THERMODYNAMICS
Chapter 4: First Law of Thermodynamics – Open Systems
Steady-Flow Energy Equation, Turbines, Compressors & Nozzles
10 Questions | Easy · Medium · Hard | Full Worked Solutions
Questions at a Glance
Q# Topic Level Key
Concept
1 Steady-Flow Energy Equation Medium Steady-Flow
2 Turbine Analysis Medium Turbine
3 Compressor Analysis Medium Compressor
4 Nozzle & Diffuser Medium Nozzle
5 Heat Exchanger Medium Heat
6 Throttling Valve Easy Throttling
7 Mixing Chamber Medium Mixing
8 Pump Work Medium Pump
9 Unsteady Filling Process Hard Unsteady
10 SFEE — Multiple Inlets/Outlets Hard SFEE
Q1. Steady-Flow Energy Equation
[Medium]
Write the steady-flow energy equation (SFEE) and identify each term. Under what conditions can
kinetic and potential energy terms be neglected?
✔ SOLUTION
STEADY-FLOW ENERGY EQUATION (per unit mass):
q – w_s = (h₂ – h₁ ) + (V₂ ² – V₁ ²)/2 + g(z₂ – z₁ )
TERMS:
q = specific heat transfer (kJ/kg) — positive INTO the system
w_s = specific shaft work (kJ/kg) — positive OUT of the system
h = specific enthalpy (kJ/kg) = u + Pv (flow work + internal energy)
(V²/2) = specific kinetic energy (kJ/kg)
gz = specific potential energy (kJ/kg)
WHEN TO NEGLECT KE: When ΔV² is small compared to Δh.
Typical: compressors, turbines, heat exchangers.
KE is IMPORTANT in: nozzles, diffusers (high-velocity flows).
WHEN TO NEGLECT PE: When elevation changes are small.
Neglected for: most devices. Important for: pumps, hydro turbines.
Thermodynamics — Practice Q&A | Page 1 of 6
, NOTE: For steady-flow, mass flow rate (m_dot) is constant: m_dot_in =
m_dot_out
KEY ANSWER: SFEE: q – w_s = Δh + ΔKE + ΔPE | Neglect KE/PE for turbines,
compressors, heat exchangers
Q2. Turbine Analysis
[Medium]
Steam enters a turbine at h₁ = 3200 kJ/kg, V₁ = 50 m/s and exits at h₂ = 2500 kJ/kg, V₂ = 180
m/s. The turbine loses 25 kJ/kg of heat to surroundings. Find the specific work output.
✔ SOLUTION
Apply SFEE (neglect potential energy change, Δz ≈ 0):
q – w_s = (h₂ – h₁ ) + (V₂ ² – V₁ ²)/2000
q = –25 kJ/kg (heat LOST, so negative for the system)
Kinetic energy terms:
KE₁ = V₁ ²/2000 = 50²/2000 = 1.25 kJ/kg
KE₂ = V₂ ²/2000 = 180²/2000 = 16.2 kJ/kg
ΔKE = 16.2 – 1.25 = 14.95 kJ/kg
SFEE:
–25 – w_s = (2500 – 3200) + 14.95
–25 – w_s = –700 + 14.95 = –685.05
–w_s = –685.05 + 25 = –660.05
w_s = 660.1 kJ/kg
The turbine produces 660.1 kJ/kg of shaft work output.
KEY ANSWER: w_s = 660.1 kJ/kg (shaft work output)
Q3. Compressor Analysis
[Medium]
Air enters a compressor at 100 kPa, 300 K (h₁ = 300.2 kJ/kg) and exits at 700 kPa, 560 K (h₂ =
567.1 kJ/kg). The compressor is adiabatic. Mass flow rate is 2 kg/s. Find power input.
✔ SOLUTION
For adiabatic compressor: q = 0
Neglect KE and PE changes (small compared to Δh).
SFEE per unit mass:
0 – w_s = h₂ – h₁
–w_s = 567.1 – 300.2 = 266.9 kJ/kg
w_s = –266.9 kJ/kg (negative → work input to system)
Power input (W_dot):
W_dot_in = m_dot × |w_s| = 2 × 266.9 = 533.8 kW
The compressor requires 533.8 kW (≈ 0.534 MW) of power.
Thermodynamics — Practice Q&A | Page 2 of 6
Chapter 4: First Law of Thermodynamics – Open Systems
Steady-Flow Energy Equation, Turbines, Compressors & Nozzles
10 Questions | Easy · Medium · Hard | Full Worked Solutions
Questions at a Glance
Q# Topic Level Key
Concept
1 Steady-Flow Energy Equation Medium Steady-Flow
2 Turbine Analysis Medium Turbine
3 Compressor Analysis Medium Compressor
4 Nozzle & Diffuser Medium Nozzle
5 Heat Exchanger Medium Heat
6 Throttling Valve Easy Throttling
7 Mixing Chamber Medium Mixing
8 Pump Work Medium Pump
9 Unsteady Filling Process Hard Unsteady
10 SFEE — Multiple Inlets/Outlets Hard SFEE
Q1. Steady-Flow Energy Equation
[Medium]
Write the steady-flow energy equation (SFEE) and identify each term. Under what conditions can
kinetic and potential energy terms be neglected?
✔ SOLUTION
STEADY-FLOW ENERGY EQUATION (per unit mass):
q – w_s = (h₂ – h₁ ) + (V₂ ² – V₁ ²)/2 + g(z₂ – z₁ )
TERMS:
q = specific heat transfer (kJ/kg) — positive INTO the system
w_s = specific shaft work (kJ/kg) — positive OUT of the system
h = specific enthalpy (kJ/kg) = u + Pv (flow work + internal energy)
(V²/2) = specific kinetic energy (kJ/kg)
gz = specific potential energy (kJ/kg)
WHEN TO NEGLECT KE: When ΔV² is small compared to Δh.
Typical: compressors, turbines, heat exchangers.
KE is IMPORTANT in: nozzles, diffusers (high-velocity flows).
WHEN TO NEGLECT PE: When elevation changes are small.
Neglected for: most devices. Important for: pumps, hydro turbines.
Thermodynamics — Practice Q&A | Page 1 of 6
, NOTE: For steady-flow, mass flow rate (m_dot) is constant: m_dot_in =
m_dot_out
KEY ANSWER: SFEE: q – w_s = Δh + ΔKE + ΔPE | Neglect KE/PE for turbines,
compressors, heat exchangers
Q2. Turbine Analysis
[Medium]
Steam enters a turbine at h₁ = 3200 kJ/kg, V₁ = 50 m/s and exits at h₂ = 2500 kJ/kg, V₂ = 180
m/s. The turbine loses 25 kJ/kg of heat to surroundings. Find the specific work output.
✔ SOLUTION
Apply SFEE (neglect potential energy change, Δz ≈ 0):
q – w_s = (h₂ – h₁ ) + (V₂ ² – V₁ ²)/2000
q = –25 kJ/kg (heat LOST, so negative for the system)
Kinetic energy terms:
KE₁ = V₁ ²/2000 = 50²/2000 = 1.25 kJ/kg
KE₂ = V₂ ²/2000 = 180²/2000 = 16.2 kJ/kg
ΔKE = 16.2 – 1.25 = 14.95 kJ/kg
SFEE:
–25 – w_s = (2500 – 3200) + 14.95
–25 – w_s = –700 + 14.95 = –685.05
–w_s = –685.05 + 25 = –660.05
w_s = 660.1 kJ/kg
The turbine produces 660.1 kJ/kg of shaft work output.
KEY ANSWER: w_s = 660.1 kJ/kg (shaft work output)
Q3. Compressor Analysis
[Medium]
Air enters a compressor at 100 kPa, 300 K (h₁ = 300.2 kJ/kg) and exits at 700 kPa, 560 K (h₂ =
567.1 kJ/kg). The compressor is adiabatic. Mass flow rate is 2 kg/s. Find power input.
✔ SOLUTION
For adiabatic compressor: q = 0
Neglect KE and PE changes (small compared to Δh).
SFEE per unit mass:
0 – w_s = h₂ – h₁
–w_s = 567.1 – 300.2 = 266.9 kJ/kg
w_s = –266.9 kJ/kg (negative → work input to system)
Power input (W_dot):
W_dot_in = m_dot × |w_s| = 2 × 266.9 = 533.8 kW
The compressor requires 533.8 kW (≈ 0.534 MW) of power.
Thermodynamics — Practice Q&A | Page 2 of 6
