Ch5Examples - Problem Statement: Nozzle Air enters a nozzle...

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Problem Statement: Nozzle Air enters a nozzle at 40.00 m/s through an area of 90.00 180.00 m/s. The density of the air at the entrance is 2.210 exits at 0.762 Schematic: Assumptions/Approximations: Nozzle is adiabatic and the flow is steady The thermal energy stored in the nozzle itself is negligible For discussion, assume constant specific heat and ideal gas behavior are appropriate Physical Laws: [Steady flow] Properties/Sources: 0.452489 1.312336 0.29 kJ/kg*K 1.01 kJ/kg*K [At 300 K] Calculations: = 0.7956 kg/s ==> 137.16 0.01 58.01 [No work or heat change and no change in elevation] -15.40 kJ/kg Reasoning/Verification/Discussion: Note that no work or heat is exchanged; only transfer of internal energy to kinetic energy; If we knew another property (T or P at a state), we would know all prope Alternatively, we could treat air as an ideal gas to yield: cm 2 . The air exits at kg/m 3 . What are the mass flow rate and the exit area? What is the change of energy in this process? ρ 1 *A 1 *V 1 = ρ 2 *A 2 *V 2 ρ 1 and ρ 2 are given v 1 = m 3 /kg v 2 = m 3 /kg R AIR = Cp AIR = ρ 2 *V 2 = kg/m 2 *s Therefore, A 2 = m 2 = cm 2 h 1 + V 1 2 /2 = h 2 + V 2 2 /2 or h = h 2 - h 1 = V 1 2 /2 - V 2 2 /2 = [Note that 1000 m 2 /s 2 = kJ/kg] AIR V 2 = 180 m/s V 1 = 40 m/s A 1 = 90 cm 2
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-15.40 kJ/kg -15.32 K = -15.32 If the entrance temperature is 300.00 K, 284.68 K = 11.53 0.634270 190.281 kPa 0.218694 62.257 kPa h 2 - h 1 = Cp AIR *(T 2 - T 1 ) = T 2 - T 1 = o C T 2 = o C P 1 = ρ 1 *R AIR *T 1 = *T 1 = P 2 = ρ 2 *R AIR *T 2 = *T 2 =
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erties kg/m 3 and
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Problem Statement: Diffuser Air enters a nozzle at 180.00 40.00 m/s through an area of 90.00 0.76 exits at 2.210 Schematic: Assumptions/Approximations: Diffuser is adiabatic and the flow is steady The thermal energy stored in the diffuser itself is negligible For discussion, assume constant specific heat and ideal gas behavior are appropriate Physical Laws: [Steady flow] Properties/Sources: 0.452489 1.312336 0.29 kJ/kg*K 1.01 kJ/kg*K Calculations: = 0.7956 kg/s ==> 137.16 0.01 58.01 [No work or heat change and no change in elevation] 15.40 kJ/kg Reasoning/Verification/Discussion: Note that no work or heat is exchanged; only transfer of kinetic energy to internal energy; If we knew another property (T or P at a state), we would know all prope Alternatively, we could treat air as an ideal gas to yield: m/s. The air exits at cm 2 . The density of the air at the entrance is kg/m 3 . What are the mass flow rate and the entrance area? What is the change of energy in this process? ρ 1 *A 1 *V 1 = ρ 2 *A 2 *V 2 ρ 1 and ρ 2 are given v 2 = m 3 /kg v 2 = m 3 /kg R AIR = Cp AIR = ρ 1 *V 1 = kg/m 2 *s Therefore, A 1 = m 2 = cm 2 h 1 + V 1 2 /2 = h 2 + V 2 2 /2 or h = h 2 - h 1 = V 1 2 /2 - V 2 2 /2 = [Note that 1000 m 2 /s 2 = kJ/kg] AIR V 1 = 180 m/s V 2 = 40 m/s A 2 = 90 cm 2
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15.32 K = 15.32 If the entrance temperature is 300.00 K, 315.32 K = 42.17 0.218694 65.608 kPa 0.634270 200.000 kPa h 2 - h 1 = Cp
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Ch5Examples - Problem Statement: Nozzle Air enters a nozzle...

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