Thermodynamics Quiz: Nozzles And Diffusers

12 questions · exam conditions

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Nozzles And DiffusersQuestion 1 of 12

Steam enters a converging nozzle at 2 MPa and 300°C with negligible velocity and exits at 1 MPa. If the nozzle operates adiabatically and reversibly, what is the approximate exit velocity? (For steam: at 2 MPa, 300°C: $h_1 = 3024$ kJ/kg; at 1 MPa, $s = 6.767$ kJ/kg·K: $h_2 = 2778$ kJ/kg)

701 m/s

492 m/s

246 m/s

986 m/s

350 m/s

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Thermodynamics Quiz

Thermodynamics Quiz: Nozzles And Diffusers

Practice Nozzles And Diffusers 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 Nozzles And Diffusers, 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

701 m/s (correct answer)
492 m/s
246 m/s
986 m/s
350 m/s

Explanation: When you encounter a nozzle problem involving steam expansion, you're dealing with steady flow through a control volume where kinetic energy changes are significant. The key principle is applying the steady flow energy equation while recognizing the process constraints. For an adiabatic, reversible (isentropic) nozzle with negligible inlet velocity, the steady flow energy equation simplifies to:

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

Since the process is isentropic, the entropy remains constant at

s_2 = s_1 = 6.767

kJ/kg·K. The problem provides the exit enthalpy at 1 MPa and this entropy:

h_2 = 2778

kJ/kg. Solving for exit velocity:

V_2 = \sqrt{2(h_1 - h_2)} = \sqrt{2(3024 - 2778) \times 1000} = \sqrt{492,000} = 701

m/s Therefore, choice A) 701 m/s is correct. Choice B) 492 m/s likely comes from forgetting the factor of 2 in the kinetic energy term or incorrectly using

V_2 = \sqrt{h_1 - h_2}

. Choice C) 246 m/s represents the enthalpy difference itself (

h_1 - h_2 = 246

kJ/kg) rather than converting it to velocity. Choice D) 986 m/s might result from unit conversion errors or using incorrect property values. Remember: always verify your units when applying the steady flow energy equation. The factor of 1000 converts kJ/kg to J/kg, ensuring dimensional consistency for velocity in m/s.

Question 2

A converging-diverging nozzle operates with air entering at 500 kPa, 400 K, and low velocity. The throat area is 10 cm² and the exit area is 25 cm². If the nozzle is designed for a back pressure of 100 kPa, what is the mass flow rate? (For air: $R = 287$ J/kg·K, $\gamma = 1.4$ )

1.84 kg/s (correct answer)
2.31 kg/s
1.65 kg/s
2.08 kg/s
1.47 kg/s

Explanation: When you encounter a converging-diverging nozzle problem, you're dealing with compressible flow where the throat represents a critical choking condition. The key insight is that for choked flow (which occurs when back pressure is sufficiently low), the mass flow rate depends only on inlet conditions and throat area. For choked flow at the throat, the Mach number equals 1, and you can use the critical flow equation:

\dot{m} = \rho^* A^* a^*

where the asterisk denotes critical conditions. The critical density and speed of sound are:

\rho^* = \rho_0 \left(\frac{2}{\gamma + 1}\right)^{\frac{1}{\gamma - 1}}

a^* = \sqrt{\gamma R T^*} = \sqrt{\gamma R T_0 \left(\frac{2}{\gamma + 1}\right)}

First, calculate inlet density:

\rho_0 = \frac{P_0}{RT_0} = \frac{500,000}{287 \times 400} = 4.36

kg/m³ Then:

\rho^* = 4.36 \times \left(\frac{2}{2.4}\right)^{2.5} = 2.67

kg/m³ And:

a^* = \sqrt{1.4 \times 287 \times 400 \times \frac{2}{2.4}} = 346

m/s Therefore:

\dot{m} = 2.67 \times 0.001 \times 346 = 1.84

kg/s Answer A (1.84 kg/s) is correct. Answer B likely uses incorrect critical ratios, C might assume incompressible flow, and D probably miscalculates the critical temperature or uses wrong area units. Remember: For choked nozzle problems, the exit conditions don't affect mass flow rate—only inlet conditions and throat geometry matter. Always check if the nozzle is choked by comparing design back pressure to critical pressure.

Question 3

Steam flows through a nozzle where the inlet stagnation enthalpy is 3200 kJ/kg and the exit static enthalpy is 2850 kJ/kg. If the inlet velocity is 120 m/s, what is the exit velocity? Assume adiabatic flow.

849 m/s
837 m/s (correct answer)
825 m/s
861 m/s
813 m/s

Explanation: When you encounter nozzle flow problems, you're dealing with the steady flow energy equation, which relates enthalpy and kinetic energy changes. For adiabatic flow through a nozzle, the key principle is that stagnation enthalpy remains constant throughout the device. The stagnation enthalpy equation is:

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

At the inlet, you can find the stagnation enthalpy using the given static enthalpy. Wait - you're actually given the inlet stagnation enthalpy directly as 3200 kJ/kg. Since stagnation enthalpy is conserved in adiabatic flow, the exit stagnation enthalpy equals 3200 kJ/kg. At the exit:

h_{0,exit} = h_{exit} + \frac{V_{exit}^2}{2}

Substituting:

3200 = 2850 + \frac{V_{exit}^2}{2}

Solving:

350 = \frac{V_{exit}^2}{2}

(converting kJ/kg to J/kg: 350,000 J/kg)

V_{exit}^2 = 700,000

V_{exit} = 837 \text{ m/s}

This confirms answer B. Answer A (849 m/s) likely results from calculation errors or unit conversion mistakes. Answer C (825 m/s) might come from incorrectly including the inlet velocity in the energy balance. Answer D (861 m/s) could result from sign errors or misapplying the energy equation. Remember: in adiabatic nozzle problems, always conserve stagnation enthalpy. The inlet velocity information here is actually unnecessary since you're given the stagnation enthalpy directly - don't let extra data confuse your approach.

Question 4

Steam expands through a nozzle from 3 MPa, 400°C to 1.5 MPa. The inlet velocity is negligible and the expansion is adiabatic with 95% efficiency. If the nozzle inlet area is 8 cm² and the mass flow rate is 1.2 kg/s, what is the inlet steam density? (At inlet: $h_1 = 3232$ kJ/kg, $v_1 = 0.0994$ m³/kg)

10.06 kg/m³ (correct answer)
15.09 kg/m³
12.55 kg/m³
8.47 kg/m³
11.32 kg/m³

Explanation: When analyzing nozzle flow problems, you're dealing with fluid mechanics fundamentals where mass flow rate connects velocity, density, and cross-sectional area through the continuity equation:

\dot{m} = \rho A V

. However, this question has a key simplification: the inlet velocity is negligible. This means you can find the inlet density directly from the given specific volume without needing the continuity equation at all. Density is simply the inverse of specific volume:

\rho = \frac{1}{v}

. With the given inlet specific volume

v_1 = 0.0994

m³/kg, the inlet density is:

\rho_1 = \frac{1}{0.0994} = 10.06 \text{ kg/m³}

This confirms answer A is correct. The wrong answers likely come from calculation errors or misapplying the given information. B (15.09 kg/m³) might result from incorrectly using the mass flow rate and area data in some combination. C (12.55 kg/m³) could come from arithmetic mistakes in the reciprocal calculation. D (8.47 kg/m³) might result from using the wrong specific volume value or making errors with unit conversions. Notice that the nozzle efficiency, exit pressure, and inlet area are red herrings for this specific question - they're not needed to find inlet density when specific volume is given. Study tip: In thermodynamics problems, always identify what's actually being asked before diving into complex calculations. Sometimes the simplest relationship (like

\rho = 1/v

) is all you need, even when lots of other data is provided.

Question 5

A steam nozzle has an efficiency of 92% and expands steam from 2.2 MPa, 380°C to 1.0 MPa. The inlet velocity is 80 m/s and the mass flow rate is 3 kg/s. What is the actual exit enthalpy? (At inlet: $h_1 = 3178$ kJ/kg; at 1.0 MPa, same entropy: $h_{2s} = 2827$ kJ/kg)

2855 kJ/kg (correct answer)
2849 kJ/kg
2861 kJ/kg
2843 kJ/kg
2867 kJ/kg

Explanation: When analyzing nozzle problems, you're dealing with isentropic efficiency, which compares actual performance to ideal (isentropic) expansion. The key insight is that real nozzles have friction losses that cause the actual exit enthalpy to be higher than the ideal case. For nozzle efficiency, the formula is:

\eta = \frac{h_1 - h_{2a}}{h_1 - h_{2s}}

Where

h_{2a}

is the actual exit enthalpy and

h_{2s}

is the isentropic exit enthalpy. Rearranging to solve for the actual exit enthalpy:

h_{2a} = h_1 - \eta(h_1 - h_{2s})

Substituting the given values:

h_{2a} = 3178 - 0.92(3178 - 2827) = 3178 - 0.92(351) = 3178 - 323 = 2855 \text{ kJ/kg}

Answer A (2855 kJ/kg) is correct as it represents the actual exit enthalpy accounting for efficiency losses. Answer B (2849 kJ/kg) likely results from calculation errors or using slightly different efficiency values. Answer C (2861 kJ/kg) might come from incorrectly adding rather than subtracting the efficiency term, or computational mistakes. Answer D (2843 kJ/kg) could result from misapplying the efficiency formula or using incorrect intermediate calculations. Notice that the inlet velocity and mass flow rate are given but not needed for this calculation - they're red herrings. Remember that in nozzle problems, efficiency always makes the actual exit enthalpy higher than the isentropic value because real processes can't achieve perfect expansion. Focus on the efficiency formula and don't get distracted by unnecessary data.

Question 6

An air diffuser has an inlet area of 0.05 m² and an exit area of 0.25 m². Air enters at 350 m/s, 80 kPa, and 280 K. If the flow is adiabatic and reversible, what is the ratio of exit stagnation pressure to inlet stagnation pressure?

1.00 (correct answer)
1.15
0.87
1.08
0.93

Explanation: When analyzing flow through a diffuser under adiabatic and reversible conditions, you're dealing with isentropic flow where entropy remains constant throughout the process. The key insight is understanding what happens to stagnation properties during such flow. In isentropic flow, stagnation pressure remains constant along a streamline. This is because stagnation pressure represents the pressure that would be achieved if the fluid were brought to rest isentropically, and since the actual process is already isentropic, no losses occur. The stagnation pressure at any point equals the static pressure plus the dynamic pressure contribution. For this diffuser, even though the static pressure increases as the air decelerates (velocity drops from the small inlet to larger exit area), the stagnation pressure remains unchanged. Therefore, the ratio of exit stagnation pressure to inlet stagnation pressure equals 1.00. Answer A (1.00) is correct because stagnation pressure is conserved in isentropic flow, making the ratio exactly unity. Answer B (1.15) and D (1.08) both suggest stagnation pressure increases, which would violate the isentropic flow condition. These likely result from incorrectly adding the static pressure rise to the stagnation pressure. Answer C (0.87) implies stagnation pressure decreases, which would only occur with losses present. This contradicts the given reversible (lossless) condition. Remember: In isentropic flow problems, stagnation properties (pressure, temperature, enthalpy) remain constant even when static properties change dramatically. Always identify whether the flow is isentropic before analyzing stagnation property changes.

Question 7

Air flows through a diffuser where the inlet conditions are 100 m/s, 100 kPa, and 300 K. If the diffuser operates adiabatically with 85% efficiency and the exit velocity is 20 m/s, what is the exit temperature? (For air: $c_p = 1.005$ kJ/kg·K)

304.0 K
303.6 K (correct answer)
308.0 K
300.4 K
307.2 K

Explanation: When you encounter a diffuser problem, remember that diffusers convert kinetic energy to internal energy (increasing temperature) by slowing down the flow. The key is applying energy conservation with isentropic efficiency. For an adiabatic diffuser, the steady flow energy equation becomes:

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

First, find the ideal exit temperature assuming 100% efficiency. The kinetic energy change is:

\frac{V_1^2 - V_2^2}{2} = \frac{(100)^2 - (20)^2}{2} = \frac{9600}{2} = 4800 \text{ J/kg}

For an ideal process:

c_p(T_{2s} - T_1) = \frac{V_1^2 - V_2^2}{2}

T_{2s} = 300 + \frac{4800}{1005} = 304.78 \text{ K}

Now apply the 85% efficiency. Diffuser efficiency is:

\eta = \frac{T_{2s} - T_1}{T_2 - T_1}

Solving for actual exit temperature:

0.85 = \frac{304.78 - 300}{T_2 - 300}

T_2 = 300 + \frac{4.78}{0.85} = 303.6 \text{ K}

This confirms answer B is correct. Answer A (304.0 K) represents the ideal case ignoring efficiency losses. Answer C (308.0 K) likely comes from incorrectly applying efficiency as

\eta = \frac{T_2 - T_1}{T_{2s} - T_1}

(inverted formula). Answer D (300.4 K) suggests a calculation error, possibly using wrong units or incorrect kinetic energy conversion. Remember: diffuser efficiency is always less than 100%, so the actual temperature rise will be less than the ideal case.

Question 8

An ideal gas flows through a converging nozzle from a large reservoir at 800 kPa and 400 K. The nozzle exhausts to atmospheric pressure (100 kPa). What is the exit Mach number? ( $\gamma = 1.3$ )

1.68 (correct answer)
1.45
1.52
1.61
1.38

Explanation: When analyzing gas flow through nozzles, you need to apply isentropic flow relations to connect upstream reservoir conditions with downstream properties. For an ideal gas expanding from high pressure to lower pressure through a converging nozzle, the flow accelerates and can become supersonic if the pressure ratio is sufficient. Start with the isentropic pressure-Mach number relationship:

\frac{P_0}{P} = \left(1 + \frac{\gamma-1}{2}M^2\right)^{\frac{\gamma}{\gamma-1}}

Here,

P_0 = 800

kPa (reservoir pressure),

P = 100

kPa (exit pressure), and

\gamma = 1.3

. The pressure ratio is

\frac{800}{100} = 8

. Substituting:

8 = \left(1 + \frac{1.3-1}{2}M^2\right)^{\frac{1.3}{0.3}} = \left(1 + 0.15M^2\right)^{4.333}

Taking the reciprocal and raising to the power

\frac{0.3}{1.3}

\left(\frac{1}{8}\right)^{0.231} = \frac{1}{1 + 0.15M^2}

This gives

0.577 = \frac{1}{1 + 0.15M^2}

, so

1 + 0.15M^2 = 1.733

Solving:

M^2 = \frac{0.733}{0.15} = 4.887

, therefore

M = 1.68

Answer A (1.68) is correct. Answer B (1.45) likely results from calculation errors in the exponent evaluation. Answer C (1.52) suggests mistakes in handling the pressure ratio algebra. Answer D (1.61) indicates errors in the final square root calculation. Remember: isentropic nozzle problems always require careful attention to the pressure ratio and systematic algebraic manipulation of the exponential relationships. Double-check your exponent calculations—they're the most common source of error.

Question 9

A rocket nozzle expands combustion gases ( $\gamma = 1.3$ , $R = 287$ J/kg·K) from a chamber at $3$ MPa and $2800$ K to vacuum conditions. The throat area is $0.05$ m² and the exit area is $0.8$ m². What is the primary limitation on the nozzle's expansion performance?

The area ratio is insufficient to achieve complete expansion to vacuum, limiting the exit velocity to approximately $85\%$ of ideal
The high temperature causes significant viscous losses in the diverging section, reducing expansion efficiency by $15-20\%$
The nozzle design is adequate for vacuum expansion, with exit conditions approaching ideal performance within $5\%$ (correct answer)
The throat area is too small relative to chamber conditions, causing choking limitations that prevent optimal expansion

Explanation: For complete expansion to vacuum (Pe = 0), the required area ratio is infinite, but practical nozzles achieve near-vacuum conditions with finite area ratios. The given area ratio is Ae/At = 0.8/0.05 = 16. For γ = 1.3 and complete expansion, this area ratio corresponds to an exit pressure ratio of approximately Pe/Pc ≈ 0.002, which gives exit pressure ≈ 6 kPa - very close to vacuum. The exit Mach number would be approximately 4.5, achieving about 95% of theoretical vacuum velocity. Choice A incorrectly suggests significant limitation. Choice B overestimates viscous effects at this area ratio. Choice D misunderstands choking physics - the throat is properly sized for the given chamber conditions.

Question 10

A convergent nozzle receives saturated steam at $1.5$ MPa and accelerates it to $400$ m/s. The mass flow rate is $2$ kg/s and the process is adiabatic. During operation, the exit pressure is measured as $1.1$ MPa. What phenomenon is most likely occurring?

Normal convergent nozzle operation with subsonic acceleration and pressure drop according to Bernoulli principles
Choking at the nozzle exit with critical flow conditions established despite the measured exit pressure
Flashing of liquid droplets to vapor in the low-pressure regions, affecting the effective flow area
Non-equilibrium expansion with superheated vapor formation due to rapid acceleration of saturated steam (correct answer)

Explanation: When saturated steam expands rapidly through a nozzle, non-equilibrium effects occur where the steam becomes superheated due to insufficient time for phase equilibrium. The measured exit pressure (1.1 MPa) and high velocity (400 m/s) suggest the steam has expanded beyond its saturation state while maintaining higher pressure than equilibrium expansion would predict. This creates superheated conditions even though the steam started saturated. Choice A is incorrect because 400 m/s suggests near-sonic or sonic conditions. Choice B is wrong because the measured pressure indicates specific expansion conditions, not choking. Choice C incorrectly describes the phase change direction - saturated steam doesn't contain liquid droplets to flash.

Question 11

A converging-diverging nozzle operates with air entering at $500$ kPa, $400$ K, and negligible velocity. The nozzle is designed for an exit pressure of $100$ kPa. If the actual exit pressure is measured as $120$ kPa with supersonic flow at the exit, what operating condition exists?

The nozzle is overexpanded with oblique shock waves forming outside the exit to compress the flow
The nozzle is underexpanded with expansion waves forming outside the exit to reduce pressure further (correct answer)
The nozzle is operating at the design point with normal pressure recovery occurring downstream
The throat area is too large, preventing the establishment of proper critical flow conditions

Explanation: The critical pressure ratio for air is approximately 0.528, giving

P^* = 500 \times 0.528 = 264

kPa at the throat. Since the actual exit pressure (120 kPa) is higher than the design exit pressure (100 kPa) but lower than the critical pressure, the flow is choked at the throat but underexpanded at the exit. The flow exits at supersonic conditions but at higher pressure than designed, so expansion waves form outside the nozzle to further expand the flow to ambient conditions. Choice A is incorrect because overexpansion would result in exit pressure below design. Choice C is wrong because 120 kPa ≠ 100 kPa design condition. Choice D is incorrect because supersonic exit flow confirms proper choking.

Question 12

A diffuser design requires air deceleration from Mach $2.5$ to Mach $0.5$ . The inlet conditions are $50$ kPa and $250$ K. Two design options are considered: (1) a single normal shock followed by subsonic diffusion, or (2) a series of oblique shocks followed by subsonic diffusion. What is the most significant advantage of option (2)?

Higher stagnation pressure recovery due to lower entropy generation across multiple weaker shock waves (correct answer)
Reduced flow separation risk because the pressure rise is distributed more gradually along the diffuser length
Lower manufacturing cost since oblique shock systems require simpler geometric configurations than normal shock systems
Better temperature control because oblique shocks produce less heating per shock compared to single normal shocks

Explanation: For a normal shock at M₁ = 2.5, the pressure recovery is approximately 72% due to significant entropy generation. A system of oblique shocks can achieve the same overall pressure rise with much higher stagnation pressure recovery (typically 85-90%) because each individual shock is weaker, generating less entropy per shock. The total entropy generation is minimized when the pressure rise is distributed among multiple shocks. Choice B is incorrect because oblique shocks actually create more complex pressure gradients. Choice C is wrong because oblique shock systems are more complex and expensive to manufacture. Choice D misses the key point about pressure recovery, and stagnation temperature actually increases the same amount regardless of shock configuration.