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1. Nusselt Number and Biot Number Are Computed in a Similar Manner (=Hd/K)

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1. Nusselt Number and Biot Number Are Computed in a Similar Manner (=Hd/K) 1. Nusselt number and Biot number are computed in a similar manner (=hD/k). What are the differences between them? When and why are each of them used? 2. During unsteady state heat transfer, can the temperature ratio (TR) = (T - T" )/(Ti - T" ) be greater than 1.0 for: a. Heating b. Cooling EXPLAIN your answer. 3. What are the quantities represented on the Heisler chart? What are their units? 4. For a counter-current heat exchanger, the following are the inlet and exit temperatures of the hot and cold fluids: Thi = 80 °C T ho = 60 °C T ci = 45 °C T co = 55 °C Without actually calculating it, EXPLAIN which of the following is a possible value for the logarithmic mean temperature difference (in °C)? a. 14.7 b. 3.8 c. 19.6 d. 26.3 5. Are the following statements true or false? a. Grashof number comes into play for forced convection b. When the Biot number is greater than 40, it can be assumed that the convective heat transfer coefficient (h) is negligible. c. In a counter-current heat exchanger, the exit temperature of the cold fluid can be higher than the inlet temperature of the hot fluid. d. When insulation is added to the outside of a pipe such that the outside radius of the pipe (including the insulation) becomes equal to the critical radius, then the heat loss from the pipe is at its maximum. e. Higher the specific heat of a product, the faster it heats up. 6. What are the units of thermal resistance? 7. When and why do we use logarithmic mean temperature difference? 8. When subjected to similar conditions, will the initial rate of cooling/heating of the center of an infinitely long cylinder be greater than, less than, or equal to that of a finite cylinder of the same radius. Explain physically and mathematically. 1. A product (cp = 4000 J/kg-K) is being heated in a 10 m long tubular heat exchanger from 20 °C to 50 °C when it is flowing at the rate of 0.1 kg/s in the inside tube. Hot water, flowing in the same direction as the product is the heating medium. Hot water (flowing at 0.15 kg/s) enters the heat exchanger at 95 °C and exits at 65 °C. The inside and outside diameters of the inside tube are 6 cm and 6.5 cm respectively while those of the outside tube are 8 and 8.5 cm respectively. The thermal conductivity of the material of the heat exchanger is 8 W/m-K and the ambient air temperature is 5 °C. In order to minimize the heat loss to the surroundings, insulation (k = 0.2 W/m-K) of thickness 1 mm is added to the outside of the outer tube. a. Determine the overall heat transfer coefficient for transfer of heat between the product and the hot water. b. Determine the convective heat transfer coefficient between the outside wall of the heat exchanger and the ambient air. Neglect the convective resistance between hot water and the inner tube of the outer wall. 3 2. A product (k = 0.5 W/m-K, ρ = 950 kg/m , cp = 4000 J/kg-K) in a cylindrical container of height 10 cm and radius 4 cm is taken out of the oven when it is at 90 °C and placed under ambient conditions (20 °C). The convective heat transfer coefficient between the container and ambient air is 25 W/m2 -K. a. Determine the center temperature of the product after 1 hour and 45 minutes. b. Outline the procedure for determining the time taken for the center temperature of the product to reach 40 °C. 3 3. A hot product (cp = 4000 J/kg-K, µ = 0.001 Pa-s, ρ = 900 kg/m , k = 0.5 W/m-K) is being conveyed in a 20 m long metal pipe of diameter 6 cm and negligible thickness at the rate of 0.5 kg/s. The inlet and exit temperatures of the hot product are 90 °C and 80 °C respectively. In order to reduce the heat loss to half its current value, insulation of 5 mm thickness is added to the outside of the pipe. If the convective heat transfer coefficient between the ambient air (which is at 10 °C) and either the pipe or the insulation is 100 W/m2 -K, determine the thermal conductivity of the insulation to be used. MAKE NECESSARY ASSUMPTIONS. 1. Nusselt number Comes into play for steady state heat transfer (forced and free convection) The thermal conductivity to be used is that of the fluid Nusselt # is used to determine convective heat transfer coefficient Biot number Comes into play for unsteady state heat transfer The thermal conductivity to be used is that of the solid object Biot # is used to determine if we use the Heisler chart or the lumped parameter analysis 2. The temperature ratio (TR) is the ratio of the difference in temperature between the center of the product & the surroundings to the difference in temperature between the initial temperature of the product and the surroundings. Irrespective of whether it is a cooling or heating process, Ti - T"" will always be higher than T - T , with the ratio becoming progressively smaller with time, till it approaches zero after a long period of time. Thus, TR cannot be greater than 1.0 for either heating or cooling. 3. Temperature ratio -- unitless 1/NBi (or k/hD) -- unitless 2 NFo (= αt/D ) -- unitless 4. Here, ∆T12 = 80 - 55 = 25 °C, ∆T = 60 - 45 = 15 °C Thus, ∆Tlm has to be a value in between these two values and the only number that matches that requirement is 19.6 °C. 5. a. False b. False c. False d. True e. False 6. Thermal resistance is given by ∆x/kA for conduction and 1/hA for convection. Thus, its units are: K/W. 7. We use logarithmic mean temperature difference when the difference in temperature between two ends of a solid are different along its length OR when the difference in temperature difference in temperature between a solid/fluid and a solid/fluid varies with position. We use logarithmic mean temperature difference in order to determine the average flow of energy from the hot end to the cold end, with ∆Tlm being the driving force. 8. For an infinite cylinder, heat transfer takes place only in the radial direction. For a finite cylinder, heat transfer takes place in both the radial and axial direction. Thus, the initial rate of heating/cooling will be higher for the finite cylinder. Mathematically, (TR)finite cylinder = (TR) infinite cylinder x (TR) infinite slab Since both the quantities on the RHS of the equation are positive and less than 1.0, we can see that the TR for the finite cylinder will be less than that for the infinite cylinder. Lower the temperature ratio, the smaller the difference in temperature between ‘T’ and ‘T" ’ and hence the faster the rate of heating/cooling. 1. The system diagram is as shown below: a. This is a forced convection problem. In this problem, Q1 is the energy gained by the product and is given by: Q1 = 0.1 (4000) (50 - 20) = 12,000 W Also, Q1 = U 1 (A lm ) (∆T lm ) 2 Here, Alm = 1.96 m (based on r i = 0.03 m, r o = 0.0325 m, L = 10 m) ∆Tlm = 37.3 °C [based on ∆T 1 = (95 - 20) = 75 °C and ∆T 2 = (65 - 50) = 15 °C] 2 Thus, U1 = 164 W/m -K b. The energy lost by the hot water is given by: Q2 = 0.15 (4194) (95 - 65) = 18,873 W [Note that cp is determined at the average temperature -- (95 + 65)/2 = 80 °C] Thus, the energy lost to the surroundings is given by: Q321 = Q - Q = 18873 - 12000 = 6,873 W Considering transfer of heat between hot water and the surroundings and neglecting the resistance to heat transfer between hot water and the inside surface of the outer tube, we get: Q3 = U 2 (A lm ) (∆T lm ) ∆Tlm = 74 °C [based on ∆T 1 = (95 - 5) = 90 °C and ∆T 2 = (65 - 5) = 60 °C] 1/(U2 A lm ) = ∆T lm /Q = (∆r/kA lmpipe ) + (∆r/kA lmins ) + (1/h o A o ) 2 For the pipe: rio = 0.04 m, r = 0.0425 m, ∆r = 0.0025 m, A lm = 2.59 m , k = 8 W/m-K 2 For insulation: rio = 0.0425 m, r = 0.0435 m, ∆r = 0.001 m, A lm = 2.7 m , k = 0.2 W/m-K 2 Also, Aoo = 2πr L = 2π(0.0435)20 = 2.73 m Thus, 74/6873 = [(0.0025)/(8)(2.59)] + [(0.001)/(0.2)(2.7)] + [1/(ho )(2.73)] 2 Solving, we get: ho = 42 W/m -K 2. a. A finite cylinder is obtained by the intersection of an infinite cylinder and an infinite slab. For infinite cylinder: D = 0.04 m X NBi = hD/k = 2.0 Thus, k/hD = 0.5 For infinite slab: D = 0.05 m X NBi = hD/k = 2.5 Thus, k/hD = 0.4 -7 2 Thus, we use the Heisler chart. α = k/(ρ cp ) = 1.32 x 10 m /s 22 NFo = αt/D = 0.52 for the infinite cylinder and N Fo = αt/D = 0.33 for the infinite slab Thus, TR = 0.4 for the infinite cylinder and TR = 0.9 for infinite slab Thus, TR for finite cylinder = (0.4) (0.9) = 0.36 = (T - T" ) / (Ti - T" ) X T = 45.2 ° C b.
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