SUPERSATURATED OR METASTABLE FLOW THROUGH A NOZZLE

The isentropic expansion of superheated steam from supply pressure p₁ to back pressure p_b 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, pv^1.3 = 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, η_n = 0.96

The h – s 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, s_f 2 = 1.3025 kJ/kg.K and s_fg2 = 6.0568 kJ/kg.K

Since s₁ > s_fg2, 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 = h₁ – h₂ = 2868 – 2691.95 = 176.05 kJ/kg

Actual enthalpy drop, (Δh)_actual = η_n (h₁ − h₂) = 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 (pv^1.3 = 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, η_n = 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, v_s3′ = 0.4 m³/kg, and x₃′ = 0.995

Figure 6.7 Superheated steam flow through a nozzle

v₃ = x_3′ v_s3′ = 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/kgAt p₂ = 1.5 bar   s₁ = s₂ = s_f 2 + x₂ s_fg26.9918 = 1.4335 + x₂ × 5.7897   x₂ = 0.96   h₂ = h_f2 + x₂h_fg2 = 467.08 + 0.96 × 2226.5 = 2604.52 kJ/kg   v₂ = v_f2 + x₂ (v_g2 − v_f2) = 0.001053 + 0.96 × (1.159 − 0.001053) = 1.113 m³/kg
2. For supersaturated flow, pv^1.3 = const. 
3. Saturation temperature at 1.5 bar = 111.37°CDegree 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 = v_f1 + x₁ (v_g1 − v_f1) = 0.001127 + 0.96 × (0.19444 − 0.001127) = 0.1867 m³/kgFor maximum discharge, Velocity of steam at throat, c₂ =  Figure 6.8 Convergent–divergent nozzle
2. For isentropic flow,s₁ = s_f1 + x₁s_fg1 = 2.1386 + 0.96 × 4.4478 = 6.4085 kJ/kgs₁ = s₂s_f1 + x₁s_fg1 = s_f2 + x₂s_fg2At p₂ = 5.774 bar6.4085 = 1.9157 + x₂ × 4.857   x₂ = 0.925   h₂ = h_f2 + x₂h_fg2 = 663.92 + 0.925 × 2091.14 = 2598.22 kJ/kg   h₁ = h_f1 + x₁h_fg1 = 762.79 + 0.96 × 2015.3 = 2697.5 kJ/kgAt p₃ = 0.8 bar, s_f3 = 1.233 kJ/kg.K, s_fg3 = 6.202 kJ/kg.K   s₂ = s₃   s_f2 + x₂ s_fg2 = s_f3 + x₃ s_fg3   1.9157 + 0.925 × 4.857 = 1.233 + x₃ × 6.202   x₃ = 0.833   h₃ = h_f3 + x₃h_fg3 = 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 h − s (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, x_3′ = 0.955, v_s3′ = 1.5 m³/kg

h₁ − h₃ = 2940−2530 = 410 kJ/kg

h₄ − h₃ = 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 = s_f3 + x₃s_fg3

 6.8212 = 1.4335 + x₃ × 5.7997

    x₃ = 0.93

    h₃ = h_f 3 + x₃h_fg3 = 467.08 + 0.93 × 2226.5 = 2537.7 kJ/kg

    (h₁ − h₃)_isen = 2748.7 − 2537.7 = 211 kJ/kg

    h₁ − h_3′ = h₁ − h₄ = 0.85 × 211 = 179.35 kJ/kg

From Mollier diagram (Fig. 6.10), x₂ = 0.965, x_3′ = 0.945, h₂ = 2662 kJ/kg

   v_s2 = 0.7 m³/kg, v_s3′ = 1.1 m³/kg, h_3′ = 2585 kJ/kg

   v₂ = x₂v_s2 = 0.965 × 0.7 = 0.676 m³/kg

   v_3′ = x_3′v_s3′ = 0.945 × 1.1 = 1.04 m³/kg

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