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.

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 p_E and the back pressure (the pressure outside the nozzle exit) p_B. As the back pressure p_B is decreased, the variation of the mass flow rate and the exit plane pressure p_E/p₀ are plotted in Fig. 6.15.

Where p_B/p₀ = 1, there is, of course, no flow, and p_E/p₀ = 1 as designated by point ‘a‘. Next, let the back pressure p_B be lowered to point b, so that p_B/p₀ is greater than the critical pressure ratio. The mass flow rate has a certain value and p_E = p_B. The exit Mach number is less than one. Next, let p_B be lowered to the critical pressure at point c. The Mach number at the exit is now unity and p_E = p_B. When p_B is decreased below the critical pressure, designated by point d, there is no further increase in the mass rate of flow, and p_E remains constant at a value equal to the critical pressure, and the exit Mach number is unity. The drop in pressure from p_E to p_B 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 p_B = p₀ and there is no flow. When p_B is decreased to the pressure indicated by point b, so that p_B/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 p_E remains constant, and the drop in pressure from p_E to p_B 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 = c_s =  = 347.2 m/s

Discharge at the exit section,

 

ṁ = 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

 

T_E = 360 × 0.9381 = 337.7 K

Sonic velocity at exit,

Velocity at exit,

Density of air at exit,

ṁ = ρ_EA_Ec_E = 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 haveFor  = 2, M*_E = 2.197,  = 0.0939, =  = 0.5089Therefore, p_E = 0.0939 × 1000 = 93.9 kPaT_E = 0.5089 × 360 = 183.2 Kc_E = M_E* c_sE = 2.197 × 271.3 = 596.1 m/sCritical pressure at throat, p* = p₀ × 0.528 = 1000 × 0.528 = 528 kPaCritical temperature, T * = 0.8333 × 360 = 300 KAt throat, 
2. From Table 6.1, we haveFor  = 2, M_E = 0.308,  = 0.936,  = 0.9812Therefore, p_E = 0.936 × 1000 = 936 kPaT_E = 0.9812 × 360 = 353.3 Kc_E = MC_sE = 0.308 × 376.8 = 116 m/sSince M = 1 at the throat, mass rate of flow is the same as in (a).