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Heat Transfer Review D. H. Willits Biological and Agricultural Engineering North Carolina State University.

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Presentation on theme: "Heat Transfer Review D. H. Willits Biological and Agricultural Engineering North Carolina State University."— Presentation transcript:

1 Heat Transfer Review D. H. Willits Biological and Agricultural Engineering North Carolina State University

2 Steady-State Ht Conduction – Composite Plane Wall (Fig 6.5)



5 Thermal Conductivity Values Interpretation of the values in Table 6.2 requires an understanding of the difference between resistivity and resistance. Resistivity = 1/k Resistance =  x/k To get resistance from resistivity, you must multiply by the thickness of the material.

6 Steady-State Ht Conduction – Composite Cylinder (Fig 6.8)


8 Steady-State Ht Conduction – Composite Sphere (Fig 6.10)

9 Steady-State Ht Conduction – Problem Consider a composite cylindrical tube with an outside diameter of 10 cm. The wall consists of two layers of different materials, A and B. The inner material A is in contact with a hot fluid and the outer material B is in contact with still air at a temperature of 30 C. Material A is stainless steel, 0.2 cm thick, and material B is 0.3 cm thick with a thermal conductivity of 0.0378 W m -1 K -1. If the outer surface temperature is 110 C, and the outside surface coefficient can be estimated as h o = 7.36 W/m 2 K, determine: a) the heat transfer through the wall, in W/m of length b) the temperature of the interface between the two materials c) the temperature of the hot fluid if the inside surface coefficient is estimated at 20 W m -2 K -1

10 Steady-State Ht Conduction – Problem Answers: a) the heat loss per unit length– b) the temperature at the interface – t interface = 158.2 C.

11 Steady-State Ht Conduction – Problem Answers: (c) the temperature of the hot fluid

12 Critical Radii for Cylinders and Spheres Radius at which maximum heat transfer occurs: Cylinder  Biot No. = 1.0 = r o h o /k Sphere  Biot No. = 2.0

13 Transient Ht Transfer Case 1: Case 2: Case 3: Heisler Charts Figs 6.11-6.13

14 Transient Ht Transfer Heisler Charts: For non-infinite geometries, TR values are multiplied together

15 Transient Ht Transfer – Problem 1 A 3 cm diameter hot dog with a length of 10 cm has an initial uniform temperature of 10 C. If it is suddenly dropped into boiling water at 100 C, determine the temperature at the center after 10 min. Assume the following values: h = 6000 W/m 2 K k = 0.5 W/m K α = 1.33 X 10 -7 m 2 /s.

16 Transient Ht Transfer – Problem 1 Intersection of cylinder and slab for cylinder: Bo = hr/k = 180 ; 1/Bo = 0.0056 (Case 2);  =  /r 2 = 0.3547; TR 1 = 0.205 (from Fig 6.12;) for slab: Bo = hL/k = 600; 1/Bo = 0.0016665 (Case 2);  =  /r 2 = 0.0.03192; TR 2 = 0.99 (from Fig. 6.11); TR = TR 1 x TR 2 = 0.20295; t c = 81.73 C

17 Transient Ht Transfer – Problem 2 An aluminum cylinder (thermal conductivity = 160 W/m K, density = 2790 kg/m 3, specific heat = 0.88 kJ/kg K) of radius r = 5 cm, length L = 0.5 m, and a uniform initial temperature of 200 C is suddenly immersed at time zero in a well-stirred fluid maintained at a constant temperature of 25 C. The heat transfer coefficient between the cylinder and the fluid is h = 300 W/m2 K. Determine the time required for the center of the cylinder to reach 50 C. What will the surface temperature be at that time?

18 Transient Ht Transfer – Problem 2 Check B o : for the cylinder: B o = hr o /k = (300W/m 2 K)(0.05m)/(160W/mK) = 0.094 This is Case 1 Note: we do not have to check B o for the slab because Case 1 says that the internal temperature gradient for the cylinder is already negligible, which says that the heat is conducted to the edge of the material faster than it can be convected away by the water. The slab case will not change that.

19 Transient Ht Transfer – Problem 2 For Case 1, use Eq 6.120 in the text: solving for  gives 6.03 min or 361.8 s The surface is the same as the center because Case 1 assumes no internal gradient.

20 Convection The basic problem is to find the appropriate h to use in Newton’s Law of Cooling. Once that is done, finding q x is fairly trivial:

21 Convection Free Horizontal cylinder, laminar flow – 1x10 4 < GrPr < 1x10 8 Horizontal cylinder, turbulent flow – 1x10 8 < GrPr < 1x10 12

22 Convection Free

23 Convection Forced Eqs 7.49 – 7.53 where

24 Convection - Problem Air at 60 C if flowing normal to a cylindrical copper tube with a diameter of 15 cm at a velocity of 2 m/s. Estimate the convective heat transfer coefficient at the surface.

25 Convection - Problem Using Eq 7.51, with properties determined at T film rho1.0604kg/m^3Density of air mu.00002005Pa*sViscosity of air cp1007J/kg*KSpecific Heat of air k.02856W/m*KThermal conductivity of air Tfilm60CFilm temperature Pr.706945028011204-Prandtl No. Nu77.6381180236944-Nusselt No. h14.7822976717114W/m^2Conv heat transfer coefficient Re15866.3341645885-Reynold's Number

26 Heat Exchangers – Basic Eqns Fig 7.1

27 Heat Exchangers – Parallel Flow

28 Heat Exchangers – Effectiveness Ratio

29 Figs 7.3 and 7.4, or

30 Radiation Heat Transfer Black Body Emissive Power where T is absolute temperature and  depends upon the unit system.

31 Radiation Heat Transfer Gray Body Emissive Power where  is the emissivity. Note: Gray bodies have constant  with wavelength.

32 Radiation Heat Transfer Gray Body Exchange

33 Radiation Heat Transfer The first problem is determining the shape factor F ij. For simple geometries, Table 8.3 may suffice. F ij is the fraction of the energy leaving i that is intercepted by j. For infinite parallel planes, the value is 1.0. For small bodies enclosed by a larger body (where the smaller body cannot see itself), the value is also 1.0.

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