Thermodynamics Quiz: Heat Exchangers And Mixing Chambers

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Heat Exchangers And Mixing ChambersQuestion 1 of 20

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?

187.2°C

165.8°C

203.4°C

174.6°C

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Thermodynamics Quiz

Thermodynamics Quiz: Heat Exchangers And Mixing Chambers

Practice Heat Exchangers And Mixing Chambers in Thermodynamics with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Heat Exchangers And Mixing Chambers, giving you a quick way to practice the rules, question types, and explanations that matter most for Thermodynamics.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

187.2°C
165.8°C (correct answer)
203.4°C
174.6°C

Explanation: 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.

Question 2

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?

It will decrease to approximately $60°C$ due to increased heat transfer
It will increase to approximately $100°C$ due to reduced residence time
It will remain at $80°C$ since the oil flow rate is unchanged
It will decrease to approximately $70°C$ due to higher water heat capacity rate (correct answer)
It cannot be determined without knowing the heat transfer coefficient

Explanation: 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

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.

Question 3

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 $150°C = 2746.0 \text{ kJ/kg}$ )

$0.68 \text{ kg/s}$ (correct answer)
$0.85 \text{ kg/s}$
$1.25 \text{ kg/s}$
$0.57 \text{ kg/s}$
$1.05 \text{ kg/s}$

Explanation: 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.

Question 4

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:

Greater than $0.75$ because the minimum heat capacity rate decreased
Less than $0.75$ because the heat capacity rate ratio changed unfavorably (correct answer)
Equal to $0.75$ because effectiveness depends only on inlet temperatures
Greater than $0.75$ because the NTU increased due to lower minimum heat capacity rate
Less than $0.75$ because the residence time for cold fluid decreased

Explanation: 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.

Question 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?

The mixing process generated entropy, causing a temperature rise above theoretical
Heat transfer to the surroundings occurred despite the adiabatic assumption (correct answer)
The process is ideal and matches theoretical predictions perfectly
Kinetic energy effects became significant during the mixing process
Pressure changes during mixing affected the final temperature significantly

Explanation: 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.

Question 6

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?

The steam flow rate is higher than specified, causing excess condensation
Heat loss from the mixing chamber to the surroundings is occurring (correct answer)
The water inlet temperature is lower than the specified $30°C$
The steam is superheated rather than saturated at the inlet condition
Incomplete mixing is occurring within the chamber volume

Explanation: 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.

Question 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?

The effectiveness will increase from $0.60$ to $0.75$ identically
The effectiveness increase will be smaller than $0.15$ due to configuration differences (correct answer)
The effectiveness increase will be larger than $0.15$ due to better heat transfer
The effectiveness will remain constant since $C_r$ is unchanged
The effectiveness cannot be determined without knowing the heat transfer area

Explanation: 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.

Question 8

In a regenerative heat exchanger, hot exhaust gas ( $\dot{m} = 2 \text{ kg/s}$ , $c_p = 1.1 \text{ kJ/kg·K}$ ) at $400°C$ heats incoming cold air ( $\dot{m} = 1.8 \text{ kg/s}$ , $c_p = 1.0 \text{ kJ/kg·K}$ ) from $25°C$ . If the cold air exits at $180°C$ , what is the hot gas exit temperature?

$274°C$ (correct answer)
$320°C$
$290°C$
$345°C$
$256°C$

Explanation: When you encounter heat exchanger problems, you're dealing with energy conservation: the heat lost by the hot fluid equals the heat gained by the cold fluid. This principle lets you find unknown temperatures when mass flow rates and specific heats are given. Start by calculating the heat gained by the cold air:

\dot{Q} = \dot{m}c_p\Delta T = 1.8 \times 1.0 \times (180-25) = 279 \text{ kW}

. Since energy is conserved, the hot gas must lose exactly this amount of heat. For the hot gas losing 279 kW:

279 = 2.0 \times 1.1 \times (400 - T_{exit})

, where

T_{exit}

is the unknown exit temperature. Solving:

279 = 2.2 \times (400 - T_{exit})

, so

400 - T_{exit} = 126.8

, giving

T_{exit} = 273.2°C

. This matches answer A)

274°C

. Answer B)

320°C

represents only a partial temperature drop, suggesting incomplete heat transfer calculation. Answer C)

290°C

might result from incorrectly assuming equal temperature changes for both fluids, ignoring their different heat capacities. Answer D)

345°C

indicates a calculation error, possibly using the wrong mass flow rate or mixing up the temperature values. The key strategy for heat exchanger problems is always to apply energy conservation systematically: calculate the known heat transfer first, then use that value to find the unknown temperature. Remember that different fluids have different heat capacities, so equal heat transfer doesn't mean equal temperature changes.

Question 9

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?

The specific volume change during the mixing process
The entropy generation rate to verify the second law compliance (correct answer)
The pressure drop through the mixing chamber
The heat transfer coefficient between the steam and water phases
The residence time required for complete phase change

Explanation: 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.

Question 10

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?

The exit temperature of the mixed air stream
The total mass flow rate of the exit stream
The exit relative humidity of the mixed air stream (correct answer)
The total energy balance across the mixing chamber
The pressure drop through the mixing chamber

Explanation: 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.

Question 11

A mixing chamber combines $0.5 \text{ kg/s}$ of steam at $200°C$ with $1.2 \text{ kg/s}$ of water at $25°C$ . Assuming the process is adiabatic and the exit pressure allows liquid water to exist, what property determines whether the mixture will be all liquid or contain some vapor?

The total mass flow rate of the combined streams entering the chamber
The pressure at the exit of the mixing chamber only
The enthalpy balance between the incoming streams and saturation properties at exit pressure (correct answer)
The temperature difference between the hot and cold inlet streams
The ratio of specific heats of the steam and liquid water phases

Explanation: When you encounter mixing chamber problems in thermodynamics, you're dealing with a steady-flow energy balance where the key question is whether the final mixture has enough energy to remain partially vaporized or will condense completely to liquid. The correct approach is option C: comparing the enthalpy balance with saturation properties. Here's why this works: In an adiabatic mixing process, energy is conserved, so the enthalpy of the exit stream equals the sum of inlet enthalpies. You calculate this by finding

h_{out} = \frac{\dot{m}_1 h_1 + \dot{m}_2 h_2}{\dot{m}_1 + \dot{m}_2}

, where the hot steam brings high enthalpy and the cold water brings low enthalpy. Then you compare this result to the saturation enthalpy of liquid water (

h_f

) at the exit pressure. If

h_{out} > h_f

, you'll have some vapor; if

h_{out} \leq h_f

, you'll get all liquid. Option A is wrong because total mass flow rate doesn't determine phase—it's the energy content per unit mass that matters. Option B misses the point because while exit pressure affects saturation properties, you still need the energy balance calculation to determine the outcome. Option D focuses on temperature difference, but temperature alone doesn't account for the latent heat effects and specific enthalpies involved in phase change processes. Remember: In mixing problems, always set up the energy balance first, then compare your result to relevant property tables at the final conditions to determine the phase state.

Question 12

An adiabatic mixing chamber operates at steady state with three inlet streams: Stream 1 ( $1 \text{ kg/s}$ at $90°C$ ), Stream 2 ( $2 \text{ kg/s}$ at $50°C$ ), and Stream 3 ( $1.5 \text{ kg/s}$ at $20°C$ ). If all streams are liquid water with $c_p = 4.18 \text{ kJ/kg·K}$ , what is the exit temperature?

$48.9°C$ (correct answer)
$53.3°C$
$45.6°C$
$51.1°C$
$47.2°C$

Explanation: When you encounter an adiabatic mixing chamber problem, you're dealing with energy conservation where no heat is transferred to or from the surroundings. The key principle is that the total enthalpy entering equals the total enthalpy leaving. For liquid water mixing, this becomes a mass-weighted average temperature calculation. Set up the energy balance:

\sum \dot{m}_{in} c_p T_{in} = \dot{m}_{out} c_p T_{out}

Since

c_p

is constant and cancels out, and total mass flow out equals total mass flow in:

T_{out} = \frac{\dot{m}_1 T_1 + \dot{m}_2 T_2 + \dot{m}_3 T_3}{\dot{m}_1 + \dot{m}_2 + \dot{m}_3}

Substituting the values:

T_{out} = \frac{(1)(90) + (2)(50) + (1.5)(20)}{1 + 2 + 1.5} = \frac{90 + 100 + 30}{4.5} = \frac{220}{4.5} = 48.9°C

This confirms answer A is correct. Answer B (53.3°C) likely results from incorrectly weighting the temperatures equally rather than by mass flow rate. Answer C (45.6°C) might come from calculation errors or mishandling the mass flow rates. Answer D (51.1°C) could result from arithmetic mistakes in the numerator or denominator. Study tip: For adiabatic mixing problems, always remember that the exit temperature is a mass-weighted average of inlet temperatures. The stream with the highest mass flow rate has the strongest influence on the final temperature, not the highest temperature stream.

Question 13

A heat exchanger operates with hot gas ( $c_p = 1.1 \text{ kJ/kg·K}$ ) entering at $300°C$ and exiting at $180°C$ . Cold air ( $c_p = 1.0 \text{ kJ/kg·K}$ ) 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?

$2.57 \text{ kg/s}$ (correct answer)
$1.80 \text{ kg/s}$
$2.20 \text{ kg/s}$
$3.00 \text{ kg/s}$
$1.50 \text{ kg/s}$

Explanation: 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·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.

Question 14

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?

$0.8$ (sum of individual effectiveness values)
$0.64$ (based on temperature change through both units) (correct answer)
$0.16$ (product of individual effectiveness values)
$0.2$ (average of the individual effectiveness values)
$0.53$ (accounting for reduced temperature differences in second unit)

Explanation: 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.

Question 15

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:

Higher effectiveness due to the multiple tube passes increasing heat transfer area
Lower effectiveness due to the mixed flow characteristics reducing temperature differences (correct answer)
Identical effectiveness since the total heat transfer area is the same
Higher effectiveness due to increased turbulence from direction changes in tubes
Lower effectiveness due to increased pressure drop reducing the driving force

Explanation: 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.

Question 16

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?

4.68 m²
5.51 m²
4.98 m² (correct answer)
6.12 m²

Explanation: 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.

Question 17

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.

89.4°C
76.8°C
95.2°C
82.1°C (correct answer)

Explanation: 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.

Question 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?

52.5°C
48.8°C
56.3°C (correct answer)
44.1°C

Explanation: 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.

Question 19

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?

0.156 (correct answer)
0.203
0.089
0.124

Explanation: 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.

Question 20

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?

52.5°C
47.0°C (correct answer)
46.8°C
50.0°C

Explanation: 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.