REFRIGERANT134 a3Ask:HEAT EXCHANGERAIRN2AIRREFRIGERANT 134a1 Bar89,6 F(AV₁) = 1m³/sMASS FLOW RATE2) HEAT TRANSFER RATEand0 P₁OF^^3 PGIVEN : KINETIC ENERGY AND POTENTIAL ENERGY = OSTEADY STATE4REFRIGERANT T₂ 71.6 F2134 a2== 0.95 BarREFRIGE RANTAIR.1349BETWEENP3 =X3=5 BarIN kg/s0.241 P4= 5 BarT4 = 68°F

Given:Inlet air condition, P1= 1 Bar, T1=89.6o F, AV1=1 m3/sOutlet air condition, P2=0.95 Bar,…

REFRIGERANT 134 a Ask 3 : HEAT EXCHANGER AIR GIVEN : KINETIC ENERGY STEADY STATE 2 AIR > 4 REFRIGERANT 134 a AND POTENTIAL 1) FLOW RATE MASS 2) HEAT TRANSFER RATE and REFRIGERANT 134 a 0 P₁ = 1 T₁ (AV₁) = 1m³/s 1 Bar @ P2 = 89,6 F = 0.95 Bar 71.6°F T₂ = OF REFRIGE RANT ENERGY = 0 AIR. 1349 BETWEEN P3 3 = 5 Bar IN kg/s 0.2 41 P4 = 5 Bar T4= 68°F

REFRIGERANT 134 a Ask 3 : HEAT EXCHANGER AIR GIVEN : KINETIC ENERGY STEADY STATE 2 AIR > 4 REFRIGERANT 134 a AND POTENTIAL 1) FLOW RATE MASS 2) HEAT TRANSFER RATE and REFRIGERANT 134 a 0 P₁ = 1 T₁ (AV₁) = 1m³/s 1 Bar @ P2 = 89,6 F = 0.95 Bar 71.6°F T₂ = OF REFRIGE RANT ENERGY = 0 AIR. 1349 BETWEEN P3 3 = 5 Bar IN kg/s 0.2 41 P4 = 5 Bar T4= 68°F

Principles of Heat Transfer (Activate Learning with these NEW titles from Engineering!)

8th Edition

ISBN:9781305387102

Author:Kreith, Frank; Manglik, Raj M.

Publisher:Kreith, Frank; Manglik, Raj M.

Chapter6: Forced Convection Over Exterior Surfaces

Section: Chapter Questions

Problem 6.12P

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