Thermodynamics Quiz: Mass Conservation Control Volumes
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Mass Conservation Control VolumesQuestion 1 of 11

A heat exchanger has hot fluid entering at m˙h,in=3.5 kg/s\dot{m}_{h,in} = 3.5 \text{ kg/s} and cold fluid entering at m˙c,in=2.8 kg/s\dot{m}_{c,in} = 2.8 \text{ kg/s}. Due to a small leak in the hot fluid side, the hot fluid exit mass flow rate is m˙h,out=3.3 kg/s\dot{m}_{h,out} = 3.3 \text{ kg/s}. Assuming steady state and that the leaked hot fluid does not mix with the cold fluid, what is the cold fluid exit mass flow rate?

m˙c,out=2.6 kg/s\dot{m}_{c,out} = 2.6 \text{ kg/s}
m˙c,out=3.0 kg/s\dot{m}_{c,out} = 3.0 \text{ kg/s}
m˙c,out=2.8 kg/s\dot{m}_{c,out} = 2.8 \text{ kg/s}
m˙c,out=3.3 kg/s\dot{m}_{c,out} = 3.3 \text{ kg/s}
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Thermodynamics Quiz

Thermodynamics Quiz: Mass Conservation Control Volumes

Practice Mass Conservation Control Volumes in Thermodynamics with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

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This quiz focuses on Mass Conservation Control Volumes, giving you a quick way to practice the rules, question types, and explanations that matter most for Thermodynamics.

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Question 1

A heat exchanger has hot fluid entering at m˙h,in=3.5 kg/s\dot{m}_{h,in} = 3.5 \text{ kg/s} and cold fluid entering at m˙c,in=2.8 kg/s\dot{m}_{c,in} = 2.8 \text{ kg/s}. Due to a small leak in the hot fluid side, the hot fluid exit mass flow rate is m˙h,out=3.3 kg/s\dot{m}_{h,out} = 3.3 \text{ kg/s}. Assuming steady state and that the leaked hot fluid does not mix with the cold fluid, what is the cold fluid exit mass flow rate?

  1. m˙c,out=2.6 kg/s\dot{m}_{c,out} = 2.6 \text{ kg/s}
  2. m˙c,out=3.0 kg/s\dot{m}_{c,out} = 3.0 \text{ kg/s}
  3. m˙c,out=2.8 kg/s\dot{m}_{c,out} = 2.8 \text{ kg/s} (correct answer)
  4. m˙c,out=3.3 kg/s\dot{m}_{c,out} = 3.3 \text{ kg/s}
Explanation: In this heat exchanger, the hot and cold fluids are separate and do not mix. Mass conservation must be applied to each fluid stream independently. For the cold fluid: m˙c,in=m˙c,out\dot{m}_{c,in} = \dot{m}_{c,out} since there are no leaks in the cold side, so m˙c,out=2.8 kg/s\dot{m}_{c,out} = 2.8 \text{ kg/s}. The leak in the hot side does not affect the cold fluid mass balance. Choice B incorrectly attempts to balance total mass. Choice C subtracts the leaked mass from cold flow. Choice D incorrectly sets cold exit equal to hot exit.

Question 2

A control volume has three inlets with mass flow rates of 2.5 kg/s, 1.8 kg/s, and 3.2 kg/s, and two outlets with mass flow rates of 4.1 kg/s and 2.9 kg/s. If the system operates under steady-state conditions, what can be concluded about the mass conservation analysis?

  1. The system violates mass conservation because the total inlet flow exceeds the total outlet flow by 0.5 kg/s (correct answer)
  2. The system satisfies mass conservation because the difference is within acceptable engineering tolerances
  3. The system violates mass conservation because the total outlet flow exceeds the total inlet flow by 0.5 kg/s
  4. The system satisfies mass conservation because steady-state conditions guarantee mass balance
  5. The system cannot be analyzed without knowing the density of the fluid at each location
Explanation: When analyzing control volumes in thermodynamics, mass conservation (continuity equation) is fundamental: under steady-state conditions, the total mass flow rate entering must exactly equal the total mass flow rate leaving. Any deviation indicates either measurement error or that the system isn't actually at steady state. Let's calculate the mass flow rates. Total inlet flow: 2.5+1.8+3.2=7.52.5 + 1.8 + 3.2 = 7.5 kg/s. Total outlet flow: 4.1+2.9=7.04.1 + 2.9 = 7.0 kg/s. The inlet flow exceeds outlet flow by 7.5−7.0=0.57.5 - 7.0 = 0.5 kg/s, which violates mass conservation. Answer A correctly identifies this violation. Answer B is wrong because there are no "acceptable engineering tolerances" for mass conservation—it's a fundamental physical law that must be satisfied exactly in steady state. The 0.5 kg/s difference represents a real imbalance that cannot be dismissed. Answer C incorrectly states that outlet flow exceeds inlet flow, when our calculation shows the opposite: inlet flow (7.5 kg/s) exceeds outlet flow (7.0 kg/s). Answer D contains a critical misconception. Steady-state conditions don't guarantee mass balance—rather, true steady state requires mass balance. If the flows don't balance, the system cannot be at steady state, regardless of what we're told. Study tip: Always verify the math first in mass conservation problems, then remember that steady state is a consequence of balanced flows, not a cause. If inlet and outlet flows don't match, question whether the system is truly at steady state or if there's measurement error.

Question 3

A control volume analysis is being performed on a system with time-varying inlet conditions. The inlet mass flow rate follows the pattern: ṁ_in(t) = 5 + 2sin(πt/10) kg/s, where t is in seconds. The outlet mass flow rate remains constant at 6.0 kg/s. What is the net change in mass within the control volume over a complete 20-second cycle?

  1. Zero, because the average inlet flow equals the constant outlet flow over the cycle
  2. -20 kg, because the outlet flow consistently exceeds the average inlet flow (correct answer)
  3. +10 kg, because the peak inlet flow provides additional mass accumulation
  4. -40 kg, reflecting the integrated effect of outlet flow exceeding inlet flow
  5. +5 kg, due to the phase relationship between inlet variation and outlet flow
Explanation: When analyzing control volumes with time-varying flows, you need to apply the conservation of mass principle by integrating flow rates over the entire time period, not just comparing average values. To find the net mass change, calculate the total mass entering minus the total mass leaving over the complete 20-second cycle. For the inlet flow m˙in(t)=5+2sin⁡(πt/10)\dot{m}_{in}(t) = 5 + 2\sin(\pi t/10), integrate over 20 seconds: ∫020(5+2sin⁡(πt/10))dt=5(20)+2∫020sin⁡(πt/10)dt\int_0^{20} (5 + 2\sin(\pi t/10)) dt = 5(20) + 2\int_0^{20} \sin(\pi t/10) dt The sine integral equals zero over a complete cycle (since ∫020sin⁡(πt/10)dt=0\int_0^{20} \sin(\pi t/10) dt = 0), giving total inlet mass = 100 kg. For the constant outlet flow: ∫0206.0dt=120\int_0^{20} 6.0 dt = 120 kg. Net change = 100 - 120 = -20 kg. Answer A incorrectly assumes that equal average flows guarantee zero net change, but this ignores the timing mismatch between inlet and outlet flows. Answer C mistakenly focuses on peak flow effects without proper integration—the oscillating component contributes zero net mass over a complete cycle. Answer D doubles the actual deficit, possibly from incorrectly calculating the integral or misunderstanding the time period. The correct answer is B: -20 kg results from the outlet flow consistently exceeding the average inlet flow throughout the cycle. Study tip: For time-varying control volume problems, always integrate the actual flow rates over the specified time period. Average flow rates alone can mislead you about the net accumulation or depletion.

Question 4

A control volume contains a compressible gas with an average density that changes from 1.2 kg/m³ to 1.8 kg/m³ over a 5-minute period. The control volume has a fixed volume of 50 m³, one inlet with constant mass flow rate of 0.8 kg/s, and one outlet with constant mass flow rate of 0.5 kg/s. Which statement correctly describes this process?

  1. The calculated mass accumulation matches the density change, confirming mass conservation is satisfied
  2. The density change indicates a violation of mass conservation due to measurement errors
  3. The process violates steady-state assumptions but satisfies overall mass conservation principles
  4. The calculated mass accumulation exceeds the observed density change, indicating mass loss (correct answer)
  5. The observed density change exceeds the calculated accumulation, suggesting mass generation
Explanation: When you encounter control volume problems with changing density, you need to apply the conservation of mass equation to check consistency between calculated mass accumulation and observed density changes. Let's calculate the expected mass accumulation. With a net mass flow rate of m˙in−m˙out=0.8−0.5=0.3 kg/s\dot{m}_{in} - \dot{m}_{out} = 0.8 - 0.5 = 0.3 \text{ kg/s} over 5 minutes (300 seconds), the calculated mass accumulation is 0.3×300=90 kg0.3 \times 300 = 90 \text{ kg}. Now let's find the actual mass change from density data. The initial mass is 1.2×50=60 kg1.2 \times 50 = 60 \text{ kg} and final mass is 1.8×50=90 kg1.8 \times 50 = 90 \text{ kg}, giving an observed mass increase of 30 kg30 \text{ kg}. Since the calculated accumulation (90 kg) exceeds the observed change (30 kg), this indicates 60 kg of mass is unaccounted for, suggesting mass loss from the system. Answer A is incorrect because the calculated and observed values don't match (90 kg vs 30 kg). Answer B wrongly attributes the discrepancy to measurement errors rather than recognizing a physical inconsistency. Answer C is incorrect because while the process isn't steady-state, mass conservation is actually violated since mass appears to be leaving the system through an unaccounted pathway. The correct answer is D because the calculated mass accumulation (90 kg) does exceed the observed density change (30 kg), indicating mass loss. Study tip: In control volume problems, always compare calculated mass flow accumulation with actual mass changes from density data. Discrepancies often reveal missing flow streams or system leaks.

Question 5

A storage tank with constant cross-sectional area receives flow from two inlet pipes and discharges through one outlet pipe. The first inlet has a constant flow rate of 200 L/min, the second inlet flow rate varies as ṁ₂(t) = 150 + 50cos(2πt/60) L/min where t is in minutes, and the outlet flow rate is constant at 320 L/min. What is the maximum rate of change of liquid level in the tank?

  1. +80 L/min occurring when the second inlet reaches its peak flow rate (correct answer)
  2. +30 L/min representing the average accumulation rate over time
  3. +50 L/min corresponding to the amplitude of the variable inlet stream
  4. -20 L/min indicating the maximum rate of level decrease
  5. Zero, since the average inlet equals the constant outlet over a complete cycle
Explanation: When analyzing fluid storage systems with multiple inlet and outlet streams, you need to apply mass conservation: the rate of liquid level change depends on the net flow rate (inflows minus outflows). The key insight is that the "maximum rate of change" occurs when the variable inlet reaches its peak value. Let's find the net flow rate. You have two constant flows: inlet 1 at 200 L/min and the outlet at 320 L/min. The second inlet varies as m˙2(t)=150+50cos⁡(2πt/60)\dot{m}_2(t) = 150 + 50\cos(2\pi t/60) L/min. The net flow rate is: Net flow=200+[150+50cos⁡(2πt/60)]−320=30+50cos⁡(2πt/60)\text{Net flow} = 200 + [150 + 50\cos(2\pi t/60)] - 320 = 30 + 50\cos(2\pi t/60) The maximum occurs when cos⁡(2πt/60)=1\cos(2\pi t/60) = 1, giving a maximum net flow of 30+50=8030 + 50 = 80 L/min. Since the tank has constant cross-sectional area, this directly equals the maximum rate of liquid level change. Answer A correctly identifies +80 L/min occurring at peak second inlet flow. Answer B (+30 L/min) represents the time-averaged net flow rate, not the maximum instantaneous rate. Answer C (+50 L/min) incorrectly uses only the amplitude of the cosine function, ignoring the baseline accumulation. Answer D (-20 L/min) has no basis in the given flow rates and would require the outlets to exceed total inlets. Study tip: For storage tank problems, always write out the complete mass balance equation including all time-dependent terms, then find when derivatives or oscillating functions reach their extreme values.

Question 6

A control volume has three streams: one inlet with mass flow rate 15.0 kg/s, and two outlets with mass flow rates ṁ₁ and ṁ₂. During steady-state operation, the ratio ṁ₁/ṁ₂ = 2.5. What is the mass flow rate of the second outlet stream?

  1. 6.0 kg/s based on the constraint ratio and mass conservation
  2. 4.3 kg/s derived from proportional flow distribution principles (correct answer)
  3. 5.0 kg/s considering equal distribution modified by the given ratio
  4. 3.8 kg/s calculated from the steady-state mass balance requirements
  5. 7.5 kg/s reflecting the primary outlet stream characteristics
Explanation: When you encounter a control volume problem with multiple streams, always start with the fundamental principle of mass conservation. For steady-state operation, the total mass flow rate entering must equal the total mass flow rate leaving the system. Given that one inlet has a mass flow rate of 15.0 kg/s and two outlets with rates ṁ₁ and ṁ₂, you can write: 15.0=m˙1+m˙215.0 = ṁ₁ + ṁ₂ Since ṁ₁/ṁ₂ = 2.5, you know that ṁ₁ = 2.5ṁ₂. Substituting this into the mass balance equation: 15.0=2.5m˙2+m˙2=3.5m˙215.0 = 2.5ṁ₂ + ṁ₂ = 3.5ṁ₂ Solving for ṁ₂: m˙2=15.0/3.5=4.29≈4.3 kg/sṁ₂ = 15.0/3.5 = 4.29 ≈ 4.3 \text{ kg/s} Option A (6.0 kg/s) incorrectly assumes ṁ₂ = 15.0/2.5, which ignores the fact that both outlets together must equal the inlet flow. Option C (5.0 kg/s) appears to use equal distribution (7.5 kg/s each) then incorrectly applies the ratio. Option D (3.8 kg/s) likely results from an algebraic error or incorrect interpretation of the ratio relationship. The correct answer is B, giving ṁ₂ = 4.3 kg/s. Study tip: For control volume problems, always write out the conservation equations first (mass, energy, momentum as needed), then substitute any given relationships. The most common mistake is misapplying ratios without considering the overall conservation constraint.

Question 7

A control volume representing a mixing chamber has three inlet streams with mass flow rates m˙1=2.5 kg/s\dot{m}_1 = 2.5 \text{ kg/s}, m˙2=1.8 kg/s\dot{m}_2 = 1.8 \text{ kg/s}, and m˙3=0.7 kg/s\dot{m}_3 = 0.7 \text{ kg/s}, and two outlet streams. If the first outlet stream has a mass flow rate of m˙4=3.2 kg/s\dot{m}_4 = 3.2 \text{ kg/s} and the control volume is at steady state with negligible accumulation, what is the mass flow rate of the second outlet stream?

  1. m˙5=1.8 kg/s\dot{m}_5 = 1.8 \text{ kg/s} (correct answer)
  2. m˙5=2.1 kg/s\dot{m}_5 = 2.1 \text{ kg/s}
  3. m˙5=1.5 kg/s\dot{m}_5 = 1.5 \text{ kg/s}
  4. m˙5=2.8 kg/s\dot{m}_5 = 2.8 \text{ kg/s}
Explanation: For steady-state control volume analysis, mass conservation requires ∑m˙in=∑m˙out\sum \dot{m}_{in} = \sum \dot{m}_{out}. Total inlet mass flow rate = 2.5+1.8+0.7=5.0 kg/s2.5 + 1.8 + 0.7 = 5.0 \text{ kg/s}. Total outlet mass flow rate must equal this: 3.2+m˙5=5.03.2 + \dot{m}_5 = 5.0, so m˙5=1.8 kg/s\dot{m}_5 = 1.8 \text{ kg/s}. Choice B results from incorrectly adding only two inlet streams (4.3−3.2=1.14.3 - 3.2 = 1.1 then miscomputing). Choice C results from subtracting the smallest inlet from the largest outlet. Choice D results from adding inlet 1 and 3 but forgetting inlet 2.

Question 8

A nozzle control volume has steam entering at section 1 and exiting at section 2. The inlet conditions are: A1=0.1 m2A_1 = 0.1 \text{ m}^2, V1=50 m/sV_1 = 50 \text{ m/s}, v1=0.5 m3/kgv_1 = 0.5 \text{ m}^3/\text{kg}. The outlet conditions are: A2=0.02 m2A_2 = 0.02 \text{ m}^2, v2=0.8 m3/kgv_2 = 0.8 \text{ m}^3/\text{kg}. If the process is steady, what is the ratio V2/V1V_2/V_1?

  1. V2/V1=10.0V_2/V_1 = 10.0
  2. V2/V1=6.25V_2/V_1 = 6.25
  3. V2/V1=8.0V_2/V_1 = 8.0 (correct answer)
  4. V2/V1=4.0V_2/V_1 = 4.0
Explanation: From steady-state mass conservation: A1V1v1=A2V2v2\frac{A_1 V_1}{v_1} = \frac{A_2 V_2}{v_2}. Rearranging: V2V1=A1v2A2v1=(0.1)(0.8)(0.02)(0.5)=0.080.01=8.0\frac{V_2}{V_1} = \frac{A_1 v_2}{A_2 v_1} = \frac{(0.1)(0.8)}{(0.02)(0.5)} = \frac{0.08}{0.01} = 8.0. Choice B results from inverting the specific volume ratio. Choice C results from using area ratio only (A1/A2=5A_1/A_2 = 5) with an error. Choice D results from using only the area ratio incorrectly.

Question 9

A mixing chamber control volume has two inlet streams of the same fluid at different temperatures. Stream 1: m˙1=4.2 kg/s\dot{m}_1 = 4.2 \text{ kg/s}, ρ1=950 kg/m3\rho_1 = 950 \text{ kg/m}^3. Stream 2: m˙2=2.8 kg/s\dot{m}_2 = 2.8 \text{ kg/s}, ρ2=850 kg/m3\rho_2 = 850 \text{ kg/m}^3. The mixed outlet stream has density ρout=920 kg/m3\rho_{out} = 920 \text{ kg/m}^3 and flows through a pipe with area Aout=0.012 m2A_{out} = 0.012 \text{ m}^2. What is the outlet velocity?

  1. Vout=0.53 m/sV_{out} = 0.53 \text{ m/s}
  2. Vout=0.63 m/sV_{out} = 0.63 \text{ m/s} (correct answer)
  3. Vout=0.48 m/sV_{out} = 0.48 \text{ m/s}
  4. Vout=0.76 m/sV_{out} = 0.76 \text{ m/s}
Explanation: First apply mass conservation: m˙out=m˙1+m˙2=4.2+2.8=7.0 kg/s\dot{m}_{out} = \dot{m}_1 + \dot{m}_2 = 4.2 + 2.8 = 7.0 \text{ kg/s}. Then use m˙out=ρoutAoutVout\dot{m}_{out} = \rho_{out} A_{out} V_{out} to find: Vout=m˙outρoutAout=7.0(920)(0.012)=7.011.04=0.634 m/s≈0.63 m/sV_{out} = \frac{\dot{m}_{out}}{\rho_{out} A_{out}} = \frac{7.0}{(920)(0.012)} = \frac{7.0}{11.04} = 0.634 \text{ m/s} \approx 0.63 \text{ m/s}. Choice A uses average of inlet densities instead of given outlet density. Choice C uses inlet density ρ1\rho_1. Choice D uses inlet density ρ2\rho_2.

Question 10

A branching junction has one inlet stream and three outlet streams. Water enters at m˙in=45 kg/s\dot{m}_{in} = 45 \text{ kg/s}. Two outlet streams have mass flow rates m˙out,1=18 kg/s\dot{m}_{out,1} = 18 \text{ kg/s} and m˙out,2=12 kg/s\dot{m}_{out,2} = 12 \text{ kg/s}. If the system operates at steady state but the third outlet valve is partially closed such that its effective flow area is reduced to 60% of its design value while maintaining the same velocity and density, what would be the mass flow rate m˙out,3\dot{m}_{out,3} compared to the required value for mass balance?

  1. m˙out,3=15 kg/s\dot{m}_{out,3} = 15 \text{ kg/s} (required) but actual flow is 9 kg/s9 \text{ kg/s}
  2. m˙out,3=15 kg/s\dot{m}_{out,3} = 15 \text{ kg/s} (required and actual)
  3. m˙out,3=25 kg/s\dot{m}_{out,3} = 25 \text{ kg/s} (required) but actual flow is 15 kg/s15 \text{ kg/s}
  4. The system cannot reach steady state with the valve restriction (correct answer)
Explanation: Mass conservation requires m˙out,3=45−18−12=15 kg/s\dot{m}_{out,3} = 45 - 18 - 12 = 15 \text{ kg/s}. However, with the valve 60% open, if velocity and density remain constant, the actual flow would be 0.6×15=9 kg/s0.6 \times 15 = 9 \text{ kg/s}. Since the required and actual flows don't match, steady state cannot be maintained - the system will accumulate mass and pressures will change until a new equilibrium is reached. Choices A and C recognize the mismatch but incorrectly assume steady state is possible. Choice B ignores the valve restriction effect.

Question 11

A storage tank is being filled through two inlet pipes and drained through one outlet pipe. The tank volume is V=50 m3V = 50 \text{ m}^3 and the fluid density is constant at ρ=800 kg/m3\rho = 800 \text{ kg/m}^3. If inlet mass flow rates are m˙in,1=12 kg/s\dot{m}_{in,1} = 12 \text{ kg/s} and m˙in,2=8 kg/s\dot{m}_{in,2} = 8 \text{ kg/s}, and the outlet mass flow rate is m˙out=15 kg/s\dot{m}_{out} = 15 \text{ kg/s}, what is the rate of change of mass within the tank?

  1. dmCVdt=5 kg/s\frac{dm_{CV}}{dt} = 5 \text{ kg/s} (mass increasing) (correct answer)
  2. dmCVdt=−5 kg/s\frac{dm_{CV}}{dt} = -5 \text{ kg/s} (mass decreasing)
  3. dmCVdt=35 kg/s\frac{dm_{CV}}{dt} = 35 \text{ kg/s} (mass increasing)
  4. dmCVdt=0 kg/s\frac{dm_{CV}}{dt} = 0 \text{ kg/s} (steady state)
Explanation: The general mass conservation equation is dmCVdt=∑m˙in−∑m˙out\frac{dm_{CV}}{dt} = \sum \dot{m}_{in} - \sum \dot{m}_{out}. Here: dmCVdt=(12+8)−15=20−15=5 kg/s\frac{dm_{CV}}{dt} = (12 + 8) - 15 = 20 - 15 = 5 \text{ kg/s}. The positive value indicates mass is accumulating in the tank. Choice B has the wrong sign (common error in applying conservation equation). Choice C incorrectly adds all flow rates without considering direction. Choice D assumes steady state incorrectly.