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 h_ox = h_oy

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 h − s 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 h − s diagram is known as the Rayleigh line. Both these lines are shown on the h − s 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 s_y − s_x ≥ 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,

 

ρ_xc_x = ρ_yc_y

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 M_y as a function of M_x 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 nozzlep_y = 4.46 × 93.9 = 512.7 kPaT_y = 1.854 × 183.2 = 339.7 Kp_oy = 0.630 × 1000 = 630 kPa
2. From Table 6.1 at point x, as the flow is isentropic to point x, Therefore, p_x = 0.2724 × 1000 = 272.4 kPaTx = 0.6897 × 360 = 248.3 KThe properties at y can be determined from the normal shock functions (Table 6.3) asp_y = 2.4583 × 272.4 = 669.6 kPaT_y = 1.320 × 248.3 = 327.8 Kp_oy = 0.9298 × 1000 = 929.8 kPa

T_ox = T_oy = 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,

p_E = 200 kPa, s_E = s₀ = 7.4089 kJ/kg.K

T_E = 178.5°C

h_E = 2826.7 kJ/kg

v_E = 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 = s_f3 + x₃s_fg3

6.7427 = 1.7766 + x₃ × 5.1193

x₃ = 0.97

h₃ = h_f3 + x₃h_fg3

   = 604.73 + 0.97 × 2133.8

   = 2674.5 kJ/kg

v₃ = v_f3 + x₃(v_g3 − v_f3) = 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 Let breadth = bThen length, l = 2.5 bArea of 6 nozzles = 6 × 2.5b × b = 15b²15b² = 2.279 × 10^–3b = 0.0123 m or 12.3 mml = 0.0308 m or 30.8 mm
2. At 4 bar, t_s = 143.65°CDegree of undercooling = 143.65 − 118.34 = 24.31°Cp_s corresponding to 118.3°C = 1.9 barDegree of super saturation =  = 2.1
3. h₁ = h₃ = 3115.3 − 2674.5 = 440.8 kJ/kgLoss of available heat drop = (h₁ − h₃) − (h₁ − h₂)(Δh)_loss = 440.8 − 437.5 = 3.3 kJ/kg
4. Increase in entropy 
5.  

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 = h_g1 = 2748.7 kJ/kg

and for p₂ = 1 bar, we find that,

h_f 2 = 417.44 kJ/kg, h_fg2 = 2258.0 kJ/kg

 

∴ h₂ = h_f2 + x₂ h_fg2

= 417.44 + 0.94 × 2258.0 = 2539.96 kJ/kg

Enthalpy drop, Δh = h₁ − h₂

= 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 = h₁ − h₂

= 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, η_n = 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, ṁ = 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 = h₁ − h₂ = 2624 − 2534 = 90 kJ/kg

Dryness fraction, x₂ = 0.92

From steam tables, v_f2 = 0.001061 m³/kg, v_g2 = 0.8857 m³/kg at p₂ = 2 bar

Specific volume at exit

v₂ = v_f2 + x₂ (v_g2 − v_f2) = 0.001061 + 0.92 × (0.8857 − 0.001061) = 0.8149 m³/kg

Now,  = h₁ − h₂

Steam mass flow rate (ṁ) 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, ṁ = 4.5 kg/s, p₃ = 1.6 bar, n = 1.135, η_n = 1 − 0.12 = 0.88

From Mollier diagram shown in Fig. 6.22,

h₁ = 2750 kJ/kg, h₃ = 2520 kJ/kg

Refer to h − s 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:

 

h₁ − h₂ = 2750 − 2650 = 100 kJ/kg

A₂ = Area of throat

v₂ = 0.964 m³/kg

From Mollier diagram, we have

 

v_g2 = 0.5321 m3/kg corresponding to p₂ = 3.468 bar

v₂ = x₂ v_g2 = 0.964 × 0.5321 = 0.513 m³/kg

For Section 2–3:

 

(h₁ − h₃) isentropic = 230 kJ/kg

h₁ − h_3‘ = (h₁ − h₃) × η_n = 230 × 0.88 = 202.4 kJ/kg

From Mollier diagram,

x_3‘ = 0.9365

v_g3 = 1.0911 m³/kg corresponding to pressure of 1.6 bar.

v₃ = x_3‘ v_g3 = 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

Using steady flow energy equation

Since w = 0

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

At 5 bar; h_f2 = 640.21 kJ/kg, h_fg2 = 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

v_g2 = 0.29 m³/kg

Δh₂ = h₁ − h₂ = 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, v_g2 = 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 + 0.1x₁n = 1.035 + 0.1 × 0.98 = 1.133Now p₁ = 10 bar,   p₂ = 10 × 0.5778 = 5.778 bar
2. From Mollier diagram, we haveh₁ = 2762 kJ/kg, h₃ = 2075 kJ/kg, h₂ = 2660 kJ/kgLoss in heat drop due to friction = 0.1 (h₁ − h₃) = 0.1 (2762 − 2075) = 68.7 kJ/kgh_3‘ − h₃ = 68.7h_3‘ = h₃ + 68.7 = 2075 + 68.7 = 2143.7 kJ/kg Figure 6.25 Mollier diagramMass flow rate, Velocity at throat, Total area of throat, x₂ = 0.953v_g2 = 0.34 m³/kg at 5.778 bar (from steam tables)Throat area per nozzle = 5 × 10 × 10⁻⁶ = 0.5 × 10⁻⁴ m²Number of nozzles =  = 7.97 = 8 (say)Exit area/nozzle =  = 7.476 × 10⁻⁴ m²Keeping the same aspect ratio of 1:2 for rectangle and let x be the smaller side2x × x = 7.476 × 10⁻⁴2x² = 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

T_2′ = 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

= T₂ − T_2′ = 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

= T₂ − T_2‘ = 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.

∴ p_2‘ = 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 = (h₁ − h₂) = 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 ṁ₁ = Mass of steam dischargedFrom Mollier diagram, h₁ = 2770 kJ/kg, h₂ = 2515 kJ/kg, and x₂ = 0.91From steam tables,Figure 6.27 Mollier diagramv_g2 = 0.885 m³/kgΔh₂ = h₁ − h₂ = 2770 − 2515 = 255 kJ/kgVelocity of steam at exit,
2. Mass of steam discharged when it is in metastable state.Let ṁ₂ = mass of steam dischargedVolume of steam at inlet,Volume of steam at exit,Furthermore, Heat drop from inlet to exit Δh₂ = h₁ − h₂ = 2770 − 2530 = 240 kJ/kgVelocity 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 barDrop 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/kgAt point 2, p₂ = 2.887 bar, h₂ = 2645 kJ/kg, and v₂ = 0.63 m³/kgAt point 3, p₃ = 0.2 bar, h₃ = 2248 kJ/kg, and v₃ = 6.6 m³/kgFigure 6.28 Mollier diagramHence, 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/kgCorresponding to h₀ = 2750 kJ/kgp₀ = 5.1 bar as shown on the chartThus, 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 chartFigure 6.29 Mollier diagramArea at throat is given by,Similarly, area at the exit from the nozzle is given by,The nozzle efficiency η_n = 0.9 and n = 1.135 for dry saturated steam.Therefore, m =  = 1.119The 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 r_p =  we getTherefore, 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/kgFigure 6.30 Mollier diagrami.e., h₁ − h’₃ = 447.8 kJ/kgand v₃′ = 6.7 m³/kgTherefore, 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 ṁ= 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 exitLet A₂ = area at the throatA₃ = area at the exitThe expansion of steam through the nozzle on the Mollier diagram is shown in Fig. 6.31. From the Mollier diagram, we find thatFigure 6.31 Mollier diagramh₁ = 2810 kJ/kg, h₂ = 2680 kJ/kg, h₃ = 2470 kJ/kg, x₂ = 0.97, and x₃ = 0.934From steam tables, we also find that the specific volume of steam at throat corresponding to 4 bar,v_f2 = 0.001084 m³/kg, v_g2 = 0.4625 m³/kgv₂ = v_f2 + x₂(v_g2 − v_f2) = 0.001084 + 0.97(0.4625 − 0.001084) = 0.449 m³/kgand specific volume of steam corresponding to 1 bar,v_f3 = 0.001043 m³/kg, v_g3 = 1.694 m³/kgv₃ = v_f3 + x₃(v_g3 − v_f3) = 0.001043 + 0.934(1.694 − 0.001043) =1.582 m³/kgHeat drop between entrance and throat,Δh₂ = h₁ − h₂ = 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,h₃ − h’₃ = 63 or h₃ = h’₃ + 63 = 2470 + 63 = 2533 kJ/kgand heat drop between entrance and exit,Δh₃ = h₁ − h₃ = 2810 − 2533 = 277 kJ/kg∴ Velocity of steam at exit,
2. Overall efficiency of the nozzleOverall 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.

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 h − s 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 h − s 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, v_s2 = 0.125 m³/kg

c₂ = [2(h₁ − h₂) × 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,