Steady Flow Energy Equation MCQ Quiz - Objective Question with Answer for Steady Flow Energy Equation - Download Free PDF

Concept:

Apply the Steady Flow Energy Equation (SFEE) for a turbine:

\( h_1 + q = h_2 + w \Rightarrow q = (h_2 + w) - h_1 \)

Calculation:

Given:

Inlet enthalpy, \( h_1 = 3000~\text{kJ/kg} \)

Outlet enthalpy, \( h_2 = 2700~\text{kJ/kg} \)

Work output, \( w = 250~\text{kJ/kg} \)

Now,

\( q = (2700 + 250) - 3000 = 2950 - 3000 = -50~\text{kJ/kg} \)

Negative sign indicates heat loss from the turbine to the surroundings.

Explanation:-

• Month C: Temperature = 27°C, Wind speed = 0.71 m/s, Humidity = 75.50%
• Month D: Temperature = 26°C, Wind speed = 0.82 m/s, Humidity = 75.22%
• Month A: Temperature = 26°C, Wind speed = 0.87 m/s, Humidity = 83.12%
• Month B: Temperature = 29°C, Wind speed = 2.91 m/s, Humidity = 72.33%

Conclusion: Month B has the highest temperature (29°C) and the highest wind speed (2.91 m/s), even though it has slightly lower humidity. These factors suggest that the highest evaporation rate will occur in Month B.

Correct Answer: Option 4 - Month B

Explanation:

Definition of Coefficient of Discharge for a Steam Nozzle

Definition: The coefficient of discharge (C_d) for a steam nozzle is defined as the ratio of the actual discharge to the theoretical discharge. It is a dimensionless number that accounts for the effects of friction, turbulence, and other factors that cause the actual flow rate to be less than the ideal flow rate. Mathematically, it can be expressed as:

C_d = (Actual Discharge) / (Theoretical Discharge)

Importance: The coefficient of discharge is a crucial parameter in the design and analysis of steam nozzles. It provides a measure of the efficiency of the nozzle in converting thermal energy into kinetic energy. A higher coefficient indicates a more efficient nozzle with less energy loss.

Concept:

The given equation:

\(\frac{C_1^2 - C_2^2}{2} = h_1 - h_2 \)

is derived from the steady flow energy equation (Bernoulli’s equation for nozzles) for adiabatic flow in steam nozzles, where:

• h₁ and h₂ are the enthalpies at the inlet and outlet, respectively.
• C₁ and C₂ are the velocities at the inlet and outlet, respectively.

Analysis of Statements:

1. The flow is adiabatic in steam nozzles:

• Correct. This equation assumes adiabatic flow, meaning there is no heat transfer to or from the fluid within the nozzle.

2. The equation considers frictional losses in the steam nozzle:

• Incorrect. The given equation does not account for frictional losses. It assumes ideal, frictionless flow. If frictional losses were considered, additional terms would be included to account for the loss of energy due to friction.

3. The equation is the steady flow energy equation for steam nozzles:

• Correct. The equation is derived from the steady flow energy equation (or Bernoulli’s equation) applied to steam nozzles.

4. There is no mechanical work done by the steam nozzle:

• Correct. In the context of this equation, it is assumed that the nozzle does not do any mechanical work on or by the fluid; it only converts enthalpy to kinetic energy or vice versa.

Conclusion

The incorrect statement about the given equation for steam nozzles is:2) The equation considers frictional losses in the steam nozzle.

Concept:

Pump:

• A pump is a mechanical device used to force a fluid (a liquid or a gas) to move forward inside a pipeline or hose.
• They are also used to produce pressure by the creation of a suction (partial vacuum), which causes the fluid to rise to a higher altitude.

The efficiency of the pump, \(\eta =\frac{Useful ~power}{Power~rating}\) 

Calculation:

Given:

Power rating = 400 W, Mass of water lifted = 500 kg, H = 30 m, time = 10 minutes = 60 × 10 = 600 seconds, g = 10 m/s²

Force, F = m × g = 500 × 10 = 5000 N

Work = Force × displacement (H) = 5000 × 30 = 150 kJ

\(Power=\frac{Work}{Time}\)

\(Power=\frac{150}{600}=0.25 ~kW=250~W\)

⇒ Useful work = 250 W

\(\eta=\frac{Useful ~power}{Power~rating}\)

\(\eta=\frac{250}{400}=0.625\) 

⇒ η = 62.5%

Concept:

Diffuser: Ideal Diffuser is a mechanical device that is used to increase pressure at the expense of kinetic energy.

Kinetic energy is higher at the inlet than that at the outlet. Therefore, the velocity at the exit will be lower than that of the inlet.

Nozzles and diffusers are commonly used in jet engines, rockets, spacecraft, and even garden sprinklers.

From the steady flow energy equation,

\(m\left( {{h_1} + \frac{{{V_1}^2}}{2} + g{Z_1}} \right) + \frac{{dQ}}{{dt}} = \;m\left( {{h_2} + \frac{{V_2^2}}{2} + g{Z_2}} \right) + \frac{{dW}}{{dt}}\)

There is neither heat nor work transfer across the boundary of the system. i.e. δQ = 0 = δW

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

For the nozzle, V₂ ≫V₁

For the diffuser, V₂ ≪V₁

i.e. the velocity of the air at exit is lower than at the inlet of the diffuser.

\({h_1} + \frac{{{V_1}^2}}{2} = \;{h_2} \Rightarrow {h_2} - {h_1} = \frac{{{V_1}^2}}{2}\)

i.e. the specific enthalpy of the air increases from inlet to exit.

A steady flow process is a process where the fluid properties do not change with time. A steady-flow process is characterized by the following:

• No properties within the control volume change with time. That is m_cv = constant; E_cv = constant
• No properties change at the boundaries with time. Thus, the fluid properties at an inlet or exit will remain the same during the whole process.
• The heat and work interactions between a steady-flow system and its surroundings do not change with time.

Δ m_cv = 0, Δ m_cv = change in mass

Explanation:

Steady Flow Energy Equation (S.F.E.E.)

\(m\left( {{h_1} + \;\frac{{v_1^2}}{2} + g{z_1}} \right) + \dot{Q} = m\left( {{h_2} + \frac{{v_2^2}}{2} + g{z_2}} \right) + \dot{W}\)

Now,

Boiler:

• Ẇ  = 0, no work is done by the boiler,
• Change in kinetic and potential energy is neglected (ΔK.E = ΔP.E = 0)

∴ Q̇ = ṁ(h₂ - h₁)

Nozzle:

• Q̇ = 0, as the nozzle is perfectly insulated.
• Ẇ = 0, no work is done by the nozzle, v₁ <<< v₂ and potential energy change is neglected

∴  \({v_2} = \sqrt {2\left( {{h_1} - {h_2}} \right)}\)

Turbine:

• Q = 0, for the adiabatic or perfectly insulated turbine,
• Change in kinetic and potential energy is neglected

∴  Ẇ = ṁ(h₁ – h₂)

Compressor:

• Q̇ = 0, for the adiabatic or perfectly insulated turbine,
• Change in kinetic and potential energy is neglected (ΔK.E = ΔP.E = 0)

∴ Ẇ = ṁ(h₂ - h₁)

Points to remember:

Concept:

Air compressor:

An air compressor is a device in which work is done on the air to raise its pressure with an appreciable increase in its density.

i.e ρ₂ > ρ₁

Volume flow rate  \((\dot{Q})\)

It is defined as the quantity of fluid (m³) flowing per second through a section. It is given by –

\(\dot{Q}=A\times{V}\)

where A = area of cross-section and V = velocity at that section.

Conservation of Mass:

For a flowing fluid, the quantity of fluid (kg) per second is constant which is given by-

\(\dot{m}=ρ_{1}A_1{V_1}=ρ_{2}A_2{V_2}\)

\(\dot{m}=ρ_{1}Q_1=ρ_{2}Q_2\)

\(∴\frac{Q_2}{Q_1}=\frac{ρ_1}{ρ_2}\)

∵ ρ₂ > ρ₁, ∴ Q₂ < Q₁ i.e. volume flow rate at the outlet decreases in case of a compressor.

Explanation:

Steady Flow Energy Equation:

When there is mass transfer across the system boundary, the system is called an open system.

When there is an involvement of heat, work and the rate of flow of mass and energy across the control surface are constant, the equation used is Steady Flow Energy Equation.

In steady flow, the ratio of heat transfer, work transfer and mass flow at inlet and outlet is the same.

\({h_1} + \frac{{V_1^2}}{2} + {z_1}g + \frac{{dQ}}{{dm}} = {h_2} + \frac{{V_2^2}}{2} + {z_2}g + \frac{{dW}}{{dm}}\;\;\;\;(1)\)

Eq (1) is the Steady flow energy equation in mass form.

\({\dot m_1}({h_1} + \frac{{V_1^2}}{2} + {z_1}g) + \dot Q = {\dot m_2}\left( {{h_2} + \frac{{V_2^2}}{2} + {z_2}g} \right) + \dot W\;\;\;\;\;(2)\)

Eq (2) is the Steady flow energy equation in rate form.

This equation can be applied to a wide variety of processes like pipeline flows, heat transfer processes, mechanical power generation in engines and turbines, flow through nozzle and diffusers etc.

Eg. nozzle and diffuser having converging and diverging passage.

Important Points

​Euler's Equation:

In fluid flow, numerous forces acting on the fluid element. When the forces due to gravity F_g and pressure force F_p are considered then it is known as Euler's equation of motion.

\(\frac{{dp}}{ρ } + gdz + vdv = 0\;\;\;\;\;(1)\)

Euler's equation is derived considering the fluid element along a streamline.

Bernoulli's equation:

Bernoulli's equation is obtained by integrating the Euler's equation of motion.

\(\frac{p}{{ρ g}} + \frac{{{v^2}}}{{2g}} + z = C\)

Following assumptions are made in the derivation of Bernoulli's equation:

1. Flow is ideal i.e inviscous.
2. Flow is steady i.e. time variation is zero.
3. Flow is incompressible i.e. ρ is constant.
4. Flow is irrotaional i.e. ωx = ωy = ωz = 0.

Laplace Equation:

If ϕ(x, y, z) represents a function, then the Laplace equation is used to check the possibility of the existence of such function which is given by:

\(\frac{\partial ^2\phi}{\partial x^2}+\frac{\partial ^2\phi}{\partial y^2}+\frac{\partial ^2\phi}{\partial z^2}=0\)

Concept:

According to Steady Flow Energy Equation:

At 1 (steam enters the whistle) and 2 (steam leaves the whistle)

\({h_1} + \frac{{V_1^2}}{2} + g{z_1} + q = {h_2} + \frac{{V_2^2}}{2} + g{z_2} + w\)

Calculation:

As initial velocity is not given so V₁ = 0 and No change in elevation (z₁ = z₂)

Q = 0 (perfectly insulated)

W = 0 (No work transfer)

Concept:

Work done by the piston = pressure × displaced volume

W = P × V

P = Pressure of the gas

\(V = Displaced\;volume = \frac{\pi }{4} \times {d^2} \times L\)

d = Diameter of the piston,L = Stroke length

Calculation:

Given:

P = 2 × 10⁵ N/m²,

d = 0.1 m, L = 0.2 m

\(V = \frac{\pi }{4} \times {0.1^2} \times 0.2\)

\(V = \frac{{3.14159}}{4} \times {0.1^2} \times 0.2 = 0.001571\;{m^3}\)

W = P × V

Explanation:

The non-flow work done by the system (only on the quasi-static process) is given by:

\({W_{non - flow}} = \mathop \smallint \limits_1^2 pdV\)

Work done in a flow process (any process):

\({W_{flow - process}} = - \mathop \smallint \limits_1^2 VdP\)

Explanation:

Steady flow process is a process where: the fluid properties can change from point to point in the control volume but remains the same at any fixed point during the whole process. A steady-flow process is characterized by the following:

• No properties within the control volume change with time. That is m_cv = constant; E_cv = constant
• No properties change at the boundaries with time. Thus, the fluid properties at an inlet or exit will remain the same during the whole process.
• The heat and work interactions between a steady-flow system and its surroundings do not change with time.

Concept:

The energy changes in a flow can be analyzed by steady flow energy equation (SFEE)

It is given as

\({h_1} + \frac{{V_1^2}}{2} + {Z_1} \times g + Q = {h_2} + \frac{{V_2^2}}{2} + {Z_2} \times g + W\)

Where h is the enthalpy of the flowing stream and V²/2 represents the kinetic energy and Zg represents the potential energy, Q and W ae the heat and work interaction respectively

Calculation:

Given,

Change in enthalpy h₁ - h₂ = 450 KJ/Kg

Now applying steady flow energy equation by the assumption (W = 0, Q = 0, ΔP.E = 0)

For ideal condition of nozzle

The initial velocity is assumed zero as it is very less compared to the exit velocity

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

\({V_2} = \sqrt {2\left( {{{\rm{h}}_1} - {{\rm{h}}_2}} \right)} = \sqrt {2 \times 450 \times 1000} = 948.68\;m/sec\;\)