Heat Transfer Finale Exam Topics

Convection, Boiling, evaporation, heat exchangers

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Heat Transfer Finale Exam Topics by Mind Map: Heat Transfer Finale Exam Topics

1. Condensation and Boiling

1.1. Condensation

1.1.1. Happens when Ts<Tsat

1.1.2. How to Solve Find Proprties @Tsat:Vapor prop (hfg,ev) @Tf:Liq prop(all other prop). Tf=Tsat+Ts/2 Q=m*hfg*. hfg*=hfg+.68Cpl*(That-Ts) Q=hA(Tsat-Ts) h(Depends On Flow), need Re

1.2. Boiling

1.2.1. NO NEED FOR Q=hA delta T

1.2.2. How to Solve 1-Find Tsat 2-Find Texcesss= Ts-Tsat 3-based on Texcess, find equation to find q 4-Find propreties to solve for q @Tsat Table A-9: Everything except for 3 things Csf and n: Table 10-3 σ: Table 10-1

1.2.3. Q=mhfg

1.3. Special kind of convection because they depend on Surface Tension σ and Latent heat (hfg)

2. Internal Forced Convection

2.1. How to solve problems

2.1.1. 1)Find equation, such as Te=Ts-(Ts-Ti)exp(hAs/mCp)

2.1.2. 2)Find As, Ac,m, h

2.1.3. 3)h(nu)-->nu(Re), depending on Re, Laminar or Turblent, Find Lt

2.1.4. 4)Find Nu, Find h, solve equation

2.1.5. If Ts is constant: use Tlm to find Q

2.1.6. If Wpump=Pl*V

2.1.7. Hints to Solve: if question mentions that thermal resistance is negligible

3. Natural Convection

3.1. Numbers:

3.1.1. Ra (Rayligh number) Same as Pe number in forced convection: determines if the flow is more controlled by convection or conduction. Ra=Gr*Pr. Nu=C*Ra^n

3.1.2. Gr(Garshof Number) Same as Re number in forced convection: Ratio of Boyauncy force/Viscous force: As Gr increases the buoyancy overcomes friction and induce flow. Whereas Re:Inertial forces/Viscous forces

3.2. How to solve

3.2.1. Hints: if mentioned:Back side or up Side

3.2.2. 1)h(nu)-->nu(Ra)

3.2.3. 2)Ra(L, Lc), depending on type of orination, find Ra, find corresponding Nu

3.2.4. Evaulate all proprieties of fluid at Tf= Ts+Tinfinte/2. B=1/Tf

3.2.5. Estimate Ts or get an estimate from Tf

4. Heat Exchangers

4.1. Equations

4.1.1. U=/ Ui =/ Uo

4.1.2. Rtot=1/UAs = 1/UiAsi,i = 1/UoAs,o UAs=UiAs,i=UoAs,o and As = As,i = As,o, hence 1/U = 1/Ui = 1/Uo

4.1.3. Q=deltaT/Rtotal = UAsDeltaT= UiAsDeltaT = UoAsDeltaT

4.1.4. 1/UAs = Σ 1/hAs + ln(Do/Di)/2*pi*k *L +ΣRf/As

4.2. How to solve

4.2.1. LMTD Method use when having all exist and inlet temperatures Q= UAsTlmF Find Tlm based on type of current Find U from eqn F from graph for complicated shell and tube or cross flow

4.2.2. NTU Method Qc=Cc (Tc, out- Tc, in) Cmin=Cp*m e=Q/Qmax e= TEMP OF C MIN/Th,in - Tc, in e, from tables depending on type of flow e=emax= 1- exp(-NTU) at c= cmin/max= zero at cmax goes to infinity Qmax=Cmin(Th,in -Tc,in) NTU=UAs/Cmin No NTU>3

5. General reviews

5.1. Constants

5.1.1. K: Thermal Conductivity: Increases with Temp for liquids but decreases with higher temp for gases

5.1.2. α:Thermal diffusivity:Heat conducted / Heat stored = K/eCp. Low α: insulating material low heat propagates through the medium

5.1.3. e=Emmisivity of the surface, resembles how perfect black body the surface is, the black body has an e=1

5.1.4. ℐ=pentratin depth = (tα)^1/2

5.2. Ideas

5.2.1. Feel cooler in the winter: body radiates 4 times for than in summer, hence more energy lost in winter due to the greater effect between the amb and body

5.2.2. Thermal contact resistance: Rc Temprature drop at interference of surfaces :

5.2.3. critical radius =: kmax/hmin, going up increases the r hence increases Q convection but we get to rc where Qconv<Qinsul. However, addiing resistance to a wall increases length which decreases Q

5.2.4. How to increase Q? 1)forced convection: increase h by adding fans 2)Increase surface area by adding fins made of Highly conductive materials

5.2.5. Effectivness for fin: e for find is < 1, fin acts as an insulator How to increase effectiveness? increase k and L and lower h. Slim and tall and metallic Best Fin length choose? @mL=1 and gives 76% of an infinity long fin heat at only 1/5 length of the super long fin

5.2.6. Lumped system: temperature with respect to time is uniform the system

5.2.7. Boundary Layer: Forms due to no slip condition and Friction

5.2.8. The thickness of the thermal boundary layer increases in the flow direction, since the effects of heat transfer are felt at greater distances from the surface further down stream.

5.2.9. As temp goes up: viscosity goes down for fluids but goes up for gases

5.2.10. Surface roughness, and the flow type and drag coeffiecnt: Surface roughness, in general, increases the drag coefficient in turbulent flow. This is especially the case for streamlined bodies. For blunt bodies such as a circular cylinder or sphere, however, an increase in the surface roughness may increase or decrease the drag coefficient depending on Reynolds number.

5.2.11. Air acts as an insulator for thermal resistance, removing air increases the thermal resistance

5.3. Numbers

5.3.1. Biot= resistance of (conduction/ convection), m=(hp/kAc)^1/2. Bi=hL/k = hV/kAs

5.3.2. Nu= hL/k. Dimensionless convection heat transfer coefficient. The Nusselt number represents the enhancement of heat transfer through a fluid layer as a result of convection relative to conduction across the same fluid layer. The larger the Nusselt number, the more effective the convection.

5.3.3. Pr: v/alpha: The relative thickness of the velocity and the thermal boundary layers is best described by the dimensionless parameter Prandtl number. The Prandtl numbers of gases are about 1, which indicates that both momentum and heat dissipate through the fluid at about the same rate. Heat diffuses very quickly in liquid metals (Pr << 1) and very slowly in oils (Pr >> 1) relative to momentum. Consequently the thermal boundary layer is much thicker for liquid metals and much thinner for oils relative to the velocity boundary layer.

5.3.4. Recritcal= 5*10^5, used to find Lc, critical length for flow over objects, when flow becomes turbulent

5.3.5. Lc: V/As characteristics length