The diffuser in a jet engine is designed to decrease the kinetic energy of the air entering the engine compressor without any work or heat interactions. Calculate the velocity at the exit of a diffuser when air at 100 kPa and&nbsp;30∘{&nbsp;C enters it with a velocity of 350 m/s and the exit state is 200 kPa and&nbsp;90∘{C.

Question

The diffuser in a jet engine is designed to decrease the kinetic energy of the air entering the engine compressor without any work or heat interactions. Calculate the velocity at the exit of a diffuser when air at 100 kPa and&amp;nbsp;30∘{&amp;nbsp;C enters it with a velocity of 350 m/s and the exit state is 200 kPa and&amp;nbsp;90∘{C.

Harr1957 · Accepted Answer

So,&amp;nbsp;E∈−Eout=△Esystem&amp;nbsp;E∈=Eout&amp;nbsp;&amp;nbsp;m(h1+v122)=m(h2+V222)&amp;nbsp;h1+V122=h2+V222&amp;nbsp;&amp;nbsp;V2=[V12+2(h1−h2)]0.5=[V12+2Cp(T1−T2)]0.5&amp;nbsp;nothing&amp;nbsp;&amp;nbsp;cp=1.007KJKg⋅K→tablea-A2b&amp;nbsp;&amp;nbsp;V2=[(350ms)2+2(1.007KjKg⋅K)(30−90)K(1000m2s21KjKg)]0.5→V2=40.7ma&amp;nbsp;

RizerMix · Accepted Answer

To solve this problem, we can use the conservation of energy equation for an adiabatic and frictionless diffuser:v222+h2=v122+h1where v1 and v2 are the velocities at the inlet and outlet of the diffuser, and h1 and h2 are the enthalpies at the inlet and outlet of the diffuser, respectively. Since the diffuser is adiabatic and frictionless, h1=h2, and we can simplify the equation to:v222=v122Solving for v2, we get:v2=v1P2P1where P1 and P2 are the pressures at the inlet and outlet of the diffuser, respectively. We can use the ideal gas law to relate the pressures to the temperatures of the air at the inlet and outlet of the diffuser:P1=ρ1RT1 and P2=ρ2RT2where ρ1 and ρ2 are the densities of the air at the inlet and outlet of the diffuser, respectively, and R is the specific gas constant for air. Substituting these equations into the expression for v2, we get:v2=v1T2T1Now, we can plug in the given values and solve for v2:v2=350363.15+90363.15+30=40.7Therefore, the velocity at the exit of the diffuser is *40.7* m/s.

Jeffrey Jordon · Accepted Answer

Answer:40.7m/sExplanation:To approach this problem is to use the conservation of mass equation for an incompressible flow:m˙=ρ1A1v1=ρ2A2v2where m˙ is the mass flow rate, ρ1 and ρ2 are the densities of the air at the inlet and outlet of the diffuser, respectively, A1 and A2 are the cross-sectional areas of the diffuser at the inlet and outlet, respectively, and v1 and v2 are the velocities at the inlet and outlet, respectively.Since the diffuser is designed to decrease the kinetic energy of the air entering the engine compressor without any work or heat interactions, we can assume that the flow is adiabatic and reversible, and that the density of the air remains constant throughout the process. Therefore, we can simplify the equation to:A1v1=A2v2We can also use the isentropic relations for an ideal gas to relate the temperatures and pressures at the inlet and outlet of the diffuser:T1=P1ρ1R and T2=P2ρ2RP1v1γ=P2v2γwhere γ is the specific heat ratio for air. Rearranging and substituting the expressions for T1 and T2 into the equation, we get:P2P1=(T2T1)γγ−1Substituting this expression into the equation A1v1=A2v2, we get:v2=v1P2P1=v1(T2T1)γ−12γPlugging in the given values, we get:v2=350(473.15393.15)0.42×1.4=40.7Therefore, the velocity at the exit of the diffuser is 40.7m/s.

Mr Solver · Accepted Answer

Answer:350m/sExplanation:m˙1=m˙2 where m˙1 is the mass flow rate at the inlet and m˙2 is the mass flow rate at the exit.The equation for conservation of energy is:h1+V122=h2+V222 where h1 and h2 are the specific enthalpies at the inlet and exit, respectively, and V1 and V2 are the velocities at the inlet and exit, respectively.Given:P1=100kPaT1=30∘CV1=350m/sP2=200kPaT2=90∘CUsing the ideal gas equation, we can find the specific volume at the inlet and exit as follows:v1=R·T1P1v2=R·T2P2 where R is the specific gas constant.Next, we can calculate the mass flow rate at the inlet using the equation:m˙1=ρ1·A1·V1 where ρ1 is the density at the inlet and A1 is the cross-sectional area of the diffuser at the inlet.Using the equation for conservation of mass, we can find the mass flow rate at the exit:m˙2=m˙1Now, we can solve for the velocity at the exit using the equation for conservation of energy:V2=2·(h1−h2)+V12To summarize, the velocity at the exit of the diffuser can be calculated using the given values and the equations mentioned above.v1=R·T1P1v2=R·T2P2 where R is the specific gas constant. For air, R=287J/(kg·K).Substituting the given values into the equations, we have:v1=287J/(kg·K)·(30+273)K100kPav2=287J/(kg·K)·(90+273)K200kPaSimplifying these expressions gives us the specific volumes:v1≈0.838m3/kgv2≈0.996m3/kgNext, we can calculate the density at the inlet using the ideal gas equation:ρ1=P1R·T1Substituting the given values:ρ1=100kPa287J/(kg·K)·(30+273)KSimplifying gives us the density:ρ1≈1.136kg/m3Now, we can calculate the mass flow rate at the inlet:m˙1=ρ1·A1·V1where A1 is the cross-sectional area of the diffuser at the inlet. Since the diffuser&#039;s geometry is not provided, we cannot calculate the exact value of A1. However, assuming a constant cross-sectional area throughout the diffuser, we can cancel out A1 when solving for the velocity at the exit.Therefore, the mass flow rate at the exit is also m˙1.Finally, we can use the equation for conservation of energy to solve for the velocity at the exit:V2=2·(h1−h2)+V12Given that the diffuser operates without any work or heat interactions, the change in specific enthalpy (h1−h2) can be considered negligible. Hence, we can neglect the first term inside the square root. Therefore, the velocity at the exit is approximately equal to the velocity at the inlet:V2≈V1Substituting the given value:V2≈350m/sThus, the velocity at the exit of the diffuser is approximately 350m/s.

madeleinejames20 · Accepted Answer

Let&#039;s denote the properties at the inlet of the diffuser with subscripts 1 and the properties at the exit with subscripts 2.The conservation of mass equation can be expressed as follows:m˙1=m˙2where m˙ represents the mass flow rate. Since we are given the velocity (v1) and the air density (ρ1) at the inlet, we can write:ρ1A1v1=ρ2A2v2where A represents the cross-sectional area.Now, let&#039;s apply the conservation of energy equation. The work done on the air during the process is zero (δW=0), and there is no heat transfer (δQ=0). Thus, the conservation of energy equation becomes:h1+v122=h2+v222where h represents the specific enthalpy of the air.Given that the pressure and temperature at the inlet and exit are known, we can determine the specific enthalpies h1 and h2 using the air properties table or specific heat relations. Then, we can solve the above equations simultaneously to find the velocity at the exit of the diffuser (v2).Now, let&#039;s calculate the solution:Given data:P1=100kPaT1=30∘Cv1=350m/sP2=200kPaT2=90∘CFirst, we need to convert the temperatures to Kelvin:T1=30+273.15=303.15KT2=90+273.15=363.15KTo find the specific enthalpy at state 1, h1, and state 2, h2, we can use the air properties table or specific heat relations. For simplicity, we&#039;ll assume air to be an ideal gas with constant specific heat:Cp=1.005kJ/kg·K (specific heat at constant pressure for air)h1=CpT1h2=CpT2Substituting the given values:h1=1.005×303.15=304.921kJ/kgh2=1.005×363.15=365.002kJ/kgNow, we can solve the conservation of mass equation to find the cross-sectional area ratio A2/A1:ρ1A1v1=ρ2A2v2Since the density ratio ρ2/ρ1 can be written as P2RT2·RT1P1, where R is the specific gas constant for air, we can rewrite the equation as:A2/A1=P1v1T2P2v2T1·RT2RT1Substituting the given values and the universal gas constant R=0.287kJ/kg·K:A2/A1=100×350×363.15200×v2×303.15·0.287×363.150.287×303.15Simplifying the equation, we have:A2/A1=7.537v2Now, let&#039;s solve the conservation of energy equation:h1+v122=h2+v222Substituting the given values:304.921+35022=365.002+v222Simplifying the equation:61000+v22=73001+v22Solving for v2, we find:v2=73001−61000=12001≈109.54m/sTherefore, the velocity at the exit of the diffuser is approximately 109.54m/s.

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