Thermodynamics Quiz: Kinetic And Potential Energy Terms

14 questions · exam conditions

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Kinetic And Potential Energy TermsQuestion 1 of 14

In the steady-flow energy equation for a control volume, under what conditions can the kinetic energy terms $\left(\sum_{in} \dot{m} \frac{V^2}{2} - \sum_{out} \dot{m} \frac{V^2}{2}\right)$ be neglected compared to the enthalpy terms?

When velocities are less than

10

m/s regardless of the specific enthalpy changes involved in the process

When the kinetic energy changes are small relative to the enthalpy changes, typically when

\frac{V^2}{2} << |\Delta h|

When the flow is incompressible since kinetic energy only matters for compressible flow situations

When the control volume is horizontal so that gravitational effects don't influence the momentum balance

When the mass flow rates are identical at all inlets and outlets regardless of velocity magnitudes

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Thermodynamics Quiz

Thermodynamics Quiz: Kinetic And Potential Energy Terms

Practice Kinetic And Potential Energy Terms 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 Kinetic And Potential Energy Terms, 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

When velocities are less than $10$ m/s regardless of the specific enthalpy changes involved in the process
When the kinetic energy changes are small relative to the enthalpy changes, typically when $\frac{V^2}{2} << |\Delta h|$ (correct answer)
When the flow is incompressible since kinetic energy only matters for compressible flow situations
When the control volume is horizontal so that gravitational effects don't influence the momentum balance
When the mass flow rates are identical at all inlets and outlets regardless of velocity magnitudes

Explanation: When analyzing the steady-flow energy equation, you need to understand when different energy terms become negligible through order-of-magnitude analysis. The key principle is that any term can be neglected when it's significantly smaller than the dominant terms in the equation. The condition

\frac{V^2}{2} << |\Delta h|

in option B correctly captures this concept. Kinetic energy per unit mass is

\frac{V^2}{2}

, while enthalpy changes

|\Delta h|

are often much larger in typical engineering processes. For example, if

V = 50

m/s, then

\frac{V^2}{2} = 1.25

kJ/kg, but phase changes or significant temperature changes involve enthalpy changes of hundreds or thousands of kJ/kg. This makes the kinetic energy terms negligible by comparison. Option A incorrectly sets an arbitrary velocity limit without considering the actual enthalpy changes. A 10 m/s flow might be significant if enthalpy changes are small, or negligible if they're large - context matters. Option C misunderstands the physics. Kinetic energy exists in both compressible and incompressible flows. Compressibility affects density changes, not whether kinetic energy should be considered. Option D confuses momentum balance with energy balance. The orientation of the control volume affects potential energy terms in the energy equation, not the relevance of kinetic energy terms. Remember: in thermodynamics problems, always compare the relative magnitudes of terms rather than applying absolute thresholds. The decision to neglect any energy term depends entirely on how it compares to the other significant energy changes in your specific process.

Question 2

For a control volume analysis of an aircraft engine at cruise altitude, air enters at $V_1 = 250$ m/s and exits as combustion gases at $V_2 = 500$ m/s. If changes in elevation are negligible and the specific enthalpy increases by $800$ kJ/kg, what percentage of the total specific energy increase is due to kinetic effects?

$10.5\%$ (correct answer)
$14.6\%$
$23.4\%$
$8.9\%$
$18.8\%$

Explanation: This problem tests your understanding of energy analysis in control volumes, specifically how to partition total energy changes between thermal and kinetic effects. When analyzing flowing systems like jet engines, you need to account for both enthalpy changes (thermal energy) and kinetic energy changes. First, calculate the specific kinetic energy change:

\Delta KE = \frac{1}{2}(V_2^2 - V_1^2) = \frac{1}{2}(500^2 - 250^2) = \frac{1}{2}(250,000 - 62,500) = 93,750

J/kg =

93.75

kJ/kg. The total specific energy increase includes both the enthalpy change and kinetic energy change:

\Delta E_{total} = \Delta h + \Delta KE = 800 + 93.75 = 893.75

kJ/kg. The percentage due to kinetic effects is:

\frac{93.75}{893.75} \times 100\% = 10.5\%

, confirming answer A. Answer B (14.6%) likely comes from incorrectly calculating the kinetic energy change or using wrong velocity values. Answer C (23.4%) probably results from calculating the ratio of kinetic energy change to enthalpy change alone (

\frac{93.75}{800}

) rather than to the total energy change. Answer D (8.9%) might stem from computational errors in the kinetic energy calculation or using incorrect conversion factors. Study tip: In control volume problems involving significant velocity changes, always calculate both thermal (enthalpy) and mechanical (kinetic) energy contributions separately, then combine them for the total. Don't forget to convert units consistently—mixing J/kg and kJ/kg is a common source of errors.

Question 3

For a control volume analysis where fluid accelerates from rest ( $V_1 = 0$ ) to $V_2 = 60$ m/s while rising $20$ m in elevation, the kinetic and potential energy terms must both be included in the energy balance. If this occurs in an adiabatic nozzle where the specific enthalpy decreases by $2.5$ kJ/kg, what fraction of the enthalpy decrease appears as kinetic energy?

$0.58$
$0.72$ (correct answer)
$0.65$
$0.43$
$0.89$

Explanation: When analyzing control volumes with significant velocity and elevation changes, you need the steady flow energy equation that accounts for all energy transformations. This problem tests your ability to apply energy conservation when enthalpy converts to both kinetic and potential energy. Start with the steady flow energy equation for an adiabatic process:

h_1 + \frac{V_1^2}{2} + gz_1 = h_2 + \frac{V_2^2}{2} + gz_2

. Since

V_1 = 0

and rearranging:

h_1 - h_2 = \frac{V_2^2}{2} + g(z_2 - z_1)

. Calculate the kinetic energy gain:

\frac{V_2^2}{2} = \frac{(60)^2}{2} = 1800

J/kg = 1.8 kJ/kg. Calculate the potential energy gain:

g\Delta z = 9.81 \times 20 = 196.2

J/kg = 0.196 kJ/kg. The enthalpy decrease is 2.5 kJ/kg, so verify energy balance:

1.8 + 0.196 = 1.996 \approx 2.0

kJ/kg (small discrepancy due to rounding). The fraction appearing as kinetic energy is:

\frac{1.8}{2.5} = 0.72

, which is answer B. Answer A (0.58) likely comes from incorrectly using the approximate 2.0 kJ/kg total instead of the given 2.5 kJ/kg. Answer C (0.65) might result from calculation errors in the kinetic energy term. Answer D (0.43) could come from confusing which energy term represents the larger fraction or arithmetic mistakes. Always double-check that your kinetic and potential energy changes sum to match the given enthalpy change, and remember that kinetic energy typically dominates at high velocities like 60 m/s.

Question 4

In the control volume analysis of a diffuser, air enters at $V_1 = 200$ m/s and exits at $V_2 = 50$ m/s. The elevation change is negligible. If the process is adiabatic and the temperature increases, which statement correctly describes the energy transformation?

Kinetic energy decreases by $18.75$ kJ/kg and this energy appears entirely as increased internal energy since enthalpy equals internal energy for ideal gases
Kinetic energy decreases by $18.75$ kJ/kg and this energy appears as increased enthalpy, which includes both internal energy and flow work effects (correct answer)
Kinetic energy decreases by $37.5$ kJ/kg and this energy appears as increased temperature with no change in enthalpy since the process is adiabatic
Kinetic energy increases by $18.75$ kJ/kg due to the acceleration in the diffuser and this energy comes from decreased internal energy
Kinetic energy decreases by $18.75$ kJ/kg but this energy is lost to friction and does not contribute to any other energy form in the system

Explanation: When analyzing diffusers using control volume analysis, you're dealing with steady-flow energy equation applications where kinetic energy converts to thermal energy. The key insight is understanding what happens to energy in adiabatic flow processes and distinguishing between internal energy and enthalpy. First, let's calculate the kinetic energy change. The kinetic energy decreases by

\frac{1}{2}(V_1^2 - V_2^2) = \frac{1}{2}(200^2 - 50^2) = \frac{1}{2}(40000 - 2500) = 18750

J/kg =

18.75

kJ/kg. In steady-flow processes like diffusers, the relevant energy equation involves enthalpy, not just internal energy. For adiabatic flow with negligible elevation changes:

h_1 + \frac{V_1^2}{2} = h_2 + \frac{V_2^2}{2}

. This means the decrease in kinetic energy directly increases the enthalpy. Enthalpy includes both internal energy (temperature effects) and flow work (

Pv

terms), making it the appropriate property for flowing systems. Option A incorrectly suggests enthalpy equals internal energy for ideal gases - this is false since

h = u + Pv

. Option C miscalculates the kinetic energy change as 37.5 kJ/kg and wrongly claims no enthalpy change in adiabatic processes. Option D incorrectly states that kinetic energy increases, when clearly the velocity decreases from 200 to 50 m/s. Remember: in steady-flow devices, always work with enthalpy rather than internal energy. The steady-flow energy equation naturally accounts for the flow work that's inherent in moving fluid through the system.

Question 5

A throttling valve reduces the pressure of steam flowing through it. The upstream conditions are $V_1 = 10$ m/s at elevation $z_1 = 2$ m, and downstream conditions are $V_2 = 35$ m/s at elevation $z_2 = 2$ m. For this throttling process, which statement about the energy balance is correct?

The enthalpy remains constant, so the kinetic energy increase of $562.5$ J/kg must come from internal energy decrease since no work is done
The total enthalpy (including kinetic energy) remains constant, so the static enthalpy decreases by $562.5$ J/kg to accommodate the kinetic energy increase (correct answer)
The enthalpy decreases by $562.5$ J/kg to provide the kinetic energy increase, while internal energy remains constant in this adiabatic process
The enthalpy increases by $562.5$ J/kg due to the pressure drop, which is balanced by the kinetic energy increase in this isentropic process
Both enthalpy and kinetic energy increase, with the total energy increase of $562.5$ J/kg coming from the flow work done by upstream pressure

Explanation: When you encounter a throttling valve problem, you're dealing with a constant enthalpy process—but you need to be careful about which enthalpy remains constant. The key insight is understanding the difference between static enthalpy and total enthalpy (also called stagnation enthalpy). For a throttling process, the total enthalpy

h_0 = h + \frac{V^2}{2} + gz

remains constant across the valve. Since the elevations are equal (

z_1 = z_2 = 2

m), the gravitational potential energy terms cancel out. Let's calculate the kinetic energy change:

\Delta KE = \frac{V_2^2 - V_1^2}{2} = \frac{35^2 - 10^2}{2} = \frac{1225 - 100}{2} = 562.5 \text{ J/kg}

Since total enthalpy is conserved:

h_1 + \frac{V_1^2}{2} = h_2 + \frac{V_2^2}{2}

This means

h_2 = h_1 - 562.5

J/kg. The static enthalpy decreases to accommodate the kinetic energy increase while keeping total enthalpy constant. Option A incorrectly states that static enthalpy remains constant—this is a common misconception. Option C wrongly suggests internal energy stays constant, but internal energy actually increases due to the irreversible throttling process. Option D incorrectly calls this an isentropic process and claims enthalpy increases, but throttling is irreversible (entropy increases) and involves energy conversion, not addition. Study tip: Always distinguish between static and total enthalpy in flow processes. Throttling conserves total enthalpy, not static enthalpy—remember this as kinetic energy trades off with static enthalpy.

Question 6

A control volume analysis is performed on a gas turbine where air enters at elevation $z_1 = 0$ m with velocity $V_1 = 150$ m/s and exits at elevation $z_2 = 10$ m with velocity $V_2 = 300$ m/s. If the specific enthalpy increases by $200$ kJ/kg, what is the change in specific kinetic energy?

$22.5$ kJ/kg
$33.75$ kJ/kg (correct answer)
$56.25$ kJ/kg
$11.25$ kJ/kg
$67.5$ kJ/kg

Explanation: When analyzing control volume problems involving turbines or compressors, you need to track changes in kinetic energy, potential energy, and enthalpy separately. The specific kinetic energy is defined as

KE = \frac{V^2}{2}

, where velocity must be in m/s to get results in J/kg. To find the change in specific kinetic energy, calculate the kinetic energy at both states: At inlet:

KE_1 = \frac{V_1^2}{2} = \frac{(150)^2}{2} = \frac{22,500}{2} = 11,250

J/kg At exit:

KE_2 = \frac{V_2^2}{2} = \frac{(300)^2}{2} = \frac{90,000}{2} = 45,000

J/kg Change in kinetic energy:

\Delta KE = KE_2 - KE_1 = 45,000 - 11,250 = 33,750

J/kg =

33.75

kJ/kg This confirms answer B is correct. A) 22.5 kJ/kg likely comes from only calculating the inlet kinetic energy (11.25 kJ/kg) and doubling it, missing the proper exit calculation. C) 56.25 kJ/kg probably results from incorrectly adding the inlet and exit kinetic energies instead of finding their difference. D) 11.25 kJ/kg is just the inlet kinetic energy alone, forgetting to account for the exit conditions entirely. Study tip: In control volume analysis, always calculate each energy term at both inlet and exit conditions before finding the change. The enthalpy increase given here is a distractor—kinetic energy depends only on velocity, not enthalpy or elevation changes.

Question 7

In a control volume analysis of a pump, water flows from a lower reservoir to an upper tank. The elevation difference is $25$ m, and the velocity increases from $2$ m/s to $8$ m/s. What fraction of the total mechanical energy change is due to kinetic energy effects?

$0.12$ (correct answer)
$0.24$
$0.36$
$0.48$
$0.88$

Explanation: When analyzing pumps in control volume problems, you need to identify all forms of mechanical energy change: kinetic energy, potential energy, and flow work. The total mechanical energy change determines how much work the pump must provide. First, calculate the kinetic energy change per unit mass:

\Delta KE = \frac{1}{2}(v_2^2 - v_1^2) = \frac{1}{2}(8^2 - 2^2) = \frac{1}{2}(64 - 4) = 30 \text{ J/kg}

Next, find the potential energy change:

\Delta PE = g \Delta z = 9.81 \times 25 = 245.25 \text{ J/kg}

The total mechanical energy change is:

\Delta E_{total} = \Delta KE + \Delta PE = 30 + 245.25 = 275.25 \text{ J/kg}

The fraction due to kinetic energy effects is:

\frac{\Delta KE}{\Delta E_{total}} = \frac{30}{275.25} = 0.109 ≈ 0.12

This confirms answer (A) 0.12. Answer (B) 0.24 would result from doubling the kinetic energy contribution, possibly from incorrectly including both initial and final kinetic energies rather than their difference. Answer (C) 0.36 might come from using incorrect velocity values or miscalculating the elevation change. Answer (D) 0.48 suggests a fundamental error in the energy balance, perhaps confusing kinetic and potential energy contributions. Study tip: In pump problems, potential energy changes typically dominate due to elevation differences, while kinetic energy effects are usually smaller contributors. Always calculate both components separately before finding their ratio.

Question 8

A steam turbine operates between two states where the inlet has specific kinetic energy of $125$ J/kg and specific potential energy of $490$ J/kg, while the outlet has specific kinetic energy of $625$ J/kg and specific potential energy of $98$ J/kg. If the specific enthalpy decreases by $300$ kJ/kg, what is the specific work output assuming no heat transfer?

$299.89$ kJ/kg (correct answer)
$300.89$ kJ/kg
$298.89$ kJ/kg
$301.11$ kJ/kg
$300.00$ kJ/kg

Explanation: When analyzing turbine performance, you need to apply the steady flow energy equation (SFEE), which accounts for all energy changes across the control volume. For a turbine with no heat transfer, the equation becomes:

h_1 + \frac{V_1^2}{2} + gz_1 = h_2 + \frac{V_2^2}{2} + gz_2 + w_{out}

To find the specific work output, rearrange to solve for

w_{out}

w_{out} = (h_1 - h_2) + \frac{V_1^2 - V_2^2}{2} + g(z_1 - z_2)

Converting to consistent units and substituting the given values:

Enthalpy change: $h_1 - h_2 = +300$ kJ/kg (since enthalpy decreases)
Kinetic energy change: $\frac{V_1^2 - V_2^2}{2} = 125 - 625 = -500$ J/kg = $-0.5$ kJ/kg
Potential energy change: $g(z_1 - z_2) = 490 - 98 = +392$ J/kg = $+0.392$ kJ/kg

Therefore:

w_{out} = 300 + (-0.5) + 0.392 = 299.892

kJ/kg ≈

299.89

kJ/kg Answer A (299.89 kJ/kg) is correct. Answer B (300.89 kJ/kg) likely results from adding the kinetic energy change instead of subtracting it. Answer C (298.89 kJ/kg) probably comes from subtracting the potential energy change incorrectly. Answer D (301.11 kJ/kg) suggests errors in both kinetic and potential energy calculations. Study tip: Always convert all energy terms to the same units before calculating, and remember that kinetic energy increases represent energy leaving the system in turbines, reducing work output.

Question 9

In a rocket nozzle operating in vacuum, combustion gases enter the control volume at $V_1 = 100$ m/s and exit at $V_2 = 2500$ m/s. The elevation change is negligible and the process is adiabatic. If the specific enthalpy decreases by $3000$ kJ/kg, what percentage error would result from neglecting the inlet kinetic energy term?

$0.17\%$ (correct answer)
$0.33\%$
$0.50\%$
$1.25\%$
$2.08\%$

Explanation: When analyzing rocket nozzles or any high-velocity flow device, you're dealing with the steady flow energy equation where kinetic energy changes can be substantial. The key question here is whether neglecting smaller energy terms introduces significant error in your calculations. To find the percentage error from neglecting inlet kinetic energy, you need to compare the complete energy equation with the simplified version. The steady flow energy equation for this adiabatic process is:

h_1 + \frac{V_1^2}{2} = h_2 + \frac{V_2^2}{2}

First, calculate the kinetic energy terms. The inlet kinetic energy is

\frac{V_1^2}{2} = \frac{(100)^2}{2} = 5,000

J/kg = 5 kJ/kg. The outlet kinetic energy is

\frac{V_2^2}{2} = \frac{(2500)^2}{2} = 3,125,000

J/kg = 3,125 kJ/kg. The complete change in kinetic energy is

3,125 - 5 = 3,120

kJ/kg. If you neglect the inlet term, you'd calculate only 3,125 kJ/kg. The percentage error is

\frac{5}{3,120} \times 100\% = 0.16\%

, which rounds to 0.17%. Answer A (0.17%) is correct. Answer B (0.33%) likely doubled the actual error. Answer C (0.50%) may have used incorrect kinetic energy calculations. Answer D (1.25%) probably compared the inlet kinetic energy to the enthalpy change rather than the total kinetic energy change. Remember: in high-velocity applications, always check whether "small" energy terms are truly negligible by calculating the actual percentage error relative to the dominant energy changes.

Question 10

In a control volume analysis of a pumping station, water is lifted from a reservoir at elevation 100 m to a storage tank at elevation 180 m. The pump inlet and outlet diameters are both 30 cm, and the volumetric flow rate is 0.2 m³/s. If the pump requires 20 kW of power input, what is the primary contribution of kinetic energy terms to the energy balance?

Kinetic energy terms are zero because inlet and outlet velocities are identical, simplifying the energy equation significantly (correct answer)
Kinetic energy terms are negligible at approximately 0.15% of the total energy input and can be safely omitted from calculations
Kinetic energy terms contribute approximately 2.8% of the power input and should be included for precise efficiency calculations
Kinetic energy terms are substantial at roughly 8.5% of the power input due to the high velocity through the pump

Explanation: Since both inlet and outlet diameters are 30 cm, the cross-sectional areas are identical:

A = \pi(0.15)^2 = 0.0707

m². The velocities are:

V = \frac{\dot{V}}{A} = \frac{0.2}{0.0707} = 2.83

m/s at both inlet and outlet. Therefore,

\Delta KE = \frac{V_{out}^2 - V_{in}^2}{2} = \frac{2.83^2 - 2.83^2}{2} = 0

. The kinetic energy terms cancel out completely, which is a common situation in pumping applications where suction and discharge line sizes are the same. Options B, C, and D incorrectly assume there is a kinetic energy contribution when there is none.

Question 11

A nozzle accelerates steam from rest in a large tank to a high-speed jet. The steam exits at 800 m/s, and the specific enthalpy decreases by 320 kJ/kg. The tank is positioned 15 m above the nozzle exit. When evaluating the relative importance of energy terms in the steady-flow energy equation, what conclusion is most appropriate?

Kinetic energy change (320 kJ/kg) exactly balances the enthalpy change, while potential energy change (0.15 kJ/kg) is negligible
Potential energy change (0.15 kJ/kg) and kinetic energy change (320 kJ/kg) together balance the enthalpy change within acceptable engineering accuracy (correct answer)
Kinetic energy change (320 kJ/kg) matches the enthalpy change indicating perfect energy conversion, but potential energy should be included for completeness
All three energy terms are significant with kinetic energy change (320 kJ/kg), enthalpy change (320 kJ/kg), and potential energy change (0.15 kJ/kg)

Explanation: Calculate kinetic energy change:

\Delta KE = \frac{V_2^2 - V_1^2}{2} = \frac{800^2 - 0^2}{2} = \frac{640000}{2} = 320000

J/kg = 320 kJ/kg. Calculate potential energy change:

\Delta PE = g(z_2 - z_1) = 9.81 \times (0 - 15) = -147.15

J/kg = -0.147 kJ/kg (approximately -0.15 kJ/kg). For the energy balance:

h_1 + \frac{V_1^2}{2} + gz_1 = h_2 + \frac{V_2^2}{2} + gz_2

. Rearranging:

h_1 - h_2 = \frac{V_2^2}{2} - \frac{V_1^2}{2} + g(z_2 - z_1) = 320 + (-0.15) = 319.85

kJ/kg. This closely matches the given enthalpy decrease of 320 kJ/kg. Option A incorrectly states that kinetic and enthalpy changes are exactly equal. Option C ignores that potential energy is already included in the balance. Option D incorrectly describes all terms as individually significant when potential energy is actually negligible.

Question 12

A centrifugal compressor draws air from atmosphere through a horizontal inlet duct and discharges it vertically upward through a 5-meter tall outlet duct. The inlet velocity is 80 m/s, outlet velocity is 120 m/s, and air density is approximately constant at 1.2 kg/m³. When applying the steady-flow energy equation to the entire compressor system including ducts, how should the kinetic and potential energy terms be incorporated?

Kinetic energy increase of 2.4 kJ/kg dominates the energy balance, while potential energy increase of 0.049 kJ/kg can be safely ignored
Kinetic energy increase of 4.0 kJ/kg must be provided by the compressor, while potential energy increase of 0.049 kJ/kg is negligible
Kinetic energy increase of 2.4 kJ/kg and potential energy increase of 0.049 kJ/kg together represent the total mechanical energy added
Kinetic energy increase of 4.0 kJ/kg and potential energy increase of 0.049 kJ/kg both contribute to the required compressor work input (correct answer)

Explanation: Calculate kinetic energy change:

\Delta KE = \frac{V_2^2 - V_1^2}{2} = \frac{120^2 - 80^2}{2} = \frac{14400 - 6400}{2} = \frac{8000}{2} = 4000

J/kg = 4.0 kJ/kg. Calculate potential energy change:

\Delta PE = g(z_2 - z_1) = 9.81 \times 5 = 49.05

J/kg = 0.049 kJ/kg. In the steady-flow energy equation for a compressor:

h_1 + \frac{V_1^2}{2} + gz_1 + w_{in} = h_2 + \frac{V_2^2}{2} + gz_2

, so

w_{in} = (h_2 - h_1) + \frac{V_2^2 - V_1^2}{2} + g(z_2 - z_1)

. Both kinetic and potential energy increases require work input from the compressor. While the potential energy term is small compared to kinetic energy (about 1.2%), it still contributes to the total work requirement. Option B incorrectly dismisses the potential energy term. Options C and D have incorrect kinetic energy values (2.4 kJ/kg instead of 4.0 kJ/kg).

Question 13

A rocket nozzle expands combustion gases from a combustion chamber (negligible velocity) to a high-speed exhaust jet at 2500 m/s. The nozzle operates horizontally, and the specific enthalpy of the gases decreases by 3.2 MJ/kg during expansion. When evaluating energy terms for nozzle performance analysis, what is the most significant consideration regarding kinetic and potential energy?

Kinetic energy change (3.125 MJ/kg) closely matches enthalpy change (3.2 MJ/kg), confirming excellent nozzle efficiency with potential energy negligible
Kinetic energy change (3.125 MJ/kg) slightly exceeds enthalpy change (3.2 MJ/kg), suggesting measurement errors or heat losses in the system
Kinetic energy change (3.125 MJ/kg) represents nearly perfect energy conversion from enthalpy, while horizontal orientation eliminates potential energy effects (correct answer)
Kinetic energy change (6.25 MJ/kg) exceeds available enthalpy change (3.2 MJ/kg), indicating impossible operating conditions requiring design revision

Explanation: Calculate kinetic energy change:

\Delta KE = \frac{V_2^2 - V_1^2}{2} = \frac{2500^2 - 0^2}{2} = \frac{6250000}{2} = 3125000

J/kg = 3.125 MJ/kg. The specific enthalpy decreases by 3.2 MJ/kg. For a horizontal nozzle, potential energy change is zero. The energy balance gives:

h_1 = h_2 + \frac{V_2^2}{2}

, so

h_1 - h_2 = \frac{V_2^2}{2}

. Theoretically, 3.2 MJ/kg of enthalpy should convert to 3.2 MJ/kg of kinetic energy, but we calculated 3.125 MJ/kg, which represents 97.7% efficiency - excellent for a rocket nozzle. Option A incorrectly states that kinetic energy matches enthalpy change when it's actually slightly less. Option B incorrectly suggests kinetic energy exceeds enthalpy change. Option C correctly identifies the near-perfect conversion and notes that horizontal orientation eliminates potential energy considerations. Option D incorrectly calculates kinetic energy as 6.25 MJ/kg (this would be

V^2

rather than

V^2/2

Question 14

A steam turbine operates between two elevations with the inlet 50 m above the outlet. Steam enters at 150 m/s and exits at 300 m/s. The specific enthalpy decreases by 800 kJ/kg across the turbine. When applying the steady-flow energy equation to determine the work output per unit mass, which statement best describes the treatment of kinetic and potential energy terms?

Potential energy change contributes +0.49 kJ/kg while kinetic energy change contributes -33.75 kJ/kg, both terms are significant compared to enthalpy change
Potential energy change contributes -0.49 kJ/kg while kinetic energy change contributes +33.75 kJ/kg, but only kinetic energy term significantly affects work calculation (correct answer)
Both potential and kinetic energy changes are less than 1% of the enthalpy change and can be neglected for practical engineering calculations
Potential energy change contributes -0.49 kJ/kg while kinetic energy change contributes -33.75 kJ/kg, requiring both terms for accurate work determination

Explanation: For potential energy:

\Delta PE = g(z_2 - z_1) = 9.81 \times (0 - 50) = -490.5

J/kg = -0.49 kJ/kg (negative because outlet is lower). For kinetic energy:

\Delta KE = \frac{V_2^2 - V_1^2}{2} = \frac{300^2 - 150^2}{2} = \frac{90000 - 22500}{2} = 33750

J/kg = +33.75 kJ/kg (positive because exit velocity is higher). The kinetic energy change (33.75 kJ/kg) is about 4.2% of the enthalpy change (800 kJ/kg), which is significant enough to include in turbine work calculations, while potential energy change is much smaller. Option A has wrong sign for potential energy. Option C underestimates the significance of kinetic energy. Option D has wrong sign for kinetic energy.