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Nozzles

1. 1. Saif al-din ali Madi Department of Mechanical Engineering/ College of Engineering/ University of Baghdad 26/11/2018 1 | P a g e [power plant Laboratory II] University of Baghdad Name: - Saif Al-din Ali -B-
2. 2. Saif al-din ali Madi Department of Mechanical Engineering/ College of Engineering/ University of Baghdad 26/11/2018 2 | P a g e TABLE OF CONTENTS Objective............................................................................I INTRODUCTION.............................................................II THEORY.........................................................................III Calculations and results..................................................V DISCUSSION..................................................................VI
3. 3. Saif al-din ali Madi Department of Mechanical Engineering/ College of Engineering/ University of Baghdad 26/11/2018 3 | P a g e Experiment Name:- 1. Objective 2. Introduction Two types of nozzles: - 1. Converging-Diverging or " De Laval" Nozzles have been widely used over the last few decades in many engineering contexts, from civil and mechanical to aerospace uses. They are designed to accelerate fluids to supersonic speeds at the nozzle exit. Nearly all rockets make use of this fact to create an effective propulsion system to reach high velocities. Whilst their operation is of a simple appearance, the combination of flows reached, be that subsonic and supersonic and the subsequent change of properties such as pressure, density and temperature make the underlying investigation of their performance more complicated than first expected. Their operation relies on the ratio between the inlet stagnation pressure P0 and outlet back pressure Pb. As this ratio Pb/Po is brought down from unity, the mass flow rate increases till a maximum value is achieved where the Mach number in the throat (see Figure 1) becomes sonic. ( Mach no = 1). This is referred to as "Choked flow". As the ratio is further decreased the flow becomes supersonic in the diverging nozzle, till the design pressure ratio is achieved (0.53 for air) with supersonic flow occurring at the nozzle exit. After this the flow becomes more complicated and normal and oblique shock waves begin to occur inside and outside of the nozzle respectively. The Purpose of this report is to gain an understanding of the nature of this flow by investigating the pressure ratios effects on the mass flow rate of air through the system and the differing pressure distributions that occur at varying lengths into the nozzle. Figure 1. Schematic of a converging-diverging nozzle 2. converging nozzle
4. 4. Saif al-din ali Madi Department of Mechanical Engineering/ College of Engineering/ University of Baghdad 26/11/2018 4 | P a g e Converging nozzles, as shown in Figure 2, are tubes with an area that decreases from the nozzle entry to the exit (or throat) of the nozzle. As the nozzle area decreases, the flow velocity increases, with the maximum flow velocity occurring at the nozzle throat. As the inlet flow velocity is increased, flow velocity at the nozzle throat keeps increasing until it reaches Mach 1. At this point, the flow at the throat gets choked, meaning that any further increase of inlet flow velocity will not increase the flow velocity at the throat. It is for this reason that converging nozzles are used to accelerate fluids in the subsonic flow regime alone and can commonly be found on all commercial jets (except for the Concord) as they travel at subsonic speeds. Figure 2. Schematic of a converging nozzle 3. divergent nozzle represents a divergent nozzle for a fluid whose peculiarity is that — decreases with the drop of pressure, i.e., specific volume increases at a faster rate than velocity with the drop of pressure. The area of cross-section should increase as the pressure decreases. Figure 3. diverging nozzle the flow conditions that can be observed in a converging nozzle: 1. No flow condition, where the back-pressure is equal to the total pressure. 2. Subsonic flow, where the flow accelerates as area decreases, and the pressure drops. 3. Subsonic flow, where there is significantly higher acceleration and the pressure drops. 4. Choked flow, where any pressure drop does not accelerate the flow. 5. Choked flow, where the flow expands after the nozzle exit (considered non- isentropic).
5. 5. Saif al-din ali Madi Department of Mechanical Engineering/ College of Engineering/ University of Baghdad 26/11/2018 5 | P a g e Figure 4. Flow conditions and regimes in a converging nozzle (theoretical predictions) converging-diverging nozzle shows the following seven profiles in the position versus pressure ratio plot. Note that the first vertical dashed line on the left of the p/pO versus distance along the nozzle plot is the location of the throat, the second vertical dashed line is the location of the nozzle exit, and the horizontal dashed line marks the choked condition. 1. Subsonic flow that never reaches choked condition. 2. Subsonic flow that reaches choked condition but does not attain supersonic velocities (considered isentropic). 3. Subsonic flow that reaches choked condition, with the resulting supersonic flow forming a normal shock, which then experiences subsonic deceleration. Here, the normal shock causes a sudden drop in velocity and an increase in back-pressure, as indicated by the sudden increase in p/pO. 4. Subsonic flow that reaches choked condition, with the resulting supersonic flow forming a normal shock after the nozzle (considered isentropic in the nozzle). 5. Over-expanded flow – the pressure at the nozzle exit is lower than the ambient pressure, causing the jet exiting the nozzle to be highly unstable with huge variations in pressure and velocity as it travels downstream. 6. Flow after the choked condition is supersonic through the nozzle, and no shock is formed. 7. Under-expanded flow – the pressure at the nozzle exit is higher than the ambient pressure and results in similar effects as over-expanded flow. Figure 5. Flow conditions and regimes in a converging-diverging nozzle (theoretical predictions) 3. Theory
6. 6. Saif al-din ali Madi Department of Mechanical Engineering/ College of Engineering/ University of Baghdad 26/11/2018 6 | P a g e The characteristic of a nozzle can be graphically represented by diagram which shows the mass flow rate through the nozzle related to the downstream pressure outlet, with fluid conditions maintained constant upstream of the nozzle. In subsonic nozzle conditions the flow rate can be expressed by: 𝑸𝒕𝒉 = 𝑨𝒖 ∗ 𝒑 𝟎 √ 𝒑 𝟎 ∗ 𝒗 ∗ √ 𝟐𝒌 𝒌 − 𝟏 ∗ [( 𝒑 𝐛 𝒑 𝟎 ) 𝟐 𝒌 − ( 𝒑 𝐛 𝒑 𝟎 ) 𝒌+𝟏 𝒌 ] 𝑨 𝒖 = 𝝅 𝟒 𝑫 𝒕𝒉 𝟐 Where: Po = Total pressures upstream the nozzle, Vo=Specific volume upstream the nozzle Au= the narrow outlet section of the nozzle (throat). P2 = Pressure downstream the nozzle, K = Specific heat ratio (taken l for temperature 170 ° C) For critical pressure downstream the nozzle the mass flow rate will be max (i.e.) the nozzle is chocked can be calculated as: 𝑸 𝒎 𝒂𝒙 = 𝑨 𝒖 𝒑 𝟎 √ 𝒑 𝟎 𝝂 𝟎 ∗ √ 𝟐𝒌 𝒌 − 𝟏 𝑸 𝐞𝐱𝐩 = 𝒌√ 𝜟𝒑 𝒗 𝒐
7. 7. Saif al-din ali Madi Department of Mechanical Engineering/ College of Engineering/ University of Baghdad 26/11/2018 7 | P a g e 4. Calculations and results 𝑨 𝒖 = 𝝅 𝟒 𝑫 𝒕𝒉 𝟐 = ( 3.14 4 ) ∗ (8 ∗ 10−3)2 = 5.03 ∗ 10−5 𝑚2 1. 𝑸𝒕𝒉 = 𝑨𝒖 ∗ 𝒑 𝟎 √ 𝒑 𝟎 ∗ 𝒗 ∗ √ 𝟐𝒌 𝒌 − 𝟏 ∗ [( 𝒑 𝐛 𝒑 𝟎 ) 𝟐 𝒌 − ( 𝒑 𝐛 𝒑 𝟎 ) 𝒌+𝟏 𝒌 ] 𝑸𝒕𝒉 = 𝟓. 𝟎𝟑 ∗ 𝟏𝟎−𝟓 ∗ 𝟒 ∗ 𝟏𝟎 𝟓 √𝟒 ∗ 𝟏𝟎 𝟓 ∗ 𝟎. 𝟒𝟔𝟒𝟐𝟓𝟏 ∗ √ 𝟐 ∗ 𝟏. 𝟏𝟒 𝟏. 𝟏𝟒 − 𝟏 ∗ [( 𝟑. 𝟓 ∗ 𝟏𝟎 𝟓 𝟒 ∗ 𝟏𝟎 𝟓 ) 𝟐 𝟏.𝟏𝟒 − ( 𝟑. 𝟓 ∗ 𝟏𝟎 𝟓 𝟒 ∗ 𝟏𝟎 𝟓 ) 𝟏.𝟏𝟒+𝟏 𝟏.𝟏𝟒 ] 𝑸𝒕𝒉 = 𝟎. 𝟎𝟑𝟏𝟏𝟐𝟏𝟖 𝐦 𝟑 /𝒉 𝑸 𝐞𝐱𝐩 = 𝒌√ 𝜟𝒑 𝒗 𝒐 𝑸 𝐞𝐱𝐩 = 𝟓. 𝟔𝟏𝟓√ 𝟗𝟏.𝟐 𝟎.𝟒𝟔𝟒𝟐𝟓 = 𝟎. 𝟎𝟕𝟖𝟔𝟗𝟗 𝒎 𝟑 𝒉𝒓 𝑹 = 𝑷𝒐/𝑷𝒃 𝑹 = 𝟒/𝟑. 𝟓 = 1.142857143 2. 𝑸𝒕𝒉 = 𝑨𝒖 ∗ 𝒑 𝟎 √ 𝒑 𝟎 ∗ 𝒗 ∗ √ 𝟐𝒌 𝒌 − 𝟏 ∗ [( 𝒑 𝐛 𝒑 𝟎 ) 𝟐 𝒌 − ( 𝒑 𝐛 𝒑 𝟎 ) 𝒌+𝟏 𝒌 ] 𝑸𝒕𝒉 = 𝟓. 𝟎𝟑 ∗ 𝟏𝟎−𝟓 ∗ 𝟒 ∗ 𝟏𝟎 𝟓 √𝟒 ∗ 𝟏𝟎 𝟓 ∗ 𝟎. 𝟒𝟔𝟒𝟐𝟓𝟏 ∗ √ 𝟐 ∗ 𝟏. 𝟏𝟒 𝟏. 𝟏𝟒 − 𝟏 ∗ [( 𝟑 ∗ 𝟏𝟎 𝟓 𝟒 ∗ 𝟏𝟎 𝟓 ) 𝟐 𝟏.𝟏𝟒 − ( 𝟑 ∗ 𝟏𝟎 𝟓 𝟒 ∗ 𝟏𝟎 𝟓 ) 𝟏.𝟏𝟒+𝟏 𝟏.𝟏𝟒 ] 𝑸𝒕𝒉 = 0.03950456 𝐦 𝟑 /𝒉 𝑸 𝐞𝐱𝐩 = 𝒌√ 𝜟𝒑 𝒗 𝒐 𝑸 𝐞𝐱𝐩 = 𝟓. 𝟔𝟏𝟓√ 𝟏𝟐𝟏.𝟔 𝟎.𝟒𝟔𝟒𝟐𝟓 = 0.090874 𝒎 𝟑 /𝒉𝒓 𝑹 = 𝑷𝒐/𝑷𝒃 𝑹 = 𝟒/𝟑 = 1.333333333
8. 8. Saif al-din ali Madi Department of Mechanical Engineering/ College of Engineering/ University of Baghdad 26/11/2018 8 | P a g e 3. 𝑸𝒕𝒉 = 𝑨𝒖 ∗ 𝒑 𝟎 √ 𝒑 𝟎 ∗ 𝒗 ∗ √ 𝟐𝒌 𝒌 − 𝟏 ∗ [( 𝒑 𝐛 𝒑 𝟎 ) 𝟐 𝒌 − ( 𝒑 𝐛 𝒑 𝟎 ) 𝒌+𝟏 𝒌 ] 𝑸𝒕𝒉 = 𝟓. 𝟎𝟑 ∗ 𝟏𝟎−𝟓 ∗ 𝟒 ∗ 𝟏𝟎 𝟓 √𝟒 ∗ 𝟏𝟎 𝟓 ∗ 𝟎. 𝟒𝟔𝟒𝟐𝟓𝟏 ∗ √ 𝟐 ∗ 𝟏. 𝟏𝟒 𝟏. 𝟏𝟒 − 𝟏 ∗ [( 𝟐. 𝟓 ∗ 𝟏𝟎 𝟓 𝟒 ∗ 𝟏𝟎 𝟓 ) 𝟐 𝟏.𝟏𝟒 − ( 𝟐. 𝟓 ∗ 𝟏𝟎 𝟓 𝟒 ∗ 𝟏𝟎 𝟓 ) 𝟏.𝟏𝟒+𝟏 𝟏.𝟏𝟒 ] 𝑸𝒕𝒉 = 0.042527341 𝐦 𝟑 /𝒉 𝑸 𝐞𝐱𝐩 = 𝒌√ 𝜟𝒑 𝒗 𝒐 𝑸 𝐞𝐱𝐩 = 𝟓. 𝟔𝟏𝟓√ 𝟏𝟒𝟒.𝟒 𝟎.𝟒𝟔𝟒𝟐𝟓 = 0.099027755 𝒎 𝟑 /𝒉𝒓 𝑹 = 𝑷𝒐/𝑷𝒃 𝑹 = 𝟒/𝟐. 𝟓 = 1.6 4. 𝑸𝒕𝒉 = 𝑨𝒖 ∗ 𝒑 𝟎 √ 𝒑 𝟎∗𝒗 ∗ √ 𝟐𝒌 𝒌−𝟏 ∗ [( 𝒑 𝐛 𝒑 𝟎 ) 𝟐 𝒌 − ( 𝒑 𝐛 𝒑 𝟎 ) 𝒌+𝟏 𝒌 ] 𝑸𝒕𝒉 = 𝟓. 𝟎𝟑 ∗ 𝟏𝟎−𝟓 ∗ 𝟒 ∗ 𝟏𝟎 𝟓 √𝟒 ∗ 𝟏𝟎 𝟓 ∗ 𝟎. 𝟒𝟔𝟒𝟐𝟓𝟏 ∗ √ 𝟐 ∗ 𝟏. 𝟏𝟒 𝟏. 𝟏𝟒 − 𝟏 ∗ [( 𝟐 ∗ 𝟏𝟎 𝟓 𝟒 ∗ 𝟏𝟎 𝟓 ) 𝟐 𝟏.𝟏𝟒 − ( 𝟐 ∗ 𝟏𝟎 𝟓 𝟒 ∗ 𝟏𝟎 𝟓 ) 𝟏.𝟏𝟒+𝟏 𝟏.𝟏𝟒 ] 𝑸𝒕𝒉 = 0.041861669 𝐦 𝟑 /𝒉 𝑸 𝐞𝐱𝐩 = 𝒌√ 𝜟𝒑 𝒗 𝒐 𝑸 𝐞𝐱𝐩 = 𝟓. 𝟔𝟏𝟓√ 𝟏𝟓𝟐 𝟎.𝟒𝟔𝟒𝟐𝟓 = 0.101600333 𝒎 𝟑 /𝒉𝒓 𝑹 = 𝑷𝒐/𝑷𝒃 𝑹 = 𝟒/𝟐 = 2
9. 9. Saif al-din ali Madi Department of Mechanical Engineering/ College of Engineering/ University of Baghdad 26/11/2018 9 | P a g e 1. 𝑸𝒕𝒉 = 𝑨𝒖 ∗ 𝒑 𝟎 √ 𝒑 𝟎 ∗ 𝒗 ∗ √ 𝟐𝒌 𝒌 − 𝟏 ∗ [( 𝒑 𝐛 𝒑 𝟎 ) 𝟐 𝒌 − ( 𝒑 𝐛 𝒑 𝟎 ) 𝒌+𝟏 𝒌 ] 𝑸𝒕𝒉 = 𝟓. 𝟎𝟑 ∗ 𝟏𝟎−𝟓 ∗ 𝟓 ∗ 𝟏𝟎 𝟓 √𝟓 ∗ 𝟏𝟎 𝟓 ∗ 𝟎. 𝟑𝟕𝟐𝟗 ∗ √ 𝟐 ∗ 𝟏. 𝟏𝟒 𝟏. 𝟏𝟒 − 𝟏 ∗ [( 𝟒. 𝟓 ∗ 𝟏𝟎 𝟓 𝟓 ∗ 𝟏𝟎 𝟓 ) 𝟐 𝟏.𝟏𝟒 − ( 𝟒. 𝟓 ∗ 𝟏𝟎 𝟓 𝟓 ∗ 𝟏𝟎 𝟓 ) 𝟏.𝟏𝟒+𝟏 𝟏.𝟏𝟒 ] 𝑸𝒕𝒉 = 0.035414399 𝐦 𝟑 /𝒉 𝑸 𝐞𝐱𝐩 = 𝒌√ 𝜟𝒑 𝒗 𝒐 𝑸 𝐞𝐱𝐩 = 𝟓. 𝟔𝟏𝟓√ 𝟏𝟐𝟏.𝟔 𝟎.𝟑𝟕𝟐𝟗 = 0.101395891 𝒎 𝟑 /𝒉𝒓 𝑹 = 𝑷𝒐/𝑷𝒃 𝑹 = 𝟓/𝟒. 𝟓 = 1.11111 2. 𝑸𝒕𝒉 = 𝑨𝒖 ∗ 𝒑 𝟎 √ 𝒑 𝟎 ∗ 𝒗 ∗ √ 𝟐𝒌 𝒌 − 𝟏 ∗ [( 𝒑 𝐛 𝒑 𝟎 ) 𝟐 𝒌 − ( 𝒑 𝐛 𝒑 𝟎 ) 𝒌+𝟏 𝒌 ] 𝑸𝒕𝒉 = 𝟓. 𝟎𝟑 ∗ 𝟏𝟎−𝟓 ∗ 𝟓 ∗ 𝟏𝟎 𝟓 √𝟓 ∗ 𝟏𝟎 𝟓 ∗ 𝟎. 𝟑𝟕𝟐𝟗 ∗ √ 𝟐 ∗ 𝟏. 𝟏𝟒 𝟏. 𝟏𝟒 − 𝟏 ∗ [( 𝟒 ∗ 𝟏𝟎 𝟓 𝟓 ∗ 𝟏𝟎 𝟓 ) 𝟐 𝟏.𝟏𝟒 − ( 𝟒 ∗ 𝟏𝟎 𝟓 𝟓 ∗ 𝟏𝟎 𝟓 ) 𝟏.𝟏𝟒+𝟏 𝟏.𝟏𝟒 ] 𝑸𝒕𝒉 = 0.04612388 𝐦 𝟑 /𝒉 𝑸 𝐞𝐱𝐩 = 𝒌√ 𝜟𝒑 𝒗 𝒐 𝑸 𝐞𝐱𝐩 = 𝟓. 𝟔𝟏𝟓√ 𝟏𝟓𝟗.𝟔 𝟎.𝟑𝟕𝟐𝟗 = 0.116163587 𝒎 𝟑 /𝒉𝒓 𝑹 = 𝑷𝒐/𝑷𝒃 𝑹 = 𝟓/𝟒 = 1.25
10. 10. Saif al-din ali Madi Department of Mechanical Engineering/ College of Engineering/ University of Baghdad 26/11/2018 10 | P a g e 3. 𝑸𝒕𝒉 = 𝑨𝒖 ∗ 𝒑 𝟎 √ 𝒑 𝟎 ∗ 𝒗 ∗ √ 𝟐𝒌 𝒌 − 𝟏 ∗ [( 𝒑 𝐛 𝒑 𝟎 ) 𝟐 𝒌 − ( 𝒑 𝐛 𝒑 𝟎 ) 𝒌+𝟏 𝒌 ] 𝑸𝒕𝒉 = 𝟓. 𝟎𝟑 ∗ 𝟏𝟎−𝟓 ∗ 𝟓 ∗ 𝟏𝟎 𝟓 √𝟓 ∗ 𝟏𝟎 𝟓 ∗ 𝟎. 𝟑𝟕𝟐𝟗 ∗ √ 𝟐 ∗ 𝟏. 𝟏𝟒 𝟏. 𝟏𝟒 − 𝟏 ∗ [( 𝟑. 𝟓 ∗ 𝟏𝟎 𝟓 𝟓 ∗ 𝟏𝟎 𝟓 ) 𝟐 𝟏.𝟏𝟒 − ( 𝟑. 𝟓 ∗ 𝟏𝟎 𝟓 𝟓 ∗ 𝟏𝟎 𝟓 ) 𝟏.𝟏𝟒+𝟏 𝟏.𝟏𝟒 ] 𝑸𝒕𝒉 = 0.051420707 𝐦 𝟑 /𝒉 𝑸 𝐞𝐱𝐩 = 𝒌√ 𝜟𝒑 𝒗 𝒐 𝑸 𝐞𝐱𝐩 = 𝟓. 𝟔𝟏𝟓√ 𝟏𝟖𝟐.𝟒 𝟎.𝟑𝟕𝟐𝟗 = 0.124184098𝒎 𝟑 /𝒉𝒓 𝑹 = 𝑷𝒐/𝑷𝒃 𝑹 = 𝟓/𝟑. 𝟓 = 1.428571429 4. 𝑸𝒕𝒉 = 𝑨𝒖 ∗ 𝒑 𝟎 √ 𝒑 𝟎 ∗ 𝒗 ∗ √ 𝟐𝒌 𝒌 − 𝟏 ∗ [( 𝒑 𝐛 𝒑 𝟎 ) 𝟐 𝒌 − ( 𝒑 𝐛 𝒑 𝟎 ) 𝒌+𝟏 𝒌 ] 𝑸𝒕𝒉 = 𝟓. 𝟎𝟑 ∗ 𝟏𝟎−𝟓 ∗ 𝟓 ∗ 𝟏𝟎 𝟓 √𝟓 ∗ 𝟏𝟎 𝟓 ∗ 𝟎. 𝟑𝟕𝟐𝟗 ∗ √ 𝟐 ∗ 𝟏. 𝟏𝟒 𝟏. 𝟏𝟒 − 𝟏 ∗ [( 𝟑 ∗ 𝟏𝟎 𝟓 𝟓 ∗ 𝟏𝟎 𝟓 ) 𝟐 𝟏.𝟏𝟒 − ( 𝟑 ∗ 𝟏𝟎 𝟓 𝟓 ∗ 𝟏𝟎 𝟓 ) 𝟏.𝟏𝟒+𝟏 𝟏.𝟏𝟒 ] 𝑸𝒕𝒉 = 0.053222816𝐦 𝟑 /𝒉 𝑸 𝐞𝐱𝐩 = 𝒌√ 𝜟𝒑 𝒗 𝒐 𝑸 𝐞𝐱𝐩 = 𝟓. 𝟔𝟏𝟓√ 𝟏𝟗𝟕.𝟔 𝟎.𝟑𝟕𝟐𝟗 = 0.129254907 𝒎 𝟑 /𝒉𝒓 𝑹 = 𝑷𝒐/𝑷𝒃 𝑹 = 𝟓/𝟑 = 1.666666667
11. 11. Saif al-din ali Madi Department of Mechanical Engineering/ College of Engineering/ University of Baghdad 26/11/2018 11 | P a g e 1. 𝑸𝒕𝒉 = 𝑨𝒖 ∗ 𝒑 𝟎 √ 𝒑 𝟎 ∗ 𝒗 ∗ √ 𝟐𝒌 𝒌 − 𝟏 ∗ [( 𝒑 𝐛 𝒑 𝟎 ) 𝟐 𝒌 − ( 𝒑 𝐛 𝒑 𝟎 ) 𝒌+𝟏 𝒌 ] 𝑸𝒕𝒉 = 𝟓. 𝟎𝟑 ∗ 𝟏𝟎−𝟓 ∗ 𝟔 ∗ 𝟏𝟎 𝟓 √𝟔 ∗ 𝟏𝟎 𝟓 ∗ 𝟎. 𝟑𝟐𝟏𝟐 ∗ √ 𝟐 ∗ 𝟏. 𝟏𝟒 𝟏. 𝟏𝟒 − 𝟏 ∗ [( 𝟓. 𝟔 ∗ 𝟏𝟎 𝟓 𝟔 ∗ 𝟏𝟎 𝟓 ) 𝟐 𝟏.𝟏𝟒 − ( 𝟓. 𝟔 ∗ 𝟏𝟎 𝟓 𝟔 ∗ 𝟏𝟎 𝟓 ) 𝟏.𝟏𝟒+𝟏 𝟏.𝟏𝟒 ] 𝑸𝒕𝒉 = 0.03500503 𝐦 𝟑 /𝒉 𝑸 𝐞𝐱𝐩 = 𝒌√ 𝜟𝒑 𝒗 𝒐 𝑸 𝐞𝐱𝐩 = 𝟓. 𝟔𝟏𝟓√ 𝟕𝟓 𝟎.𝟑𝟐𝟏𝟐 = 0.085801054𝒎 𝟑 /𝒉𝒓 𝑹 = 𝑷𝒐/𝑷𝒃 𝑹 = 𝟔/𝟓. 𝟔 = 1.071428571 2. 𝑸𝒕𝒉 = 𝑨𝒖 ∗ 𝒑 𝟎 √ 𝒑 𝟎 ∗ 𝒗 ∗ √ 𝟐𝒌 𝒌 − 𝟏 ∗ [( 𝒑 𝐛 𝒑 𝟎 ) 𝟐 𝒌 − ( 𝒑 𝐛 𝒑 𝟎 ) 𝒌+𝟏 𝒌 ] 𝑸𝒕𝒉 = 𝟓. 𝟎𝟑 ∗ 𝟏𝟎−𝟓 ∗ 𝟔 ∗ 𝟏𝟎 𝟓 √𝟔 ∗ 𝟏𝟎 𝟓 ∗ 𝟎. 𝟑𝟐𝟏𝟐 ∗ √ 𝟐 ∗ 𝟏. 𝟏𝟒 𝟏. 𝟏𝟒 − 𝟏 ∗ [( 𝟓. 𝟒 ∗ 𝟏𝟎 𝟓 𝟔 ∗ 𝟏𝟎 𝟓 ) 𝟐 𝟏.𝟏𝟒 − ( 𝟓. 𝟒 ∗ 𝟏𝟎 𝟓 𝟔 ∗ 𝟏𝟎 𝟓 ) 𝟏.𝟏𝟒+𝟏 𝟏.𝟏𝟒 ] 𝑸𝒕𝒉 = 0.041800254 𝐦 𝟑 /𝒉 𝑸 𝐞𝐱𝐩 = 𝒌√ 𝜟𝒑 𝒗 𝒐 𝑸 𝐞𝐱𝐩 = 𝟓. 𝟔𝟏𝟓√ 𝟗𝟎 𝟎.𝟑𝟐𝟏𝟐 = 0.093990346 𝒎 𝟑 /𝒉𝒓 𝑹 = 𝑷𝒐/𝑷𝒃 𝑹 = 𝟔/𝟓. 𝟒 = 1.111111111
12. 12. Saif al-din ali Madi Department of Mechanical Engineering/ College of Engineering/ University of Baghdad 26/11/2018 12 | P a g e 3. 𝑸𝒕𝒉 = 𝑨𝒖 ∗ 𝒑 𝟎 √ 𝒑 𝟎 ∗ 𝒗 ∗ √ 𝟐𝒌 𝒌 − 𝟏 ∗ [( 𝒑 𝐛 𝒑 𝟎 ) 𝟐 𝒌 − ( 𝒑 𝐛 𝒑 𝟎 ) 𝒌+𝟏 𝒌 ] 𝑸𝒕𝒉 = 𝟓. 𝟎𝟑 ∗ 𝟏𝟎−𝟓 ∗ 𝟔 ∗ 𝟏𝟎 𝟓 √𝟔 ∗ 𝟏𝟎 𝟓 ∗ 𝟎. 𝟑𝟐𝟏𝟐 ∗ √ 𝟐 ∗ 𝟏. 𝟏𝟒 𝟏. 𝟏𝟒 − 𝟏 ∗ [( 𝟓. 𝟏 ∗ 𝟏𝟎 𝟓 𝟔 ∗ 𝟏𝟎 𝟓 ) 𝟐 𝟏.𝟏𝟒 − ( 𝟓. 𝟏 ∗ 𝟏𝟎 𝟓 𝟔 ∗ 𝟏𝟎 𝟓 ) 𝟏.𝟏𝟒+𝟏 𝟏.𝟏𝟒 ] 𝑸𝒕𝒉 = 0.049192115 𝐦 𝟑 /𝒉 𝑸 𝐞𝐱𝐩 = 𝒌√ 𝜟𝒑 𝒗 𝒐 𝑸 𝐞𝐱𝐩 = 𝟓. 𝟔𝟏𝟓√ 𝟏𝟐𝟓 𝟎.𝟑𝟐𝟏𝟐 = 0.110768685𝒎 𝟑 /𝒉𝒓 𝑹 = 𝑷𝒐/𝑷𝒃 𝑹 = 𝟔/𝟓. 𝟏 = 1.176470588 4. 𝑸𝒕𝒉 = 𝑨𝒖 ∗ 𝒑 𝟎 √ 𝒑 𝟎 ∗ 𝒗 ∗ √ 𝟐𝒌 𝒌 − 𝟏 ∗ [( 𝒑 𝐛 𝒑 𝟎 ) 𝟐 𝒌 − ( 𝒑 𝐛 𝒑 𝟎 ) 𝒌+𝟏 𝒌 ] 𝑸𝒕𝒉 = 𝟓. 𝟎𝟑 ∗ 𝟏𝟎−𝟓 ∗ 𝟔 ∗ 𝟏𝟎 𝟓 √𝟔 ∗ 𝟏𝟎 𝟓 ∗ 𝟎. 𝟑𝟐𝟏𝟐 ∗ √ 𝟐 ∗ 𝟏. 𝟏𝟒 𝟏. 𝟏𝟒 − 𝟏 ∗ [( 𝟒. 𝟗 ∗ 𝟏𝟎 𝟓 𝟔 ∗ 𝟏𝟎 𝟓 ) 𝟐 𝟏.𝟏𝟒 − ( 𝟒. 𝟗 ∗ 𝟏𝟎 𝟓 𝟔 ∗ 𝟏𝟎 𝟓 ) 𝟏.𝟏𝟒+𝟏 𝟏.𝟏𝟒 ] 𝑸𝒕𝒉 = 0.052882165𝐦 𝟑 /𝒉 𝑸 𝐞𝐱𝐩 = 𝒌√ 𝜟𝒑 𝒗 𝒐 𝑸 𝐞𝐱𝐩 = 𝟓. 𝟔𝟏𝟓√ 𝟏𝟒𝟓 𝟎.𝟑𝟐𝟏𝟐 = 0.119301525𝒎 𝟑 /𝒉𝒓 𝑹 = 𝑷𝒐/𝑷𝒃 𝑹 = 𝟔/𝟒. 𝟗 = 1.224489796
13. 13. Saif al-din ali Madi Department of Mechanical Engineering/ College of Engineering/ University of Baghdad 26/11/2018 13 | P a g e 1. 𝑸𝒕𝒉 = 𝑨𝒖 ∗ 𝒑 𝟎 √ 𝒑 𝟎 ∗ 𝒗 ∗ √ 𝟐𝒌 𝒌 − 𝟏 ∗ [( 𝒑 𝐛 𝒑 𝟎 ) 𝟐 𝒌 − ( 𝒑 𝐛 𝒑 𝟎 ) 𝒌+𝟏 𝒌 ] 𝑸𝒕𝒉 = 𝟓. 𝟎𝟑 ∗ 𝟏𝟎−𝟓 ∗ 𝟕 ∗ 𝟏𝟎 𝟓 √𝟕 ∗ 𝟏𝟎 𝟓 ∗ 𝟎. 𝟑𝟐𝟏𝟐 ∗ √ 𝟐 ∗ 𝟏. 𝟏𝟒 𝟏. 𝟏𝟒 − 𝟏 ∗ [( 𝟔. 𝟓 ∗ 𝟏𝟎 𝟓 𝟔 ∗ 𝟏𝟎 𝟓 ) 𝟐 𝟏.𝟏𝟒 − ( 𝟔. 𝟓 ∗ 𝟏𝟎 𝟓 𝟔 ∗ 𝟏𝟎 𝟓 ) 𝟏.𝟏𝟒+𝟏 𝟏.𝟏𝟒 ] 𝑸𝒕𝒉 = 0.03500503 𝐦 𝟑 /𝒉 𝑸 𝐞𝐱𝐩 = 𝒌√ 𝜟𝒑 𝒗 𝒐 𝑸 𝐞𝐱𝐩 = 𝟓. 𝟔𝟏𝟓√ 𝟕𝟓 𝟎.𝟐𝟕𝟓𝟐 = 0.092687731 𝒎 𝟑 /𝒉𝒓 𝑹 = 𝑷𝒐/𝑷𝒃 𝑹 = 𝟕 ∗ 𝟔. 𝟓 = 1.076923077 2. 𝑸𝒕𝒉 = 𝑨𝒖 ∗ 𝒑 𝟎 √ 𝒑 𝟎 ∗ 𝒗 ∗ √ 𝟐𝒌 𝒌 − 𝟏 ∗ [( 𝒑 𝐛 𝒑 𝟎 ) 𝟐 𝒌 − ( 𝒑 𝐛 𝒑 𝟎 ) 𝒌+𝟏 𝒌 ] 𝑸𝒕𝒉 = 𝟓. 𝟎𝟑 ∗ 𝟏𝟎−𝟓 ∗ 𝟕 ∗ 𝟏𝟎 𝟓 √𝟕 ∗ 𝟏𝟎 𝟓 ∗ 𝟎. 𝟑𝟐𝟏𝟐 ∗ √ 𝟐 ∗ 𝟏. 𝟏𝟒 𝟏. 𝟏𝟒 − 𝟏 ∗ [( 𝟔. 𝟐 ∗ 𝟏𝟎 𝟓 𝟔 ∗ 𝟏𝟎 𝟓 ) 𝟐 𝟏.𝟏𝟒 − ( 𝟔. 𝟐 ∗ 𝟏𝟎 𝟓 𝟔 ∗ 𝟏𝟎 𝟓 ) 𝟏.𝟏𝟒+𝟏 𝟏.𝟏𝟒 ] 𝑸𝒕𝒉 = 0.051562331 𝐦 𝟑 /𝒉 𝑸 𝐞𝐱𝐩 = 𝒌√ 𝜟𝒑 𝒗 𝒐 𝑸 𝐞𝐱𝐩 = 𝟓. 𝟔𝟏𝟓√ 𝟗𝟎 𝟎.𝟐𝟕𝟓𝟐 = 0.101534322𝒎 𝟑 /𝒉𝒓 𝑹 = 𝑷𝒐/𝑷𝒃 𝑹 = 𝟕 ∗ 𝟔. 𝟐 = 1.129032258
14. 14. Saif al-din ali Madi Department of Mechanical Engineering/ College of Engineering/ University of Baghdad 26/11/2018 14 | P a g e 3. 𝑸𝒕𝒉 = 𝑨𝒖 ∗ 𝒑 𝟎 √ 𝒑 𝟎 ∗ 𝒗 ∗ √ 𝟐𝒌 𝒌 − 𝟏 ∗ [( 𝒑 𝐛 𝒑 𝟎 ) 𝟐 𝒌 − ( 𝒑 𝐛 𝒑 𝟎 ) 𝒌+𝟏 𝒌 ] 𝑸𝒕𝒉 = 𝟓. 𝟎𝟑 ∗ 𝟏𝟎−𝟓 ∗ 𝟕 ∗ 𝟏𝟎 𝟓 √𝟕 ∗ 𝟏𝟎 𝟓 ∗ 𝟎. 𝟑𝟐𝟏𝟐 ∗ √ 𝟐 ∗ 𝟏. 𝟏𝟒 𝟏. 𝟏𝟒 − 𝟏 ∗ [( 𝟓. 𝟗 ∗ 𝟏𝟎 𝟓 𝟔 ∗ 𝟏𝟎 𝟓 ) 𝟐 𝟏.𝟏𝟒 − ( 𝟓. 𝟗 ∗ 𝟏𝟎 𝟓 𝟔 ∗ 𝟏𝟎 𝟓 ) 𝟏.𝟏𝟒+𝟏 𝟏.𝟏𝟒 ] 𝑸𝒕𝒉 = 0.05840326 𝐦 𝟑 /𝒉 𝑸 𝐞𝐱𝐩 = 𝒌√ 𝜟𝒑 𝒗 𝒐 𝑸 𝐞𝐱𝐩 = 𝟓. 𝟔𝟏𝟓√ 𝟏𝟐𝟓 𝟎.𝟐𝟕𝟓𝟐 = 0.119659346 𝒎 𝟑 /𝒉𝒓 𝑹 = 𝑷𝒐/𝑷𝒃 𝑹 = 𝟕 ∗ 𝟓. 𝟗 = 1.186440678 4. 𝑸𝒕𝒉 = 𝑨𝒖 ∗ 𝒑 𝟎 √ 𝒑 𝟎 ∗ 𝒗 ∗ √ 𝟐𝒌 𝒌 − 𝟏 ∗ [( 𝒑 𝐛 𝒑 𝟎 ) 𝟐 𝒌 − ( 𝒑 𝐛 𝒑 𝟎 ) 𝒌+𝟏 𝒌 ] 𝑸𝒕𝒉 = 𝟓. 𝟎𝟑 ∗ 𝟏𝟎−𝟓 ∗ 𝟕 ∗ 𝟏𝟎 𝟓 √𝟕 ∗ 𝟏𝟎 𝟓 ∗ 𝟎. 𝟑𝟐𝟏𝟐 ∗ √ 𝟐 ∗ 𝟏. 𝟏𝟒 𝟏. 𝟏𝟒 − 𝟏 ∗ [( 𝟓. 𝟔 ∗ 𝟏𝟎 𝟓 𝟔 ∗ 𝟏𝟎 𝟓 ) 𝟐 𝟏.𝟏𝟒 − ( 𝟓. 𝟔 ∗ 𝟏𝟎 𝟓 𝟔 ∗ 𝟏𝟎 𝟓 ) 𝟏.𝟏𝟒+𝟏 𝟏.𝟏𝟒 ] 𝑸𝒕𝒉 = 0.063522547 𝐦 𝟑 /𝒉 𝑸 𝐞𝐱𝐩 = 𝒌√ 𝜟𝒑 𝒗 𝒐 𝑸 𝐞𝐱𝐩 = 𝟓. 𝟔𝟏𝟓√ 𝟏𝟒𝟓 𝟎.𝟐𝟕𝟓𝟐 = 0.12887706 𝒎 𝟑 /𝒉𝒓 𝑹 = 𝑷𝒐/𝑷𝒃 𝑹 = 𝟕 ∗ 𝟓. 𝟔 = 1.25
15. 15. Saif al-din ali Madi Department of Mechanical Engineering/ College of Engineering/ University of Baghdad 26/11/2018 15 | P a g e Experimental 1 state Po bar Pb (bar) To ( c) ∆P (mmHg) Tb Vo (m3/kg) 1 3 2.5 145 0.12 137 0.464 2 3 2 145 0.16 130 0.464 3 3 1.5 145 0.19 125 0.464 4 3 1 145 0.2 120 0.464 2 1 4 3.5 150 0.16 144 0.3729 2 4 3 150 0.21 132.5 0.3729 3 4 2.5 150 0.24 132 0.3729 4 4 2 150 0.26 130 0.3729 Theoretical 1 1 5 4.6 165 75 0.3212 2 5 4.4 165 90 0.3212 3 5 4.1 165 125 0.3212 4 5 3.9 165 145 0.3212 2 1 6 5.5 168 75 0.275 2 6 5.2 168 90 0.2752 3 6 4.9 168 125 0.2752 4 6 4.6 168 145 0.2752
16. 16. Saif al-din ali Madi Department of Mechanical Engineering/ College of Engineering/ University of Baghdad 26/11/2018 16 | P a g e Experimental Q_exp R Q_th 1 0.078699279 1.142857143 0.031121877 0.0908741 1.333333333 0.039504565 0.099027755 1.6 0.042527341 0.101600333 2 0.041861669 2 0.101395891 1.111111111 0.035414399 0.116163587 1.25 0.04612388 0.124184098 1.428571429 0.051420707 0.129254907 1.666666667 0.053222816 Theoretical 1 0.085801054 1.071428571 0.03500503 0.093990346 1.111111111 0.041800254 0.110768685 1.176470588 0.049192115 0.119301525 1.224489796 0.052882165 2 0.092687731 1.076923077 0.042127888 0.101534322 1.129032258 0.051562331 0.119659346 1.186440678 0.05840326 0.12887706 1.25 0.063522547 Experimental 1 ∆P*760 Pb (bar) +1 Po bar +1 91.2 3.5 4 121.6 3 4 144.4 2.5 4 152 2 4 2 121.6 4.5 5 159.6 4 5 182.4 3.5 5 197.6 3 5 Theoretical 1 75 5.6 6 90 5.4 6 125 5.1 6 145 4.9 6 2 75 6.5 7 90 6.2 7 125 5.9 7 145 5.6 7
17. 17. Saif al-din ali Madi Department of Mechanical Engineering/ College of Engineering/ University of Baghdad 26/11/2018 17 | P a g e 5. DISCUSSION 1. 0 0.02 0.04 0.06 0.08 0.1 0.12 0 0.5 1 1.5 2 2.5 Q R Experimental 1 Q_exp Q_th 0 0.02 0.04 0.06 0.08 0.1 0.12 0.14 0 0.5 1 1.5 2 Q R Experimental 2 Q_exp Q_th
18. 18. Saif al-din ali Madi Department of Mechanical Engineering/ College of Engineering/ University of Baghdad 26/11/2018 18 | P a g e In all the drawings, it turns out that the condition is direct and the pitch of the gradient varies depending on the situation 0 0.02 0.04 0.06 0.08 0.1 0.12 0.14 1.05 1.1 1.15 1.2 1.25 Q R Theoretical 1 Q_exp Q_th 0 0.02 0.04 0.06 0.08 0.1 0.12 0.14 1.05 1.1 1.15 1.2 1.25 1.3 Q R Theoretical 2 Q_exp Q_th
19. 19. Saif al-din ali Madi Department of Mechanical Engineering/ College of Engineering/ University of Baghdad 26/11/2018 19 | P a g e What is the effect of the downstream pressure nozzle on the flow rate values? It has great influence on the amount of outflow and the type of flow as per the drawing