Daily Heat Loss Cost Analysis for Concrete Slabs and Systems
School
University of British Columbia**We aren't endorsed by this school
Course
CHBE 352
Subject
Mechanical Engineering
Date
Dec 11, 2024
Pages
18
Uploaded by SuperHumanPheasant169
PROBLEM 1.5KNOWN:Dimensions, thermal conductivity and surface temperatures of a concrete slab. Efficiency of gas furnace and cost of natural gas. FIND:Daily cost of heat loss. SCHEMATIC:Furnace, = 0.90ηfNatural gas,C= $0.01/MJgConcrete, k = 1.4 W/m-Kt = 0.2 mL = 11 mW = 8 mT= 17C 1oT= 10C 2oq Warm airASSUMPTIONS:(1) Steady state, (2) One-dimensional conduction, (3) Constant properties. ANALYSIS:The rate of heat loss by conduction through the slab is ()()12TT7 CqkLW1.4W / m K 11m8m4312 Wt0.20m−°==⋅×=<The daily cost of natural gas that must be combusted to compensate for the heat loss is ()()gd6fqC4312W$0.02/ MJCt24h /d3600s/ h$8.28/d0.910 J / MJη×=∆=×=×<COMMENTS:The loss could be reduced by installing a floor covering with a layer of insulation between it and the concrete.
PROBLEM 1.18KNOWN:Length, diameter and calibration of a hot wire anemometer. Temperature of air stream. Current, voltage drop and surface temperature of wire for a particular application. FIND:Air velocity SCHEMATIC:AirT= 75CsoHot wire (V ~ h)2L = 20 mm, D = 0.5 mmE = 5V, I = 100 mAT = 25C oooV, ASSUMPTIONS:(1) Steady-state conditions, (2) Negligible heat transfer from the wire by natural convection or radiation. ANALYSIS:If all of the electric energy is transferred by convection to the air, the following equality must be satisfied ()PEIhA TTelecs==∞where ()52ADL0.0005m0.02m3.1410m .ππ−==×=×Hence, ()()EI5V0.1A2h318 W/mK52A TTs3.1410m50 C×===−−∞×()25252V6.2510h6.2510318 W/mK6.3 m/s−−=×=×⋅=<COMMENTS:The convection coefficient is sufficiently large to render buoyancy (natural convection) and radiation effects negligible.
PROBLEM 1.41 KNOWN: Blood inlet and outlet temperatures and flow rate. Dimensions of tubing.FIND: Required rate of heat addition and estimate of kinetic and potential energy changes. SCHEMATIC: ASSUMPTIONS: (1) Steady-state conditions, (2) Incompressible liquid with negligible kinetic and potential energy changes, (3) Blood has properties of water. PROPERTIES: Table A.6, Water (T≈ 300 K): cp,f =4179 J/ kg∙ K, ρf = 1/vf = 997 kg/m3. ANALYSIS: From an overall energy balance, Equation 1.12e,poutinq = mc(T- T)where 3-63-3fm = ρ= 997 kg/m× 200 m/min × 10m /m 60 s/min = 3.32 × 10kg/s∀Thus -3ooq = 3.32 × 10kg/s × 4179 J/kgK × (37C - 10C) = 375 W⋅< The velocity in the tube is given by -63-3-3cV = /A= 200 m/min × 10m /m (60 s/min × 6.4 × 10m × 1.6 × 10m)= 0.33 m/s∀The change in kinetic energy is 2-32-41122m(V- 0) = 3.32 × 10kg/s × × (0.33 m/s)= 1.8 × 10W<The change in potential energy is -32mgz = 3.32 × 10kg/s × 9.8 m/s× 2 m = 0.065 W<COMMENT: The kinetic and potential energy changes are both negligible relative to the thermal energy change. 1.6 mm6.4 mmTout= 37°C2 mẠ= 200 mℓ/min, Tin= 10°CBlood1.6 mm6.4 mmTout= 37°C2 mẠ= 200 mℓ/min, Tin= 10°CBlood
PROBLEM 2.21KNOWN:Steady-state temperature distribution in a cylindrical rod having uniform heat generation of 731610W/mq=×. FIND:(a) Steady-state centerline and surface heat transfer rates per unit length, ′qr.(b) Initial time rate of change of the centerline and surface temperatures in response to a change in the generation rate from 8312qto q= 10W/m .SCHEMATIC:ASSUMPTIONS:(1) One-dimensional conduction in the r direction, (2) Uniform generation, and (3) Steady-state for 731610W/mq=×. ANALYSIS:(a) From the rate equations for cylindrical coordinates, ′′ = −qkTrq = -kATrrr∂∂∂∂.Hence, ()rTqk 2rLr∂π∂= −or ′ = −qkrTrr2π∂∂(1) where ∂T/∂r may be evaluated from the prescribed temperature distribution, T(r). At r = 0, the gradient is (∂T/∂r) = 0. Hence, from Equation (1) the heat rate is ()rq00.′=<At r = ro,the temperature gradient is()()()oo55o2r=r3r=rTK2 5.2610r2 5.26100.030mrmT31.610K/m.r∂∂∂∂=−×=−×=−×Continued ... ro= 0.030 mT(r) = 900 – 5.26∙105r2= 6∙107W/m3
PROBLEM 2.21 (Cont.)Hence, the heat rate at the outer surface (r = ro) per unit length is()[]()3roqr230 W/mK0.030m31.610K/mπ′=−⋅−×()5roqr1.78510W/m.′=×<(b) Transient (time-dependent) conditions will exist when the generation is changed, and for the prescribed assumptions, the temperature is determined by the following form of the heat equation, Equation 2.26 2p1TTkrqcrrrt∂∂∂r∂∂∂+=Hence 2pT11Tkrq.tcrrr∂∂∂∂r∂∂=+However, initially (at t = 0), the temperature distribution is given by the prescribed form, T(r) = 800 - 5.26×105r2,and()51Tkkrr -10.5210rrrrrr∂∂∂∂∂∂=×⋅()5k21.0410rr=−×⋅5230 W/m K -21.0410K/m=⋅×()7316.3110W/m the original q=q.=×Hence, everywhere in the wall, 7833T16.311010W/mt1100 kg/m800 J/kgK∂∂=−×+×⋅or 41.91 K/sTt∂=∂< COMMENTS:(1) The value of (∂T/∂t) will decrease with increasing time, until a new steady-state condition is reached and once again (∂T/∂t) = 0. (2) By applying the energy conservation requirement, Equation 1.12c, to a unit length of the rod for the steady-state condition,.′−′+′=EEEinoutgen0Hence( )()()2rro1oq0qrqr.π′′−=−
PROBLEM 3.4 KNOWN:Thermal conductivities and thicknesses of original wall, insulation layer, and glass layer. Interior and exterior air temperatures and convection heat transfer coefficients. FIND:Heat flux through original and retrofitted walls. SCHEMATIC: ASSUMPTIONS: (1) One-dimensional conduction, (2) Steady-state conditions, (3) Constant properties, (4) Negligible contact resistances. ANALYSIS: The original wall with convection inside and outside can be represented by the following thermal resistance network, where the resistances are each for a unit area: Thus the heat flux can be expressed as ,,22222 C0 C41.3 W/m10.030 m1115 W/mK0.1 W/mK30 W/mKiosisoTTqLhkh∞∞−°−°′′ ===++++⋅⋅⋅<The retrofitted wall has three layers. The thermal circuit can be represented as follows: Thus the heat flux can be expressed as ,,2221122 C0 C14.0 W/m10.030 m0.030 m0.005 m15 W/mK0.1 W/mK0.029 W/mK1.4 W/mK25 W/mKiogsiisigoTTqLLLhkkkh∞∞−′′ =++++°−°==++++⋅⋅⋅⋅⋅<COMMENTS:The heat flux has been reduced to approximately one-third of the original value because of the increased resistance, which is mainly due to the insulation layer. 1/hi Ls/ks 1/ho T∞,iT∞,oT∞,o1/hi Ls/ks 1/ho T∞,iLi/ki Lg/kg
PROBLEM 3.44KNOWN:Diameter of electrical wire. Thickness and thermal conductivity of rubberized sheath. Contact resistance between sheath and wire. Convection coefficient and ambient air temperature. Maximum allowable sheath temperature. FIND:Maximum allowable power dissipation per unit length of wire. Critical radius of insulation. SCHEMATIC:ASSUMPTIONS:(1) Steady-state, (2) One-dimensional radial conduction through insulation, (3) Constant properties, (4) Negligible radiation exchange with surroundings. ANALYSIS:The maximum insulation temperature corresponds to its inner surface and is independent of the contact resistance. From the thermal circuit, we may write ()()in,iin,igcondconvin,oin,iin,oTTTTEqRRn r/ r/ 2 k1/ 2 rhππ∞∞−−′′===′′++where in,iin,oin,irD / 20.0015m, rrt0.0035m,===+=and in,imaxTT50 C==yields the maximum allowable power dissipation. Hence, ()()()g,max25020C30 CE7.37 W / mn 2.33311.043.03 m K / W20.13W / m K20.0035m 15 W / mKππ−°°′===+⋅+×⋅⋅<The critical insulation radius is also unaffected by the contact resistance and is given by cr2k0.13W / m Kr0.0087m8.7mmh15W / mK⋅====⋅<Hence, rin,o< rcrand g,maxE′could be increased by increasing rin,oup to a value of 8.7 mm (t = 7.2 mm). COMMENTS:The contact resistance affects the temperature of the wire, and for g,maxqE′′=7.37 W / m,=the outer surface temperature of the wire is Tw,o= Tin,i+t,cq R50 C′′=()7.37 W / m+()()42310mK / W /0.003m50.2 C.π−×⋅= °Hence, the temperature change across the contact resistance is negligible.
PROBLEM 3.65KNOWN:Plane wall with internal heat generation which is insulated at the inner surface and subjected to a convection process at the outer surface. FIND:Maximum temperature in the wall. SCHEMATIC:ASSUMPTIONS:(1) Steady-state conditions, (2) One-dimensional conduction with uniform volumetric heat generation, (3) Inner surface is adiabatic.ANALYSIS: The temperature at the inner surface is given by Eq. 3.48 and is the maximum temperature within the wall, 2osTqL / 2k+T .=The outer surface temperature follows from Eq. 3.51, s62s3TTqL/hWT92 C+0.4100.2m/400W/mK=92 C+200 C=292 C.m∞=+=××⋅It follows that ()263oT0.410 W/m0.2m/ 230W/m K+292 C=×××⋅oT267 C+292 C=559 C.=<COMMENTS:The heat flux leaving the wall can be determined from knowledge of h, Tsand T∞using Newton’s law of cooling. ()()22convsqh TT400W/mK 29292C=80kW/m .∞′′=−=⋅−This same result can be determined from an energy balance on the entire wall, which has the form goutEE0−=where goutconvEqAL and EqA.′′==Hence, 632convqqL=0.410 W/m0.2m=80kW/m .′′=××
PROBLEM 4.2 KNOWN:Two-dimensional rectangular plate subjected to prescribed uniform temperature boundary conditions. FIND:Temperatures along the mid-plane at y = 0.25, 0.5, and 0.75 m considering the first five non-zero terms; assess error resulting from using only first three terms. SCHEMATIC:ASSUMPTIONS:(1) Two-dimensional, steady-state conduction, (2) Constant properties. ANALYSIS:From Section 4.2, the temperature distribution is ()()()()n 1121n 111sinh ny LTT2nxx, ysinTTnLsinh nW Lθππθππ+=−+−≡=⋅−∑. (1,4.19) Considering now the point (x,y) = (1.0,0.5) and recognizing x/L = 1/2, y/L = 1/4 and W/L = 1/2, ()()()()n 1121n 111sinh n4TT2n1,0.5sinTTn2sinh n2θππθππ+=−+−≡=⋅−∑. (2) When n is even (2, 4, 6 ...), the corresponding term is zero; hence we need only consider n = 1, 3, 5, 7 and 9 as the first five non-zero terms. ()()()()()sinh4sinh 342231,0.52sinsin2sinh232sinh 32ππππθπππ=++()()()()()()sinh 54sinh 74sinh 94252729sinsinsin52sinh 5272sinh 7292sinh 92πππππππππ++()[]21,0.50.7550.0630.0080.0010.0000.445θπ=−+−+=()()()()211T 1,0.51,0.5TTT0.445 150505094.5 Cθ=−+=−. <Repeating the calculation for y = 0.25 and 0.75 m, the only thing that changes is y/L = 1/8 and 3/8 respectively in the sinh term in the numerator. The results are (repeating the above result for completeness): ()()()(1,0.25)0.212, T 1,0.2571.2 C,(1,0.50)0.445, T 1,0.2594.5 C,(1,0.75)0.711, T 1,0.75121 C.θθθ======<Continued...
PROBLEM 4.2 (Cont.) If only the first three terms of the series, Eq. (2), are considered, the results are: ()()()(1,0.25)0.212, T 1,0.2571.2 C,(1,0.50)0.446, T 1,0.2594.6 C,(1,0.75)0.719, T 1,0.75122 C.θθθ======<The worst error is for y = 0.75 m, with only a 0.6% difference. COMMENTS:The number of terms needed for an accurate result depends on the location, with more terms need near those corners where there is a discontinuous change in temperature.
PROBLEM 4.15KNOWN:Dimensions and boundary temperatures of a steam pipe embedded in a concrete casing. FIND:Heat loss per unit length. SCHEMATIC:ASSUMPTIONS:(1) Steady-state conditions, (2) Negligible steam side convection resistance, pipe wall resistance and contact resistance (T1= 400K), (3) Constant properties. PROPERTIES:Table A-3, Concrete (300K): k = 1.4 W/m⋅K. ANALYSIS:The heat rate can be expressed as ()1-212qSk TSk TT= ∆=-From Table 4.1, the shape factor is 2LS.1.08 wnDπ=Hence, ()122 k TTqq1.08 wLnDπ-′ ==()21.4W/m K400300 Kq767 W/m.1.081.75mn0.6mπ×⋅×-′ ==×<COMMENTS:Having neglected the steam side convection resistance, the pipe wall resistance, and the contact resistance, the foregoing result overestimates the actual heat loss.
PROBLEM 4.38KNOWN:Two-dimensional cylindrical configuration with prescribed radial (∆r) and angular (∆φ) spacings of nodes. FIND:Finite-difference equations for nodes 2, 3 and 1. SCHEMATIC:ASSUMPTIONS:(1) Steady-state conditions, (2) Two-dimensional conduction in cylindrical coordinates (r,φ), (3) Constant properties. ANALYSIS:The method of solution is to define the appropriate control volume for each node, to identify relevant processes and then to perform an energy balance. (a) Node 2. This is an interiornode with control volume as shown above. The energy balance is inabcdEqqqq0.′′′′=+++=Using Fourier’s law for each process, find()()()()()()()()5232iii212iiTTTT3krrkr2rrrTTTT1krrkr0.2rrrφφφφ−−+∆∆+∆+∆+ ∆∆−−++∆∆+∆=∆+ ∆∆Canceling terms and regrouping yields, ()()()()()()()()22i2i531ii22iirr1312rrTrrTTTrrT0.rr22rrφφ∆∆−+ ∆+++∆++++∆=+ ∆∆+ ∆∆(b) Node 3. The adiabatic surface behaves as a symmetry surface. We can utilize the result of Part (a) to write the finite-difference equation by inspection as ()()()()()()()22i3i62ii22iir2r1312rrTrrTTrrT0.rr22rrφφ∆∆−+ ∆+++∆+⋅++∆=+ ∆∆+ ∆∆(c) Node 1. The energy balance is abcdqqqq0.′′′′+++=Substituting,()()()()4121iiTTTT3krrkr22rrrφφ−−∆+∆+∆+∆+ ∆∆()()()i1i1TT1krrhrTT022rφ∞−∆++∆+∆−=∆<This expression could now be rearranged.
PROBLEM 4.58 KNOWN:Long bar of square cross section, three sides of which are maintained at a constant temperature while the fourth side is subjected to a convection process. FIND:(a) The mid-point temperature and heat transfer rate between the bar and fluid; a numerical technique with grid spacing of 0.2 m is suggested, and (b) Reducing the grid spacing by a factor of 2, find the midpoint temperature and the heat transfer rate. Also, plot temperature distribution across the surface exposed to the fluid. SCHEMATIC: ASSUMPTIONS:(1) Steady-state, two-dimensional conduction, (2) Constant properties. ANALYSIS:(a) Considering symmetry, the nodal network is shown above. The matrix inversion method of solution will be employed. The finite-difference equations are: Nodes 1, 3, 5- Interior nodes, Eq. 4.29; written by inspection. Nodes 2, 4, 6- Also can be treated as interior points, considering symmetry. Nodes 7, 8- On a plane with convection, Eq. 4.42; noting that h∆x/k = 10 W/m2⋅K ×0.2 m/2W/m⋅K = 1, find Node 7: (2T5+ 300 + T8) + 2×1⋅100 - 2(1+2)T7= 0 Node 8: (2T6+ T7+ T7) + 2×1⋅100 - 2(1+2)T8= 0 The solution matrix [T] can be found using a stock matrix program using the [A] and [C] matrices shown below to obtain the solution matrix [T] (Eq. 4.48). Alternatively, the set of equations could be entered into the IHT workspace and solved for the nodal temperatures. 41100000600292.224010000300289.210411000300279.7012401000272.2ACT00104110300254.5000124010240.100002061500198.100000226200179.4−−−−−−−===−−−−−−−From the solution matrix, [T], find the mid-point temperature as T4= 272.2°C <Continued...
PROBLEM 4.58 (Cont.) The heat rate by convection between the bar and fluid is given as, ()convabcq2 qqq′′′′=++()()()()()()conv87q2 hx 2TThxTThx 2300T∞∞∞′=∆−+∆−+∆−() ()()()2convq2 10W mK0.2m 2179.41002 198.1100300100K′=⋅×−+−+−convq952W m′=. <(b) Reducing the grid spacing by a factor of 2, the nodal arrangement will appear as shown. The finite-difference equation for the interior and centerline nodes were written by inspection and entered into the IHT workspace. The IHT Finite-Difference Equations Toolfor 2-D, SS conditions, was used to obtain the FDE for the nodes on the exposed surface. The midpoint temperature T13and heat rate for the finer mesh are T13= 271.0°C q′= 834 W/m <COMMENTS:The midpoint temperatures for the coarse and finer meshes agree closely, T4= 272°C vs. T13= 271.0°C, respectively. However, the estimate for the heat rate is substantially influenced by the mesh size; q′= 952 vs. 834 W/m for the coarse and finer meshes, respectively.
PROBLEM 5.7KNOWN:Diameter and initial temperature of steel balls cooling in air. FIND:Time required to cool to a prescribed temperature. SCHEMATIC:D = 0. 01 mh = 25 W/m2• KASSUMPTIONS:(1) Negligible radiation effects, (2) Constant properties. ANALYSIS:Applying Eq. 5.10 to a sphere (Lc= ro/3), ()()2och r/ 325 W/mK 0.005 m/3hLBi0.001.kk40 W/mK⋅====⋅Hence, the temperature of the steel remains approximately uniform during the cooling process, and the lumped capacitance method may be used. From Eqs. 5.4 and 5.5, ()3ppii2sD/ 6 cVcTTTTtlnlnhATTTThDρ pρp∞∞∞∞−−==−−()327800 kg/m0.01 m600 J/kgK1150325tln450325625 W/mK⋅−=−×⋅t589 s 0.164 h==<COMMENTS:Due to the large value of Ti, radiation effects are likely to be significant during the early portion of the transient. The effect is to shorten the cooling time.
PROBLEM 5.38KNOWN:Thickness, properties and initial temperature of steel slab. Convection conditions. FIND:Heating time required to achieve a minimum temperature of 550°C in the slab. SCHEMATIC:L = 0.05 mSteel, T = 20io0 Cρ = 7830 kg/m3c = 550 J/kg-Kk = 48 W/m-KT = 800oCooh = 250 W/m-K2T , h ooCombustiongasesASSUMPTIONS:(1) One-dimensional conduction, (2) Negligible radiation effects, (3) Constant properties. ANALYSIS:With a Biot number of hL/k = (250 W/m2⋅K ×0.05m)/48 W/m⋅K = 0.260, a lumped capacitance analysis should not be performed. At any time during heating, the lowest temperature in the slab is at the midplane, and from the one-term approximation to the transient thermal response of a plane wall, Eq. (5.44), we obtain ()()()2oo11i550800CTT0.417C expFoTT200800Cθζ∗∞∞−°−====−−°With 10.488 radζ≈and 1C1.0396≈from Table 5.1 and 52k /c1.11510m/ s,αρ−==×()()221t / Lln 0.4010.914ζα−==−()()22225210.914 0.05m0.914Lt861s0.4881.11510m/ sζα−===×<COMMENTS:The surface temperature at t = 861s may be obtained from Eq. (5.43b), where ()()o1cos0.417 cos 0.488 rad0.368.xθθζ∗∗∗===Hence, ()()siTL,861sTT0.368 TT∞∞≡=+−800 C221 C579 C.=°−°=Assuming a surface emissivity of ε= 1 and surroundings that are at surTT800 C,∞==the radiation heat transfer coefficient corresponding to this surface temperature is ()()222rssurssurhTTTT205 W / mK.εs=++=⋅Since this value is comparable to the convection coefficient, radiation is not negligible and the desired heating will occur well before t = 861s.
PROBLEM 5.102KNOWN:Very thick plate, initially at a uniform temperature, Ti, is suddenly exposed to a convection cooling process (T∞,h). FIND:Temperatures at the surface and a 45mm depth after 3 minutes using finite-difference method with space and time increments of 15mm and 18s. SCHEMATIC:ASSUMPTIONS:(1) One-dimensional transient conduction, (2) Plate approximates semi-infinite medium, (3) Constant properties. ANALYSIS:The grid network representing the plate is shown above. The finite-difference equation for node 0 is given by Eq. 5.90 for one-dimensional conditions or Eq. 5.85, ()()p+1pp010T2 FoTBiT12 Fo2 BiFo T .∞=+⋅+−−⋅(1) The numerical values of Fo and Bi are ()()62222-3t5.610m/s18sFo0.448x0.015m100 W/mK1510mh xBi0.075.k20 W/mKα−∆××===∆⋅××∆===⋅Recognizing that T∞= 15°C, Eq. (1) has the form p+1pp001T0.0359 T0.897 T1.01.=++(2) It is important to satisfy the stability criterion, Fo (1+Bi) ≤1/2. Substituting values, 0.448 (1+0.075) = 0.482 ≤1/2, and the criterion is satisfied. The finite-difference equation for the interior nodes, m = 1, 2…, follows from Eq. 5.78, ()()p+1pppmmm+1m-1TFo TT12Fo T.=++−(3) Recognizing that the stability criterion, Fo ≤1/2, is satisfied with Fo = 0.448, ()p+1pppmmm+1m-1T0.448 TT0.104T.=++(4) Continued …
PROBLEM 5.102 (Cont.)The time scale is related to p, the number of steps in the calculation procedure, and ∆t, the time increment, tp t.=∆(5) The finite-difference calculations can now be performed using Eqs. (2) and (4). The results are tabulated below. p t(s) T0T1T2T3 T4T5T6T7(K) 0 0 325 325 325 325 325 325 325 325 1 18 304.2 324.7 325 325 325 325 325 325 2 36 303.2 315.3 324.5 325 325 325 325 325 3 54 294.7 313.7 320.3 324.5 325 325 325 325 4 72 293.0 307.8 318.9 322.5 324.5 325 325 325 5 90 287.6 305.8 315.2 321.5 323.5 324.5 325 325 6 108 285.6 301.6 313.5 319.3 322.7 324.0 324.5 325 7 126 281.8 299.5 310.5 317.9 321.4 323.3 324.2 8 144 279.8 296.2 308.6 315.8 320.4 322.5 9 162 276.7 294.1 306.0 314.3 319.0 10 180 274.8 291.3 304.1 312.4 Hence, find ()()101003T0, 180sT275 C T45mm, 180sT312 C.====<COMMENTS:(1) The above results can be readily checked against the analytical solution represented in Fig. 5.8 (see also Eq. 5.63). For x = 0 and t = 180s, find ()()()1/ 21/ 22-621/ 2x02t100 W/mK 5.6010m/s180sht0.16k20 W/mKαα=⋅××==⋅for which the figure gives iiTT0.15TT∞−=−so that, ()()()()iiT0, 180s0.15 TTT0.15 15325C325 CT0, 180s278 C.∞=−+=−=For x = 45mm, the procedure yields T(45mm, 180s) = 316°C. The agreement with the numerical solution is nearly within 1%.