Option 4 : 949 m/s

**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 V^{2}/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 h_{1 }- h_{2 }= 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

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

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

Diffuser increases the pressure of a fluid at the expense of its:

Option 1 : kinetic energy

**Explanation:**

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

Kinetic energy is higher at the inlet than that of 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.

Option 3 : 58 kg/s

__Concept:__

Mass flow rate (ṁ) = ρ × A × V

Where ρ = density

Assuming air to be ideal gas; which follows the equation

P v = m R T

\(P = \frac{m}{v}RT\)

\(⇒ \rho = \frac{P}{{RT}}\)

__Calculation:__

Given, A = 0.5 m^{2}, V = 100 m/s

Here we have to find out density (ρ)

Here P = 100 kPa

R = 0.287 KJ/kg K

T = 27°C = 300 K

\(⇒ \rho = \frac{{100}}{{0.287 \times 300}} = 1.16\frac{{kg}}{{{m^3}}}\)

ṁ = 1.16 × 0.5 × 100 = 58 kg/s

Option 1 : It is a duct which decreases the velocity and increases pressure.

__Explanation:__

**Diffuser:**

- A diffuser is defined as "a device for
**reducing the velocity**of a fluid passing through a system while**increasing the static pressure**". - The fluid's static pressure rise as it passes through a duct is commonly referred to as
**pressure recovery.**

This is because the energy in the flow is conserved.

Bernoulli's principle describes this property of a steady, incompressible, and inviscid flow, which states that in this flow,

\({\frac{1}{2}\rho V^2~+~p~+~\rho gz~=~constant}\)

The contraction or enlargement of the section in a duct leads to a respective increase or decrease of velocity to conserve the mass flow.

The section of a diffuser grows along the stream, causing the flow to slow down.

Assuming the height is the same, the previous equation will result in (index 1 for upstream and index 2 for downstream):

\(\frac{1}{2}\rho V_1^2~+~p_1~=~\frac{1}{2}\rho V_2^2~+~p_2\) (V1 > V2)

This implies pressure has to increase to conserve energy, i. e. p2 has to be higher than p1.

__Additional Information__

**Nozzle**:

- A nozzle is used to increase the discharge velocity while simultaneously lowering the pressure of a fluid passing through it.

Nozzles used in steam turbines convert ______

Option 3 : enthalpy of steam into kinetic energy

__Explanation: __

Steam nozzles:

A steam nozzle is a passage of varying cross-sections, which converts the heat energy of steam into kinetic energy. The main use of a steam nozzle is in a steam turbine, which produces a jet of steam with a high velocity.

- The smallest section of the nozzle is called the throat.
- The steam enters the nozzle with high pressure and negligible velocity. But leaves the nozzle with a high velocity and small pressure.
- The pressure, at which the steam leaves the nozzle, is known as back-pressure.
- No heat is supplied or rejected by the steam during flow through a nozzle. Therefore it is considered as an adiabatic flow, and the corresponding expansion is considered as an adiabatic expansion.

The following are the three type of nozzles:

(1) Convergent nozzle:

When the cross-section of a nozzle decreases continuously from the entrance to exit, it is called a convergent nozzle.

(2) Divergent nozzle:

When the cross-section of a nozzle increases continuously from the entrance to exit, it is called a divergent nozzle.

(3) Convergent-divergent nozzle:

When the cross-section of a nozzle first decreases from its entrance to the throat and then increases from its throat to exit, it is called a convergent-divergent nozzle. This type of nozzle is widely used these days in various types of steam turbines.

__Additional Information__

Process |
Properties |

Adiabatic process |
δQ = 0 |

Polytropic process |
δQ ≠ 0 |

Throttling |
h = constant |

Free expansion |
δQ = 0, δW = 0, du = 0 |

where, δQ = Change in heat, δW = Change in work, du = Change in internal energy, h = enthalpy.

Option 2 : isentropic flow

__Explanation:__

An adiabatic flow is a condition where the overall heat transfer interaction across the boundary of the system and surrounding is zero.

- If the system boundary is perfectly insulated with the surrounding then the heat interaction will be zero.
- A nozzle is a steady-state steady flow device to create a high-velocity fluid stream at the expense of its pressure.
- The nozzle is perfectly insulated so the flow is adiabatic and also there is no moving part so shaft work is also zero.

From the steady flow energy equation,

\({h_1} + \frac{{C_1^2}}{2} + {z_1}g + \frac{{\partial Q}}{{dm}} = {h_2} + \frac{{C_2^2}}{2} + {z_2}g + \frac{{\partial W}}{{dm}}\)

As the change in potential energy is also zero hence SFEE is given by,

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

Option 2 : 0.93 to 0.98

__Explanation:__

Coefficient of discharge is the ratio of actual discharge to the theoretical discharge.

Coefficient of discharge for various devices are:

⇒ Venturimeter – 0.95 to 0.98

⇒ Orifice meter – 0.62 to 0.65

⇒ Nozzle meter – 0.93 to 0.98

∴ Coefficient of discharge for nozzle meter lies between venturi meter and orifice meter.

Although the maximum range of coefficient of discharge for nozzle meter and venturi meter seems to be same, the average value of the coefficient of discharge for nozzle meter is less than venturi mete

Steam flows through a nozzle at a mass flow rate of \(\dot m = 0.1\) kg/s with a heat loss of 5 kW. The enthalpies at inlet and exit are 2500 kJ/kg and 2350 kJ/kg, respectively. Assuming negligible velocity at inlet (C_{1} ≈ 0), the velocity (C_{2}) of steam (in m/s) at the nozzle exit is __________ (correct to two decimal places).

**Concept:**

Steady Flow Energy Equation for steady flow devices,

\({h_1} + \frac{{C_1^2}}{2} + {z_1}g + \frac{{\partial Q}}{{dm}} = {h_2} + \frac{{C_2^2}}{2} + {z_2}g + \frac{{\partial W}}{{dm}}\)

Where, h_{1} is the enthalpy,

\(\frac{C^2}{2}\) is the kinetic energy

Zg is the potential energy

Q and W are the heat and work transfer

**Calculation:**

For nozzle, Z_{1} = Z_{2}, dw/dm = 0, C_{1} = 0

\({h_1}+ \frac{{\partial Q}}{{dm}} = {h_2} + \frac{{C_2^2}}{2} \),given Q̇ = - 5 kW.

\(2500 - \frac{5}{{0.1}}= 2350 + \frac{{C_2^2}}{{2000}}\ { \Rightarrow \frac{{C_2^2}}{{2000}} = 100}\)

C_{2} = 447.21 m/s

The temperature and pressure of air in a large reservoir are 400 K and 3 bar respectively. A converging–diverging nozzle of exit area 0.005 m^{2} is fitted to the wall of the reservoir as shown in the figure. The static pressure of air at the exit section for isentropic flow through the nozzle is 50 kPa. The characteristic gas constant and the ratio of specific heats of air are 0.287 kJ/kgK and 1.4 respectively.

Option 4 : 2.06

**Concept:**

The mass flow rate through the nozzle is, m = (ρAV)_{i} = (ρAV)_{o}

The velocity of flow through nozzle is find from steady flow energy equation,

\(h_i+\frac{V_i^2}{2}+Q=h_o+\frac{V_o^2}{2}+W\)

where, V_{i} = 0 , W = 0 ,Q = 0 ...adiabatic

Therefore, \(h_i-h_o=\frac{V_o^2}{2}\Rightarrow C_p(T_i-T_o)=\frac{V_o^2}{2}\)

**Calculation:**

**Given:**

T_{1} = 400 K, P_{1} = 300 kPa, P_{2} = 50 kPa, R = 0.289 kJ/kgK, γ = 1.4, A_{2} = 0.005 m^{2}, ρ = 0.727 kg/m^{3}

The process followed from entrance to exit is isentropic process, therefore

\(\frac{{{T_2}}}{{{T_1}}} = {\left( {\frac{{{P_2}}}{{{P_1}}}} \right)^{\frac{{γ - 1}}{γ }}} \Rightarrow {T_2} = 400{\left( {\frac{{50}}{{300}}} \right)^{\frac{{0.4}}{{1.4}}}} = 239.5~K\)

Therefore, for exit velocity,

1.005 (400 - 239.5) × 2000 = V_{o}^{2}

V_{o} = 567.98 m/s

The mass flow rate, m = 0.727 × 0.005 × 567.98 = **2.06 kg/sec **

Option 2 : produce high velocity jet of steam

__Explanation:__

**Steam Nozzle:**

- A
**steam nozzle**is a passage of**varying cross-sections**that converts the heat energy of steam into kinetic energy.**The main application of the steam nozzle is in a steam turbine which produces a jet of steam with high velocity.** - The increase of velocity of the steam jet at the exit of the nozzle is obtained due to a decrease in enthalpy (total heat content) of the steam.
- The nozzle is so shaped that it will perform this conversion of energy with minimum loss.

**Types of Steam Nozzle:**

**Convergent Nozzle:**

In a **convergent nozzle**, the nozzle cross-sectional area **decreases continuously from its entrance to exit**. It is used in a case **where back pressure (exit pressure) is equal to or more than the critical pressure ratio.**

**Divergent Nozzle:**

The cross-sectional area of the **divergent nozzle increases continuously from its entrance to exit.** It is used in a case **where the back pressure is less than the critical pressure ratio.**

**Convergent-Divergent Nozzle:**

In this condition, the cross-sectional area first decreases from its entrance to the throat and then again increases from throat to the exit. This case is used in the case where the back pressure is less than the critical pressure. It is widely used in many types of steam turbines in the modern-day turbine.

__Additional Information__

The relation among area, velocity, and Mach number for the nozzle is given by,

\(\frac{{dA}}{A} = \frac{{dV}}{V}\left( {{M^2} - 1} \right)\)

- When M < 1, i.e. the inlet velocity is subsonic, as flow area decreases, the pressure decreases, and velocity increases. So for subsonic flow, a convergent portion becomes a nozzle and a divergent portion becomes a diffuser.
- When M > 1, i.e. when the inlet velocity is supersonic, as flow area decreases, pressure increases, and velocity decreases, and as the flow area increases, pressure decreases and velocity increases. So for supersonic flow, a convergent passage is a diffuser and a divergent portion is a nozzle.
- If A = Constant, So M = 1, Occurs only at the throat and nowhere else and this happens only when the discharge is maximum.

Option 3 : Will remain constant till a point and then gradually decrease

**Concept:**

Given:

Back pressure, P_{b }= 0.5 MPa, Adiabatic index, \(\gamma = 1.4\)

For expansion process, critical pressure ratio

\(\frac{{{{\rm{P}}_{\rm{c}}}}}{{{{\rm{P}}_{\rm{b}}}}} = {\left[ {\frac{2}{{{\rm{\gamma }}\;+\;1}}} \right]^{\frac{{\rm{\gamma }}}{{\left( {{\rm{\gamma }}\;-\;1} \right)}}}}\)

\(\frac{{{{\rm{P}}_{\rm{c}}}}}{{0.5}} = {\left[ {\frac{2}{{1.4\; + \;1}}} \right]^{\frac{{1.4}}{{\left( {1.4\; -\; 1} \right)}}}}\)

∴ P_{c} = 0.264 MPa

Critical pressure, P_{c} = 0.264 MPa

The temperature and pressure of air in a large reservoir are 400 K and 3 bar respectively. A converging–diverging nozzle of exit area 0.005 m^{2} is fitted to the wall of the reservoir as shown in the figure. The static pressure of air at the exit section for isentropic flow through the nozzle is 50 kPa. The characteristic gas constant and the ratio of specific heats of air are 0.287 kJ/kgK and 1.4 respectively.

Option 3 : 0.727

__ Concept__:

The relation between pressure and temperature for the isentropic process.

\(\frac{{{T_2}}}{{{T_2}}} = {\left( {\frac{{{P_2}}}{{{P_1}}}} \right)^{\frac{{γ - 1}}{γ }}}\)

**Calculation:**

**Given:**

T_{1} = 400 K, P_{1} = 300 kPa, P_{2} = 50 kPa, R = 0.289 kJ/kgK

γ = 1.4, A_{2} = 0.005 m^{2}

The process followed from entrance to exit is isentropic process, therefore

\(\frac{{{T_2}}}{{{T_1}}} = {\left( {\frac{{{P_2}}}{{{P_1}}}} \right)^{\frac{{γ - 1}}{γ }}} \Rightarrow {T_2} = 400{\left( {\frac{{50}}{{300}}} \right)^{\frac{{0.4}}{{1.4}}}} = 239.5~K\)

From the perfect gas equation

\(\rho = \frac{P}{{RT}}~or~{\rho _2} = \frac{{{P_2}}}{{R{T_2}}} = \frac{{50}}{{0.287 \times239.5}} \Rightarrow {\rho _2} = 0.727~\frac{{kg}}{{{m^3}}}\)

**Statement (I) :** Mass flow through the convergent nozzle is maximum when the exit Mach number is 1.

**Statement (II) :** The divergent section is added to convergent nozzle to increase the exit Mach number and not to increase the mass flow rate

Option 2 : Both Statement (I) and Statement (II) are individually true but Statement (II) is NOT the correct explanation of Statement (I)

__Concept:__

A nozzle is a passage of smoothly varying cross-section by means of which the pressure energy of working fluid is converted into kinetic energy.

- A nozzle whose flow area decreases in the flow direction is called a Convergent nozzle.
- It is designed such that a drop in pressure from inlet to outlet accelerates the flow.
- The location of the smallest flow area of a nozzle is called the throat.
- The ratio of the pressure at the section where sonic velocity is attained to the inlet pressure of a nozzle is called the critical pressure ratio.
- It happens at Mach number equals 1.

Mach number < 1 → subsonic velocity

Mach number = 1 → sonic velocity

Mach number > 1 → supersonic velocity

- If a convergent nozzle is operating under choked condition, the exit Mach number is unity.
- To determine whether a nozzle is choked or not, we calculate the actual pressure ratio and then compare this with the critical pressure ratio.
- If the actual pressure ratio > critical pressure ratio, the nozzle is said to be choked.

__Convergent-divergent nozzle:__

- Convergent - Divergent nozzles are used to
**increase the flow of gas to supersonic speeds**(as in the case of rockets). - Their cross-sectional area first decreases and then increases. The area where the diameter is minimum is called the throat.
- As the gas enters the converging section, its velocity increases, considering the mass flow rate to be constant.
- As the gas passes through the throat, it attains sonic velocity (Mach number = 1).
- As the gas passes through the divergent nozzle, the velocity increases to supersonic (Mach number >1)
- The flow rate is maximum for a given nozzle if the flow is sonic at the throat. This condition is achieved by managing the back pressure.
- For the compressible fluid flow, the Mach number is an important dimensionless parameter. On the basis of the Mach number, the flow is defined.
- The addition of divergent to convergent does not alter the mass flow rate as the flow is choked.

**Statement (I) :** In an isentropic nozzle flow, discharge reaches a maximum value when the throat pressure reaches the critical value.

Option 2 : Both Statement (I) and Statement (II) are individually true but Statement (II) is NOT the correct explanation of Statement (I)

__Concept:__

A nozzle is a device that increases fluid velocity while causing pressure drop of the fluid i.e. dP > 0,

dP < 0.

The equation for a nozzle is given by

\(\frac{{dA}}{A} = \frac{{dP}}{{\rho {{\vec V}^2}}}\left( {1 - {M^2}} \right)\)

Where A is the cross-sectional area of the nozzle, P is the fluid pressure, is the fluid velocity, M is the Mach number.

Various conditions are described as below:

- Subsonic: M < 1, dP(1 – M
^{2}) < 0, dA < 0 - Sonic: M = 1, dP(1 – M
^{2}) = 0, dA = 0 - Supersonic: M > 1, dP(1 – M
^{2}) < 0, dA > 0

As depicted from the above picture, to accelerate the subsonic flow, the nozzle flow area must first decrease in the flow direction. The **flow area reaches a minimum at the point where the Mach number is unity**. To continue to accelerate the flow to supersonic conditions, the flow area must increase.

The minimum flow area is called the **throat of the nozzle**.

Now if we plot mass flow rate vs. static to stagnation pressure ratio, we get a plot like this

The plot shows that there is a value of P/P_{o} that makes the **mass flow rate a maximum**.

The pressure ratio that makes the **mass flow rate a maximum** is the same pressure ratio at which the Mach number is unity at the flow cross-sectional area. This value of pressure ratio is called **critical pressure ratio** for nozzle flow or we can say mass flow rate and hence discharge reaches a maximum value when the throat pressure reaches the critical value.

So, from the above discussion, it is clear that both the statements are individually true but statement II is not the correct explanation of statement I.