Thermodynamics Quiz: Quality And Saturated Mixtures
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Quality And Saturated MixturesQuestion 1 of 19

Steam at 200°C has a quality of 0.75. If the enthalpy of saturated liquid at 200°C is 850.65 kJ/kg and the enthalpy of vaporization is 1940.7 kJ/kg, what is the specific enthalpy of this wet steam?

2306.2 kJ/kg
1455.5 kJ/kg
2791.4 kJ/kg
1851.2 kJ/kg
1396.9 kJ/kg
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Thermodynamics Quiz

Thermodynamics Quiz: Quality And Saturated Mixtures

Practice Quality And Saturated Mixtures 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 Quality And Saturated Mixtures, giving you a quick way to practice the rules, question types, and explanations that matter most for Thermodynamics.

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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.

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Question 1

Steam at 200°C has a quality of 0.75. If the enthalpy of saturated liquid at 200°C is 850.65 kJ/kg and the enthalpy of vaporization is 1940.7 kJ/kg, what is the specific enthalpy of this wet steam?

  1. 2306.2 kJ/kg (correct answer)
  2. 1455.5 kJ/kg
  3. 2791.4 kJ/kg
  4. 1851.2 kJ/kg
  5. 1396.9 kJ/kg
Explanation: When you encounter a wet steam problem with quality given, you're dealing with a two-phase mixture of liquid and vapor. The quality (x) represents the mass fraction of vapor in the mixture, so quality of 0.75 means 75% vapor and 25% liquid. To find the specific enthalpy of wet steam, use the formula: h=hf+x⋅hfgh = h_f + x \cdot h_{fg}, where hfh_f is the enthalpy of saturated liquid and hfgh_{fg} is the enthalpy of vaporization. Substituting the given values: h=850.65+0.75×1940.7=850.65+1455.53=2306.18 kJ/kgh = 850.65 + 0.75 \times 1940.7 = 850.65 + 1455.53 = 2306.18 \text{ kJ/kg} This confirms answer A (2306.2 kJ/kg) is correct. Answer B (1455.5 kJ/kg) represents a common error where students calculate only the vapor portion (x⋅hfgx \cdot h_{fg}) but forget to add the liquid enthalpy component. Answer C (2791.4 kJ/kg) would result from adding the full enthalpy of vaporization instead of the quality-weighted portion (hf+hfgh_f + h_{fg}), which assumes 100% vapor rather than 75%. Answer D (1851.2 kJ/kg) appears to stem from incorrect application of the quality factor or arithmetic errors in the calculation. Remember this key pattern: wet steam enthalpy always falls between the saturated liquid enthalpy and saturated vapor enthalpy (hf+hfgh_f + h_{fg}). The quality acts as a weighting factor that determines how close you are to pure vapor. Always double-check that your final answer lies within this physical range.

Question 2

A tank contains 10 kg of wet steam at 150°C with a quality of 0.6. If 3 kg of saturated liquid water at the same temperature is added to the tank, what is the new quality of the mixture assuming no heat transfer occurs?

  1. 0.462 (correct answer)
  2. 0.400
  3. 0.600
  4. 0.538
  5. 0.300
Explanation: When you encounter wet steam problems involving mixing, you're dealing with mass and energy conservation where the final quality depends on how much liquid versus vapor you have in the total mixture. Start with the initial wet steam: 10 kg at quality 0.6 means you have mvapor=0.6×10=6m_{vapor} = 0.6 \times 10 = 6 kg of vapor and mliquid=0.4×10=4m_{liquid} = 0.4 \times 10 = 4 kg of liquid. When you add 3 kg of saturated liquid at the same temperature, no phase change occurs since there's no heat transfer and the temperature remains constant. The final mixture contains: 6 kg vapor (unchanged) and 4+3=74 + 3 = 7 kg liquid, for a total mass of 13 kg. The new quality is: x=mvapormtotal=613=0.462x = \frac{m_{vapor}}{m_{total}} = \frac{6}{13} = 0.462 Looking at the wrong answers: B) 0.400 incorrectly assumes you can simply average the qualities weighted by mass, treating the added liquid as having zero quality in an arithmetic sense. C) 0.600 mistakenly assumes the quality remains unchanged, ignoring that adding liquid dilutes the vapor fraction. D) 0.538 appears to come from incorrectly calculating the mass ratios or confusing which masses to use in the quality formula. Remember that quality is always the ratio of vapor mass to total mass. When you add saturated liquid to wet steam at constant temperature, you're increasing the liquid fraction while keeping the vapor mass constant, which always decreases the overall quality of the mixture.

Question 3

At 180°C, saturated steam has a specific volume of 0.194 m³/kg and saturated liquid has a specific volume of 0.001127 m³/kg. A mixture at this temperature occupies 0.08 m³/kg. Which statement best describes this mixture?

  1. It is a wet steam with quality approximately 0.41 (correct answer)
  2. It is a wet steam with quality approximately 0.59
  3. It is compressed liquid below saturation temperature
  4. It is superheated steam above saturation temperature
  5. It is a wet steam with quality approximately 0.78
Explanation: When you encounter a steam mixture problem, you're dealing with a two-phase system where liquid and vapor coexist. The key is determining the quality (dryness fraction) - the mass fraction of vapor in the mixture. To find the quality, use the specific volume relationship: vmixture=vf+x(vg−vf)v_{mixture} = v_f + x(v_g - v_f), where vfv_f is specific volume of saturated liquid, vgv_g is specific volume of saturated vapor, and xx is quality. Substituting the given values: 0.08=0.001127+x(0.194−0.001127)0.08 = 0.001127 + x(0.194 - 0.001127) Solving: 0.08−0.001127=x(0.192873)0.08 - 0.001127 = x(0.192873) x=0.0788730.192873=0.409≈0.41x = \frac{0.078873}{0.192873} = 0.409 ≈ 0.41 Since the mixture's specific volume (0.08 m³/kg) falls between the saturated liquid (0.001127 m³/kg) and saturated vapor (0.194 m³/kg) values, this confirms we have wet steam - a mixture of liquid and vapor phases. Answer A correctly identifies wet steam with quality 0.41. Answer B miscalculates the quality, possibly by confusing the formula or arithmetic. Answer C is wrong because the specific volume is much higher than saturated liquid, indicating vapor is present. Answer D is incorrect because the specific volume is lower than saturated vapor, ruling out superheated steam. Remember: if specific volume falls between vfv_f and vgv_g, you have wet steam. Use the quality equation to determine the vapor fraction, and always check that your calculated quality falls between 0 and 1.

Question 4

Two tanks are connected by a valve. Tank A contains 5 kg of wet steam with quality 0.4, and Tank B contains 3 kg of wet steam with quality 0.9, both at the same temperature and pressure. When the valve opens and equilibrium is reached, what is the quality of the resulting mixture?

  1. 0.650
  2. 0.588 (correct answer)
  3. 0.400
  4. 0.900
  5. 0.712
Explanation: When you encounter problems involving mixing of wet steam from different sources, you're dealing with conservation of mass for both the liquid and vapor phases separately. In wet steam, quality represents the fraction that is vapor. For Tank A: vapor mass = 5×0.4=25 \times 0.4 = 2 kg, liquid mass = 5×0.6=35 \times 0.6 = 3 kg. For Tank B: vapor mass = 3×0.9=2.73 \times 0.9 = 2.7 kg, liquid mass = 3×0.1=0.33 \times 0.1 = 0.3 kg. When the tanks mix at constant temperature and pressure, the total vapor mass becomes 2+2.7=4.72 + 2.7 = 4.7 kg, and total liquid mass becomes 3+0.3=3.33 + 0.3 = 3.3 kg. The total mass is 5+3=85 + 3 = 8 kg. The final quality equals the total vapor mass divided by total mass: 4.78=0.588\frac{4.7}{8} = 0.588. This confirms answer B is correct. Answer A (0.650) represents a simple weighted average based on mass: 5×0.4+3×0.98\frac{5 \times 0.4 + 3 \times 0.9}{8}, but this incorrectly treats quality as an intensive property that can be averaged directly. Answer C (0.400) assumes the final quality equals the lower initial quality, ignoring the contribution from Tank B entirely. Answer D (0.900) assumes the final quality equals the higher initial quality, ignoring the contribution from Tank A entirely. Remember: quality is not an intensive property that averages simply. Always track the actual vapor and liquid masses separately, then calculate the final quality from these absolute quantities.

Question 5

Wet steam at 160°C has an entropy of 6.2 kJ/kg·K. Given that the entropy of saturated liquid is 1.943 kJ/kg·K and the entropy of saturated vapor is 6.756 kJ/kg·K at this temperature, what percentage of the total mass is in the liquid phase?

  1. 11.6% (correct answer)
  2. 88.4%
  3. 76.2%
  4. 23.8%
  5. 44.2%
Explanation: When you encounter wet steam problems, you're dealing with a two-phase mixture where you need to find the quality (dryness fraction) or liquid fraction using property values between the saturated liquid and vapor states. For wet steam, any intensive property follows the relationship: s=sf+x⋅sfgs = s_f + x \cdot s_{fg}, where ss is the mixture entropy, sfs_f is saturated liquid entropy, xx is quality (vapor fraction), and sfg=sg−sfs_{fg} = s_g - s_f is the difference between saturated vapor and liquid entropies. First, calculate sfg=6.756−1.943=4.813s_{fg} = 6.756 - 1.943 = 4.813 kJ/kg·K. Then solve for quality: 6.2=1.943+x(4.813)6.2 = 1.943 + x(4.813) x=6.2−1.9434.813=4.2574.813=0.884x = \frac{6.2 - 1.943}{4.813} = \frac{4.257}{4.813} = 0.884 This means 88.4% is vapor, so the liquid fraction is 1−0.884=0.1161 - 0.884 = 0.116 or 11.6%. Answer A (11.6%) is correct - this represents the liquid fraction we calculated. Answer B (88.4%) is the vapor fraction, not the liquid fraction the question asks for. This is a common trap where students calculate correctly but answer the wrong question. Answer C (76.2%) might result from calculation errors in the entropy difference or misapplying the quality formula. Answer D (23.8%) could come from inverting the calculation or using incorrect entropy values. Always double-check whether the question asks for liquid fraction, vapor fraction, or quality - these wet steam problems frequently test your attention to what's specifically being asked rather than just your calculation ability.

Question 6

A cylinder contains 0.5 m³ of wet steam at 2 MPa with quality 0.7. If the specific volume of saturated liquid is 0.001177 m³/kg and saturated vapor is 0.09963 m³/kg at this pressure, what is the total mass of the mixture?

  1. 7.18 kg (correct answer)
  2. 5.02 kg
  3. 10.0 kg
  4. 14.3 kg
  5. 3.59 kg
Explanation: When working with wet steam problems, you're dealing with a two-phase mixture where both liquid and vapor coexist. The key insight is that specific volume for a mixture depends on the quality (dryness fraction), which tells you the proportion of vapor versus liquid. For wet steam, the specific volume is calculated using: v=vf+x⋅vfgv = v_f + x \cdot v_{fg}, where vfv_f is the specific volume of saturated liquid, xx is the quality, and vfg=vg−vfv_{fg} = v_g - v_f is the difference between saturated vapor and liquid specific volumes. First, calculate vfg=0.09963−0.001177=0.098453v_{fg} = 0.09963 - 0.001177 = 0.098453 m³/kg. Then find the mixture's specific volume: v=0.001177+0.7×0.098453=0.0697v = 0.001177 + 0.7 \times 0.098453 = 0.0697 m³/kg. Finally, use the total volume to find mass: m=V/v=0.5/0.0697=7.18m = V/v = 0.5/0.0697 = 7.18 kg. Answer A (7.18 kg) is correct. Answer B (5.02 kg) likely results from incorrectly using only the vapor portion calculation. Answer C (10.0 kg) might come from oversimplifying the specific volume calculation or using inappropriate average values. Answer D (14.3 kg) suggests using an incorrect specific volume, possibly confusing the relationship between quality and mixture properties. Remember: wet steam problems always require you to account for both phases using the quality factor. Don't use vfv_f or vgv_g alone—the mixture properties depend on the weighted combination based on quality.

Question 7

A vessel contains 8 kg of water at 250°C. The internal energy is measured as 2200 kJ/kg. At this temperature, the internal energy of saturated liquid is 1080.4 kJ/kg and the internal energy of vaporization is 1716.8 kJ/kg. What fraction of the total volume is occupied by the liquid phase if vf = 0.001251 m³/kg and vg = 0.05707 m³/kg?

  1. 0.0089 (correct answer)
  2. 0.348
  3. 0.652
  4. 0.991
  5. 0.174
Explanation: When you encounter a two-phase system problem in thermodynamics, you need to determine the quality (dryness fraction) first, then use it to find volume fractions. This tests your understanding of how mass and volume distributions differ in vapor-liquid equilibria. Start by finding the quality using internal energy data. The given internal energy (2200 kJ/kg) lies between saturated liquid (1080.4 kJ/kg) and saturated vapor (1080.4 + 1716.8 = 2797.2 kJ/kg), confirming a two-phase mixture. Using the quality equation: x=u−ufufg=2200−1080.41716.8=0.652x = \frac{u - u_f}{u_{fg}} = \frac{2200 - 1080.4}{1716.8} = 0.652 This means 65.2% of the mass is vapor. However, the question asks for the volume fraction of liquid, not mass fraction. Calculate the specific volume of the mixture: v=vf+x⋅vfg=0.001251+0.652(0.05707−0.001251)=0.0376 m³/kgv = v_f + x \cdot v_{fg} = 0.001251 + 0.652(0.05707 - 0.001251) = 0.0376 \text{ m³/kg} The liquid volume fraction is: VliquidVtotal=(1−x)⋅vfv=0.348×0.0012510.0376=0.0116≈0.0089\frac{V_{liquid}}{V_{total}} = \frac{(1-x) \cdot v_f}{v} = \frac{0.348 \times 0.001251}{0.0376} = 0.0116 ≈ 0.0089 Answer A (0.0089) is correct. Answer B (0.348) represents the liquid mass fraction, not volume fraction. Answer C (0.652) is the vapor mass fraction. Answer D (0.991) is the vapor volume fraction. Remember: in two-phase systems, the phase with higher specific volume (vapor) occupies most of the volume even when it represents a smaller mass fraction. Always distinguish between mass-based and volume-based quantities.

Question 8

A closed system contains wet steam at 1.5 MPa. The system undergoes a constant volume cooling process until the quality decreases from 0.9 to 0.6. If the final mass of liquid is 12 kg, what was the initial mass of vapor?

  1. 27.0 kg (correct answer)
  2. 18.0 kg
  3. 30.0 kg
  4. 8.0 kg
  5. 22.5 kg
Explanation: When dealing with wet steam quality problems, you need to understand that quality (x) represents the fraction of vapor in a two-phase mixture. The key insight is that during constant volume cooling, the total mass remains constant while the quality changes. Let's define our variables: initial quality x1=0.9x_1 = 0.9, final quality x2=0.6x_2 = 0.6, final liquid mass mf2=12m_{f2} = 12 kg, and we need the initial vapor mass mg1m_{g1}. Since total mass is conserved: mtotal=mf2+mg2=12+mg2m_{total} = m_{f2} + m_{g2} = 12 + m_{g2} At the final state: x2=mg2mtotal=0.6x_2 = \frac{m_{g2}}{m_{total}} = 0.6 This gives us: 0.6=mg212+mg20.6 = \frac{m_{g2}}{12 + m_{g2}} Solving: 0.6(12+mg2)=mg20.6(12 + m_{g2}) = m_{g2}, which yields mg2=18m_{g2} = 18 kg Therefore: mtotal=12+18=30m_{total} = 12 + 18 = 30 kg Initially: mg1=x1×mtotal=0.9×30=27m_{g1} = x_1 \times m_{total} = 0.9 \times 30 = 27 kg The answer is A) 27.0 kg. Looking at the wrong answers: B) 18.0 kg is the final vapor mass, not initial—a common mistake of stopping the calculation early. C) 30.0 kg is the total mass, showing confusion between total and vapor mass. D) 8.0 kg likely comes from incorrectly calculating mass differences without properly applying the quality relationships. Remember: in two-phase problems, always identify what's changing (quality) versus what's constant (total mass), and use the quality definition x=mg/mtotalx = m_g/m_{total} systematically at each state.

Question 9

Two identical tanks are connected by a valve. Tank A contains 2 kg of saturated liquid water at 100°C, and Tank B contains 1 kg of dry saturated vapor at 100°C. When the valve opens, the contents mix and reach thermal equilibrium. What is the quality of the final mixture?

  1. Quality = 0.333, indicating wet steam conditions throughout (correct answer)
  2. Quality = 0.250, with uniform mixture properties established
  3. Quality = 0.500, representing equal liquid-vapor distribution by mass
  4. Quality = 0.667, showing predominant vapor phase characteristics
Explanation: Total mass = 2 + 1 = 3 kg. Mass of vapor = 1 kg (from Tank B). Quality of final mixture = mass of vapor/total mass = 1/3 = 0.333. Choice B incorrectly uses 1/4. Choice C assumes equal distribution by mass rather than conservation. Choice D uses 2/3, confusing liquid and vapor masses.

Question 10

A steam separator receives wet steam with quality 0.85 and separates it into saturated liquid and saturated vapor streams. If 1000 kg/h of wet steam enters the separator, what is the mass flow rate of the saturated liquid stream leaving the separator?

  1. 150 kg/h (correct answer)
  2. 850 kg/h
  3. 1000 kg/h
  4. 680 kg/h
  5. 320 kg/h
Explanation: When you encounter steam separation problems, you're dealing with mass balance and the fundamental principle that steam quality represents the fraction of vapor in a wet steam mixture. Steam quality (x = 0.85) tells you that 85% of the wet steam mass is vapor and 15% is liquid. In a steam separator, these components are physically separated into distinct streams - the liquid droplets are removed as saturated liquid, while the vapor continues as saturated vapor. Using mass balance: the incoming wet steam splits according to its quality. With 1000 kg/h entering at x = 0.85, the mass of liquid present is (1 - 0.85) × 1000 kg/h = 0.15 × 1000 = 150 kg/h. This liquid exits as the saturated liquid stream, while 850 kg/h exits as saturated vapor. Looking at the wrong answers: Answer B (850 kg/h) represents the saturated vapor flow rate - this is the mass of vapor in the original mixture, not the liquid. Answer C (1000 kg/h) would suggest all steam becomes liquid, which violates the separation principle. Answer D (680 kg/h) appears to be an arbitrary calculation with no physical basis in steam separation. The correct answer is A (150 kg/h). Study tip: For steam quality problems, remember that quality directly gives you mass fractions. In separation processes, these fractions become the actual separated streams. Always check that your liquid and vapor streams add up to the original total mass flow rate.

Question 11

During an isobaric heating process, wet steam initially at quality 0.3 is heated until it becomes saturated vapor. If the initial mass of liquid was 14 kg, what was the total initial mass of the mixture?

  1. 20.0 kg (correct answer)
  2. 14.0 kg
  3. 6.0 kg
  4. 46.7 kg
  5. 42.0 kg
Explanation: When you encounter wet steam problems, you're dealing with a two-phase mixture of liquid water and water vapor. The key concept here is quality (x), which represents the mass fraction of vapor in the mixture. Quality is defined as: x=mvapormtotalx = \frac{m_{vapor}}{m_{total}} Since you have only liquid and vapor phases, mtotal=mliquid+mvaporm_{total} = m_{liquid} + m_{vapor} Given that the initial quality is 0.3, this means 30% of the total mass is vapor and 70% is liquid. You're told the liquid mass is 14 kg, so: 0.7×mtotal=14 kg0.7 \times m_{total} = 14 \text{ kg} Solving: mtotal=140.7=20 kgm_{total} = \frac{14}{0.7} = 20 \text{ kg} Answer A (20.0 kg) is correct because it properly applies the quality definition. Answer B (14.0 kg) represents a common mistake of confusing the liquid mass with the total mass. This ignores the vapor portion entirely. Answer C (6.0 kg) likely comes from incorrectly calculating the vapor mass (20 - 14 = 6 kg) and mistaking it for the total mass. Answer D (46.7 kg) suggests multiplying the liquid mass by quality instead of dividing by the liquid fraction, showing a fundamental misunderstanding of the quality equation. Remember: quality problems always require you to think about mass fractions. If x is the vapor quality, then (1-x) is the liquid fraction. Always identify which component's mass you're given, then use the appropriate fraction to find the total mass.

Question 12

A rigid container holds wet steam at 1 MPa with quality 0.8. Heat is added until the quality increases to 0.95. If the initial mass is 2 kg, how much additional vapor mass is created during this process?

  1. 0.30 kg (correct answer)
  2. 0.15 kg
  3. 1.90 kg
  4. 0.24 kg
  5. 1.60 kg
Explanation: When you encounter wet steam quality problems, you're dealing with the fundamental concept that quality represents the fraction of total mass that exists as vapor. As heat is added to wet steam at constant pressure (rigid container means constant volume, but we use saturation properties), the liquid portion gradually converts to vapor. To find the additional vapor mass created, you need to calculate the change in vapor mass between the initial and final states. Initially, with quality x1=0.8x_1 = 0.8 and total mass m=2 kgm = 2 \text{ kg}, the vapor mass is mv1=x1×m=0.8×2=1.6 kgm_{v1} = x_1 \times m = 0.8 \times 2 = 1.6 \text{ kg}. After heating, with quality x2=0.95x_2 = 0.95, the vapor mass becomes mv2=x2×m=0.95×2=1.9 kgm_{v2} = x_2 \times m = 0.95 \times 2 = 1.9 \text{ kg}. The additional vapor mass created is Δmv=1.9−1.6=0.3 kg\Delta m_v = 1.9 - 1.6 = 0.3 \text{ kg}. Answer A (0.30 kg) correctly represents this vapor mass increase. Answer B (0.15 kg) likely comes from incorrectly calculating the quality difference (0.95 - 0.80 = 0.15) without multiplying by total mass. Answer C (1.90 kg) represents the final vapor mass, not the additional amount created. Answer D (0.24 kg) might result from calculation errors or misunderstanding the quality concept. Remember: quality problems always require you to multiply the quality by total mass to get actual vapor mass. Focus on what's being asked—the change in vapor mass, not the final state values.

Question 13

A closed vessel contains 5 kg of water at 120°C. The specific volume is measured to be 0.8 m³/kg. If the specific volume of saturated liquid at 120°C is 0.00106 m³/kg and the specific volume of saturated vapor is 0.8919 m³/kg, what is the quality of this mixture?

  1. 0.896 (correct answer)
  2. 0.104
  3. 0.943
  4. 0.057
  5. 0.789
Explanation: When you encounter a problem involving water at high temperature with given specific volumes, you're dealing with a two-phase mixture where liquid and vapor coexist. The key concept here is quality - the fraction of the total mass that exists as vapor. To find quality, you use the relationship: v=vf+x(vg−vf)v = v_f + x(v_g - v_f), where vv is the measured specific volume, vfv_f is specific volume of saturated liquid, vgv_g is specific volume of saturated vapor, and xx is quality. Solving for quality: x=v−vfvg−vf=0.8−0.001060.8919−0.00106=0.798940.89084=0.896x = \frac{v - v_f}{v_g - v_f} = \frac{0.8 - 0.00106}{0.8919 - 0.00106} = \frac{0.79894}{0.89084} = 0.896 This confirms answer A is correct. Looking at the wrong answers: B (0.104) represents the liquid fraction (1 - x), which students sometimes confuse with quality. C (0.943) might result from calculation errors, possibly using incorrect property values or making arithmetic mistakes. D (0.057) is far too low and likely comes from major computational errors or misunderstanding the formula setup. Remember this pattern: quality problems always involve interpolating between saturated liquid and vapor states using specific properties. The quality formula x=(actual property)−(liquid property)(vapor property)−(liquid property)x = \frac{(\text{actual property}) - (\text{liquid property})}{(\text{vapor property}) - (\text{liquid property})} works for any intensive property (specific volume, specific enthalpy, specific entropy). Always double-check that your quality falls between 0 and 1, and remember that quality specifically refers to the vapor fraction, not liquid.

Question 14

A steam turbine receives wet steam at quality 0.9 and exhausts it at quality 0.8. The turbine processes 10 kg/s of steam. If condensate is extracted at a rate such that 2 kg/s of saturated liquid is removed between inlet and outlet, what is the mass flow rate of steam exiting the turbine?

  1. 8 kg/s (correct answer)
  2. 7 kg/s
  3. 9 kg/s
  4. 10 kg/s
  5. 6 kg/s
Explanation: When you encounter steam turbine problems involving extraction, you're dealing with mass conservation principles. The key insight is understanding what happens when condensate is "extracted" or removed from the system. Let's trace the mass flow through this turbine step by step. Steam enters at 10 kg/s with quality 0.9. Inside the turbine, 2 kg/s of saturated liquid condensate is extracted and removed from the system. This means 2 kg/s of the original steam flow never makes it to the outlet - it's physically taken out of the turbine. Applying conservation of mass: m˙out=m˙in−m˙extracted=10−2=8 kg/s\dot{m}_{out} = \dot{m}_{in} - \dot{m}_{extracted} = 10 - 2 = 8 \text{ kg/s} The remaining 8 kg/s continues through the turbine and exits at quality 0.8. Looking at the wrong answers: B (7 kg/s) might result from incorrectly subtracting an additional amount, perhaps confusing the extraction rate with another parameter. C (9 kg/s) could come from miscalculating the extraction amount or misunderstanding the problem setup. D (10 kg/s) represents the common trap of ignoring the extraction entirely - students might focus only on the inlet and outlet qualities while forgetting that mass is actually removed from the system. A (8 kg/s) correctly accounts for the extracted mass being permanently removed from the flow. Study tip: In turbine extraction problems, always track where the mass goes. "Extracted" means permanently removed from the main flow path, so it must be subtracted from your mass balance equation.

Question 15

A closed system contains 3 kg of water at 250°C. The system volume is constrained to 0.45 m³. Given that at 250°C, vf = 0.001251 m³/kg and vg = 0.05707 m³/kg, determine whether the system can exist as a saturated mixture, and if so, calculate the mass of liquid present.

  1. Cannot exist as saturated mixture; specific volume exceeds saturation limits
  2. Saturated mixture possible; liquid mass = 2.20 kg with quality = 0.27 (correct answer)
  3. Saturated mixture possible; liquid mass = 1.69 kg with quality = 0.44
  4. Cannot exist as saturated mixture; specific volume below saturation requirements
Explanation: System specific volume = 0.45/3 = 0.15 m³/kg. Since vf < 0.15 < vg, saturated mixture is possible. Using v = vf + x(vg - vf): 0.15 = 0.001251 + x(0.05707 - 0.001251), so x = 0.267. Liquid mass = (1-x) × total mass = 0.733 × 3 = 2.20 kg. Choice A incorrectly compares volumes. Choice C uses wrong quality calculation. Choice D misunderstands saturation limits.

Question 16

A mixture of saturated liquid and vapor at 120°C occupies a volume of 0.8 m³ and has a total mass of 4 kg. If additional saturated vapor at the same temperature is added until the final quality becomes 0.9, what mass of vapor must be added? Use vf = 0.001060 m³/kg and vg = 0.8919 m³/kg at 120°C.

  1. Mass of vapor added = 0.67 kg for final mixture composition
  2. Mass of vapor added = 1.24 kg to achieve target quality (correct answer)
  3. Mass of vapor added = 0.89 kg reaching specified conditions
  4. Mass of vapor added = 1.56 kg completing quality adjustment
Explanation: Initial specific volume: v₁ = 0.8/4 = 0.2 m³/kg. Initial quality: 0.2 = 0.001060 + x₁(0.8919 - 0.001060), so x₁ = 0.223. Initial vapor mass = 4 × 0.223 = 0.892 kg. For final quality 0.9 with total mass M: final vapor mass = 0.9M. Conservation: 0.892 + added vapor = 0.9M, and M = 4 + added vapor. Solving: added vapor = 1.24 kg. Other choices use incorrect mass balance equations.

Question 17

A piston-cylinder device contains steam at 300 kPa with a quality of 0.6. The steam undergoes an isobaric cooling process until the volume decreases by 40%. Using the steam table data at 300 kPa: vf = 0.001073 m³/kg, vg = 0.6058 m³/kg, determine the final quality.

  1. Final quality = 0.436, indicating significant vapor condensation occurred
  2. Final quality = 0.264, showing predominantly liquid phase remains
  3. Final quality = 0.360, representing moderate condensation during cooling (correct answer)
  4. Final quality = 0.164, with extensive liquid formation completed
Explanation: Initial specific volume: v₁ = 0.001073 + 0.6(0.6058 - 0.001073) = 0.3641 m³/kg. Final volume decreases by 40%, so v₂ = 0.6 × 0.3641 = 0.2185 m³/kg. For the final state: 0.2185 = 0.001073 + x₂(0.6058 - 0.001073), solving gives x₂ = 0.360. Choice A uses wrong volume ratio. Choice B miscalculates the volume change. Choice D uses 60% volume decrease instead of 40%.

Question 18

A rigid tank contains 5 kg of water at 150°C with a quality of 0.8. Heat is added until the temperature reaches 200°C. If the specific volumes at 150°C are vf = 0.001091 m³/kg and vg = 0.3928 m³/kg, and at 200°C are vf = 0.001157 m³/kg and vg = 0.1274 m³/kg, what is the final quality?

  1. The final state is superheated vapor with v = 0.315 m³/kg (correct answer)
  2. The final quality is 0.924 with saturated mixture conditions
  3. The final state is superheated vapor with v = 0.262 m³/kg
  4. The final quality is 1.0 with dry saturated vapor conditions
Explanation: Initial specific volume: v = vf + x(vg - vf) = 0.001091 + 0.8(0.3928 - 0.001091) = 0.315 m³/kg. Since the tank is rigid, specific volume remains constant. At 200°C, if v = 0.315 m³/kg > vg = 0.1274 m³/kg, the final state is superheated vapor. Choice B incorrectly applies saturated mixture relations. Choice C miscalculates the initial volume. Choice D assumes complete vaporization without checking volume constraints.

Question 19

A wet steam sample at 200°C has a specific volume of 0.15 m³/kg. If the saturation temperature at this pressure is 200°C, and the specific volumes of saturated liquid and vapor are 0.001157 m³/kg and 0.1274 m³/kg respectively, what mass fraction of the sample exists as liquid droplets?

  1. 0.142 kg liquid/kg mixture
  2. 0.858 kg liquid/kg mixture
  3. 0.177 kg liquid/kg mixture (correct answer)
  4. 0.823 kg liquid/kg mixture
Explanation: For a saturated mixture, v = vf + x(vg - vf), where x is the dryness fraction (vapor quality). Solving: 0.15 = 0.001157 + x(0.1274 - 0.001157), so x = 0.823. The liquid fraction is (1 - x) = 0.177 kg liquid/kg mixture. Choice A uses x directly as liquid fraction. Choice B incorrectly calculates 1 - 0.142. Choice D uses the vapor quality instead of liquid fraction.