Heat Exchangers & Mixing Chambers - Thermodynamics

1. A mixing chamber receives superheated steam and subcooled liquid water. The steam enters at $3 \text{ kg/s}$, $200°C$, and $500 \text{ kPa}$. The water enters at $8 \text{ kg/s}$, $40°C$, and $500 \text{ kPa}$. If the exit pressure is $500 \text{ kPa}$ and the exit state is saturated liquid, which thermodynamic property is most critical for determining the feasibility of this process?
   Answer: The entropy generation rate to verify the second law compliance When analyzing mixing processes in thermodynamics, you need to consider both energy conservation (first law) and the fundamental requirement that all real processes must increase total entropy (second law). While the first law helps you find final states, only the second law tells you whether a proposed process can actually occur. The correct answer is B because entropy generation determines process feasibility. For this mixing process to be physically possible, the total entropy of the exit stream must exceed the total entropy of the inlet streams. You'd calculate this by finding the specific entropies of superheated steam at 200°C/500 kPa and subcooled liquid at 40°C/500 kPa, then comparing the mass-weighted sum of inlet entropies to the exit entropy of saturated liquid at 500 kPa. If entropy decreases, the process violates the second law and cannot occur spontaneously. Answer A focuses on specific volume change, which affects chamber design but doesn't determine thermodynamic feasibility. A process can be feasible regardless of volume changes as long as entropy increases. Answer C mentions pressure drop, but this problem states constant pressure throughout, making pressure drop irrelevant to feasibility analysis. Answer D addresses heat transfer coefficients, which affect the rate of mixing and equipment design but not whether the process can thermodynamically occur. Heat transfer rates are kinetic considerations, not thermodynamic limitations. Remember: whenever you see "feasibility" in thermodynamics problems, immediately think second law compliance. Energy conservation tells you what happens; entropy generation tells you what can happen.
2. A regenerative heat exchanger operates with hot combustion gases ($c_p = 1.15 kJ/kg·K$) entering at 450°C with a mass flow rate of 3.5 kg/s, and cold air ($c_p = 1.005 kJ/kg·K$) entering at 20°C with a mass flow rate of 4.0 kg/s. If the heat exchanger has an effectiveness of 0.78 and operates in counterflow, what is the exit temperature of the hot gases?
   Answer: 165.8°C Heat capacity rates: $C_h = 3.5 \times 1.15 = 4.025 kW/K$, $C_c = 4.0 \times 1.005 = 4.02 kW/K$. $C_{min} = 4.02 kW/K$. Maximum heat transfer: $q_{max} = C_{min}(T_{h,in} - T_{c,in}) = 4.02 \times (450-20) = 1728.6 kW$. Actual heat transfer: $q = 0.78 \times 1728.6 = 1348.3 kW$. Hot gas outlet temperature: $T_{h,out} = T_{h,in} - q/C_h = 450 - 1348.3/4.025 = 165.8°C$. Choice A uses the cold air heat capacity rate for hot gas temperature calculation. Choice C incorrectly applies the effectiveness to temperature difference directly. Choice D uses average heat capacity rate instead of the correct hot gas value.
3. A shell-and-tube heat exchanger with one shell pass and two tube passes has a correction factor F = 0.85. The log-mean temperature difference for a counterflow configuration would be 45°C. If the overall heat transfer coefficient is 850 W/m²·K and the required heat transfer rate is 180 kW, what minimum surface area is needed?
   Answer: 4.98 m² For shell-and-tube exchangers, the effective LMTD is: $\Delta T_{eff} = F \times \Delta T_{LMTD,cf} = 0.85 \times 45 = 38.25°C$. Using the heat transfer equation: $q = UA\Delta T_{eff}$, therefore $A = q/(U\Delta T_{eff}) = 180,000/(850 \times 38.25) = 4.98 m²$. Choice A neglects the correction factor entirely and uses the counterflow LMTD directly. Choice B incorrectly applies the correction factor as a multiplier rather than accounting for the reduced driving temperature difference. Choice D uses an incorrect heat transfer coefficient calculation.
4. A shell-and-tube heat exchanger with one shell pass and two tube passes (1-2 configuration) has the same fluid flow rates and properties as a single-pass counter-flow unit. Compared to the counter-flow exchanger, the 1-2 configuration will have:
   Answer: Lower effectiveness due to the mixed flow characteristics reducing temperature differences When analyzing heat exchanger effectiveness, you need to understand how flow configuration affects the temperature driving force for heat transfer. The key insight is that effectiveness depends on how well the configuration maintains temperature differences between hot and cold fluids throughout the exchanger. In a true counter-flow arrangement, the hottest cold fluid encounters the hottest hot fluid, and the coldest cold fluid encounters the coldest hot fluid. This configuration maximizes the temperature difference at every point, leading to optimal heat transfer effectiveness. The 1-2 shell-and-tube configuration creates mixed flow characteristics that reduce this advantage. In the shell side, there's cross-flow mixing, and the two tube passes mean that some of the cold fluid that has already been partially heated in the first pass flows parallel (rather than counter) to the hot fluid in the second pass. This reduces the average temperature difference compared to pure counter-flow, resulting in lower effectiveness. Answer A incorrectly assumes that multiple passes automatically improve performance by increasing area - but the question states both exchangers have the same heat transfer area. Answer C misses the crucial point that identical area doesn't guarantee identical performance when flow patterns differ. Answer D incorrectly focuses on turbulence benefits, but the primary effect of the 1-2 configuration is the detrimental impact on temperature driving forces, not enhanced turbulence. Remember: counter-flow is the gold standard for heat exchanger effectiveness. Any deviation from pure counter-flow typically reduces performance, even with clever multi-pass designs.
5. In an adiabatic mixing process, $3 \text{ kg/s}$ of liquid water at $15°C$ is mixed with $1 \text{ kg/s}$ of liquid water at $75°C$. If the actual exit temperature is measured as $29°C$, which of the following best explains this result?
   Answer: Heat transfer to the surroundings occurred despite the adiabatic assumption When you encounter adiabatic mixing problems, start by calculating the theoretical exit temperature assuming perfect adiabatic conditions, then compare it to the actual result to diagnose what's really happening. For ideal adiabatic mixing, energy balance gives us: $\dot{m}_1 c_p T_1 + \dot{m}_2 c_p T_2 = (\dot{m}_1 + \dot{m}_2) c_p T_{exit}$. Substituting the given values: $(3)(15°C) + (1)(75°C) = (4)T_{theoretical}$, which yields $T_{theoretical} = 30°C$. However, the actual measured temperature is only $29°C$ - a full degree lower than the theoretical prediction. This temperature deficit indicates that energy was lost from the system, violating the adiabatic assumption. Answer B correctly identifies that heat transfer to the surroundings occurred despite the intended adiabatic conditions. A is incorrect because entropy generation in irreversible processes doesn't create energy - it would actually require energy dissipation, potentially lowering temperature further. C is clearly wrong since $29°C \neq 30°C$. D misunderstands the physics - kinetic energy effects in liquid mixing are negligible compared to thermal energy, and wouldn't account for this systematic energy loss. The key insight is that when actual temperatures fall short of theoretical adiabatic predictions, suspect heat loss to surroundings. Perfect adiabatic conditions are difficult to achieve in practice, especially with temperature differences and extended mixing times. Study tip: In thermodynamics problems, always calculate the ideal case first, then use deviations from theory to diagnose real-world effects like heat transfer, friction, or other irreversibilities.
6. An open feedwater heater operates with 5 kg/s of liquid water at 40°C and 800 kPa entering through one inlet, while 0.8 kg/s of steam at 800 kPa and 90% quality enters through another inlet. If the mixing process is adiabatic and steady, and the outlet is saturated liquid at 800 kPa, what is the rate of entropy generation for this process?
   Answer: 0.147 kW/K At 800 kPa: $T_{sat} = 170.4°C$, $s_f = 2.046 kJ/kg·K$, $s_{fg} = 4.614 kJ/kg·K$, $s_f@170.4°C = 2.046 kJ/kg·K$. For liquid water at 40°C: $s_1 ≈ 0.573 kJ/kg·K$. For steam at 90% quality: $s_2 = s_f + x s_{fg} = 2.046 + 0.9 \times 4.614 = 6.199 kJ/kg·K$. Outlet saturated liquid: $s_3 = 2.046 kJ/kg·K$. Entropy generation rate: $\dot{S}_{gen} = \dot{m}_3 s_3 - \dot{m}_1 s_1 - \dot{m}_2 s_2 = 5.8 \times 2.046 - 5 \times 0.573 - 0.8 \times 6.199 = 0.147 kW/K$. Choices B, C, and D use incorrect entropy values or wrong mass flow rate calculations in the entropy balance.
7. A heat exchanger effectiveness-NTU analysis shows that increasing the NTU from $1.5$ to $3.0$ for a counter-flow configuration with $C_r = 0.5$ increases the effectiveness from $0.60$ to $0.75$. If the same change is made to a parallel-flow heat exchanger with identical $C_r$, what can be expected?
   Answer: The effectiveness increase will be smaller than $0.15$ due to configuration differences When analyzing heat exchanger effectiveness-NTU relationships, you need to understand how flow configuration fundamentally affects performance limits and improvement potential. Counter-flow and parallel-flow heat exchangers behave differently as NTU increases. Counter-flow configurations can theoretically achieve higher effectiveness values because the temperature profiles allow for more efficient heat transfer - the cold fluid outlet can actually exceed the hot fluid outlet temperature when $C_r < 1$. Parallel-flow arrangements are inherently limited because both fluids flow in the same direction, preventing the cold fluid from ever getting hotter than the hot fluid outlet. The key insight is that parallel-flow heat exchangers approach their effectiveness limits more quickly than counter-flow units. Since the counter-flow case improved from 0.60 to 0.75 (an increase of 0.15), the parallel-flow case starting from the same initial effectiveness will show a smaller improvement because it's closer to its theoretical maximum. Option A incorrectly assumes flow configuration doesn't matter - this ignores fundamental differences in temperature profiles between configurations. Option C suggests parallel-flow would perform better, which contradicts established heat transfer principles since counter-flow is always superior or equal for given conditions. Option D misunderstands that effectiveness depends on both NTU and configuration, not just $C_r$. Study tip: Remember that counter-flow always outperforms parallel-flow, and parallel-flow effectiveness saturates at lower NTU values. When comparing configurations, always consider which one is closer to its theoretical limit.
8. In a mixing chamber, 0.5 kg/s of steam at 200°C and 300 kPa mixes adiabatically with 1.2 kg/s of liquid water at 25°C and 300 kPa. Assuming the process is steady-state and kinetic and potential energy effects are negligible, what is the approximate quality of the mixture at the exit if it reaches saturation conditions at 300 kPa?
   Answer: 0.156 For steady-state adiabatic mixing: $\dot{m}_1 h_1 + \dot{m}_2 h_2 = \dot{m}_3 h_3$. At 300 kPa: $T_{sat} = 133.5°C$, $h_f = 561.4 kJ/kg$, $h_{fg} = 2163.8 kJ/kg$. Steam at 200°C, 300 kPa: $h_1 ≈ 2865 kJ/kg$. Liquid water at 25°C: $h_2 ≈ 104.8 kJ/kg$. Energy balance: $0.5 \times 2865 + 1.2 \times 104.8 = 1.7 \times h_3$, so $h_3 = 918.1 kJ/kg$. For wet mixture: $h_3 = h_f + x h_{fg}$, therefore $x = (918.1 - 561.4)/2163.8 = 0.156$. Choice B assumes incorrect saturation properties. Choice C uses wrong mass flow rates. Choice D incorrectly applies the energy balance equation.
9. An adiabatic mixing chamber receives two streams of air at 150 kPa. Stream 1 has a mass flow rate of 0.8 kg/s at 80°C and 40% relative humidity, while Stream 2 has a mass flow rate of 1.2 kg/s at 25°C and 60% relative humidity. Assuming ideal gas behavior for dry air and using $c_p = 1.005 kJ/kg·K$ for dry air, what is the approximate dry-bulb temperature of the mixed stream?
   Answer: 47.0°C For adiabatic mixing of moist air, considering primarily the sensible heat of dry air (moisture effects are secondary for temperature estimation): $\dot{m}_1 c_p T_1 + \dot{m}_2 c_p T_2 = (\dot{m}_1 + \dot{m}_2) c_p T_3$. Simplifying: $T_3 = (\dot{m}_1 T_1 + \dot{m}_2 T_2)/(\dot{m}_1 + \dot{m}_2) = (0.8 \times 80 + 1.2 \times 25)/(0.8 + 1.2) = (64 + 30)/2.0 = 47.0°C$. Note: This approximation neglects latent heat effects from moisture, which would slightly affect the final temperature. Choice B incorrectly weights by relative humidity. Choice C uses wrong mass flow calculations. Choice D assumes equal mass flow rates.
10. In a direct-contact mixing chamber, superheated steam at 400°C and 500 kPa with a mass flow rate of 0.3 kg/s mixes with subcooled liquid water at 30°C and 500 kPa with a mass flow rate of 1.8 kg/s. Assuming the mixing process is adiabatic and the exit pressure is 500 kPa, determine the exit temperature if the mixture reaches thermal equilibrium.
    Answer: 82.1°C For adiabatic mixing: $\dot{m}_1 h_1 + \dot{m}_2 h_2 = \dot{m}_3 h_3$. Superheated steam at 400°C, 500 kPa: $h_1 ≈ 3270 kJ/kg$. Subcooled liquid at 30°C: $h_2 ≈ 125.7 kJ/kg$. Total mass flow rate: $\dot{m}_3 = 0.3 + 1.8 = 2.1 kg/s$. Energy balance: $0.3 \times 3270 + 1.8 \times 125.7 = 2.1 \times h_3$, giving $h_3 = 575.4 kJ/kg$. Since $h_3 < h_f@500kPa$ (640.1 kJ/kg), the mixture is subcooled liquid. Using $h ≈ c_p T$: $T_3 ≈ 575.4/4.18 ≈ 82.1°C$. Choice A assumes incorrect specific heat values. Choice B neglects the energy contribution from superheated steam. Choice C incorrectly assumes the mixture is saturated.
11. In a closed feedwater heater, 8 kg/s of liquid water at 50°C enters the tube side while 1.5 kg/s of steam at 600 kPa and 95% quality enters the shell side for condensation. The condensate leaves as saturated liquid at 600 kPa, and the feedwater is heated but remains subcooled. If the process is steady and adiabatic, what is the exit temperature of the feedwater?
    Answer: 112.9°C At 600 kPa: $T_{sat} = 158.8°C$, $h_f = 670.4 kJ/kg$, $h_{fg} = 2756.8 kJ/kg$. Steam enthalpy: $h_{steam} = h_f + x h_{fg} = 670.4 + 0.95 \times 2756.8 = 3290.9 kJ/kg$. Condensate leaves as saturated liquid: $h_{condensate} = 670.4 kJ/kg$. Heat released by steam: $q = 1.5 \times (3290.9 - 670.4) = 3930.8 kW$. For feedwater: $q = \dot{m} c_p (T_{out} - T_{in})$, so $3930.8 = 8 \times 4.18 \times (T_{out} - 50)$. Solving: $T_{out} = 50 + 3930.8/(8 \times 4.18) = 112.9°C$. Choice A uses incorrect steam quality. Choice B assumes the feedwater reaches saturation temperature. Choice D neglects part of the latent heat release.
12. In a direct-contact mixing chamber, saturated steam at $150°C$ is mixed with subcooled water at $25°C$ to produce saturated liquid water at $100°C$. If the water flow rate is $5 \text{ kg/s}$, what is the required steam flow rate? (Use: $h_f$ at $25°C = 104.8 \text{ kJ/kg}$, $h_f$ at $100°C = 419.1 \text{ kJ/kg}$, $h_g$ at $150°C = 2746.0 \text{ kJ/kg}$)
    Answer: $0.68 \text{ kg/s}$ When analyzing mixing chambers, you're dealing with a steady-flow energy balance where mass and energy must both be conserved. The key insight is that the enthalpy entering the system equals the enthalpy leaving the system. Set up your conservation equations systematically. For mass conservation: $\dot{m}_{steam} + \dot{m}_{water} = \dot{m}_{out}$, so $\dot{m}_{steam} + 5 = \dot{m}_{out}$. For energy conservation: $\dot{m}_{steam} \cdot h_{steam} + \dot{m}_{water} \cdot h_{water} = \dot{m}_{out} \cdot h_{out}$. Substituting the given enthalpies: $\dot{m}_{steam} \times 2746.0 + 5 \times 104.8 = (\dot{m}_{steam} + 5) \times 419.1$ Expanding: $2746.0\dot{m}_{steam} + 524 = 419.1\dot{m}_{steam} + 2095.5$ Solving: $2326.9\dot{m}_{steam} = 1571.5$, giving $\dot{m}_{steam} = 0.675 \text{ kg/s}$, which rounds to 0.68 kg/s. Answer A (0.68 kg/s) is correct based on this energy balance. Answer B (0.85 kg/s) likely results from incorrectly using saturated liquid enthalpy for the steam instead of saturated vapor. Answer C (1.25 kg/s) suggests a fundamental error in setting up the energy balance, possibly confusing inlet and outlet conditions. Answer D (0.57 kg/s) might come from arithmetic errors or using wrong enthalpy values. Strategy tip: Always identify your inlet and outlet states clearly, use the correct enthalpy values for each phase (liquid vs. vapor), and double-check that your mass balance matches your energy balance setup.
13. A heat exchanger operates with hot gas ($c_p = 1.1 \text{ kJ/kg}\cdot\text{K}$) entering at $300°C$ and exiting at $180°C$. Cold air ($c_p = 1.0 \text{ kJ/kg}\cdot\text{K}$) enters at $50°C$ and exits at $120°C$. If the hot gas flow rate is $1.5 \text{ kg/s}$, and the measured heat transfer rate is $180 \text{ kW}$, what is the cold air flow rate?
    Answer: $2.57 \text{ kg/s}$ When you encounter heat exchanger problems, you're dealing with energy conservation between two fluid streams. The key principle is that heat lost by the hot fluid equals heat gained by the cold fluid (assuming no losses to surroundings). For any fluid stream, the heat transfer rate is $\dot{Q} = \dot{m} \cdot c_p \cdot \Delta T$, where $\dot{m}$ is mass flow rate, $c_p$ is specific heat, and $\Delta T$ is temperature change. Let's verify the energy balance using the hot gas: $\dot{Q}_{hot} = 1.5 \text{ kg/s} \times 1.1 \text{ kJ/kg}\cdot\text{K} \times (300-180)°C = 1.5 \times 1.1 \times 120 = 198 \text{ kW}$. This differs slightly from the given 180 kW, but we'll use the measured value of 180 kW as it accounts for real-world losses. For the cold air gaining this heat: $180 = \dot{m}_{cold} \times 1.0 \times (120-50)$ $180 = \dot{m}_{cold} \times 1.0 \times 70$ $\dot{m}_{cold} = 180/70 = 2.57 \text{ kg/s}$ This confirms answer A is correct. Answer B (1.80 kg/s) would result from incorrectly using 100°C as the temperature difference. Answer C (2.20 kg/s) might come from calculation errors or wrong temperature values. Answer D (3.00 kg/s) could result from using 60°C as the temperature difference instead of 70°C. Always use the measured heat transfer rate when provided, as it accounts for real losses. Double-check your temperature differences—subtract inlet from outlet for the fluid gaining heat.
14. In an adiabatic mixing chamber, $4 \text{ kg/s}$ of air at $40°C$ and $60\%$ relative humidity mixes with $2 \text{ kg/s}$ of air at $10°C$ and $90\%$ relative humidity. Assuming the exit pressure is $101.3 \text{ kPa}$, which property requires psychrometric analysis rather than simple mass and energy balances?
    Answer: The exit relative humidity of the mixed air stream When you encounter adiabatic mixing problems in thermodynamics, you can handle most properties through straightforward conservation principles, but humidity properties require special psychrometric analysis due to the complex behavior of water vapor in air mixtures. For adiabatic mixing, you can directly apply conservation of mass and energy. The exit temperature (A) comes from energy balance: $\dot{m}_1 h_1 + \dot{m}_2 h_2 = \dot{m}_{exit} h_{exit}$, where enthalpies can be determined from temperature and humidity data. The total mass flow rate (B) is simply $4 + 2 = 6 \text{ kg/s}$ by mass conservation. The energy balance (D) itself is a fundamental conservation equation. However, the exit relative humidity (C) cannot be found through simple mass and energy balances alone. Relative humidity depends on both the actual moisture content and the saturation moisture content at the exit temperature. While you can find the total moisture content by adding the absolute humidity contributions from each stream, determining what percentage this represents of the maximum possible moisture (saturation) at the exit temperature requires psychrometric properties. You need saturation pressure data, humidity ratio relationships, and psychrometric charts or equations to convert between absolute and relative humidity values. Options A, B, and D all follow directly from conservation laws, while C requires you to understand the relationship between moisture content and saturation conditions at the final temperature. Study tip: In mixing problems, remember that intensive properties like relative humidity require psychrometric analysis, while extensive properties typically follow conservation principles.
15. Two identical heat exchangers are connected in series, with the hot fluid flowing through both units and the cold fluid also flowing through both units in the same direction. Each heat exchanger has an effectiveness of $0.4$ when operating individually. What is the overall effectiveness of the combined system?
    Answer: $0.64$ (based on temperature change through both units) When analyzing heat exchangers in series, you need to track how the fluid temperatures change progressively through each unit, rather than simply adding or averaging effectiveness values. For two identical heat exchangers in series with $\varepsilon = 0.4$ each, start by understanding that effectiveness represents the fraction of maximum possible heat transfer achieved. In the first heat exchanger, if the initial temperature difference is $\Delta T_0$, the actual temperature change will be $0.4 \Delta T_0$, leaving a remaining temperature difference of $0.6 \Delta T_0$ entering the second unit. The second heat exchanger operates on this reduced temperature difference with the same effectiveness of $0.4$, so it achieves an additional temperature change of $0.4 \times 0.6 \Delta T_0 = 0.24 \Delta T_0$. The total temperature change is $0.4 \Delta T_0 + 0.24 \Delta T_0 = 0.64 \Delta T_0$, giving an overall effectiveness of $0.64$. Choice A ($0.8$) incorrectly assumes you can simply add effectiveness values, ignoring that the second unit works with a reduced driving force. Choice C ($0.16$) mistakenly multiplies the effectiveness values as if they were independent probabilities. Choice D ($0.2$) incorrectly averages the values, which has no physical basis for heat exchanger performance. Remember: for heat exchangers in series, each subsequent unit operates with the reduced temperature difference left by the previous unit. Always trace the temperature changes step by step rather than using shortcuts that ignore the sequential nature of the process.
16. A steam-heated mixing chamber receives $0.2 \text{ kg/s}$ of saturated steam at $120°C$ and $4 \text{ kg/s}$ of water at $30°C$. The chamber is designed to produce hot water at $85°C$. If the actual exit temperature is only $78°C$, what is the most likely cause?
    Answer: Heat loss from the mixing chamber to the surroundings is occurring When analyzing mixing chamber problems in thermodynamics, you need to apply both mass and energy conservation principles. If the actual exit temperature is lower than the designed temperature, something is preventing the system from achieving its intended energy balance. Let's examine what should happen: The high-energy saturated steam at 120°C should condense and mix with the cooler water, releasing latent heat to achieve the target 85°C. When the actual temperature (78°C) falls short, energy is being lost from the system. Option B correctly identifies heat loss to the surroundings as the culprit. In real mixing chambers, thermal energy can escape through chamber walls, piping connections, and other surfaces. This lost energy means less thermal energy remains to heat the water mixture, resulting in a lower exit temperature than predicted by ideal calculations. Option A is incorrect because higher steam flow would actually increase the energy input, raising the exit temperature above the design value, not lowering it. Option C suggests lower inlet water temperature, but this would require the steam to provide even more energy to reach 85°C - the actual result shows insufficient energy, not excess demand. Option D proposes superheated steam, which would contain more thermal energy than saturated steam, again leading to higher exit temperatures. Study tip: In thermodynamics problems, when actual performance falls short of theoretical predictions, always consider heat losses first. Real systems rarely achieve perfect insulation, and energy losses to surroundings are the most common cause of reduced thermal performance in heat exchangers and mixing equipment.
17. A shell-and-tube heat exchanger operates with oil flowing through the tubes and water flowing through the shell. The oil inlet temperature is $120°C$ and outlet temperature is $80°C$. The water inlet temperature is $25°C$ and outlet temperature is $45°C$. If the water flow rate is doubled while keeping all other conditions constant, what will happen to the oil outlet temperature?
    Answer: It will decrease to approximately $70°C$ due to higher water heat capacity rate Heat exchanger problems require you to think about energy balance and heat capacity rates. When operating conditions change, the system adjusts to maintain energy conservation between the hot and cold fluids. When the water flow rate doubles, its heat capacity rate ($\dot{m}c_p$) doubles, meaning it can absorb more thermal energy per unit time. Since energy must be conserved, the oil must give up more heat to satisfy this increased capacity. With the oil flow rate unchanged, the only way to transfer more heat is for the oil to exit at a lower temperature, creating a larger temperature difference driving force. Using energy balance principles: $\dot{Q}_{oil} = \dot{Q}_{water}$. Originally, the oil loses $40°C$ while water gains $20°C$. When water's heat capacity rate doubles, it can absorb the same heat with only a $10°C$ rise (from $25°C$ to $35°C$). For energy balance, the oil must now cool more significantly - to approximately $70°C$ - to provide the necessary heat transfer. Choice A ($60°C$) overestimates the temperature drop. Choice B ($100°C$) incorrectly suggests the oil outlet temperature increases, which violates energy conservation. Choice C ($80°C$) reflects the misconception that unchanged oil flow rate means unchanged outlet temperature, ignoring the energy balance requirement. Study tip: In heat exchanger problems, always apply energy balance first. When one fluid's heat capacity rate changes, the other fluid must adjust its temperature change to maintain $\dot{Q}_{hot} = \dot{Q}_{cold}$. The fluid with unchanged flow rate adjusts through larger temperature differences.
18. A counterflow heat exchanger operates with hot water entering at 90°C and cold water entering at 15°C. The hot water flow rate is 2 kg/s and the cold water flow rate is 3 kg/s. If the heat exchanger effectiveness is 0.75 and the specific heat of water is 4.18 kJ/kg·K, what is the outlet temperature of the hot water?
    Answer: 56.3°C First, determine the minimum heat capacity rate: $C_{min} = \min(\dot{m}_h c_p, \dot{m}_c c_p) = \min(2 \times 4.18, 3 \times 4.18) = 8.36 kJ/s·K$. The maximum possible heat transfer is $q_{max} = C_{min}(T_{h,in} - T_{c,in}) = 8.36 \times (90-15) = 627 kJ/s$. The actual heat transfer is $q = \varepsilon q_{max} = 0.75 \times 627 = 470.25 kJ/s$. Since the hot water has the minimum heat capacity rate, $T_{h,out} = T_{h,in} - q/C_{h} = 90 - 470.25/8.36 = 56.3°C$. Choice A uses the wrong heat capacity rate in the denominator. Choice B incorrectly assumes equal temperature changes. Choice D uses the cold water inlet temperature in the calculation.
19. A parallel-flow heat exchanger has an overall heat transfer coefficient of 1200 W/m²·K and a surface area of 15 m². Hot oil enters at 150°C with a heat capacity rate of 8000 W/K, while cold water enters at 20°C with a heat capacity rate of 6000 W/K. What is the Number of Transfer Units (NTU) and the theoretical maximum heat transfer rate?
    Answer: NTU = 3.0, $q_{max} = 780 kW$ $C_{min} = \min(8000, 6000) = 6000 W/K$. $NTU = UA/C_{min} = (1200 \times 15)/6000 = 3.0$. The maximum heat transfer rate occurs when the fluid with minimum heat capacity rate undergoes the maximum possible temperature change: $q_{max} = C_{min}(T_{h,in} - T_{c,in}) = 6000 \times (150-20) = 780 kW$. Choice B incorrectly uses the average heat capacity rate for NTU calculation. Choice C correctly calculates NTU but uses the wrong heat capacity rate for $q_{max}$. Choice D uses both incorrect calculations from choices B and C.
20. A counter-flow heat exchanger has an effectiveness of $0.75$ when operating with hot water ($\dot{m}_h = 2 \text{ kg/s}$) and cold water ($\dot{m}_c = 1.5 \text{ kg/s}$). The hot water enters at $90°C$ and cold water enters at $20°C$. If the cold water flow rate is increased to $3.0 \text{ kg/s}$ while maintaining the same inlet temperatures and overall heat transfer coefficient, the new effectiveness will be:
    Answer: Less than $0.75$ because the heat capacity rate ratio changed unfavorably When analyzing heat exchanger effectiveness changes, you need to understand how the heat capacity rate ratio and NTU (Number of Transfer Units) affect performance. Effectiveness depends on both the heat capacity rate ratio $C_r = C_{min}/C_{max}$ and the NTU value. Initially, calculate the heat capacity rates: $C_h = \dot{m}_h c_p = 2 \times 4180 = 8360 \text{ W/K}$ and $C_c = 1.5 \times 4180 = 6270 \text{ W/K}$. Since $C_c < C_h$, we have $C_{min} = 6270 \text{ W/K}$ and $C_r = 6270/8360 = 0.75$. When the cold water flow rate doubles to 3.0 kg/s, $C_c$ becomes $3 \times 4180 = 12540 \text{ W/K}$. Now $C_h < C_c$, so $C_{min} = 8360 \text{ W/K}$ and $C_r = 8360/12540 = 0.67$. The NTU also changes because $\text{NTU} = UA/C_{min}$, and $C_{min}$ increased from 6270 to 8360 W/K, making NTU smaller. For counter-flow heat exchangers, effectiveness decreases when the heat capacity rate ratio moves away from unity (optimal is $C_r = 1$). Moving from 0.75 to 0.67 represents this unfavorable change, and the reduced NTU further decreases effectiveness. Answer B correctly identifies that the heat capacity rate ratio changed unfavorably. Answer A incorrectly states that $C_{min}$ decreased (it increased). Answer C is wrong because effectiveness depends on flow rates, not just inlet temperatures. Answer D incorrectly claims NTU increased when it actually decreased. Study tip: Remember that heat exchanger effectiveness is maximized when heat capacity rates are balanced ($C_r = 1$) and when NTU is high. Changes that move away from these conditions typically reduce effectiveness.

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