Fundamentals of Heat and Mass Transfer 7th Edition Bergman Solutions Manual

Fundamentals of heat and mass transfer 7th edition Bergman Solutions Manual

KNOWN: Temperature, size and orientation of Surfaces A and B in a two-dimensional geometry. Thermal conductivity dependence on temperature. FIND: Temperature gradient 8T/8y at surface A. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) No volumetric generation, (3) Two-dimensional conduction. ANALYSIS: At Surface A, kA = ko + aTA = 10 W/m-K - 10-3 W/m-K2 x 273 K = 9.73 W/m-K while at Surface B, kB = ko + aTB = 10 W/m-K - 10-3 W/m-K2 x 373 K = 9.63 W/m-K. For steady-state conditions, Ein = Eout which may be written in terms of Fourier’s law as

COMMENTS: (1) If the thermal conductivity is not temperature-dependent, then the temperature gradient at A is 15 K/m. (2) Surfaces A and B are both isothermal. Hence, 8T /dx\ = 8T /dy\ = 0.

Fundamentals of heat and mass transfer 7th edition by Theodore L. Bergman, Adrienne S. Lavine, Frank P. Incropera, David P. DeWitt solutions manual This is Solutions manual for Fundamentals of Heat and Mass Transfer Bergman Lavine Incropera DeWitt 7th edition a the solutions manual for original book, easily to download in pdf. Download clean no errors text formating: https://solutionsmanualbank.com/download/fundamentals-of-heat-and-mass-transfer-7thedition-incropera-solutions-manual/

KNOWN: A rod of constant thermal conductivity k and variable cross-sectional area Ax(x) = Aoeax where Ao and a are constants. FIND: (a) Expression for the conduction heat rate, qx(x); use this expression to determine the temperature distribution, T(x); and sketch of the temperature distribution, (b) Considering the presence of volumetric heat generation rate, q = qo exp (-ax), obtain an expression for qx(x) when the left face, x = 0, is well insulated. SCHEMATIC:

0

L

ASSUMPTIONS: (1) One-dimensional conduction in the rod, (2) Constant properties, (3) Steady- state conditions. ANALYSIS: Perform an energy balance on the control volume, A(x)-dx,

x

That is, the product of the cross-sectional area and the temperature gradient is a constant, independent of x. Hence, with T(0) > T(L), the temperature distribution is exponential, and as shown in the sketch above. Separating variables and integrating Eq. (3), the general form for the temperature distribution can be determined,

Ao exp (ax)•

dT dx

dT = QAQ 1 exp (-ax) dx

Cl

KNOWN: Electrical heater sandwiched between two identical cylindrical (30 mm dia. x 60 mm length) samples whose opposite ends contact plates maintained at To. FIND: (a) Thermal conductivity of SS316 samples for the prescribed conditions (A) and their average temperature, (b) Thermal conductivity of Armco iron sample for the prescribed conditions (B), (c) Comment on advantages of experimental arrangement, lateral heat losses, and conditions for which ATi ^ AT2. SCHEMATIC:

ASSUMPTIONS: (1) One-dimensional heat transfer in samples, (2) Steady-state conditions, (3) Negligible contact resistance between materials. PROPERTIES: Table A.2, Stainless steel 316 (T=400 K) : kss = 15.2 W/m • K; Armco iron (T=380 K) : kiron = 67.2 W/m • K.

The total temperature drop across the length of the sample is AT1(L/Ax) = 25°C (60 mm/15 mm) = 1000C. Hence, the heater temperature is Th = 177°C. Thus the average temperature of the sample is T= (To + Th)/2 = 127°C=400 K.

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We compare the calculated value of k with the tabulated value (see above) at 400 K and note the good agreement. (b) For Case B, we assume that the thermal conductivity of the SS316 sample is the same as that found in Part (a). The heat rate through the Armco iron sample is

KNOWN: Geometry and steady-state conditions used to measure the thermal conductivity of an aerogel sheet. FIND: (a) Reason the apparatus of Problem 2.17 cannot be used, (b) Thermal conductivity of the aerogel, (c) Temperature difference across the aluminum sheets, and (d) Outlet temperature of the coolant. SCHEMATIC:

Heater Cootant Tc, = 25°C

leads

in (typ)

mc = 10 kg/min

ASSUMPTIONS: (1) Steady-state conditions, (2) Constant properties, (3) One-dimensional heat transfer. PROPERTIES: Table A.1, pure aluminum [T = (Ti + Tc,i)/2 = 40°C = 313 K]: ka = 239 W/m-K. Table A.6, liquid water (25°C = 298 K): cp = 4180 J/kg-K. ANALYSIS: (a) The apparatus of Problem 2.17 cannot be used because it operates under the assumption that the heat transfer is one-dimensional in the axial direction. Since the aerogel is expected to have an extremely small thermal conductivity, the insulation used in Problem 2.17 will likely have a higher thermal conductivity than aerogel. Radial heat losses would be significant, invalidating any measured results. (b) The electrical power is Eg = V(I) = 10V x 0.125 A = 1.25 W

KNOWN: Dimensions of and temperature difference across an aircraft window. Window materials and cost of energy. FIND: Heat loss through one window and cost of heating for 130 windows on 8-hour trip. SCHEMATIC:

Ti

k —H

L = 0.01 m

ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction in the xdirection, (3) Constant properties. PROPERTIES: Table A.3, soda lime glass (300 K): kg = 1.4 W/m-K. ANALYSIS: From Eq. 2.1, , A dT , ,(T 1 -T 2 ) q x = -kA— = k a b—--------— x dx L For glass, W qx g = 1.4 -------- x 0.3 m x 0.3 m x x,g m - K

80°C 0.01m

= 1010W

The cost associated with heat loss through N windows at a rate of R = $1/kW^h over a t = 8 h flight time is Cgg ,g= Nq x g R t = 130 x 1010 W x 1 —— x 8h x 1kW =$1050 kW • h 1000W

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Repeating the calculation for the polycarbonate yields qx,p = 151W, Cp = $157

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while for aerogel, Ox,a = 101 W, Ca=$10

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COMMENT: Polycarbonate provides significant savings relative to glass. It is also lighter (pp = 1200 kg/m3) relative to glass (pg = 2500 kg/m3). The aerogel offers the best thermal performance and is very light (pa = 2 kg/m3) but would be relatively expensive.

KNOWN: Volume of unknown metal of high thermal conductivity. Known heating rate. FIND: (a) Differential equation that may be used to determine the temperature response of the metal to heating. (b) If the model can be used to identify the metal, based upon matching the predicted and measured thermal responses. SCHEMATIC:

Unknown metal of volume, V

ASSUMPTIONS: (1) Negligible spatial temperature gradients, (2) Constant properties. PROPERTIES: Table A.1 (T = 300 K): Aluminum; p = 2702 kg/m3, cp = 903 J/kg-K, Gold; p = 19300 kg/m3, cp = 129 J/kg-K, Silver; p = 10500 kg/m3, cp = 235 J/kg-K. ANALYSIS: (a) An energy balance on the control volume yields Est = Ein which may be written dT q q1 T. dT pVcn — = q or — = —-— = ----------- --p dt dt pVcp V pcp The thermal response, dT/dt, may be measured. Alternatively, the expression may be integrated to find T(t) given the initial temperature, volume, heat rate, and product of the density and specific heat. (b) For a known metal volume, V, the thermal response to constant heating is determined by the product of the density and specific heat, pcp. This product is listed below for each of the three candidate materials.

Material _____ o (kg/m3) _____ cp (J/kg-K) Aluminum Gold Silver

2702 19300 10500

903 129 235

ocP (J/m3-K) 2.44 x 106 2.49 x 106 2.47 x 106

Because the product of the densities and specific heats are so similar for these four candidate materials, in general this approach cannot be used to distinguish which material is being heated. < COMMENTS: (1) By neglecting spatial temperature gradients, the proposed approach is based only on thermodynamics principles. Therefore, it is limited in its usefulness relative to alternative schemes discussed in the text such as in Problem 2.23. (2) For many metals, the product of the density and specific heat lies within a relatively narrow band.

KNOWN: Temperatures of various materials. FIND: (a) Graph of thermal conductivity, k, versus temperature, T, for pure copper, 2024 aluminum and AISI 302 stainless steel for 300 < T < 600 K, (b) Graph of thermal conductivity, k, for helium and air over the range 300 < T < 800 K, (c) Graph of kinematic viscosity, v, for engine oil, ethylene glycol, and liquid water for 300 < T < 360 K, (d) Graph of thermal conductivity, k, versus volume fraction, p, of a water-Al2O3 nanofluid for 0 < p < 0.08 and T = 300 K. Comment on the trends for each case. ASSUMPTION: (1) Constant nanoparticle properties. ANALYSIS: (a) Using the IHT workspace of Comment 1 yields Thermal Conductivity of Cu, 2024 Al, and 302 ss 500 400 Copper