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A hot fluid passes through a thin-walled tube of \(10-\mathrm{mm}\) diameter and \(1 . \mathrm{m}\) length. and a coolant at \(T_{\mathrm{w}}=25^{\circ} \mathrm{C}\) is in cross flow over the tube. When the flow rate is \(\dot{m}=18 \mathrm{~kg} / \mathrm{h}\) and the inlet temperature is \(T_{m u}=85^{\circ} \mathrm{C}\). the outlet temperature is \(T_{\operatorname{mor}}=78^{\circ} \mathrm{C}\). Assuming fully developed flow and thermal conditions in the tube, determine the outlet temperarure, \(T_{\text {man }}\) if the flow rate is increased by a factor of 2 . That is, \(m=36 \mathrm{~kg} / \mathrm{h}\), with all other conditions the same, The thermophysical properties of the hot fluid are $$ \rho=1079 \mathrm{~kg} / \mathrm{m}^{3}, \quad c_{p}=2637 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}, \quad \mu=0.0034 \mathrm{~N} \text { - } $$ \(\mathrm{sm}^{2}\), and \(k=0.261 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\).

Step by step solution

01

- Determine Heat Transfer Rate for Original Conditions

The heat transfer rate can be calculated using the equation: \( q = \dot{m} \cdot c_p \cdot (T_{in} - T_{out}) \). Here, \( \dot{m} = 18 \frac{\text{kg}}{\text{h}} = 0.005 \frac{\text{kg}}{\text{s}} \), \( c_p = 2637 \frac{\text{J}}{\text{kg} \cdot \text{K}} \), \( T_{in} = 85^\circ C \), and \( T_{out} = 78^\circ C \). Substituting these values, we have: \( q = 0.005 \cdot 2637 \cdot (85 - 78) \). Calculate \( q \) to determine the heat transfer rate.

02

- Calculate Heat Transfer for Increased Flow Rate

When the mass flow rate is doubled, \( \dot{m} = 36 \frac{\text{kg}}{\text{h}} = 0.01 \frac{\text{kg}}{\text{s}} \). The heat transfer rate must remain approximately the same since the external conditions are unchanged. Assume \( q' \approx q \). Thus, \( q' = \dot{m}_{new} \cdot c_p \cdot (T_{in} - T_{out,new}) \). Set \( q' = q \) from Step 1 and solve for \( T_{out,new} \).

03

- Solve for New Outlet Temperature

Rearrange the equation from Step 2 to solve for \( T_{out,new} \): \( T_{out,new} = T_{in} - \frac{q}{\dot{m}_{new} \cdot c_p} \). Insert the values \( q \) from Step 1 and \( \dot{m}_{new} = 0.01 \), \( c_p = 2637 \) to find \( T_{out,new} \).

04

- Result and Interpretation

Substituting the calculated heat transfer rate \( q \) from Step 1 into the equation from Step 3, compute \( T_{out,new} \). Verify if this temperature is consistent with physical expectations noting that higher flow rates generally reduce the outlet temperature. Provide the solution as the final outlet temperature for the increased flow rate scenario.

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Key Concepts

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

Thermophysical Properties

Understanding the thermophysical properties of a substance is key in analyzing heat transfer scenarios. These properties describe the behavior of a fluid as it undergoes changes in temperature. Key thermophysical properties include:

• Density (\( \rho \)): This property, measured in \( \text{kg/m}^3 \), indicates the mass per unit volume of a substance. In this case, the hot fluid has a density of \( 1079 \text{ kg/m}^3 \).
• Specific Heat Capacity (\( c_p \)): Represented in \( \text{J/kg} \cdot \text{K} \), this is the amount of heat required to change the temperature of a unit mass by one degree Kelvin. Here, \( c_p \) is \( 2637 \text{ J/kg} \cdot \text{K} \) for the hot fluid.
• Viscosity (\( \mu \)): Measured in \( \text{N} \cdot \text{s/m}^2 \), it indicates a fluid's resistance to flow. The hot fluid has a viscosity of \( 0.0034 \text{ N} \cdot \text{s/m}^2 \).
• Thermal Conductivity (\( k \)): This property, measured in \( \text{W/m} \cdot \text{K} \), indicates a material's ability to conduct heat. For our fluid, it is \( 0.261 \text{ W/m} \cdot \text{K} \).

By understanding these properties, engineers can predict how efficiently a fluid will transfer heat when heated or cooled, which is crucial when designing systems involving heat exchange.

Mass Flow Rate

The mass flow rate is a critical concept in heat transfer, especially in systems involving fluid flows. It measures the mass of the fluid passing through a given cross-sectional area per unit of time. In our exercise, the mass flow rate (\( \dot{m} \)) initially is \( 18 \text{ kg/h} \), which is then doubled.

• To work with mass flow rate conveniently in calculations, it is often converted from hours to seconds, yielding here \( 0.005 \text{ kg/s} \) initially.
• A higher mass flow rate generally means more fluid is available to carry heat away, which can affect the outlet temperature of the system.
• When the mass flow rate was doubled to \( 0.01 \text{ kg/s} \), there was a drop in outlet temperature due to the increased heat capacity of the added mass, which carries away more heat.

The relationship shows why controlling mass flow rate is essential for managing the temperature in thermal systems.

Outlet Temperature Calculation

Calculating the outlet temperature of a fluid in heat-exchange systems is pivotal for system effectiveness. This requires understanding of how changes in conditions, like mass flow rate, impact the temperature.

• The outlet temperature can be found using the equation: \[ T_{out,new} = T_{in} - \frac{q}{\dot{m}_{new} \cdot c_p} \] where \( T_{in} \) is the inlet temperature, \( q \) is the heat transfer rate from the original conditions, \( \dot{m}_{new} \) is the new mass flow rate, and \( c_p \) is the specific heat capacity.
• By doubling the mass flow rate, assuming unchanged external conditions, the outlet temperature \( T_{out,new} \) will be lower than the original outlet temperature due to higher fluid mass absorbing the heat.
• This calculation illustrates how control over fluid flow and heat transfer can manage the thermal output of a system, making it a powerful tool in engineering applications where precise temperature control is necessary.

Understanding and executing these calculations allow engineers to ensure systems operate within desired thermal limits.