Example: evaluate the thermal conductivity of hybrid nanofluid consist of Cu, TiO2, H20, 91-0.02 for Cu, 91-0.03 for TiO₂ Density, specific heat and thermal conductivity of base fluids and nanoparticles Cu H₂O Properties/constituents 997.1 8933 Density, p (kg/m³) 4179 385 Specific heat, Cp (/kg K) Thermal conductivity, k (W/mK) 0.613 401 TiO2 4250 686.2 8.9538

Introduction to Chemical Engineering Thermodynamics
8th Edition
ISBN:9781259696527
Author:J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart
Publisher:J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart
Chapter1: Introduction
Section: Chapter Questions
Problem 1.1P
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Hybrid nanofluid
Consist of mixed more than one nanopartical with working fluid such as CuO
A1203, TiO2 with water and the Thermo physical properties of hybrid nanofluid
depend on the type of nanoparticle.
For example hybrid nanofluid consist of two of nanofluid
Thermo physical properties of hybrid nanofluid
Hybrid Nanofluid
Phnf = Pfl
= p;(1-42) [(1-4) + 4 (B;})]
(PCp)nag = (PCp), (1—4₂) [(1–4₂) + 4,
Hj
Mined (1-01)25(1-02) 25
Ps +$2ps2
Kinf
köf
=
(pcp)st
(pcp)f
ksz + (S-1)kbf-(S-1) ₂ (kbf-ksz)
k2 +(S-1)ky + d₂ (kb-ksz)
+$₂ (PCp) sz
Where
kbf_ks1 + (S-1)kg-(S-1) 1 (ky-ks1)
ksi + (S−1)kƒ + ₁ (kj-ks1)
=
kf
Example: evaluate the thermal conductivity of hybrid nanofluid consist of Cu,
TiO2 ,H20, 9₁-0.02 for Cu, 9₁-0.03 for TiO₂
Density, specific heat and thermal conductivity of base fluids and nanoparticles
H₂O
Cu
Properties/constituents
997.1
8933
Density, p (kg/m³)
4179
385
Specific heat, Cp (/kg K)
Thermal conductivity, k (W/mK)
0.613
401
TiO₂
4250
686.2
8.9538
Transcribed Image Text:Hybrid nanofluid Consist of mixed more than one nanopartical with working fluid such as CuO A1203, TiO2 with water and the Thermo physical properties of hybrid nanofluid depend on the type of nanoparticle. For example hybrid nanofluid consist of two of nanofluid Thermo physical properties of hybrid nanofluid Hybrid Nanofluid Phnf = Pfl = p;(1-42) [(1-4) + 4 (B;})] (PCp)nag = (PCp), (1—4₂) [(1–4₂) + 4, Hj Mined (1-01)25(1-02) 25 Ps +$2ps2 Kinf köf = (pcp)st (pcp)f ksz + (S-1)kbf-(S-1) ₂ (kbf-ksz) k2 +(S-1)ky + d₂ (kb-ksz) +$₂ (PCp) sz Where kbf_ks1 + (S-1)kg-(S-1) 1 (ky-ks1) ksi + (S−1)kƒ + ₁ (kj-ks1) = kf Example: evaluate the thermal conductivity of hybrid nanofluid consist of Cu, TiO2 ,H20, 9₁-0.02 for Cu, 9₁-0.03 for TiO₂ Density, specific heat and thermal conductivity of base fluids and nanoparticles H₂O Cu Properties/constituents 997.1 8933 Density, p (kg/m³) 4179 385 Specific heat, Cp (/kg K) Thermal conductivity, k (W/mK) 0.613 401 TiO₂ 4250 686.2 8.9538
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