter side wall film material and the fouling lay, on both the tube-side and shell-side surfaces, considering the respective heat transfer areas. i. Analyse the impact of increased fouling on either the shell-side or tube-side on the value of the overall heat transfer coefficient and the subsequent effect on the required heat transfer surface area to maintain a specified heat duty. ii. Deduce the relationship for the Logarithmic Mean Temperature Difference (LMTD) for a pure counter-flow heat exchanger by performing a heat balance over a differential element of the exchanger length, assuming a constant U and constant specific heats for both fluids. + When dealing with multi-pass heat exchangers, the deviation from true counter-flow behaveur requires a correction to the LMTD. i. Obtain the expression for the LMTD correction factor (F), defined as the ratio of the actual mean temperature difference in the multi-pass exchanger to the LMTD of a counter-flow exchanger operating between the same inlet and outlet temperatures, by considering the temperature effectiveness (P) and the heat capacity rate ratio (R). ii. Reduce the expression for the heat transfer rate Q to the form Q = U * A * F « LMTD counterfiow, where A is the total heat transfer area, and discuss how the value of F is determined from design charts based on P and R for a specific exchanger configuration (e.g., 1-shell pass, 2-tube passes). Ro 4 Rei = a 2 ha; \\ ; Ns Y, Rome i07 " EX "hts Ne Re Q 2. A solid steel sphere with a diameter of 5 cm, initially at a uniform temperature of 450 °C, is suddenly immersed in a large oil bath maintained at a constant temperature of 50 °C. The convection heat transfer coefficient between the sphere surface and the oil is 120 W/(m2-K). The properties of the steel are: density p = 7800 kg/m? specific heat capacity C, = 480 J/(kg'K), and thermal conductivity k= 50 W/mK). +. Calculate the Biot number (Bi) for this system to justify the validity of the lumped capacitance method, and subsequently determine the time required for the center of the sphere to cool down to a temperature of 150 °C. «=. Based on the governing differential equation for the lumped capacitance method, reduce the expression for the temperature-time history, T(t), to a dimensionless form, Fae = exp(~Bi Fo), where Fo is the Fourier number, by appropriately defining the dimensionless time and position variables.

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Act like a helpful tutor and exlain me step-by-step solution with explanation and final answer:Uploaded Image1. The design of a shell and tube heat exchanger necessitates a rigorous evaluation of the overall heat transfer coefficient, which is a composite parameter accounting for all thermal resistances in the system. The diagram below illustrates the cross-section of a single tube, showing the inner fluid film, the inner fouling layer, cuter fouling layer are tube wall, and outer fluid film, with their respective thermal resistances and the temperature profile across them. «For a counter-flow shell and tube heat exchanger operating under steady-state conditions, with constant fluid properties and a uniform overall heat transfer coefficient U, derive the expression for U based on the summation of the individual thermal resistances offered by the inner and outer fluid film, the tube wall material, and the fouling layers on both the tube-side and shell-side surfaces, considering the ozter side wall film material and the fouling lay, on both the tube-side and shell-side surfaces, considering the respective heat transfer areas. i. Analyse the impact of increased fouling on either the shell-side or tube-side on the value of the overall heat transfer coefficient and the subsequent effect on the required heat transfer surface area to maintain a specified heat duty. ii. Deduce the relationship for the Logarithmic Mean Temperature Difference (LMTD) for a pure counter-flow heat exchanger by performing a heat balance over a differential element of the exchanger length, assuming a constant U and constant specific heats for both fluids. + When dealing with multi-pass heat exchangers, the deviation from true counter-flow behaveur requires a correction to the LMTD. i. Obtain the expression for the LMTD correction factor (F), defined as the ratio of the actual mean temperature difference in the multi-pass exchanger to the LMTD of a counter-flow exchanger operating between the same inlet and outlet temperatures, by considering the temperature effectiveness (P) and the heat capacity rate ratio (R). ii. Reduce the expression for the heat transfer rate Q to the form Q = U * A * F « LMTD counterfiow, where A is the total heat transfer area, and discuss how the value of F is determined from design charts based on P and R for a specific exchanger configuration (e.g., 1-shell pass, 2-tube passes). Ro 4 Rei = a 2 ha; \\ ; Ns Y, Rome i07 " EX "hts Ne Re Q 2. A solid steel sphere with a diameter of 5 cm, initially at a uniform temperature of 450 °C, is suddenly immersed in a large oil bath maintained at a constant temperature of 50 °C. The convection heat transfer coefficient between the sphere surface and the oil is 120 W/(m2-K). The properties of the steel are: density p = 7800 kg/m? specific heat capacity C, = 480 J/(kg'K), and thermal conductivity k= 50 W/mK). +. Calculate the Biot number (Bi) for this system to justify the validity of the lumped capacitance method, and subsequently determine the time required for the center of the sphere to cool down to a temperature of 150 °C. «=. Based on the governing differential equation for the lumped capacitance method, reduce the expression for the temperature-time history, T(t), to a dimensionless form, Fae = exp(~Bi Fo), where Fo is the Fourier number, by appropriately defining the dimensionless time and position variables.

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Asked by: Francis
Created at: 2026-02-09 07:01:28

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Let's break down the **step-by-step solution** for **Question 2 (a)**, which involves calculating the **Biot number** and the **cooling time** for a steel sphere immersed in oil. --- # **Step-by-Step Solution** --- ## **Given Data** - **Diameter of sphere, \( d \) = 5 cm = .05 m** - **Radius, \( r_ \) = d/2 = .025 m** - **Initial temperature, \( T_{initial} \) = 450°C** - **Oil bath temperature, \( T_{fluid} \) = 50°C** - **Convective heat transfer coefficient, \( h \) = 120 W/(m²·K)** - **Density, \( \rho \) = 780 kg/m³** - **Specific heat, \( C_p \) = 480 J/(kg·K)** - **Thermal conductivity, \( k \) = 50 W/(m·K)** - **Target temperature, \( T \) = 150°C** --- ## **(a) Calculate the Biot Number (Bi)** ### **Biot Number Formula** \[ Bi = \frac{h L_c}{k} \] Where **\(L_c\)** is the characteristic length for a sphere: \[ L_c = \frac{V}{A} = \frac{\frac{4}{3}\pi r_^3}{4\pi r_^2} = \frac{r_}{3} \] So, \[ L_c = \frac{.025}{3} = .00833~\text{m} \] ### **Plug in the Values** \[ Bi = \frac{120 \times .00833}{50} = \frac{1}{50} = .02 \] ### **Interpretation** - **If Bi < .1**, lumped capacitance method is valid. - Here, **Bi = .02 < .1**: **Lumped capacitance method is justified**. --- ## **(a) Time required to cool to 150°C** ### **Lumped Capacitance Equation** \[ \frac{T(t)-T_\text{fluid}}{T_\text{initial}-T_\text{fluid}} = \exp\left(-\frac{hA}{\rho V C_p} t\right) \] ### **Calculate Surface Area (A) and Volume (V)** - \( A = 4\pi r_^2 = 4\pi (.025)^2 = .00785~\text{m}^2 \) - \( V = \frac{4}{3}\pi r_^3 = \frac{4}{3}\pi (.025)^3 = 6.54 \times 10^{-5}~\text{m}^3 \) ### **Find the Time (t) for T = 150°C** 1. **Set up the equation:** \[ \frac{T(t) - 50}{450 - 50} = \exp\left(-\frac{hA}{\rho V C_p} t\right) \] \[ \frac{150 - 50}{400} = .25 \] \[ .25 = \exp\left(-\frac{hA}{\rho V C_p} t\right) \] 2. **Take the natural log on both sides:** \[ \ln(.25) = -\frac{hA}{\rho V C_p} t \] \[ -1.3863 = -\frac{hA}{\rho V C_p} t \] \[ 1.3863 = \frac{hA}{\rho V C_p} t \] \[ t = \frac{1.3863 \cdot \rho V C_p}{hA} \] 3. **Plug in the values:** - \( \rho = 780~\text{kg/m}^3 \) - \( V = 6.54 \times 10^{-5}~\text{m}^3 \) - \( C_p = 480~\text{J/(kg·K)} \) - \( h = 120~\text{W/(m}^2\text{·K)} \) - \( A = .00785~\text{m}^2 \) \[ t = \frac{1.3863 \cdot 780 \cdot 6.54 \times 10^{-5} \cdot 480}{120 \cdot .00785} \] Calculate numerator: - \( 780 \cdot 6.54 \times 10^{-5} = .51012 \) - \( .51012 \cdot 480 = 244.8576 \) - \( 244.8576 \cdot 1.3863 = 339.594 \) Calculate denominator: - \( 120 \cdot .00785 = .942 \) \[ t = \frac{339.594}{.942} = 360.6~\text{seconds} \] --- ## **Final Answers** ### **Biot Number:** \[ \boxed{Bi = .02} \] Lumped capacitance method is justified. ### **Cooling Time to 150°C:** \[ \boxed{t = 361~\text{seconds}} \] (Rounded to the nearest second) --- **If you want the solution for part (b), let me know!**

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