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Hot brine from a geothermal well passes through \(1203 \mathrm{~cm}-\mathrm{O} . \mathrm{D} ., 1 \mathrm{~mm}-\mathrm{wall}-\) thickness, \(16 \mathrm{~m}\)-long titanium alloy \((k=14 \mathrm{~W} / \mathrm{m} \mathrm{K})\) tubes. Refrigerant-113 boils on the exterior of the tubes at \(150^{\circ} \mathrm{C}\), with an outside heat transfer coefficient of \(20,000 \mathrm{~W} / \mathrm{m}^{2} \mathrm{~K}\). The hot brine enters the tubes at \(210^{\circ} \mathrm{C}\), and the bulk velocity in each tube is \(2 \mathrm{~m} / \mathrm{s}\). Calculate the following quantities. (i) \(\operatorname{Re}_{D}\), the Reynolds number inside a tube (ii) \(h_{c, i}\), the inside heat transfer coefficient (iii) \(U\), the overall heat transfer coefficient (iv) \(\mathrm{N}_{\mathrm{tu}}\), the number of transfer units of the boiler (v) \(\varepsilon\), the exchanger effectiveness (vi) \(T_{H, \text { out }}\), the hot stream outlet temperature (vii) \(\dot{Q}\), the heat transferred in the boiler (viii) \(\dot{m}_{C}\), the R-113 vapor production rate For the brine, take \(k=0.6 \mathrm{~W} / \mathrm{m} \mathrm{K}, \rho=900 \mathrm{~kg} / \mathrm{m}^{3}, c_{p}=4000 \mathrm{~J} / \mathrm{kg} \mathrm{K}\), and \(v=0.20 \times 10^{-6} \mathrm{~m}^{2} / \mathrm{s} ;\) for R-113, take \(h_{\mathrm{fg}}=0.2 \times 10^{6} \mathrm{~J} / \mathrm{kg}\).

Step by step solution

01

Calculate the Reynolds Number (Re_D)

The Reynolds number (Re) is a dimensionless quantity used to predict flow patterns in different fluid flow situations. It is given by the formula \( \operatorname{Re}_{D} = \frac{\rho V D}{\mu} \), where \( \rho \) is the density, \( V \) is the velocity, \( D \) is the diameter of the tube, and \( \mu \) is the dynamic viscosity (which can also be expressed using kinematic viscosity \( u \) as \( \mu = \rho u \)).Given:\- \( \rho = 900 \, \text{kg/m}^3 \)- \( V = 2 \, \text{m/s} \)- \( D = 1200 \, \text{mm} - 2 \, \text{mm} = 0.201 \, \text{m} \)\ (correcting for inner diameter)- \( u = 0.20 \times 10^{-6} \, \text{m}^2/\text{s} \)Substitute the known values:\[ \operatorname{Re}_{D} = \frac{900 \times 2 \times 0.201}{0.20 \times 10^{-6}} = \frac{362.4}{0.20 \times 10^{-6}} = 1.812 \times 10^6 \]

02

Calculate Inside Heat Transfer Coefficient (h_c,i)

For calculating the inside heat transfer coefficient, \( h_{c,i} \), we use the Dittus-Boelter equation for turbulent flow, given as \( h = 0.023 \frac{k}{D} \operatorname{Re}_{D}^{0.8} \operatorname{Pr}^{n} \), where \( n=0.3 \) for heating of fluid, \( k \) is the fluid thermal conductivity, and \( \operatorname{Pr} \) is the Prandtl number.Given:- \( k = 0.6 \, \text{W/m} \cdot \text{K} \)- \( \operatorname{Pr} = \frac{c_p \mu}{k} = \frac{4000 \times 0.20 \times 10^{-6}}{0.6} = 1.333 \)- \( D = 0.201 \, \text{m} \) (inner diameter of the tube)Using the Reynolds number from Step 1 and substituting:\[ h_{c,i} = 0.023 \frac{0.6}{0.201} (1.812 \times 10^6)^{0.8} (1.333)^{0.3} \]Calculate it:\[ h_{c,i} = 0.023 \times 2.985 \times 5.757 \times 1.105 = 5022.13 \, \text{W/m}^2 \cdot \text{K} \]

03

Calculate Overall Heat Transfer Coefficient (U)

The overall heat transfer coefficient, \( U \), can be calculated using the formula:\[ \frac{1}{U} = \frac{1}{h_{c,i}} + \frac{t}{k_w} + \frac{1}{h_{c,o}} \]where:- \( h_{c,i} \) is the inside heat transfer coefficient- \( t = 1 \, \text{mm} = 0.001 \, \text{m} \) (wall thickness of the tube)- \( k_w = 14 \, \text{W/m} \cdot \text{K} \) (thermal conductivity of titanium alloy)- \( h_{c,o} = 20000 \, \text{W/m}^2 \cdot \text{K} \) (outside heat transfer coefficient)Substitute the values:\[ \frac{1}{U} = \frac{1}{5022.13} + \frac{0.001}{14} + \frac{1}{20000} \]\[ U = \frac{1}{0.00019952 + 0.00007143 + 0.000050} = 3931.4 \, \text{W/m}^2 \cdot \text{K} \]

04

Calculate Number of Transfer Units (NTU)

The Number of Transfer Units, \( \mathrm{N}_{\mathrm{tu}} \), is defined by the formula \( \mathrm{N}_{\mathrm{tu}} = \frac{UA}{C_\text{min}} \), where \( A \) is the heat transfer area and \( C_\text{min} \) is the minimum heat capacity rate.- Heat transfer area, \( A = \pi DL = \pi \times 0.201 \times 16 \approx 10.097 \, \text{m}^2 \)- Heat capacity rate of hot stream, \( C_{H} = \dot{m}_{H} c_{p,H} \)- \( C_{H} \) can be calculated assuming a mass flow rate, \( \dot{m}_{H} = \rho V A_{\text{cross-section}} = 900 \times 2 \times \frac{\pi (0.201)^2}{4} = 57.009 \, \text{kg/s} \) Given \( c_{p,H} = 4000 \, \text{J/kg K} \):\[ C_{H} = 57.009 \times 4000 = 228036 \text{ W/K} \] (\( C_\text{min} = C_H \))\[ \mathrm{N}_{\mathrm{tu}} = \frac{3931.4 \times 10.097}{228036} = 0.1737 \]

05

Calculate Heat Exchanger Effectiveness (蔚)

Effectiveness \( \epsilon \) is defined based on NTU for a counter flow heat exchanger:\[ \epsilon = 1 - e^{-\mathrm{N}_{\mathrm{tu}}} \]Calculate \( \epsilon \):\[ \epsilon = 1 - e^{-0.1737} = 0.1592 \]

06

Calculate Hot Stream Outlet Temperature (T_H,out)

To find the outlet temperature of the hot stream, use the effectiveness definition:Given \( T_{H,in} = 210^{\circ} \text{C} \) and \( T_C = 150^{\circ} \text{C} \), the effectiveness formula is:\[ \epsilon = \frac{T_{H,in} - T_{H,out}}{T_{H,in} - T_{C}} \]Rearrange the formula:\[ T_{H,out} = T_{H,in} - \epsilon (T_{H,in} - T_C) = 210 - 0.1592 \times (210 - 150) \]Calculate \( T_{H,out} \):\[ T_{H,out} = 210 - 9.55 = 200.45^{\circ} \text{C} \]

07

Calculate Heat Transferred (Q虈)

The heat transferred \( \dot{Q} \) can be calculated as:\[ \dot{Q} = C_{H} \times (T_{H,in} - T_{H,out}) \]Substitute known values:\[ \dot{Q} = 228036 \times (210 - 200.45) = 228036 \times 9.55 = 2.178 \times 10^{6} \text{ W} \]

08

Calculate R-113 Vapor Production Rate (峁乢C)

The vapor production rate of the refrigerant can be found using the heat transferred and the latent heat of vaporization:\[ \dot{m}_C = \frac{\dot{Q}}{h_{fg}} \]Given \( h_{fg} = 0.2 \times 10^6 \text{ J/kg} \):\[ \dot{m}_{C} = \frac{2.178 \times 10^6}{0.2 \times 10^6} = 10.89 \text{ kg/s} \]

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Key Concepts

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

Reynolds Number

The Reynolds Number is a vital concept when analyzing flow within heat exchangers. It helps determine whether the flow will be laminar or turbulent, which in turn affects the heat transfer processes. The equation for the Reynolds Number is given by \( \operatorname{Re}_{D} = \frac{\rho V D}{\mu} \), where:

• \( \rho \) is the fluid's density.
• \( V \) is the fluid's velocity.
• \( D \) is the diameter of the tube.
• \( \mu \) is the dynamic viscosity.

In the case of our geothermal heat exchanger, the Reynolds Number calculated was approximately \( 1.812 \times 10^6 \). This indicates a turbulent flow since the number is much greater than 4000, the typical threshold for turbulence. Turbulent flows have a more chaotic behavior conducive to mixing, which enhances the rate of heat transfer.

Heat Transfer Coefficient

The Heat Transfer Coefficient, particularly the 'inside heat transfer coefficient' \( h_{c,i} \), describes the heat transferred per unit surface area per unit temperature difference. It is a critical parameter as it determines how effectively heat is being exchanged between the fluid and the tube walls. The Dittus-Boelter equation is suitable for calculating this coefficient in turbulent flow situations:\[h = 0.023 \left(\frac{k}{D}\right) \operatorname{Re}_{D}^{0.8} \operatorname{Pr}^{n}\]Where:

• \( k \) is the thermal conductivity of the fluid.
• \( \operatorname{Pr} \) is the Prandtl number, indicative of the fluid's property ratio.
• \( n \) is 0.3 for heating processes.

In our system, \( h_{c,i} \) was calculated to be 5022.13 \( \text{W/m}^2 \cdot \text{K} \). A higher heat transfer coefficient means more efficient heat exchange, useful in thermal design optimizations.

Overall Heat Transfer Coefficient

The Overall Heat Transfer Coefficient, \( U \), is a comprehensive measure integrating all resistances to heat transfer both inside and outside the heat exchanger tube walls. It plays an integral role in thermal system design, impacting the heat exchanger's performance:\[\frac{1}{U} = \frac{1}{h_{c,i}} + \frac{t}{k_w} + \frac{1}{h_{c,o}}\]Considerations for \( U \) include:

• \( h_{c,i} \): Inside heat transfer coefficient.
• \( t \): Tube wall thickness.
• \( k_w \): Thermal conductivity of the tube material.
• \( h_{c,o} \): Outside heat transfer coefficient.

The calculated \( U \) value of 3931.4 \( \text{W/m}^2 \cdot \text{K} \) suggests that the heat exchanger is moderately efficient, with scope for improvement by reducing any of the thermal resistances, like increasing \( h_{c,o} \) or selecting a material with higher \( k_w \).

Number of Transfer Units

The Number of Transfer Units, \( \mathrm{N}_{\mathrm{tu}} \), provides insight into the heat exchanger's capacity and efficiency by representing the ratio of the heat transfer to the minimum energy capacity rate:\[\mathrm{N}_{\mathrm{tu}} = \frac{UA}{C_\text{min}}\]Where:

• \( U \) is the overall heat transfer coefficient.
• \( A \) is the heat transfer area.
• \( C_\text{min} \) is the minimum heat capacity rate.

A \( \mathrm{N}_{\mathrm{tu}} \) value calculated to be 0.1737 implies a low potential effectiveness in transferring heat under specific operation conditions. Generally, a higher NTU indicates better heat exchange performance in the heat exchanger.

Exchanger Effectiveness

The Exchanger Effectiveness \( \varepsilon \) is a performance metric that measures how well a heat exchanger transfers the available energy between two fluids. For a counter-flow exchanger, the effectiveness is defined as \[\varepsilon = 1 - e^{-\mathrm{N}_{\mathrm{tu}}}\]In our example, the effectiveness was found to be 0.1592. This generally low effectiveness indicates that a relatively small portion of the maximum possible heat transfer is occurring.
For improving \( \varepsilon \), strategies may involve altering the flow arrangement or increasing the overall heat transfer coefficient \( U \). Efficient operation of a heat exchanger directly impacts energy savings and operational costs in industrial applications.