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Use the Kedzierski (2003) refrigerant/lubricant mixture pool boiling model to predict the boiling heat transfer coefficient (hm) for a range of superheats (47,8 K to 40 K) and Ts=277.6 K: 5.9×107(1-x₂)PhAT, k₂ (1) Where h 9m T-T r x,To PL-Px 5.9×10 (1-x)ph AT, KL (W/mK) 0.139 0.7551 p (1-x₂) 18.75AP (1-x)_18.75×10-¹[m]p, (1-x₂) Xb Prv XbPn x,T,o, XbPrv Assume that λ = 1.34 for xb=0.005 and that λ = 0.3 for xb = 0.02. The properties of the refrigerant (R123) at the film temperature are: R123 hig Or (J/kg) (N/m) 179692.3 0.01764 The properties of the mineral oil (lubricant) are: York-C V (cSt) 60 PL (kg/m³) 917.8 Prv (kg/m³) 2.701 V (m²/s) 6 x 10-5 OL (N/m) 0.026

Use the Kedzierski (2003) refrigerant/lubricant mixture pool boiling model to predict the boiling heat transfer coefficient (hm) for a range of superheats (47,8 K to 40 K) and Ts=277.6 K: 5.9×107(1-x₂)PhAT, k₂ (1) Where h 9m T-T r x,To PL-Px 5.9×10 (1-x)ph AT, KL (W/mK) 0.139 0.7551 p (1-x₂) 18.75AP (1-x)_18.75×10-¹[m]p, (1-x₂) Xb Prv XbPn x,T,o, XbPrv Assume that λ = 1.34 for xb=0.005 and that λ = 0.3 for xb = 0.02. The properties of the refrigerant (R123) at the film temperature are: R123 hig Or (J/kg) (N/m) 179692.3 0.01764 The properties of the mineral oil (lubricant) are: York-C V (cSt) 60 PL (kg/m³) 917.8 Prv (kg/m³) 2.701 V (m²/s) 6 x 10-5 OL (N/m) 0.026

Elements Of Electromagnetics
Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
ChapterMA: Math Assessment
Section: Chapter Questions
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Use the Kedzierski (2003) refrigerant/lubricant mixture pool boiling model to predict the boiling heat transfer coefficient (hm) for a range of superheats (AT, = 8 K to 40 K) and Ts=277.6 K: 5.9×10 (1-x₂)ph AT, k₂ (1-e) x,To, Where 1 hm = T-T, Г x,T,o, PLP% 5.9x10' (1-x₂)ph AT, 0.7551 (1-x₂) 18.75AP, (1-x₂) 18.75×10-¹ [m]p, (1-x) Xp Prv XbPr XbPrv KL (W/mK) 0.139 Assume that λ = 1.34 for xb=0.005 and that λ = 0.3 for xb = 0.02. The properties of the refrigerant (R123) at the film temperature are: R123 hig Or (J/kg) (N/m) 179692.3 0.01764 The properties of the mineral oil (lubricant) are: York-C V (cSt) 60 PL (kg/m³) 917.8 Prv (kg/m³) 2.701 VL (m²/s) 6 x 10-5 OL (N/m) 0.026
Transcribed Image Text:Use the Kedzierski (2003) refrigerant/lubricant mixture pool boiling model to predict the boiling heat transfer coefficient (hm) for a range of superheats (AT, = 8 K to 40 K) and Ts=277.6 K: 5.9×10 (1-x₂)ph AT, k₂ (1-e) x,To, Where 1 hm = T-T, Г x,T,o, PLP% 5.9x10' (1-x₂)ph AT, 0.7551 (1-x₂) 18.75AP, (1-x₂) 18.75×10-¹ [m]p, (1-x) Xp Prv XbPr XbPrv KL (W/mK) 0.139 Assume that λ = 1.34 for xb=0.005 and that λ = 0.3 for xb = 0.02. The properties of the refrigerant (R123) at the film temperature are: R123 hig Or (J/kg) (N/m) 179692.3 0.01764 The properties of the mineral oil (lubricant) are: York-C V (cSt) 60 PL (kg/m³) 917.8 Prv (kg/m³) 2.701 VL (m²/s) 6 x 10-5 OL (N/m) 0.026
3.) Calculate the benefit of adding AIO nanoparticles to the x = 0.005 and the x = 0.02 mass fraction mixtures using the Kedzierski (2012) model: 9ap = 1 + 9PL 1.45×10 [s m¹] No VPX Ą, 3/2 Dm (9.)³¹² P₁ (Pm - P₁) 8 (1-x)² Assume that for x = 0.005 (Nap/As)G= 3 x 1020 m² and for x = 0.02 (Nnp/As)G= 4 x 10²⁰ m². The properties of the AlO nanoparticles are: AIO nanoparticles Dup (nm) 10 Pnp (kg/m³) 3600 np Plot - versus q "PL where q "PL is varied from 5000 W/m² to 30000 W/m². Remember that q'n 9PL = q "PL/1 W/m². Which mass fraction benefits the most from the addition of nanoparticles. Explain why? Use the results of part 1 of this problem for the refrigerant/lubricant without nanoparticles in your explanation.
Transcribed Image Text:3.) Calculate the benefit of adding AIO nanoparticles to the x = 0.005 and the x = 0.02 mass fraction mixtures using the Kedzierski (2012) model: 9ap = 1 + 9PL 1.45×10 [s m¹] No VPX Ą, 3/2 Dm (9.)³¹² P₁ (Pm - P₁) 8 (1-x)² Assume that for x = 0.005 (Nap/As)G= 3 x 1020 m² and for x = 0.02 (Nnp/As)G= 4 x 10²⁰ m². The properties of the AlO nanoparticles are: AIO nanoparticles Dup (nm) 10 Pnp (kg/m³) 3600 np Plot - versus q "PL where q "PL is varied from 5000 W/m² to 30000 W/m². Remember that q'n 9PL = q "PL/1 W/m². Which mass fraction benefits the most from the addition of nanoparticles. Explain why? Use the results of part 1 of this problem for the refrigerant/lubricant without nanoparticles in your explanation.
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