Flow through a nozzle

Objectives:

1. To study the pressure distribution through a nozzle at different ratios across it.
   
2. To learn, study and understand how a nozzle works.

Apparatus:
    Air is admitted to a cast iron pressure chest by way of adjustable values. A nozzle of highly finished brass is screwed into seating in the base of the chest and the air expands through the nozzle.

To enable the pressure distribution through the nozzle to be plotted, a search tube or probe of stainless steel may be traversed along the axis of the nozzle. A small cross-hole in the search tube connects with a high-grade pressure gauge, which registers the pressure at any point in the nozzle. The search tube is traversed by rotating a calibrated dial and pressures are usually recorded at intervals of 2.5mm. A pointer moves with the search tube past a replica of the nozzle profile in order to indicate the point in the nozzle at which the pressure is being measured. The nozzle discharges into a vertical pipe of large bore fitted with the throttling valve for controlling the downstream pressure. Other instruments include a second pressure gauge for recording the pressure in the chest and a thermometer for indicating the temperature of the air.

Procedure:

1. Open the back pressure valve (inlet valve) whilst keeping the probe in position no. 1.
   
2. Set the inlet pressure to 300KPa and check that the inlet pressure and the chest pressure are to be equal.
   
3. Chest pressure is to be observed throughout the experiment and re-adjusted to initial setting when necessary.
   
4. We start recording the probe pressure at position no. 1 and then we go throughout all the other positions till reaching final position (38) measuring the pressure at each position.
   
5. Finally we repeat our procedures different back pressure values (400KPa and 600 KPa)
   

Sample of calculations:

1. Pressure ratio when P₀ = 300KPa 220/300 = 0.733
   
2. Vt when P₀ = 400KPa = (2*1.4*287*(18+273))/1.4-1=584619 584619*(1-(0.75)^0.4/1.4)= 46130.8584 46130.8584^1/2= 214.781
   
3. Mt when P₀ = 500KPa= 9.16(500)(0.76)^(1/1.4)*[2*1.4/(1.4-1)(287)(291)*(1-(.76)^(0.4/1.4))]^1/2= 9.47
   

Discussions & Conclusions:

1. The pressure decreases with the nozzle’s diameter decreasing
   
2. A nozzle increases velocity
   
3. Velocity keeps increasing till it chocks, if reaches velocity of sound (A) it cant increase anymore
   
4. If VA which is the nozzle V increases. If VA then V decreases
   
5. The negative value declares that the probe is out of the nozzle 
   
6. The relation between the pressure ratio and the mass flow is direct
   
7. The relation between the pressure ratio and the velocity of the throat is inverse.