# Chapter 6 Flow Through Steam Nozzles – Thermal Engineering

## Flow Through Steam Nozzles

##### 6.1 ❐ INTRODUCTION

The steam nozzle is a passage of varying cross-section by means of which a part of the enthalpy of steam is converted into kinetic energy as the steam expands from a higher pressure to a lower pressure. Therefore, a nozzle is a device designed to increase the velocity of steam. Steam nozzles are of three types, namely convergent nozzle, divergent nozzle, and convergent–divergent nozzle. If the cross-section of the nozzle decreases continuously from the entrance to exit, then it is called a convergent nozzle, as shown in Fig. 6.1(a). If the cross-section increases, then it is called a divergent nozzle, as shown in Fig. 6.1(b). If the cross-section of the nozzle first decreases and then increases, it is called a convergent–divergent nozzle, as shown in Fig. 6.1(c). The least area of cross-section of the nozzle is called the throat. The divergent section has to be long as the divergent angle is limited to about 7° in order to prevent separation at the wall.

The main purpose of steam nozzles is to produce a high velocity jet of steam which is used in steam turbine injectors for pumping feed water into boilers and to maintain high vacuum in power plant condensers or steam jet refrigeration condensers.

Figure 6.1 Types of nozzles: (a) Convergent nozzle (b) Divergent nozzle (c) Convergent-Divergent nozzle

##### 6.2 ❐ CONTINUITY EQUATION

Consider the flow of steam through a nozzle.

Let = steady state mass flow rate of steam, kg/s

A = cross-sectional area of nozzle normal to the direction of steam flow at any section, m2

v = specific volume of steam at the same section, m3/kg

c = velocity of steam across the section, m/s

Then for steady flow of steam through the nozzle,

Equation (6.1) is the continuity equation for steam flow through the nozzle.

##### 6.3 ❐ VELOCITY OF FLOW OF STEAM THROUGH NOZZLES

1. The steam flows through the nozzle without any work or heat transfer.
2. The frictional forces are neglected.

Consider the flow of steam through a nozzle, as shown in Fig. 6.2. Applying the energy equation to sections 1 and 2 at entrance and exit of nozzle, we have

Usually c1 = 0,

where (Δh)isen = h1h2 = isentropic enthalpy change, kJ/kg

Figure 6.2 Flow through a nozzle

#### 6.3.1 Flow of Steam Through the Nozzle

Let pvn = constant

where n = 1.035 + 0.1 x1 for wet steam

= 1.135 for saturated steam

= 1.3 for superheated steam

As the steam pressure drops while passing through the nozzle, its enthalpy is reduced. This reduction of enthalpy of steam must be equal to the increase in kinetic energy. Hence,

where p1, p2 = pressure of steam at entry and exit, respectively, v1, v2 = specific volume at entry and exit, respectively.

If c1 << c2, then

##### 6.4 ❐ MASS FLOW RATE OF STEAM

The mass flow of steam per second is flowing with velocity c2 through a cross-sectional area A2, and specific volume v2 is:

Now, v2 = v1

Also, from Eq. (6.3), we have

##### 6.5 ❐ CRITICAL PRESSURE RATIO

From Eq. (6.4), we have

The mass flow rate per unit area is maximum at the throat because it has minimum area of cross-section. Therefore, will be maximum when is maximum

Equation (6.5) gives the critical pressure ratio at which the discharge through the nozzle is maximum.

Substituting for from Eq. (6.5), we get

where cs2 = velocity of sound at exit of the nozzle.

where cs1 = velocity of sound at entry of nozzle.

##### 6.6 ❐ MAXIMUM DISCHARGE

From Eq. (6.5), maximum mass flow rate of steam can be obtained as

As p2 is gradually reduced, the discharge gradually increases and becomes maximum as critical pressure is approached, as shown in Fig. 6.3.

Figure 6.3 Discharge v’s pressure ratio in a nozzle

##### 6.7 ❐ EFFECT OF FRICTION ON EXPANSION OF STEAM

The exit velocity of steam for a given pressure drop is reduced due to the following reasons:

1. Friction between nozzle surface and steam
2. Internal fluid friction in the steam
3. Shock losses

Most of these losses occur beyond the throat in the divergent section of the nozzle as the length and the velocity of steam is much higher there. The effects of these friction losses are as follows:

1. The expansion does not remain isentropic and the enthalpy drop is reduced, resulting in lower exit velocity.
2. The final dryness fraction of steam is increased as a part of kinetic energy gets converted into heat due to friction and is absorbed by steam, which increases the enthalpy.
3. The specific volume of steam is increased due to frictional reheating.

The effect of friction of steam flow through a nozzle is shown in hs (Mollier) diagram in Fig. 6.4. The point A represents the initial condition of steam and the point E represents the throat of a convergent–divergent nozzle. AB represents the isentropic expansion without friction, and AC represents the expansion with friction. Isentropic enthalpy drop is (hAhC) and actual enthalpy drop with friction is (hAhc). If the actual enthalpy drop as percentage of theoretical enthalpy drop is known, then point C can be located. However, expansion must end at same pressure as at B. The horizontal line drawn through C to cut the back pressure line pb at point D represents the final condition of steam. The hs diagram shows that the dryness fraction of steam at point D is greater than that at point B and the specific volume of steam also increases.

Most of the friction occurs in the divergent part of the nozzle and actual expansion is represented by the line AED. AE represents the expansion in the convergent part, whereas ED represents in the divergent part of the nozzle. Lines AEB and AED represent expansion of steam for initially saturated steam. Likewise, A1E1B1 and A1E1D1 are for initially superheated steam.

Figure 6.4 Effect of friction on expansion of steam in a convergent–divergent nozzle

##### 6.8 ❐ NOZZLE EFFICIENCY

Nozzle efficiency, ηn is a factor that takes into account the effect of friction during expansion of steam in the nozzle. It is defined as:

The exit velocity of steam considering friction is:

where K =

##### 6.9 ❐ SUPERSATURATED OR METASTABLE FLOW THROUGH A NOZZLE

The isentropic expansion of superheated steam from supply pressure p1 to back pressure pb can be represented on the Mollier diagram by line AE, as shown in Fig. 6.5. During expansion, change of phase must start at point B where the pressure line p2 meets the saturation line. However, in nozzles, under certain conditions, this phenomenon of condensation does not occur at point B as the time available is very short due to very high velocity of steam (nearly sonic) through the nozzle. The equilibrium between the vapour phase and liquid is, therefore, delayed and the vapour continues to expand in dry state even beyond point B upon point B1. The pressure at B1 can be found by extending the superheated constant pressure line p3 up to B1. The steam during the expansion BB1 remains dry and condensation is suppressed.

The vapours between pressure p2 and p3 are said to be supersaturated or supercooled and such a flow in nozzles is called supersaturated or metastable flow of steam. A limit to the supersaturated state was observed by Wilson and the line drawn on the Mollier chart through the observed points is known as Wilson line. For all practical purposes, this line has become the saturation line.

The flow is also called supercooled flow because at any pressure between p2 and p3, the temperature of the vapour is always lower than the saturation temperature corresponding to that pressure. The difference in this temperature is called the degree of undercooling.

When the expansion reaches at point B, on the Wilson line, the condensation occurs at constant enthalpy, and the pressure remaining constant, as shown by horizontal line BC. Further isentropic expansion to the exit pressure is represented by CD. The ratio of saturation pressures corresponding to the temperatures at B and B1 is called the degree of supersaturation.

Figure 6.5 Super-saturated flow of steam in a nozzle

Velocity of steam at the end of expansion,

Specific volume, v2 = v1

Temperature, T2 = T1

and A2 =

The superheated expansion law, pv1.3 = constant, is followed in the supersaturated flow.

#### Example 6.1

Steam expands from 2.5 bar to 1 bar in a nozzle. The initial velocity of steam is 80 m/s and initial temperature is 200°C. Taking nozzle efficiency as 96%, find the exit velocity.

Solution

Given that p1 = 2.5 bar, p2 = 1 bar, c1 = 80 m/s, t1 = 200°C, ηn = 0.96

The hs diagram is shown in Fig. 6.6.

From steam tables for superheated steam, we have

h1 = 2868 kJ/kg at p1 = 2.5 bar and 200°C

s1 = 7.3593 kJ/kg.K

At p2 = 1 bar, sf 2 = 1.3025 kJ/kg.K and sfg2 = 6.0568 kJ/kg.K

Since s1 > sfg2, steam is superheated. At 1 bar and s1 = 7.3593 kJ/kg.K, the enthalpy for superheated steam by interpolation is:

Figure 6.6 Steam flow through a nozzle

Enthalpy drop = h1h2 = 2868 – 2691.95 = 176.05 kJ/kg

Actual enthalpy drop, (Δh)actual = ηn (h1h2) = 0.96 × 176.05 = 169.0 kJ/kg

#### Example 6.2

Superheated steam enters a convergent–divergent nozzle at 20 bar and 300°C. The exit pressure is 4.5 bar. Assuming frictionless flow up to the throat (pv1.3 = const.) and a nozzle efficiency of 90%, determine (a) the flow rate for a throat area of 30 cm2 and (b) exit area.

Solution

Given that p1 = 20 bar, t1 = 300°C, p3 = 4.5 bar, ηn = 0.9, A2 = 30 cm2

Critical pressure ratio, = 0.5457

Throat pressure, p2 = 20 × 0.5457 = 10.91 bar

From Mollier diagram (Fig. 6.7), we get

h1 = 3025 kJ/kg, h2 = 2880 kJ/kg, v2 = 0.2 m3/kg, h3 = 2700 kJ/kg, vs3′ = 0.4 m3/kg, and x3′ = 0.995

Figure 6.7 Superheated steam flow through a nozzle

v3 = x3′ vs3′ = 0.995 × 0.4 = 0.398 m3/kg

#### Example 6.3

Dry saturated steam at 3 bar is expanded through a convergent nozzle to 1.5 bar. The exit area is 2.5 cm2. Calculate the exit velocity and mass flow rate, assuming (a) isentropic expansion, (b) supersaturated flow, and (c) degree of under-cooling at exit.

Solution

1. At p1 = 3 bar

s1 = 6.992 kJ/kg.K, h1 = 2724.3 kJ/kg, v1 = 0.6058 m3/kg

At p2 = 1.5 bar

s1 = s2 = sf 2 + x2 sfg2

6.9918 = 1.4335 + x2 × 5.7897

x2 = 0.96

h2 = hf2 + x2hfg2 = 467.08 + 0.96 × 2226.5 = 2604.52 kJ/kg

v2 = vf2 + x2 (vg2 vf2) = 0.001053 + 0.96 × (1.159 − 0.001053) = 1.113 m3/kg

2. For supersaturated flow, pv1.3 = const.
3. Saturation temperature at 1.5 bar = 111.37°C

Degree of under cooling at exit = 111.37 − 73.45 = 37.92°C.

#### Example 6.4

Steam at a pressure of 10 bar, 0.96 dry is expanded through a convergent–divergent nozzle and leaves the nozzle at 0.3 bar.

1. Calculate the velocity of steam at throat for maximum discharge. Take n = 1.134.
2. Calculate the exit area and steam discharge if the throat area is 1.5 cm2. Assume isentropic flow and ignore friction losses.

Solution

The nozzle is shown in Fig. 6.8.

1. v1 = vf1 + x1 (vg1vf1) = 0.001127 + 0.96 × (0.19444 − 0.001127) = 0.1867 m3/kg

For maximum discharge,

Velocity of steam at throat, c2 =

Figure 6.8 Convergent–divergent nozzle

2. For isentropic flow,

s1 = sf1 + x1sfg1 = 2.1386 + 0.96 × 4.4478 = 6.4085 kJ/kg

s1 = s2

sf1 + x1sfg1 = sf2 + x2sfg2

At p2 = 5.774 bar

6.4085 = 1.9157 + x2 × 4.857

x2 = 0.925

h2 = hf2 + x2hfg2 = 663.92 + 0.925 × 2091.14 = 2598.22 kJ/kg

h1 = hf1 + x1hfg1 = 762.79 + 0.96 × 2015.3 = 2697.5 kJ/kg

At p3 = 0.8 bar, sf3 = 1.233 kJ/kg.K, sfg3 = 6.202 kJ/kg.K

s2 = s3

sf2 + x2 sfg2 = sf3 + x3 sfg3

1.9157 + 0.925 × 4.857 = 1.233 + x3 × 6.202

x3 = 0.833

h3 = hf3 + x3hfg3 = 391.7 + 0.833 × 2274.1 = 2286 kJ/kg

#### Example 6.5

A convergent–divergent nozzle is supplied with steam at 10 bar and 250°C. The divergent portion of the nozzle is 4 cm long and throat diameter is 6 mm. Find the semi-cone angle of the divergent section so that steam may leave the nozzle at 1.2 bar. The frictional loss in the nozzle is 10 percent of the total enthalpy drop. Assume that the frictional loss occurs only in the divergent part of the nozzle.

Solution

Assuming maximum discharge, the throat pressure,

For superheated steam, n = 1.3

The nozzle and the hs (Mollier) diagram is shown in Fig. 6.9. Locate point ‘1’ on the Mollier diagram corresponding to p1 = 10 bar and 250°C. Draw a vertical line from point ‘1’ to cut the p2 line at point ‘2’ and p3 line at point 3.

Length (1 − 4) = 0.9 × Length (1 − 3)

Draw horizontal line 4 − 3′ to cut p3 line at point 3′. Point 3′ gives the exit condition of steam.

From the Mollier diagram, we have

p1 = 10 bar: h1 = 2940 kJ/kg

p2 = 5.457 bar: h2 = 2830 kJ/kg, v2 = 0.4 m3/kg

p3 = 1.2 bar: h3 = 2530 kJ/kg, x3′ = 0.955, vs3′ = 1.5 m3/kg

h1h3 = 2940−2530 = 410 kJ/kg

h4h3 = 0.1 × 410 = 41 kJ/kg

Figure 6.9 Superheated steam flow through a convergent–divergent nozzle

#### Example 6.6

Dry saturated steam at pressure of 5 bar flows through a convergent–divergent nozzle at the rate of 4 kg/s and discharges at a pressure of 1.5 bar. The loss due to friction occurs only in the diverging portion of the nozzle and its magnitude is 15% of the total isentropic enthalpy drop. Assume the isentropic index of expansion n = 1.135. Determine the area of cross-section at the throat and exit of the nozzle.

Solution

Refer to Fig. 6.10.

For maximum discharge,

From steam table, we get

h1 = 2748.7 kJ/kg, s1 = 6.8212 kJ/kg.K

Now, s1 = s3 = sf3 + x3sfg3

6.8212 = 1.4335 + x3 × 5.7997

x3 = 0.93

h3 = hf 3 + x3hfg3 = 467.08 + 0.93 × 2226.5 = 2537.7 kJ/kg

(h1h3)isen = 2748.7 − 2537.7 = 211 kJ/kg

h1h3 = h1h4 = 0.85 × 211 = 179.35 kJ/kg

From Mollier diagram (Fig. 6.10), x2 = 0.965, x3′ = 0.945, h2 = 2662 kJ/kg

vs2 = 0.7 m3/kg, vs3′ = 1.1 m3/kg, h3′ = 2585 kJ/kg

v2 = x2vs2 = 0.965 × 0.7 = 0.676 m3/kg

v3′ = x3′vs3′ = 0.945 × 1.1 = 1.04 m3/kg

Figure 6.10 Dry saturated steam flow through a convergent–divergent nozzle

##### 6.10 ❐ ISENTROPIC, ONE-DIMENSIONAL STEADY FLOW THROUGH A NOZZLE

A nozzle with both converging and diverging section is shown in Fig. 6.11. For the control volume shown, the following relations can be written:

First law:

Property relation

Continuity equation:

ρAc = = const.

By logarithmic differentiation, we get

Combining Eqs (6.11) and (6.12), we have

Substituting this in Eq. (6.13), we have

Since the flow is isentropic,

and therefore,

Figure 6.11 One-dimensional isentropic flow through a nozzle

where cs = velocity of sound

M = Mach number =

γ = ratio of specific heats = cp/cv

This is a very significant equation, and from it, we can draw the following conclusions about the proper shape for nozzles and diffusers:

1. For a nozzle, dp < 0. Therefore,

for a subsonic nozzle, M < 1, dA < 0, and the nozzle is converging;

for a supersonic nozzle, M > 1, dA > 0, and the nozzle is diverging.

2. For a diffuser, dp > 0. Therefore,

for a subsonic diffuser, M < 1, dA > 0, and the diffuser is diverging;

for a supersonic diffuser, M > 1, dA < 0, and the diffuser is converging.

3. When M = 1, dA = 0, which means that some velocity can be achieved only at the throat of a nozzle or diffuser. These conclusions are summarised in Fig. 6.12.

Figure 6.12 Required area changes for (a) nozzles and (b) diffusers

#### 6.10.1 Relationship between Actual and Stagnation Properties

The relation between enthalpy h, stagnation enthalpy h0, and kinetic energy is:

For an ideal gas with constant specific heat, Eq. (6.15) can be written as:

Since = γRT, where γ = cp0/cv0

For an isentropic process,

Values of are given as a function of M in Table 6.1 for the value of γ = 1.40.

Table 6.1 One-dimensional insentropic compressible-flow functions for an ideal gas with constant specific heat and molecular weight and γ = 1.4

The conditions at the throat of the nozzle can be found by putting M = 1 at the throat. The properties at the throat are denoted as an asterisk (*) and are referred to as critical properties.

Table 6.2 gives these ratios for various values of k.

Table 6.2 Critical properties for isentropic flow of an ideal gas

##### 6.11 ❐ MASS RATE OF FLOW THROUGH AN ISENTROPIC NOZZLE

From the continuity equation, we have

Substituting Eq. (6.17) in Eq. (6.22), the flow per unit area can be expressed in terms of stagnation pressure, stagnation temperature, Mach number, and gas properties.

At the throat, M = 1, and therefore, the flow per unit area at the throat, can be found by setting M = 1 in Eq. (6.23).

The area ratio can be obtained by dividing Eq. (6.24) by Eq. (6.23).

Figure 6.13 Area ratio as a function of Mach number for isentropic nozzle

The values of A/A* as a function of M are given in Table 6.1. Fig. 6.13 depicts the variation of A/A* with M, which shows that a subsonic nozzle is converging and a supersonic nozzle is diverging.

#### 6.11.1 Effect of Varying the Back Pressure on Mass Flow Rate

Consider first a convergent nozzle as shown in Fig. 6.14, which also shows the pressure ratio p/p0 along the length of the nozzle. The conditions upstream are the stagnation conditions, which are assumed to be constant. The pressure at the exit plane of the nozzle is designated pE and the back pressure (the pressure outside the nozzle exit) pB. As the back pressure pB is decreased, the variation of the mass flow rate and the exit plane pressure pE/p0 are plotted in Fig. 6.15.

Where pB/p0 = 1, there is, of course, no flow, and pE/p0 = 1 as designated by point 'a'. Next, let the back pressure pB be lowered to point b, so that pB/p0 is greater than the critical pressure ratio. The mass flow rate has a certain value and pE = pB. The exit Mach number is less than one. Next, let pB be lowered to the critical pressure at point c. The Mach number at the exit is now unity and pE = pB. When pB is decreased below the critical pressure, designated by point d, there is no further increase in the mass rate of flow, and pE remains constant at a value equal to the critical pressure, and the exit Mach number is unity. The drop in pressure from pE to pB takes place outside the nozzle exit. Under these conditions, the nozzle is said to be choked, which means that for given stagnation conditions, the nozzle is passing the maximum possible mass flow.

Figure 6.14 Pressure ratio as a function of back pressure for a convergent nozzle

Figure 6.15 Mass rate of flow and exit pressure as a function of back pressure for a convergent nozzle

Figure 6.16 Nozzle pressure ratio as a function of back pressure for a convergent–divergent nozzle

Consider next a convergent–divergent nozzle as shown in Fig. 6.16. Point ‘a’ designated the condition when pB = p0 and there is no flow. When pB is decreased to the pressure indicated by point b, so that pB/p0 is less than unity but considerably greater than that the critical pressure ratio, the velocity increases in the convergent section, but M < 1 at the throat. Therefore, the diverging section acts as a subsonic diffuser in which the pressure increases and velocity decreases. Point ‘c’ designated the back pressure at which M = 1 at the throat, but the diverging section acts as a subsonic diffuser (with M = 1 at the inlet) in which the pressure increases and velocity decreases. Point ‘d’ designates one other back pressure that permits isentropic flow, and in this case, the diverging section acts as a supersonic nozzle with a decrease in pressure and an increase in velocity. Between the back pressure designated by points c and d, an isentropic solution is not possible, and shock waves will be present. When the back pressure is decreased below that designated by point d, the exit pressure pE remains constant, and the drop in pressure from pE to pB takes place outside the nozzle. This is designated by point e.

#### Example 6.7

A convergent nozzle has an exit area of 500 mm2. Air enters the nozzle with a stagnation pressure of 1000 kPa and a stagnation temperature of 360 K. Determine the mass rate of flow for back pressures of 800 kPa, 528 kPa, and 300 kPa, assuming isentropic flow. For air, k = 1.4.

Solution

Critical pressure ratio, = 0.528

p* = 1000 × 0.528 = 528 kPa

Therefore, for a back pressure of 528 kPa, M = 1 at the nozzle exit and the nozzle is choked.

Decreasing the back pressure below 528 kPa will not increase the flow.

For a back pressure of 528 kPa,

At the exit, c = cs = = 347.2 m/s

Discharge at the exit section,

= ρ*Ac = 6.1324 × 500 × 10–6 × 347.2 = 1.0646 kg/s

For a back pressure of 800 kPa, = 0.8

From Table 6.1 for by interpolation

TE = 360 × 0.9381 = 337.7 K

Sonic velocity at exit,

Velocity at exit,

Density of air at exit,

= ρEAEcE = 8.2542 × 500 × 10−6 × 211.1 = 0.8712 kg/s

For a back pressure less than the critical pressure (528 kPa), the nozzle is choked and the mass rate of flow is the same as that for the critical pressure. Therefore, for an exhaust pressure of 300 kPa, the mass flow rate is 1.0646 kg/s.

#### Example 6.8

A converging–diverging nozzle has an exit area to throat area ratio of 2. Air enters this nozzle with a stagnation pressure of 1000 kPa and a stagnation temperature of 360 K. The throat area is 500 mm2. Determine the mass rate of flow, exit pressure, exit temperature, exit Mach number, and exit velocity for the following conditions:

1. Sonic velocity at the throat, diverging section acting as a nozzle.
2. Sonic velocity at the throat, diverging section acting as a diffuser.

Solution

1. From Table 6.1, we have

For = 2, M*E = 2.197, = 0.0939, = = 0.5089

Therefore, pE = 0.0939 × 1000 = 93.9 kPa

TE = 0.5089 × 360 = 183.2 K

cE = ME* csE = 2.197 × 271.3 = 596.1 m/s

Critical pressure at throat, p* = p0 × 0.528 = 1000 × 0.528 = 528 kPa

Critical temperature, T * = 0.8333 × 360 = 300 K

At throat,

2. From Table 6.1, we have

For = 2, ME = 0.308, = 0.936, = 0.9812

Therefore, pE = 0.936 × 1000 = 936 kPa

TE = 0.9812 × 360 = 353.3 K

cE = MCsE = 0.308 × 376.8 = 116 m/s

Since M = 1 at the throat, mass rate of flow is the same as in (a).

##### 6.12 ❐ NORMAL SHOCK IN AN IDEAL GAS FLOWING THROUGH A NOZZLE

A shock wave involves an extremely rapid and abrupt change of state. In a normal shock, this change of state takes place across a plane normal to the direction of flow. Figure 6.17 shows a control surface that includes such a normal shock. Let subscripts x and y denote the conditions upstream and downstream of shock, respectively, and assuming steady-state, steady-flow with no heat and work transfer across the control surface, then the various relations are as follows:

First law:

or hox = hoy

Figure 6.17 One-dimensional normal shock

Continuity equation:

Momentum equation:

Second law:

The energy and continuity equations can be combined to establish an equation that when plotted on the hs diagram is called the Fanno line. Similarly, the momentum and continuity equations can be combined to establish an equation the plot of which on the hs diagram is known as the Rayleigh line. Both these lines are shown on the hs diagram in Fig. 6.18. The point of maximum entropy on each line, points ‘a’ and ‘b’, corresponds to M = 1. The lower part of each line corresponds to supersonic flow, whereas the upper part corresponds to subsonic flow.

The two points where all three equations are satisfied are points x and y, where x being in the supersonic region and y in the subsonic region. Since sysx ≥ 0, the normal shock can proceed only from x to y. This means that the velocity changes from supersonic (M > 1) before the shock to subsonic (M < 1) after the shock.

Assuming constant specific heats, the energy Eq. (6.26) gives,

That is, there is no change in stagnation temperature across a normal shock. Using Eq. (6.16), we have

and substituting into Eq. (6.30), we have

Figure 6.18 End states for a one-dimensional normal shock on an enthalpy–entropy diagram

The continuity equation is,

ρxcx = ρycy

But

Combining energy Eq. (6.31) and continuity Eq. (6.32) gives the equation of the Fanno line.

The momentum and continuity equations can be combined as follows to give the equation of the Rayleigh line.

Combining Eqs (6.33) and (6.34), we get

Table 6.3 gives the normal shock functions, which includes My as a function of Mx for γ = 1.4.

Table 6.3 One-dimensional normal shock functions for an ideal gas with constant specific heat and molecular weight and γ = 1.4

#### Example 6.9

For the convergent–divergent nozzle of Example 6.8 in which the diverging section acts as a supersonic nozzle (Fig. 6.19), a normal shock stands in the exit plane of the nozzle. Determine the static pressure and temperature and the stagnation pressure (a) just downstream of the normal shock and (b) at a point where M = 1.4.

Solution

1. From Table 6.3, we have

Figure 6.19 Convergent–divergent nozzle

py = 4.46 × 93.9 = 512.7 kPa

Ty = 1.854 × 183.2 = 339.7 K

poy = 0.630 × 1000 = 630 kPa

2. From Table 6.1 at point x, as the flow is isentropic to point x,

Therefore, px = 0.2724 × 1000 = 272.4 kPa

Tx = 0.6897 × 360 = 248.3 K

The properties at y can be determined from the normal shock functions (Table 6.3) as

py = 2.4583 × 272.4 = 669.6 kPa

Ty = 1.320 × 248.3 = 327.8 K

poy = 0.9298 × 1000 = 929.8 kPa

Tox = Toy = 360 K, as there is no change in stagnation temperature across a normal shock. From y to E, the diverging section acts as a diffuser.

#### Example 6.10

Steam at stagnation pressure of 800 kPa and a stagnation temperature of 350°C expands in a nozzle to 200 kPa. When the mass flow rate is 3 kg/s, determine the throat area and exit area for isentropic flow.

Solution

Critical pressure ratio at the throat, = 0.545

p* = 0.545 × 800 = 436 kPa

s* = s0 = 7.4089 kJ/kg. K

h0 = 3161.7 kJ/kg

T * = 268.7°C

h* = 3001.4 kJ/kg

At the nozzle exit,

pE = 200 kPa, sE = s0 = 7.4089 kJ/kg.K

TE = 178.5°C

hE = 2826.7 kJ/kg

vE = 1.0284 m3/kg

#### Example 6.11

5 kg/s of steam at 30 bar and 350°C is supplied to a group of 6 nozzles in a wheel diameter maintained at 4 bar. Calculate the following:

1. The dimensions of the nozzles of rectangular cross-sectional flow area with aspect ratio of 2.5:1. The expansion may be considered metastable and friction is neglected.
2. Degree of undercooling and supersaturation
3. Loss in available heat drop due to irreversibility
4. Increase in entropy
5. Ratio of mass flow rate with metastable expansion to that if expansion is in thermal equilibrium.

Solution

The h-s diagram is shown in Fig. 6.20.

h1 = 3115.3 kJ/kg

v1 = 0.09053 m3/kg

s1 = 6.74257 kJ/kg.K

s3 = s1 = sf3 + x3sfg3

6.7427 = 1.7766 + x3 × 5.1193

x3 = 0.97

h3 = hf3 + x3hfg3

= 604.73 + 0.97 × 2133.8

= 2674.5 kJ/kg

v3 = vf3 + x3(vg3vf3) = 0.0010840.97 × (0.4625 − 0.001084)

= 0.45 m3/kg

Figure 6.20 Mollier diagram for superheated steam

1. For supersaturated steam, n = 1.3

Then length, l = 2.5 b

Area of 6 nozzles = 6 × 2.5b × b = 15b2

15b2 = 2.279 × 10–3

b = 0.0123 m or 12.3 mm

l = 0.0308 m or 30.8 mm

2. At 4 bar, ts = 143.65°C

Degree of undercooling = 143.65 − 118.34 = 24.31°C

ps corresponding to 118.3°C = 1.9 bar

Degree of super saturation = = 2.1

3. h1 = h3 = 3115.3 − 2674.5 = 440.8 kJ/kg

Loss of available heat drop = (h1h3) − (h1h2)

h)loss = 440.8 − 437.5 = 3.3 kJ/kg

4. Increase in entropy

#### Example 6.12

Dry saturated steam at 5 bar with negligible velocity expands isentropically in a convergent nozzle to 1 bar and dryness fraction 0.94. Determine the velocity of steam leaving the nozzle.

Solution

Given that p1 = 5 bar; p2 = 1 bar; x2 = 0.94

For p1 = 5 bar, enthalpy of dry saturated steam from steam tables,

h1 = hg1 = 2748.7 kJ/kg

and for p2 = 1 bar, we find that,

hf 2 = 417.44 kJ/kg, hfg2 = 2258.0 kJ/kg

h2 = hf2 + x2 hfg2

= 417.44 + 0.94 × 2258.0 = 2539.96 kJ/kg

Enthalpy drop, Δh = h1h2

= 2748.75 − 2539.96 = 208.74 kJ/kg

Velocity of steam leaving the nozzle

#### Example 6.13

Dry saturated steam at a pressure of 15 bar enters in a nozzle and is discharged at a pressure of 1.5 bar. Find the final velocity of steam when initial velocity of steam is negligible. If 15% of the heat drop is lost in friction, then find the percentage reduction in the final velocity.

Solution

For p1 = 15 bar from steam tables,

h1 = 2792.1 kJ/kg

and for p2 = 1.5 bar,

h2 = 2693.5 kJ/kg

Enthalpy drop, Δh = h1h2

= 2792.19 − 2693.5 = 98.6 kJ/kg

Final velocity of steam,

Percentage reduction in the final enthalpy = 15% = 0.15

Nozzle coefficient or nozzle efficiency, ηn = 1 − 0.15 = 0.85

Final velocity of steam,

Percentage reduction in final velocity

#### Example 6.14

Steam approaches a nozzle with velocity of 250 m/s at a pressure of 3.5 bar and dryness fraction 0.95. If the isentropic expansion in the nozzle proceeds till the pressure at exit is 2 bar, then determine the change of enthalpy and dryness fraction of steam. Calculate also the exit velocity from the nozzle and the area of exit of nozzle for flow of 0.75 kg/s.

Solution

Given that c1 = 250 m/s, p1 = 3.5 bar, x1 = 0.95, p2 = 2 bar, = 0.75 kg/s

From Mollier diagram, as shown in Fig. 6.21,

Figure 6.21 Mollier diagram

h1 = 2624 kJ/kg

h2 = 2534 kJ/kg

Δh = h1h2 = 2624 − 2534 = 90 kJ/kg

Dryness fraction, x2 = 0.92

From steam tables, vf2 = 0.001061 m3/kg, vg2 = 0.8857 m3/kg at p2 = 2 bar

Specific volume at exit

v2 = vf2 + x2 (vg2vf2) = 0.001061 + 0.92 × (0.8857 − 0.001061) = 0.8149 m3/kg

Now, = h1h2

Steam mass flow rate () is given by,

#### Example 6.15

Dry saturated steam at a pressure of 6 bar flows through a convergent–divergent nozzle at a rate of 4.5 kg/s and discharges at a pressure of 1.6 bar. The loss due to friction occurs only in the diverging portion of nozzle and its magnitude is 12% of total isentropic enthalpy drop. Assuming the isentropic index of expansion, n = 1.135, determine the cross-sectional area at throat and exit of nozzles.

Solution

Given that p1 = 6 bar, = 4.5 kg/s, p3 = 1.6 bar, n = 1.135, ηn = 1 − 0.12 = 0.88

From Mollier diagram shown in Fig. 6.22,

h1 = 2750 kJ/kg, h3 = 2520 kJ/kg

Refer to hs diagram (Mollier chart) as shown in Fig. 6.22.

Let suffix 1, 2 and 3 denote entrance, throat, and exit.

We have

p2 = 6 × 0.578 = 3.468 bar and

h2 = 2650 kJ/kg

Figure 6.22 Mollier diagram

For Section 1–2:

h1h2 = 2750 − 2650 = 100 kJ/kg
A2 = Area of throat
v2 = 0.964 m3/kg

From Mollier diagram, we have

vg2 = 0.5321 m3/kg corresponding to p2 = 3.468 bar
v2 = x2 vg2 = 0.964 × 0.5321 = 0.513 m3/kg

For Section 2–3:

(h1h3) isentropic = 230 kJ/kg
h1h3' = (h1h3) × ηn = 230 × 0.88 = 202.4 kJ/kg

From Mollier diagram,

x3' = 0.9365

vg3 = 1.0911 m3/kg corresponding to pressure of 1.6 bar.

v3 = x3' vg3 = 0.9365 × 1.0911 = 1.022 m3/kg

#### Example 6.16

Dry saturated steam enters a nozzle at a pressure of 10 bar and velocity of 100 m/s. The discharge pressure is 5 bar and discharge velocity is 500 m/s. Heat loss from the nozzle is 5 kJ/kg. Determine the final dryness fraction of steam.

Solution

Given that p1 = 10 bar, c1 = 100 m/s, p2 = 5 bar, c2 = 500 m/s, q = −5 kJ/kg

Since w = 0

At p1 = 10 bar, x1 = 1.0, h1 = 2778.1 kJ/kg

At 5 bar; hf2 = 640.21 kJ/kg, hfg2 = 2108.5 kJ/kg

#### Example 6.17

Steam enters a group of nozzles of a steam turbine at 12 bar and 220°C and leaves at 1.2 bar. The steam turbine develops 220 kW with a specific steam consumption of 13.5 kg/kWh. If the diameter of nozzles at throat is 7 mm, then calculate the number of nozzles.

Solution

Given that p1 = 12 bar, t1 = 220°C, p3 = 1.2 bar, power developed = 220 KW, m3 = 13.5 kg/kWh, and

d2 = 7 mm

Refer to Mollier diagram, as shown in Fig. 6.23.

The steam is initially superheated.

n = 1.3 and p2/p1 = 0.5457

p2 = 0.5457 p1 = 0.5457 × 12 = 6.548 bar

From Mollier diagram, we have

h1 = 2860 kJ/kg, h2 = 2750 kJ/kg, x2 = 0.992

Figure 6.23 Mollier diagram

vg2 = 0.29 m3/kg

Δh2 = h1h2 = 2860 − 2750 = 110 kJ/kg

A2 = Area of throat

Mass flow rate per nozzle

Total mass flow rate,

#### Example 6.18

Dry saturated steam at a pressure of 8 bar enters a convergent–divergent nozzle and leaves it at a pressure of 1.5 bar. If the flow is isentropic and the corresponding expansion index is 1.135, then find the ratio of cross-sectional area at exit and throat for maximum discharge.

Solution

Given that p1 = 8 bar, p3 = 1.5 bar, n = 1.135

Let A2 = Area of cross-section at throat

A3 = Area of cross-section at exit

m = mass of steam discharged per second

Refer to Mollier diagram as shown in Fig. 6.24.

Figure 6.24 Mollier diagram

For dry saturated steam, n = 1.135

From Mollier chart, h1 = 2775 kJ/kg

h2 = 2650 kJ/kg
h3 = 2465 kJ/kg
x2 = 0.965, x3 = 0.902

From steam tables, vg2 = 0.405 m3/kg, v'g2 = 1.159 m3/kg

Ratio of cross-sectional area at exit and throat =

#### Example 6.19

Steam at a pressure of 10 bar and dryness fraction of 0.98 is discharged through convergent–divergent nozzle to a back pressure of 0.1 bar. The mass flow rate is 10 kg/kWh. If the power developed is 220 kW, then determine the following:

1. Pressure at throat
2. Dimensions at exit of the nozzle if the throat is of rectangular cross-section of 5 mm × 10 mm.

The loss due to friction is 10% of the overall isentropic enthalpy drop in the divergent portion.

Solution

Refer to the Mollier diagram as shown in Fig. 6.25.

1. The isentropic index of expansion for wet steam:

n = 1.035 + 0.1x1

n = 1.035 + 0.1 × 0.98 = 1.133

Now p1 = 10 bar,

p2 = 10 × 0.5778 = 5.778 bar

2. From Mollier diagram, we have

h1 = 2762 kJ/kg, h3 = 2075 kJ/kg, h2 = 2660 kJ/kg

Loss in heat drop due to friction = 0.1 (h1h3) = 0.1 (2762 − 2075) = 68.7 kJ/kg

h3'h3 = 68.7
h3' = h3 + 68.7 = 2075 + 68.7 = 2143.7 kJ/kg

Figure 6.25 Mollier diagram

Mass flow rate,

Velocity at throat,

Total area of throat,

x2 = 0.953
vg2 = 0.34 m3/kg at 5.778 bar (from steam tables)

Throat area per nozzle = 5 × 10 × 10−6 = 0.5 × 10−4 m2

Number of nozzles = = 7.97 = 8 (say)

Exit area/nozzle = = 7.476 × 10−4 m2

Keeping the same aspect ratio of 1:2 for rectangle and let x be the smaller side

2x × x = 7.476 × 10−4
2x2 = 7.476 cm2

Therefore, exit rectangle is 19.33 mm × 38.66 mm.

#### Example 6.20

The dry saturated steam is expanded in a nozzle from pressure of 10 bar to pressure of 5 bar. If the expansion is supersaturated, then find (a) the degree of under cooling, and (b) the degree of supersaturation.

Solution

Given that p1 = 10 bar and p2 = 5 bar

From steam tables corresponding to p1 = 10 bar, saturation temperature of steam.

T1 = 179.91°C = 452.91 K

T2 = Temperature at which supersaturation occurs

From steam tables corresponding to a pressure of 5 bar, we find saturation temperature

T2 = 151.86°C

∴ Degree of undercooling

= T2T2 = 151.86 − 112.88 = 38.98°C

#### Example 6.21

Steam enters a nozzle in a dry saturated condition and expands from a pressure of 2 bar to 1 bar. It is observed that supersaturated flow is taking place and steam flow reverts to normal flow at 1 bar. What is the degree of undercooling and increase in entropy and loss in available heat drop due to irreversibility?

Solution

For supersaturated flow,

From steam tables at 2 bar,

T1 = (120.23 + 273) = 393.23 K

Hence, = 335 K

From steam tables corresponding to 1 bar, T2 (saturation temperature)

= 99.62°C = 99.62 + 273 = 372.62 K

Degree of undercooling

= T2T2' = 372.62 − 335 = 37.62 K

Refer to the Mollier diagram as shown in Fig. 6.26.

Corresponding to 335 K, saturation pressure from steam tables = 0.22 bar.

p2' = 0.22 bar

Figure 6.26 Mollier diagram

Degree of supersaturation = = 4.545

For supersaturated flow, the formula for superheated steam is used.

Thus,

where n = 1.3

v1 = 0.8857 m3/kg from steam tables

Isentropic heat drop = (h1h2) = 117.5 kJ/kg (from Mollier diagram)

Therefore, loss in availability = 117.5 − 113.5 = 4 kJ/kg

Increase in entropy = = 0.01073 kJ/kg.K

#### Example 6.22

Find the percentage increase in discharge from a convergent–divergent nozzle expanding steam from 8.75 bar dry to 2 bar, when (a) the expansion is taking place under thermal equilibrium and (b) the steam is in metastable state during part of its expansion.

Take area of nozzle as 2500 mm2.

Solution

Given that p1 = 8.75 bar, p2 = 2 bar, A2 = 2500 mm2 = 2500 × 10–6 m2

Refer to the Mollier diagram shown in Fig. 6.27.

1. Mass of steam discharged when expansion is under thermal equilibrium.

Let 1 = Mass of steam discharged

From Mollier diagram, h1 = 2770 kJ/kg, h2 = 2515 kJ/kg, and x2 = 0.91

From steam tables,

Figure 6.27 Mollier diagram

vg2 = 0.885 m3/kg
Δh2 = h1h2 = 2770 − 2515 = 255 kJ/kg

Velocity of steam at exit,

2. Mass of steam discharged when it is in metastable state.

Let 2 = mass of steam discharged

Volume of steam at inlet,

Volume of steam at exit,

Furthermore,

Heat drop from inlet to exit

Δh2 = h1h2 = 2770 − 2530 = 240 kJ/kg

Velocity of steam at exit,

∴ Percentage increase in discharge

#### Example 6.23

Steam expands through a nozzle from 5 bar and dry saturated to a back pressure of 0.2 bar. Mass flow is 2 kg/s. Calculate the exit and the throat areas under the following conditions:

1. Isentropic expansion with negligible velocity
2. Isentropic expansion with initial velocity of 100 m/s
3. Friction loss at pressure amounts to 10% of the total heat drop up to that pressure and initial velocity negligible

Solution

1. Pressure at throat is given by

p2 = 0.5774 × p1 = 0.5774 × 5 = 2.887 bar

Drop a vertical from point 1 to cut pressure for 2.887 bar at point 2 and the pressure line for 0.2 bar, i.e., the back pressure at point 3. Then from the Mollier diagram (Fig. 6.28), the following values are obtained:

At point 1, h1 = 2745 kJ/kg

At point 2, p2 = 2.887 bar, h2 = 2645 kJ/kg, and v2 = 0.63 m3/kg

At point 3, p3 = 0.2 bar, h3 = 2248 kJ/kg, and v3 = 6.6 m3/kg

Figure 6.28 Mollier diagram

Hence, the area at throat for a mass flow of 2 kg/s is given by,

Similarly, area at exit for a mass flow of 2 kg/s is given by,

2. Refer Fig. 6.29.

Hence, h0 = 5.0 + h1 = 5 + 2745 = 2750 kJ/kg

Corresponding to h0 = 2750 kJ/kg

p0 = 5.1 bar as shown on the chart

Thus, the problem is resolved to the condition of p0 = 5.1 bar and h0 = 2750 kJ and velocity of zero at entry to the nozzle.

Therefore, pressure at throat is given by,

And at p2 = 2.945 bar

h2 = 2653 kJ/kg, v2 = 0.61 m3/kg, from the Mollier chart

p3 = 0.2 bar

h3 = 2248 kJ/kg, v3 = 6.6 m3/kg, from the Mollier chart

Figure 6.29 Mollier diagram

Area at throat is given by,

Similarly, area at the exit from the nozzle is given by,

The nozzle efficiency ηn = 0.9 and n = 1.135 for dry saturated steam.

Therefore, m = = 1.119

The critical pressure ratio is given by,

And the discharge per unit area is given by,

Area at throat, A2 =

Using the same formula and putting the value of rp = we get

Therefore, area at exit is given by,

The area at the exit may also be found with the help of Mollier diagram (Fig. 6.30). Isentropic enthalpy drop as found in part (a) = 497 kJ/kg.

Therefore, the actual heat drop = 0.9 × (2745 − 2248) kJ/kg

= 497 + 0.9 = 447.3 kJ/kg

Figure 6.30 Mollier diagram

i.e., h1h'3 = 447.8 kJ/kg

and v3′ = 6.7 m3/kg

Therefore, area at exit is given by,

#### Example 6.24

A convergent–divergent nozzle is required to discharge 2 kg of steam per second. The nozzle is supplied with steam at 7 bar and 180°C and discharge takes place against a back pressure of 1 bar. The expansion up to throat is isentropic and the frictional resistance between the throat and the exit is equivalent to 63 kJ/kg of steam. Taking approach velocity of 75 m/s and throat pressure of 4 bar, estimate: (a) suitable areas for the throat and exit and (b) overall efficiency of the nozzle based on the enthalpy drop between the actual inlet pressure and temperature and the exit pressure.

Solution

Given that = 2 kg/s; p1 = 7 bar, T1 = 180°C, p3 = 1 bar, frictional resistance = 63 kJ/kg of steam, c1 = 75 m/s, p2 = 4 bar

1. Suitable areas for the throat and exit

Let A2 = area at the throat

A3 = area at the exit

The expansion of steam through the nozzle on the Mollier diagram is shown in Fig. 6.31. From the Mollier diagram, we find that

Figure 6.31 Mollier diagram

h1 = 2810 kJ/kg, h2 = 2680 kJ/kg, h3 = 2470 kJ/kg, x2 = 0.97, and x3 = 0.934

From steam tables, we also find that the specific volume of steam at throat corresponding to 4 bar,

vf2 = 0.001084 m3/kg, vg2 = 0.4625 m3/kg

v2 = vf2 + x2(vg2vf2) = 0.001084 + 0.97(0.4625 − 0.001084) = 0.449 m3/kg

and specific volume of steam corresponding to 1 bar,

vf3 = 0.001043 m3/kg, vg3 = 1.694 m3/kg

v3 = vf3 + x3(vg3vf3) = 0.001043 + 0.934(1.694 − 0.001043) =1.582 m3/kg

Heat drop between entrance and throat,

Δh2 = h1h2 = 2810 − 2680 = 130 kJ/kg

∴ Velocity of steam at throat,

Since there is a frictional resistance of 63 kJ/kg of steam between the throat and the exit,

h3h'3 = 63 or h3 = h'3 + 63 = 2470 + 63 = 2533 kJ/kg

and heat drop between entrance and exit,

Δh3 = h1h3 = 2810 − 2533 = 277 kJ/kg

∴ Velocity of steam at exit,

2. Overall efficiency of the nozzle

Overall efficiency of the nozzle

#### Example 6.25

Air is expanded reversibly and adiabatically in a nozzle from 13 bar and 150°C to a pressure of 6 bar. The inlet velocity of the nozzle is very small and the process occurs under steady-state flow conditions. Calculate the exit velocity of the nozzle.

[IES, 1992]

Solution

Given that p1 = 13 bar, T1 = 273 + 180 = 423 K, p2 = 6 bar, and c1 = 0

SFEE for unit mass flow rate is:

Figure 6.32 Flow through steam nozzle

Since the air expands reversibly and adiabatically from state 1 to 2, as shown in Fig. 6.32, we have q = 0, w = 0, and z1 = z2

Exit velocity of the nozzle,

#### Example 6.26

A steam nozzle receives steam at 40 bar and 400°C at an initial velocity of 40 m/s. The final pressure of steam is 10 bar. The mass flow rate of steam is 2 kg/s. The nozzle efficiency is 90%. The cross-section of the nozzle is circular. The angle of divergence is 6°. Calculate the throat and exit diameters and the length of the divergent portion. Show the representation of process on hs diagram and sketch the nozzle and label the dimensions calculated.

[IES, 2005]

Figure 6.33 Flow through convergent-divergent steam nozzle

Solution

The convergent–divergent nozzle is shown in Fig. 6.33 and hs diagram in Fig. 6.34.

From steam tables and Mollier diagram, we have

h1 =3213.5 kJ/kg for p1 = 40 bar and t = 400°C

p2 = 0.5457, p1 = 0.5457 × 40 = 21.828 bar, for superheated steam

h2 = 3025 kJ/kg, vs2 = 0.125 m3/kg

c2 = [2(h1h2) × 103 + ]0.5

= [2(3213.5 − 3025) × 103 + 402]0.5 = 615.3 m/s

Figure 6.34 h − s diagram

In Fig. 6.34,

#### Summary for Quick Revision

1. A steam nozzle may be defined as a passage of varying cross-section by means of which a part of enthalpy of steam is converted into kinetic energy as the steam expands from a higher pressure to a lower pressure.
2. Continuity equation:
3. Velocity of steam through a nozzle:

For

where h1 and h2 are in kJ/kg

4. Isentropic flow through a nozzle
1. pvn= constant

where n = 1.035 + 0.1 x1 from wet steam

= 1.135 for saturated steam

= 1.3 for superheated steam

2. Velocity, c2 =
3. Mass of steam discharged,
4. Critical pressure ratio, for maximum discharge.
5. Velocity at exit for maximum discharge,
6. Maximum discharge, ()max = A2
5. Most of the friction losses occur beyond the throat in the divergent section of the nozzle.
6. Nozzle efficiency,
7. Exit velocity considering friction in nozzle,
8. Supersaturated (or meta stable) flow:

In steam nozzles, the condensation of steam beyond the saturation line is suppressed or delayed due to very high velocity of steam and the vapours continue to expand in dry state even beyond the saturation line. The steam vapours are said to be supersaturated or supercooled in this region.

Velocity of steam at the end of expansion,

9. Wilson line represents the limit to the supersaturated state on the Mollier chart.
10. Isentropic, one-dimensional steady flow through a nozzle:
1. where

2. For a nozzle, dp < 0.

Subsonic nozzle: M < 1, dA < 0, and nozzle is converging.

Supersonic nozzle: M > 1, dA > 0, and nozzle is diverging.

3. For a diffuser, dp > 0.

Subsonic diffuser, M < 1, dA > 0, and diffuser is diverging.

Supersonic diffuser, M > 1, dA < 0, and diffuser is converging.

4. When M = 1, dA = 0, which occurs at the throat of nozzle and diffuser.
11. Stagnation properties:

12. Properties at the throat, M = 1
13. Mass rate of flow through isentropic nozzle.

At the throat, M = 1

14. Area ratio,
15. A shock wave involves an extremely rapid and abrupt change of state.
1. In a normal shock, the abrupt change of state takes place across a plane normal to the direction of flow.
2. The normal shock proceeds from supersonic to subsonic region.
3. There is no change in stagnation temperature across a normal shock, i.e., Tox = Toy.

#### Multiple-choice Questions

1. A nozzle has velocity head at outlet of 10 m. If it is kept vertical, the height reached by the stream is
1. 100 m
2. 10 m
2. The effect of friction on flow of steam through a nozzle is to
1. decrease the mass flow rate and to increase the wetness at the exit
2. increase the mass flow rate and to increase the exit temperature
3. decrease the mass flow rate and to decrease the wetness of the steam
4. increase the exit temperature, without any effect on the mass flow rate
3. For adiabatic expansion with friction through a nozzle, the following remains constant
1. Entropy
2. Static enthalpy
3. Stagnation enthalpy
4. Stagnation pressure
4. The critical pressure ratios for flow of dry saturated steam and superheated steam through a nozzle are, respectively,
1. 0.5279 and 0.550
2. 0.577 and 0.550
3. 0.577 and 0.546
4. 0.5279 and 0.546
5. At which location of a converging–diverging nozzle, does the shock-boundary layer interaction take place?
1. Converging portion
2. Throat
3. Inlet
4. Diverging portion
6. The effect of friction in a steam nozzle is to
1. increase velocity and increase dryness fraction
2. increase velocity and decrease dryness fraction
3. decrease velocity and increase dryness fraction
4. decrease velocity and decrease dryness fraction
7. Consider the following statements:

When dry saturated or slightly superheated steam expands through a nozzle,

1. The coefficient of discharge is greater than unity.
2. It is dry up to Wilson’s line.
3. Expansion is isentropic throughout.

Of these statements

1. I, II, and III are correct
2. I and II are correct
3. I and III are correct
4. II and III are correct
8. The total and static pressures at the inlet of a steam nozzle are 186 kPa and 178 kPa, respectively. If the total pressure at the exit is 180 kPa and static pressure is 100 kPa, then the loss of energy per unit mass in the nozzle will be
1. 78 kPa
2. 8 kPa
3. 6 kPa
4. 2 kPa
9. Under ideal conditions, the velocity of steam at the outlet of a nozzle for a heat drop of 400 kJ/kg will be approximately
1. 1200 m/s
2. 900 m/s
3. 60 m/s
4. the same as the sonic velocity
10. Consider the following statements:
1. De Laval nozzle is a subsonic nozzle.
2. Supersonic nozzle is a converging passage
3. Subsonic diffuser is a diverging passage

Which of these statements is/are correct?

1. I and II
2. II and III
3. I alone
4. III alone
11. In a steam nozzle, inlet pressure of superheated steam is 10 bar. The exit pressure is decreased from 3 bar to 1 bar. The discharge rate will
1. remain constant
2. decrease
3. increase slightly
4. increase or decrease depending on whether the nozzle is convergent or convergent–divergent
12. Air from a reservoir is to be passed through a supersonic nozzle so that the jet will have a Mach number of 2. If the static temperature of the jet is not to be less than 27°C, the minimum temperature of air in the reservoir should be
1. 48.6°C
2. 167°C
3. 267°C
4. 367°C
13. Consider the following statements:

For supersaturated flow through a steam nozzle, the

1. Enthalpy drop reduces further
2. Exit temperature increases
3. Flow rate increases

Which of these statements are correct?

1. I, II, and III
2. I and II
3. II and III
4. I and III
14. The critical pressure ratio for maximum discharge through a nozzle is given by
15. Consider the following statements in relation to a convergent–divergent steam nozzle operating under choked conditions:
1. In the convergent portion, steam velocity is less than sonic velocity.
2. In the convergent portion, steam velocity is greater than sonic velocity.
3. In the divergent portion, the steam velocity is less than sonic velocity.
4. In the divergent portion, the steam velocity is greater than sonic velocity.

Which of the above statements are correct?

1. I and III
2. I and IV
3. II and III
4. II and IV
16. For maximum discharge through a convergent nozzle, the pressure ratio should be (where n is the index for expansion)
17. Wilson line is associated with which one of the following?
1. Total steam consumption with respect to power output
2. Supersonic flow of steam through a nozzle
3. Nozzle flow with friction
4. Supersaturated flow of steam through a nozzle
18. In a steam nozzle, to increase the velocity of steam above sonic velocity by expanding steam below critical pressure
1. a vacuum pump is added
2. ring diffusers are used
3. divergent portion of the nozzle is necessary
4. abrupt change in cross-section is needed
19. Which one of the following is the correct statement? To get supersonic velocity of steam at nozzle exit with a large pressure drop across it, the duct must
1. converge from inlet to exit
2. diverge from inlet to exit
3. first converge to the throat and then diverge till exit
4. remain constant in cross-section
20. Which one of the following is the correct expression for the critical pressure ratio of a nozzle?
21. What is the critical pressure ratio for isentropic nozzle flow with ratio of specific heats as 1.5?
1. (0.8)3
2. (0.8)0.6
3. (1.25)0.33
4. (1.25)3
22. A compressible fluid flows through a passage as shown in Fig. 6.35. The velocity of the fluid at the point A is 400 m/s.

Figure 6.35

Which one of the following is correct?

At point B, the fluid experiences

1. an increase in velocity and decrease in pressure
2. a decrease in velocity and increase in pressure
3. a decrease in velocity and pressure
4. an increase in velocity and pressure
23. If the cross-section of a nozzle is increasing in the direction of flow in supersonic flow, then in the downstream direction
1. both pressure and velocity will increase
2. both pressure and velocity will decrease
3. pressure and increase but velocity will decrease
4. pressure will decrease but velocity will increase
24. If the velocity of propagation of small disturbances in air at 27°C is 330 m/s, then at a temperature of 54°C, its speed would be
1. 660 m/s
2. 300 × m/s
3. 330/ m/s
4. 330 × m/s
25. For one-dimensional isentropic flow in a diverging passage, if the initial static pressure is p1 and the initial Mach number is M1 (M1 < 1), then for the downstream flow
1. M2 < M1; p1 < p2
2. M2 < M1; p2 > p1
3. M2 > M1; p2 > p1
4. M2 > M1; p2 < p1
26. Which of the following diagrams correctly depict the behaviour of compressible fluid flow in the given geometries (Fig. 6.36)?

Figure 6.36

Codes:

1. 1 and 4
2. 2 and 4
3. 2 and 3
4. 1 and 3
27. Figure 6.37 represents a schematic view of the arrangement of supersonic wind tunnel section. A normal shock can exist without affecting the test conditions

Figure 6.37

1. Between Sections 4 and 5
2. At Section 4
3. Between Sections 4 and 3
4. Between Sections 1 and 2
28. The Mach number for nitrogen flowing at 195 m/s when the pressure and temperature in the undisturbed flow are 690 kN/m abs and 93°C, respectively, will be
1. 0.25
2. 0.50
3. 0.66
4. 0.75
29. The Mach number at inlet of gas turbine diffuser is 0.3. The shape of the diffuser would be
1. converging
2. diverging
3. stagnation enthalpy
4. stagnation pressure
30. The identical convergent–divergent nozzles A and B are connected in series, as shown in Fig. 6.38 to carry a compressible fluid.

Figure 6.38

Which one of the following statements regarding the velocities at the throats of the nozzles is correct?

1. Sonic and supersonic velocities exist at the throats of the nozzles A and B, respectively.
2. Sonic velocity can exist at throats of both nozzles A and B.
3. Sonic velocity will always exist at the throat of nozzle A, whereas subsonic velocity will exist at throat of nozzle B.
4. Sonic velocity exists at the throat of nozzle B, whereas subsonic velocity exists at the throat of nozzle A.
31. Semi-angle of a Mach cone is
1. sin–1 M

where M is the Mach number.

32. Which of the following statements are correct?
1. Mach wave is a very weak shock wave.
2. Entropy change across a shock wave is nearly zero.
3. Total pressure behind a shock wave is less than that ahead of it.
4. Mach number behind a normal shock is less than one.

Select the correct answer using the codes given below:

Codes:

1. I, II, and III
2. I, III, and IV
3. I, II, and IV
4. II, III, and IV
33. Introduction of a Pitot tube in a supersonic flow would produce
1. normal shock at the tube nose
2. curved shock at the tube nose
3. normal shock at the upstream of the tube nose
4. curved shock at the upstream of the tube nose
34. In the Fanno line shown in Fig. 6.39,

Figure 6.39

1. subsonic flow proceeds along PQR
2. supersonic flow proceeds along PQR
3. subsonic flow proceeds along PQ and supersonic flow proceeds along RQ
4. subsonic flow proceeds along PQ and supersonic flow proceeds along QR
35. Given k = ratio of specific heats, for Rayleigh line, the temperature is maximum at a Mach number of
1. 1/k
2. k
36. What is/are the effect(s) of supersaturation in nozzle flow?
1. It increases the mass flow
2. It increases friction in the nozzle
3. It increases exit velocity
4. It reduces dryness fraction of steam.

Select the correct answer using the code given below:

1. I only
2. I and III
3. I, II and IV
4. III only
37. Consider the following statements pertaining to isentropic flow:
1. To obtain stagnation enthalpy, the flow need not be decelerated isentropically but should be decelerated adiabatically.
2. The effect of friction in an adiabatic flow is to reduce the stagnation pressure and increase entropy.
3. A constant area tube with rough surfaces can be used as a subsonic nozzle.

Of these statements:

1. I, II, and III are correct
2. I and II are correct
3. I and III are correct
4. II and III are correct
38. Consider the following statements:

A convergent–divergent nozzle is said to be choked when:

1. Critical pressure is attained at the throat.
2. Velocity at throat becomes sonic.
3. Exit velocity becomes supersonic.

Of these statements

1. I, II, and III are correct
2. I and II are correct
3. II and III correct
4. I and III are correct
39. In flow through a convergent nozzle, the ratio of back pressure to the inlet pressure is given by the relation

If the back pressure is lower than pB given in the above equation, then:

1. The flow in the nozzle is supersonic
2. A shock wave exists inside the nozzle
3. The gases expand outside the nozzle and a shock wave appears outside the nozzle
4. A shock wave appears at the nozzle exit
40. Consider the following statements:

Across the normal shock, the fluid properties change in such a manner that the:

1. Velocity of flow is subsonic
2. Pressure increases
3. Specific volume decreases
4. Temperature decreases

Of these statements:

1. II, III, and IV are correct
2. I, II, and IV are correct
3. I, III, and IV are correct
4. I, II, and III are correct
41. Match List I (Property ratios as the critical and stagnation conditions) with List II (values of ratios) and select the correct answer using the codes given below the lists:

Codes:

A B C D

1.   2  1  4  4
2.   1  2  3  4
3.   2  1  3  4
4.   1  2  4  3
42. For oblique shock, the downstream Mach number
1. is always more than unity
2. is always less than unity
3. may be less or more than unity
4. can never be unity
43. Fanno line flow is a flow in a constant area duct
1. with friction and heat transfer but in the absence of work
2. with friction and heat transfer and accompanied by work
3. with friction but in the absence of heat transfer and work
4. without friction but accompanied by heat transfer and work
44. Reyleigh line flow is a flow in a constant area duct
1. with friction but without heat transfer
2. without friction but with heat transfer
3. with both friction and heat transfer
4. without either friction or heat transfer
45. The plot of the pressure ratio along the length of the convergent–divergent nozzle is shown in Fig. 6.40. The sequence of the flow conditions labelled (1), (2), (3), and (4) in the figure is, respectively

Figure 6.40

1. Supersonic, sonic, subsonic, and supersonic
2. Sonic, supersonic, subsonic, and supersonic
3. Subsonic, supersonic, sonic, and subsonic
4. Subsonic, sonic, supersonic, and subsonic
46. Acoustic velocity in an elastic gaseous medium is proportional to
1. absolute temperature
2. stagnation temperature
3. square root of absolute temperature
4. square root of stagnation temperature
47. Which of the following statement(s) is/are relevant to critical flow through a steam nozzle?
1. Flow rate through the nozzle is minimum.
2. Flow rate through the nozzle is maximum.
3. Velocity at the throat is supersonic.
4. Velocity at the throat is sonic.

Select the correct answer using the codes given below:

Codes:

1. I alone
2. I and III
3. II and IV
4. IV alone
48. Which of the following statements is/are true in case of one-dimensional flow of perfect gas through a converging–diverging nozzle?
1. The exit velocity is always supersonic.
2. The exit velocity can be subsonic or supersonic.
3. If the flow is isentropic, then the exit velocity must be supersonic.
4. If the exit velocity is supersonic, then the flow must be isentropic.

Select the correct answer from the codes given below:

Codes:

1. II and IV
2. II, III, and IV
3. I, III, and IV
4. II alone
49. In a normal shock, in a gas
1. the stagnation pressure remains the same on both sides of the shock
2. the stagnation density remains the same on both sides of the shock
3. the stagnation temperature remains the same on both sides of the shock
4. the Mach number remains the same on both sides of the shock
50. A normal shock
1. causes a disruption and reversal of flow pattern
2. may occur only in a diverging passage
3. is more severe than an oblique shock
4. moves with a velocity equal to the sonic velocity
51. In a normal shock, in a gas
1. both the downstream flow and the upstream flow are supersonic
2. only the upstream flow is supersonic
3. the downstream flow is sonic
4. the upstream flow is subsonic
52. Which of the following parameters decrease across a normal shock wave?
1. Mach number
2. Static pressure
3. Stagnation pressure
4. Static temperature

Select the correct answer using the code given below:

1. Only I and III
2. Only II and IV
3. I, II, and III
4. II, III, and IV
53. When are shock waves formed in air compressors?
1. Mach number < 0.9
2. Mach number > 0.9
3. Mach number = 2
4. Mach number changes suddenly from one value to another
54. It is recommended that the diffuser angle should be kept less than 18° because
1. pressure decrease in flow direction and flow separation may occur
2. pressure decreases in flow direction and flow may become turbulent
3. pressure increases in flow direction and flow separation may occur
4. pressure increases in flow direction and flow may become turbulent
55. A converging diverging nozzle is connected to a gas pipeline. At the inlet of the nozzle (converging section) the Mach number is 2. It is observed that there is a shock in the diverging section. What is the value of the Mach number at the throat?
1. < 1
2. Equal to 1
3. > 1
4. ≥ 1
56. A nozzle is discharging steam through critical pressure ratio. When the back pressure is further decreased, the nozzle flow rate will
1. decrease
2. increase
3. remain unaltered
4. first increase to a maximum and then will decrease
57. In isentropic flow between two points
1. the stagnation pressure decreases in the direction of flow
2. the stagnation temperature and stagnation pressure decrease with increase in the velocity
3. the stagnation temperature and stagnation pressure may vary
4. the stagnation temperature and stagnation pressure remain constant
58. Steam flows at the rate of 10 kg/s through a supersonic nozzle. Throat diameter is 50 mm. Density ratio and velocity ratio with reference to throat and exit are respectively 2.45 and 0.8. What is the diameter at the exit?
1. 122.5 mm
2. 58 mm
3. 70 mm
4. 62.5 mm
59. Which of the following is caused by the occurrence of a normal shock in the diverging section of a convergent-divergent nozzle?
1. Velocity jump
2. Pressure jump
3. Velocity drop
4. Pressure drop

Select the correct answer using the codes given below:

1. I only
2. I and II
3. II and III
4. I and IV
60. Total enthalpy of stream at the inlet of a nozzle is 2800 kJ while static enthalpy at the exit is 2555 kJ. What is the steam velocity at the exit if expansion is isentropic?
1. 70 m/s
2. 245 m/s
3. 450 m/s
4. 700 m/s
61. What is/are the effect(s) of supersaturation in nozzle flow?
1. It increases the mass flow
2. It increases friction in the nozzle
3. It increases exit velocity
4. It reduces dryness fraction of steam.

Select the correct answer using the code given below:

1. I only
2. I and III
3. I, II, and IV
4. III only

#### Explanatory Notes

1. 1. (b) Velocity head = = h = 10 cm
2. 8. (c) Pressure loss = poipoe = 186 – 180 = 6 kPa
3. 9. (b) Velocity of steam at outlet of nozzle = = 894.4 ≃ 900 m/s
1. 11. (a) Critical pressure ratio for superheated steam = 0.5457

p2 = 10 × 0.5457 = 5.457 bar

Exit pressure of 3 bar is less than p2. With further decrease of pressure from 3 bar to 1 bar, discharge rate will remain the same.

2. 12. (a)

= 1 + 0.5 (1.4 − 1) × 22 = 1.8

Tr = 27 × 1.8 = 48.6°C

3. 21. (a) Critical pressure ratio =
4. 24. (d)
5. 28. (b) For nitrogen,

Velocity of sound in nitrogen,

6. 36. (d) Flight Mach number of Aircraft = = 2
7. 58. (c) At ct ρt = Ae ce ρe

= 0.8 × 2.45 × 502 = 4900

de = 70 mm

8. 60. (d) ce = = 700 m/s

#### Review Questions

1. Define a nozzle.
2. Differentiate between a nozzle and a diffuser.
3. What is a throat of nozzle?
4. What is the value of n in pvn = constant for saturated steam?
5. Write the formula for critical pressure ratio in terms of index of expansion.
6. Define Mach number.
7. What are the reasons for reduction in velocity of steam in the nozzle?
8. What are the effects of friction losses in the nozzle?
9. Define nozzle efficiency.
10. What is metastable steam flow?
11. What is Wilson line?
12. Write the conditions for a nozzle to be subsonic and supersonic.
13. Define stagnation enthalpy and stagnation pressure.
14. Define back pressure.
15. What is a normal shock?
16. Explain Fanno line and Rayleigh line.

#### Exercises

6.1 Steam is expanded in a set of nozzles from 10 bar and 200°C to 5 bar. What type of nozzle is it? Neglecting the initial velocity find minimum area of the nozzle required to allow a flow of 3 kg/s under the given conditions. Assume isentropic expansion of steam.

[Ans. Convergent, 21 cm2]

6.2 Steam expands in a nozzle from 4 bar and 200°C to 1 bar. If the velocity of steam at entry is 60 m/s and nozzle efficiency is 92%, determine the exit velocity.

[Ans. 687.7 m/s]

6.3 Dry saturated steam enters a convergent–divergent nozzle at 11 bar and exit at a pressure of 2 bar. The flow is frictionless and isentropic, following the law pv1.135 = constant. Calculate (a) the exit velocity of steam and (b) the ratio of cross-section areas at exit and throat.

[Ans. 774.6 m/s, 1.62]

6.4 A convergent–divergent nozzle is to be designed in which steam at 14 bar and 275°C is to be expanded to 1.05 bar. Calculate the necessary throat and exit diameter of the nozzle for steam discharge of 500 kg/h. Assume that 12% of the total heat drop is lost due to friction only in the divergent part of the nozzle.

[Ans. 9.9 mm, 17.2 mm]

6.5 Steam at 15 bar, 350°C enters a nozzle of the convergent–divergent type and leaves at 1 bar. The throat diameter is 6 mm and length of the diverging part is 80 mm. Determine the cone angle of the divergent part. Assume that 12 per cent of the total available enthalpy drop is lost in friction in the divergent part only.

[Ans. 4° 22′]

6.6 A convergent–divergent nozzle is required to discharge 2 kg/s of stream. The nozzle is supplied with steam at 6.9 bar, 180°C and discharge takes place against a back pressure of 0.98 bar. Expansion up to throat is isentropic and the frictional resistance between the throat and the exit is equivalent to 62.76 kJ/kg of steam. Taking approach velocity of 75 m/s and throat pressure 3.9 bar, estimate: (a) suitable areas for the throat and exit and (b) overall efficiency of the nozzle based on the enthalpy drop between the actual inlet pressure, temperature and the exit pressure.

[Ans. 17.66 cm2, 41.67 cm2, 82.25%]

6.7 The nozzles in the stage of an impulse turbine receive steam at 17.5 bar, 300°C and pressure in the wheel chamber is 10.5 bar. If there are 16 nozzles, then find the cross-sectional area at the exit of each nozzle for a total discharge to be 280 kg/min. Assume nozzle efficiency of 90%.

[Ans. 1.36 cm2]

6.8 Steam at a pressure of 10 bar and 95% dry is supplied through a convergent–divergent nozzle to the wheel chamber where the pressure is maintained at 0.12 bar. The mass flow rate through the nozzle is 8.16 kg/kWh and the work developed by the wheel is 110 kW. Determine the following:

1. Pressure of the throat for maximum discharge
2. Number of nozzles required if throat diameter of each nozzle is 5 mm
3. The exit diameter of nozzle if 10% of overall enthalpy drop overcomes friction in the divergent portion

[Ans. 5.93 bar, 9, 18.1 mm]

6.9 Five kilogram of steam per minute passes through a convergent–divergent nozzle. The pressure and temperature of steam supplied to the nozzle is 10 bar and 200°C, respectively. The discharge pressure is 0.1 bar. The expansion is supersaturated up to throat and in thermal equilibrium afterwards. Calculate (a) the area of nozzle at exit, (b) the maximum degree of super-saturation, and (c) the degree of under cooling at the throat. Assume pv1.3 = constant for super-saturated flow.

[Ans. 8.4 cm2, 1.65, 18°C]

6.10 Supersaturated expansion occurs in a nozzle supplied with steam at 20 bar and 325°C. Exit pressure is 3.6 bar. For a flow rate of 7.5 kg/s, determine (a) the throat and exit areas and (b) the degree of undercooling at the exit.

[Ans. 28.9 cm2, 42.8 cm2, 10.9°C]

6.11 A converging nozzle has an exit area of 0.001 m2. Air enters the nozzle with negligible velocity at a pressure of 1.0 MPa and a temperature of 360 K. For isentropic flow of an ideal gas with k = 1.4, determine the mass flow rate and the exit Mach number for back pressure of (a) 500 kPa and (b) 784 kPa.

[Ans. 2.13 kg/s, 1.79 kg/s]

6.12 Air flows through a convergent–divergent nozzle. At some section in the nozzle the pressure is 2 bar, the velocity is 171.1 m/s and the temperature is 1200 K. The area of flow at this section is 1000 cm2. Assuming isentropic flow conditions, determine the following:

1. Stagnation temperature and stagnation pressure.
2. Sonic velocity and Mach number at this section.
3. Velocity, Mach number, and area of flow at the exit section if the pressure at that section is 1.013 bar.
4. Pressure, temperature, velocity, and area of flow at the throat of the nozzle.

Assume cp = 1.005 kJ/kg.K and cv = 0.718 kJ/kg.K

[ESE, 1983]

6.13 A set of steam nozzles in an impulse turbine stage is supplied with steam at 20 bar and 230°C. The mass flow of steam is 60 kg/s. The steam is expanded from rest to a back pressure of 14 bar with an efficiency of 98%. The mean diameter of nozzle disc is 800 mm and the nozzle angle is 20 degrees. Assuming that steam is admitted all around the periphery of the nozzle disc, determine:

1. whether a convergent or a convergent–divergent nozzle is to be used,
2. the velocity of flow of steam at the exit of the nozzle,
3. the area of flow normal to the axis of the stage, and
4. the height of the nozzles.

Assume adiabatic index for superheated steam to be 1.3. What would be the mass flow through the set of nozzles when the back pressure is reduced to 9 bar keeping the upstream conditions same as before?

[ESE, 1984]

6.14 Air is isentropically expanded in a convergent–divergent nozzle from an initial pressure of 5 bar and 25°C to a back pressure of 1.5 bar. The velocity of the air entering the nozzle is 100 m/s. The mass flow rate of the air is 2 kg/s. Determine (a) Mach number at inlet to the nozzle, (b) pressure at the throat, (c) area of flow at the throat, and (d) the area of flow at the exit of the nozzle. Assume for air the ratio of specific heats to be 1.4 and R to be 0.287 kJ/kg.K.

[ESE, 1986]

6.15 Air enters a constant-area combustion chamber at a Mach number of 0.25 with a total pressure of 5.5 bar. Combustion gas leaves the chamber with a Mach number of 0.35. Neglecting friction and the mass of the fuel, determine the drop in total pressure of air in the combustion chamber. Assume cp/cv = 1.35.

[ESE, 1995]

1. b
2. c
3. c
4. c
5. d
6. c
7. d
8. c
9. b
10. d
11. a
12. a
13. d
14. b
15. b
16. b
17. d
18. c
19. b
20. c
21. a
22. d
23. d
24. d
25. a
26. c
27. d
28. b
29. b
30. b
31. c
32. a
33. a
34. d
35. a
36. d
37. a
38. b
39. c
40. d
41. a
42. c
43. c
44. b
45. d
46. c
47. c
48. a
49. c
50. c
51. b
52. a
53. b
54. c
55. b
56. c
57. a
58. c
59. d
60. d
61. b