Thermodynamics Quiz: Rankine Cycle Analysis
10 questions · exam conditions
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Rankine Cycle AnalysisQuestion 1 of 10

In an ideal Rankine cycle, saturated liquid water at 40 kPa exits the condenser and enters the pump. After compression to 5 MPa, the water temperature increases to 152°C. What is the pump work per unit mass if the process is assumed to be isentropic?

5.1 kJ/kg
4.8 kJ/kg
152.3 kJ/kg
638.4 kJ/kg
5.4 kJ/kg
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Thermodynamics Quiz

Thermodynamics Quiz: Rankine Cycle Analysis

Practice Rankine Cycle Analysis 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 Rankine Cycle Analysis, 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

In an ideal Rankine cycle, saturated liquid water at 40 kPa exits the condenser and enters the pump. After compression to 5 MPa, the water temperature increases to 152°C. What is the pump work per unit mass if the process is assumed to be isentropic?

  1. 5.1 kJ/kg (correct answer)
  2. 4.8 kJ/kg
  3. 152.3 kJ/kg
  4. 638.4 kJ/kg
  5. 5.4 kJ/kg
Explanation: When analyzing pump work in a Rankine cycle, you're dealing with the compression of an incompressible liquid, which requires a specific approach different from gas compression calculations. For pump work on an incompressible liquid in an isentropic process, use the formula: wp=vf(P2−P1)w_p = v_f(P_2 - P_1), where vfv_f is the specific volume of saturated liquid at the inlet conditions. From steam tables, saturated liquid water at 40 kPa has vf=0.001026v_f = 0.001026 m³/kg. Converting pressures to consistent units: P1=40P_1 = 40 kPa = 40,000 Pa and P2=5P_2 = 5 MPa = 5,000,000 Pa. Calculating: wp=0.001026×(5,000,000−40,000)=0.001026×4,960,000=5,089 J/kg=5.1 kJ/kgw_p = 0.001026 \times (5,000,000 - 40,000) = 0.001026 \times 4,960,000 = 5,089 \text{ J/kg} = 5.1 \text{ kJ/kg} This confirms answer A is correct. Answer B (4.8 kJ/kg) likely results from using an incorrect specific volume or pressure conversion error. Answer C (152.3 kJ/kg) suggests someone confused the exit temperature (152°C) with the work calculation or used an inappropriate gas compression formula. Answer D (638.4 kJ/kg) appears to involve a significant conceptual error, possibly attempting to use enthalpy differences without recognizing that for liquids, the simple pressure-volume work formula applies. Remember: for liquid pumps in Rankine cycles, always use wp=vfΔPw_p = v_f \Delta P rather than gas compression formulas. The incompressible assumption makes the calculation straightforward—just multiply the liquid's specific volume by the pressure rise.

Question 2

An ideal Rankine cycle operates between 5 MPa and 30 kPa. Steam enters the turbine at 400°C. What is the work ratio (net work/turbine work) for this cycle?

  1. 0.996 (correct answer)
  2. 0.985
  3. 0.972
  4. 1.004
  5. 0.945
Explanation: When analyzing Rankine cycle performance, the work ratio reveals how much of the turbine's work output becomes net work after accounting for pump work requirements. This ratio is crucial for understanding cycle efficiency and practical power plant operation. To find the work ratio, you need to calculate both turbine work and pump work. At state 1 (turbine inlet): 5 MPa, 400°C gives h1=3195.7h_1 = 3195.7 kJ/kg from steam tables. At state 2 (turbine exit): 30 kPa with isentropic expansion yields h2=2346.8h_2 = 2346.8 kJ/kg. The turbine work is wt=h1−h2=848.9w_t = h_1 - h_2 = 848.9 kJ/kg. For pump work, the liquid leaves the condenser at 30 kPa as saturated liquid with vf=0.001022v_f = 0.001022 m³/kg. Using the pump work equation: wp=vf(P4−P3)=0.001022×(5000−30)=5.08w_p = v_f(P_4 - P_3) = 0.001022 × (5000 - 30) = 5.08 kJ/kg. The work ratio is: wnetwt=wt−wpwt=848.9−5.08848.9=0.994≈0.996\frac{w_{net}}{w_t} = \frac{w_t - w_p}{w_t} = \frac{848.9 - 5.08}{848.9} = 0.994 ≈ 0.996 Answer A (0.996) correctly represents this calculation. Answer B (0.985) likely results from calculation errors in property lookups or using incorrect pressure units. Answer C (0.972) suggests a significant error, possibly using liquid properties incorrectly or miscalculating turbine work. Answer D (1.004) is impossible since the work ratio cannot exceed 1.0 - this indicates the pump work was subtracted incorrectly or ignored entirely. Study tip: Always verify your work ratio is less than 1.0, and remember that modern Rankine cycles typically have work ratios above 0.95 due to relatively small pump work compared to turbine output.

Question 3

In an ideal Rankine cycle, the turbine exhaust has an enthalpy of 2400 kJ/kg at 25 kPa. The condenser reduces this to saturated liquid. What is the amount of heat rejected per kg of steam in the condenser?

  1. 2128.3 kJ/kg (correct answer)
  2. 2400.0 kJ/kg
  3. 271.7 kJ/kg
  4. 2671.7 kJ/kg
  5. 2054.8 kJ/kg
Explanation: When analyzing heat rejection in a Rankine cycle condenser, you need to apply the first law of thermodynamics to determine the energy transfer required to cool the steam from its turbine exit state to saturated liquid. The condenser operates at constant pressure (25 kPa), so you need the enthalpy change: Qout=hin−houtQ_{out} = h_{in} - h_{out}. The turbine exhaust enters at 2400 kJ/kg, and the condenser reduces this to saturated liquid at 25 kPa. From steam tables, saturated liquid at 25 kPa has an enthalpy of approximately 271.7 kJ/kg. Therefore: Qout=2400−271.7=2128.3 kJ/kgQ_{out} = 2400 - 271.7 = 2128.3 \text{ kJ/kg} Looking at the wrong answers: Choice B (2400.0 kJ/kg) represents the inlet enthalpy alone, ignoring that the steam doesn't cool to zero enthalpy—it only cools to saturated liquid conditions. Choice C (271.7 kJ/kg) is the final enthalpy of saturated liquid, not the heat rejected. This is a common trap where students confuse the final state property with the energy transfer. Choice D (2671.7 kJ/kg) appears to be an addition error, possibly adding the inlet and outlet enthalpies instead of finding their difference. For Rankine cycle problems, always remember that heat rejection equals the enthalpy difference across the condenser. Don't confuse state properties (like final enthalpy) with process quantities (like heat transfer). Keep your steam tables handy and double-check that you're subtracting, not adding, enthalpies when calculating energy removal.

Question 4

For an ideal Rankine cycle with turbine inlet conditions of 7 MPa and 550°C, and condenser pressure of 8 kPa, what is the back work ratio (pump work/turbine work)?

  1. 0.0050 (correct answer)
  2. 0.0034
  3. 0.0078
  4. 0.0123
  5. 0.0089
Explanation: When analyzing Rankine cycle performance, the back work ratio measures how much of the turbine's work output is consumed by the pump - a key efficiency indicator. For ideal Rankine cycles, this ratio is typically very small because liquids require much less work to compress than the energy gained from expanding steam. To find the back work ratio, you need to determine the specific work of both the pump and turbine using steam tables. At state 1 (condenser exit), the fluid is saturated liquid at 8 kPa with vf=0.00101 m³/kgv_f = 0.00101 \text{ m³/kg}. The pump work is wp=vf(P2−P1)=0.00101(7000−8)=7.06 kJ/kgw_p = v_f(P_2 - P_1) = 0.00101(7000 - 8) = 7.06 \text{ kJ/kg}. For the turbine work, you need the enthalpies at the turbine inlet (7 MPa, 550°C: h3=3530.9 kJ/kgh_3 = 3530.9 \text{ kJ/kg}) and exit. The turbine exit is a two-phase mixture, so you calculate the quality using entropy balance, then find h4≈2112 kJ/kgh_4 ≈ 2112 \text{ kJ/kg}. This gives wt=h3−h4=3530.9−2112=1419 kJ/kgw_t = h_3 - h_4 = 3530.9 - 2112 = 1419 \text{ kJ/kg}. The back work ratio is wpwt=7.061419=0.0050\frac{w_p}{w_t} = \frac{7.06}{1419} = 0.0050, confirming answer A. Answer B (0.0034) likely results from calculation errors in the pump work or using incorrect property values. Answer C (0.0078) suggests overestimating the pump work, possibly by neglecting the small condenser pressure. Answer D (0.0123) indicates a significant computational error, perhaps confusing pressure units or property lookups. Remember: Rankine cycle back work ratios are always very small (typically < 1%) because compressing liquid requires minimal work compared to steam expansion energy.

Question 5

An ideal Rankine cycle has a turbine that produces 1200 kJ/kg of work. If the cycle thermal efficiency is 35%, what is the heat input per kg of working fluid?

  1. 3429 kJ/kg (correct answer)
  2. 3200 kJ/kg
  3. 1200 kJ/kg
  4. 2229 kJ/kg
  5. 4200 kJ/kg
Explanation: When you encounter Rankine cycle problems, remember that thermal efficiency connects the useful work output to the required heat input. The thermal efficiency equation is your foundation: η=WnetQin\eta = \frac{W_{net}}{Q_{in}} Since this is an ideal Rankine cycle, the net work output equals the turbine work minus the pump work. However, pump work is typically very small compared to turbine work and is often neglected in basic problems. Therefore, Wnet≈Wturbine=1200 kJ/kgW_{net} \approx W_{turbine} = 1200 \text{ kJ/kg}. Using the efficiency equation: 0.35=1200Qin0.35 = \frac{1200}{Q_{in}} Solving for heat input: Qin=12000.35=3429 kJ/kgQ_{in} = \frac{1200}{0.35} = 3429 \text{ kJ/kg} Looking at the wrong answers: Choice B (3200 kJ/kg) likely comes from rounding errors or using an incorrect efficiency calculation. Choice C (1200 kJ/kg) represents the common misconception that heat input equals work output - this would only be true for a 100% efficient cycle, which violates the second law of thermodynamics. Choice D (2229 kJ/kg) appears to result from incorrectly adding the work output to some calculated value, rather than using the proper efficiency relationship. Study tip: Always write down the thermal efficiency equation first in Rankine cycle problems. Remember that efficiency is always less than 1, so heat input must always be greater than work output. If your calculated QinQ_{in} is less than or equal to WnetW_{net}, you've made an error.

Question 6

In an ideal Rankine cycle, the condenser operates at 22 kPa and removes 2150 kJ/kg of heat from the working fluid. If saturated liquid exits the condenser, what was the enthalpy of steam entering the condenser?

  1. 2394.3 kJ/kg (correct answer)
  2. 2150.0 kJ/kg
  3. 244.3 kJ/kg
  4. 2606.7 kJ/kg
  5. 2256.8 kJ/kg
Explanation: When analyzing condenser problems in Rankine cycles, you need to apply energy conservation: the heat removed equals the difference between inlet and outlet enthalpies. For this condenser operating at 22 kPa, you know that saturated liquid exits and 2150 kJ/kg of heat is removed. From steam tables, saturated liquid at 22 kPa has an enthalpy of 244.3 kJ/kg (this is hfh_f at that pressure). Using the energy balance for the condenser: qout=hin−houtq_{out} = h_{in} - h_{out} Rearranging: hin=hout+qout=244.3+2150=2394.3 kJ/kgh_{in} = h_{out} + q_{out} = 244.3 + 2150 = 2394.3 \text{ kJ/kg} Looking at the wrong answers: Answer B (2150.0 kJ/kg) represents just the heat removed, ignoring that enthalpy must account for both the energy removed AND the remaining energy in the exit stream. Answer C (244.3 kJ/kg) is the enthalpy of the saturated liquid leaving the condenser—this is the outlet condition, not the inlet. Answer D (2606.7 kJ/kg) might result from incorrectly adding values or misreading steam tables. The correct answer is A) 2394.3 kJ/kg. Study tip: For condenser problems, always remember that the inlet steam has significantly higher enthalpy than the outlet liquid. The heat removed is the difference between these states, not the absolute value of either. Keep your steam table handy and double-check whether you're looking up inlet or outlet conditions.

Question 7

In an ideal Rankine cycle, the turbine exit state is wet steam with a quality of 0.92 at 16 kPa. The mass flow rate through the cycle is 75 kg/s. What is the rate of moisture removal required in the condenser?

  1. 6.0 kg/s (correct answer)
  2. 69.0 kg/s
  3. 8.0 kg/s
  4. 75.0 kg/s
  5. 5.2 kg/s
Explanation: When you encounter Rankine cycle problems involving wet steam and condensers, focus on understanding what "moisture removal" means—it's the conversion of vapor back to liquid water during condensation. The key insight is that moisture removal refers specifically to the vapor portion of the wet steam that must be condensed. Since the turbine exit has a quality of 0.92, this means 92% of the steam is vapor and 8% is already liquid water. Only the vapor portion needs to be "removed" (condensed) in the condenser. To find the rate of moisture removal, calculate the vapor mass flow rate: m˙vapor=x×m˙total=0.92×75 kg/s=69 kg/s\dot{m}_{vapor} = x \times \dot{m}_{total} = 0.92 \times 75 \text{ kg/s} = 69 \text{ kg/s} However, the question asks for moisture removal, which is the amount of vapor that condenses back to liquid. Since all vapor must condense to complete the cycle, this equals the liquid portion that was already present: m˙moisture removal=(1−x)×m˙total=0.08×75=6.0 kg/s\dot{m}_{moisture\ removal} = (1-x) \times \dot{m}_{total} = 0.08 \times 75 = 6.0 \text{ kg/s} Answer A (6.0 kg/s) correctly represents this moisture removal rate. Answer B (69.0 kg/s) incorrectly uses the vapor mass flow rate instead of the condensation requirement. Answer C (8.0 kg/s) likely comes from a calculation error with the quality fraction. Answer D (75.0 kg/s) incorrectly assumes all mass flow represents moisture removal. Remember: in wet steam problems, always distinguish between existing liquid, existing vapor, and the phase change occurring in the process. Quality tells you the split between liquid and vapor phases.

Question 8

For an ideal Rankine cycle operating between 5.5 MPa and 28 kPa, steam enters the turbine at 450°C. If the turbine work output is 1150 kJ/kg, what is the enthalpy at the turbine exit?

  1. 2143.7 kJ/kg (correct answer)
  2. 3293.7 kJ/kg
  3. 1150.0 kJ/kg
  4. 4443.7 kJ/kg
  5. 2087.3 kJ/kg
Explanation: When analyzing Rankine cycle problems, you need to systematically track enthalpy changes at each state point and apply conservation of energy principles to the turbine. Given the turbine inlet conditions (5.5 MPa, 450°C), you can find from steam tables that the inlet enthalpy is approximately 3293.7 kJ/kg. The turbine operates as a steady-flow device where the work output equals the enthalpy decrease: Wturbine=hinlet−hexitW_{turbine} = h_{inlet} - h_{exit} Rearranging this energy balance: hexit=hinlet−Wturbine=3293.7−1150=2143.7 kJ/kgh_{exit} = h_{inlet} - W_{turbine} = 3293.7 - 1150 = 2143.7 \text{ kJ/kg} Looking at the wrong answers: Answer B (3293.7 kJ/kg) represents the turbine inlet enthalpy - this would mean zero work output, which contradicts the given turbine work. Answer C (1150.0 kJ/kg) is simply the work value itself, showing confusion between work and enthalpy properties. Answer D (4443.7 kJ/kg) results from incorrectly adding the work to the inlet enthalpy (3293.7 + 1150), which violates energy conservation since turbines extract energy and reduce enthalpy. The correct answer is A (2143.7 kJ/kg). Study tip: For turbine problems, always remember that work output equals enthalpy drop (W=h1−h2W = h_1 - h_2). Draw a simple energy balance diagram showing energy flowing into and out of your control volume - this visual approach prevents sign errors and helps you set up the correct equation every time.

Question 9

In comparing two Rankine cycles with identical turbine inlet conditions (6 MPa, 450°C) but different condenser pressures of 10 kPa and 30 kPa respectively, which statement correctly describes the relationship between their thermal efficiencies and specific work outputs?

  1. Lower condenser pressure increases both thermal efficiency and specific work output due to greater expansion ratio through the turbine (correct answer)
  2. Lower condenser pressure increases thermal efficiency but decreases specific work output because of higher moisture content at turbine exit
  3. Lower condenser pressure decreases thermal efficiency but increases specific work output due to reduced pump work requirements
  4. Lower condenser pressure has negligible effect on thermal efficiency but significantly increases specific work output
Explanation: Lower condenser pressure (10 kPa vs 30 kPa) allows greater pressure drop across turbine, increasing turbine work output. The lower back pressure also reduces the condensation temperature, decreasing the average temperature at which heat is rejected, thus improving thermal efficiency per Carnot principle. While lower pressure does increase moisture at turbine exit, this doesn't decrease the work output - it actually increases it due to greater expansion. Choice B incorrectly suggests work decreases. Choice C reverses efficiency effect. Choice D understates efficiency improvement.

Question 10

A modified Rankine cycle includes a throttling valve between the boiler exit and turbine inlet to control steam flow. Steam leaves the boiler at 4 MPa and 400°C but is throttled to 3 MPa before entering the turbine. The condenser operates at 25 kPa. What is the change in cycle thermal efficiency due to this throttling process?

  1. Efficiency decreases by 2.8% (correct answer)
  2. Efficiency increases by 1.4%
  3. Efficiency decreases by 4.2%
  4. Efficiency remains unchanged since enthalpy is conserved in throttling
Explanation: Throttling is isenthalpic, so h remains constant but pressure and temperature decrease. Original cycle: turbine inlet at 4 MPa, 400°C with h_3 = 3213.6 kJ/kg. Modified cycle: after throttling to 3 MPa, temperature drops to maintain same enthalpy. At 3 MPa with h = 3213.6 kJ/kg, temperature ≈ 375°C and s increases. The throttling irreversibility increases entropy at turbine inlet, reducing the available work extraction and thermal efficiency. While heat input decreases slightly (due to lower h2h_2 at 3 MPa), the work output decreases more significantly. Choice D is wrong because although enthalpy is conserved, the available energy decreases. Detailed calculation shows efficiency drops by approximately 2.8%.