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3 Pages

### Thermodynamics HW Solutions 70

Course: PHY 4803, Fall 2010
School: UNF
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Word Count: 308

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1 Chapter Basics of Heat Transfer 1-132 A 0.2-L glass of water at 20C is to be cooled with ice to 5C. The amounts of ice or cold water that needs to be added to the water are to be determined. Assumptions 1 Thermal properties of the ice and water are constant. 2 Heat transfer to the water is negligible. Properties The density of water is 1 kg/L, and the specific heat of water at room temperature is C = 4.18 kJ/kgC...

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1 Chapter Basics of Heat Transfer 1-132 A 0.2-L glass of water at 20C is to be cooled with ice to 5C. The amounts of ice or cold water that needs to be added to the water are to be determined. Assumptions 1 Thermal properties of the ice and water are constant. 2 Heat transfer to the water is negligible. Properties The density of water is 1 kg/L, and the specific heat of water at room temperature is C = 4.18 kJ/kgC (Table A-9). The heat of fusion of ice at atmospheric pressure is 333.7 kJ/kg,. Analysis The mass of the water is Ice, 0C m w = V = (1kg/L )(0.2 L ) = 0.2kg We take the ice and the water as our system, and disregard any heat and mass transfer. This is a reasonable assumption since the time period of the process is very short. Then the energy balance can be written as E in E out 1 24 43 = Net energy transfer by heat, work, and system 1 mass E 24 43 [mC(0 C T ) o 0 = U (U )ice + (U )water = 0 Change in internal, kinetic, potential, etc. energies 1 solid ( + mhif + mC T2 0 o C )] liquid ice + [mC (T2 T1 )]water = 0 Noting that T1, ice = 0C and T2 = 5C and substituting m[0 + 333.7 kJ/kg + (4.18 kJ/kgC)(5-0)C] + (0.2 kg)(4.18 kJ/kgC)(5-20)C = 0 It gives m = 0.0354 kg = 35.4 g Cooling with cold water can be handled the same way. All we need to do is replace the terms for ice by the ones for cold water at 0C: (U )coldwater + (U )water = 0 [mC (T2 T1 )]coldwater + [mC (T2 T1 )]water = 0 Substituting, [mcold water (4.18 kJ/kgC)(5 - 0)C] + (0.2 kg)(4.18 kJ/kgC)(5-20)C = 0 It gives m = 0.6 kg = 600 g Discussion Note that this is 17 times the amount of ice needed, and it explains why we use ice instead of water to cool drinks. 1-70 Water 0.2 L 20C
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UNF - PHY - 4803
Chapter 1 Basics of Heat Transfer1-133&quot;GIVEN&quot;V=0.0002 &quot;[m^3]&quot;T_w1=20 &quot;[C]&quot;T_w2=5 &quot;[C]&quot;&quot;T_ice=0 [C], parameter to be varied&quot;T_melting=0 &quot;[C]&quot;&quot;PROPERTIES&quot;rho=density(water, T=25, P=101.3) &quot;at room temperature&quot;C_w=CP(water, T=25, P=101.3) &quot;at room
UNF - PHY - 4803
Chapter 1 Basics of Heat Transfer1-134E A 1-short ton (2000 lbm) of water at 70F is to be cooled in a tank by pouring 160 lbm of ice at25F into it. The final equilibrium temperature in the tank is to be determined. The melting temperature andthe heat o
UNF - PHY - 4803
Chapter 1 Basics of Heat Transfer1-136 Somebody takes a shower using a mixture of hot and cold water. The mass flow rate of hot waterand the average temperature of mixed water are to be determined.Assumptions The hot water temperature changes from 80C
UNF - PHY - 4803
Chapter 1 Basics of Heat Transfer1-137 The glass cover of a flat plate solar collector with specified inner and outer surface temperatures isconsidered. The fraction of heat lost from the glass cover by radiation is to be determined.Assumptions 1 Stead
UNF - PHY - 4803
Chapter 1 Basics of Heat Transfer1-138 The range of U-factors for windows are given. The range for the rate of heat loss through thewindow of a house is to be determined.Assumptions 1 Steady operating conditions exist. 2 Heat losses associated with the
UNF - PHY - 4803
Chapter 1 Basics of Heat Transfer1-139&quot;GIVEN&quot;A=1.2*1.8 &quot;[m^2]&quot;T_1=20 &quot;[C]&quot;T_2=-8 &quot;[C]&quot;&quot;U=1.25 [W/m^2-C], parameter to be varied&quot;&quot;ANALYSIS&quot;Q_dot_window=U*A*(T_1-T_2)U [W/m2.C]1.251.752.252.753.253.754.254.755.255.756.25Qwindow [W]75.6
UNF - PHY - 4803
Chapter 1 Basics of Heat Transfer1-140 . . . 1-144 Design and Essay Problems1-77
UNF - PHY - 4803
Chapter 2 Heat Conduction EquationChapter 2HEAT CONDUCTION EQUATIONIntroduction2-1C Heat transfer is a vector quantity since it has direction as well as magnitude. Therefore, we mustspecify both direction and magnitude in order to describe heat trans
UNF - PHY - 4803
Chapter 2 Heat Conduction Equation2-6C Heat transfer to a hot dog can be modeled as two-dimensional since temperature differences (and thusheat transfer) will exist in the radial and axial directions (but there will be symmetry about the center lineand
UNF - PHY - 4803
Chapter 2 Heat Conduction Equation&amp;g=&amp;GV wire=&amp;G(D / 4) L2= 3.412 Btu/h 73 = 7.820 10 Btu/h ft1W[(0.08 / 12 ft) / 4](15 / 12 ft) 1000 W2Similarly, heat flux on the outer surface of the wire as a result of this heat generation is determ
UNF - PHY - 4803
Chapter 2 Heat Conduction Equation2-15E&quot;GIVEN&quot;E_dot=1000 &quot;[W]&quot;L=15 &quot;[in]&quot;&quot;D=0.08 [in], parameter to be varied&quot;&quot;ANALYSIS&quot;g_dot=E_dot/V_wire*Convert(W, Btu/h)V_wire=pi*D^2/4*L*Convert(in^3, ft^3)q_dot=E_dot/A_wire*Convert(W, Btu/h)A_wire=pi*D*L*Co
UNF - PHY - 4803
Chapter 2 Heat Conduction Equation2-16 The rate of heat generation per unit volume in the uranium rods is given. The total rate of heatgeneration in each rod is to be determined.g = 7107 W/m3Assumptions Heat is generated uniformly in the uranium rods.
UNF - PHY - 4803
Chapter 2 Heat Conduction Equationvariable, g is the heat generation per unit volume, k is the thermal conductivity, is the thermal diffusivity,and t is the time.2-6
UNF - PHY - 4803
Chapter 2 Heat Conduction Equation2-21 We consider a thin element of thickness x in a large plane wall (see Fig. 2-13 in the text). Thedensity of the wall is , the specific heat is C, and the area of the wall normal to the direction of heattransfer is
UNF - PHY - 4803
Chapter 2 Heat Conduction Equation2-22 We consider a thin cylindrical shell element of thickness r in a long cylinder (see Fig. 2-15 in thetext). The density of the cylinder is , the specific heat is C, and the length is L. The area of the cylindernorm
UNF - PHY - 4803
Chapter 2 Heat Conduction Equation2-23 We consider a thin spherical shell element of thickness r in a sphere (see Fig. 2-17 in the text). Thedensity of the sphere is , the specific heat is C, and the length is L. The area of the sphere normal to thedir
UNF - PHY - 4803
Chapter 2 Heat Conduction Equation(a) Heat transfer is transient, (b) it is one-dimensional, (c) there is no heat generation, and (d) the thermalconductivity is constant.2-27 For a medium in which the heat conduction equation is given in its simplest b
UNF - PHY - 4803
Chapter 2 Heat Conduction Equation2-29 We consider a thin ring shaped volume element of width z and thickness r in a cylinder. Thedensity of the cylinder is and the specific heat is C. In general, an energy balance on this ring elementduring a small ti
UNF - PHY - 4803
Chapter 2 Heat Conduction Equation2-30 Consider a thin disk element of thickness z and diameter D in a long cylinder (Fig. P2-30). Thedensity of the cylinder is , the specific heat is C, and the area of the cylinder normal to the direction ofheat trans
UNF - PHY - 4803
Chapter 2 Heat Conduction Equation2-31 For a medium in which the heat conduction equation is given by2Tx2+2Ty2=1 T: t(a) Heat transfer is transient, (b) it is two-dimensional, (c) there is no heat generation, and (d) the thermalconductivity
UNF - PHY - 4803
Chapter 2 Heat Conduction Equation2-39C We try to avoid the radiation boundary condition in heat transfer analysis because it is a non-linearexpression that causes mathematical difficulties while solving the problem; often making it impossible toobtain
UNF - PHY - 4803
Chapter 2 Heat Conduction Equation2-43 A spherical shell of inner radius r1 , outer radius r2 , and thermal conductivity k is considered. Theouter surface of the shell is subjected to radiation to surrounding surfaces at Tsurr . Assuming no convectiona
UNF - PHY - 4803
Chapter 2 Heat Conduction Equation2-46E A 1.5-kW resistance heater wire is used for space heating. Assuming constant thermal conductivityand one-dimensional heat transfer, the mathematical formulation (the differential equation and theboundary conditio
UNF - PHY - 4803
Chapter 2 Heat Conduction Equation2-48 Water flows through a pipe whose outer surface is wrapped with a thin electric heater that consumes300 W per m length of the pipe. The exposed surface of the heater is heavily insulated so that the entireheat gene
UNF - PHY - 4803
Chapter 2 Heat Conduction EquationAssumptions 1 Heat transfer is given to be transient and one-dimensional. 2 Thermal conductivity is givento be variable. 3 There is no heat generation in the medium. 4 The outer surface at r = r0 is subjected toconvect
UNF - PHY - 4803
Chapter 2 Heat Conduction EquationSolution of Steady One-Dimensional Heat Conduction Problems2-52C Yes, this claim is reasonable since in the absence of any heat generation the rate of heat transferthrough a plain wall in steady operation must be const
UNF - PHY - 4803
Chapter 2 Heat Conduction Equation2-56 A large plane wall is subjected to specified temperature on the left surface and convection on the rightsurface. The mathematical formulation, the variation of temperature, and the rate of heat transfer are to bed
UNF - PHY - 4803
Chapter 2 Heat Conduction Equation2-57 The top and bottom surfaces of a solid cylindrical rod are maintained at constant temperatures of20C and 95C while the side surface is perfectly insulated. The rate of heat transfer through the rod is tobe determi
UNF - PHY - 4803
Chapter 2 Heat Conduction Equation2-58&quot;GIVEN&quot;L=0.15 &quot;[m]&quot;D=0.05 &quot;[m]&quot;T_1=20 &quot;[C]&quot;T_2=95 &quot;[C]&quot;&quot;k=1.2 [W/m-C], parameter to be varied&quot;&quot;ANALYSIS&quot;A=pi*D^2/4Q_dot=k*A*(T_2-T_1)/Lk [W/m.C]122436485106127148169190211232253274295316337
UNF - PHY - 4803
Chapter 2 Heat Conduction Equation2-59 The base plate of a household iron is subjected to specified heat flux on the left surface and tospecified temperature on the right surface. The mathematical formulation, the variation of temperature inthe plate,
UNF - PHY - 4803
Chapter 2 Heat Conduction Equation2-60 The base plate of a household iron is subjected to specified heat flux on the left surface and tospecified temperature on the right surface. The mathematical formulation, the variation of temperature inthe plate,
UNF - PHY - 4803
Chapter 2 Heat Conduction Equation2-61&quot;GIVEN&quot;Q_dot=800 &quot;[W]&quot;L=0.006 &quot;[m]&quot;A_base=160E-4 &quot;[m^2]&quot;k=20 &quot;[W/m-C]&quot;T_2=85 &quot;[C]&quot;&quot;ANALYSIS&quot;q_dot_0=Q_dot/A_baseT=q_dot_0*(L-x)/k+T_2 &quot;Variation of temperature&quot;&quot;x is the parameter to be varied&quot;00.0006667
UNF - PHY - 4803
Chapter 2 Heat Conduction EquationAssumptions 1 Heat conduction is steady and one-dimensional since the pipe is long relative to itsthickness, and there is thermal symmetry about the center line. 2 Thermal conductivity is constant. 3 Thereis no heat ge
UNF - PHY - 4803
Chapter 2 Heat Conduction Equation2-27
UNF - PHY - 4803
Chapter 2 Heat Conduction Equation2-63 A spherical container is subjected to specified temperature on the inner surface and convection on theouter surface. The mathematical formulation, the variation of temperature, and the rate of heat transfer areto
UNF - PHY - 4803
Chapter 2 Heat Conduction EquationC1r (T T )dT&amp;= k (4r 2 ) 2 = 4kC1 = 4k 2 1 Q = kArkdxr1 2 r1 hr2(2.1 m)(0 25)C= 4 (30 W/m C)= 23,460 W2.130 W/m C12 (18 W/m 2 C)(2.1 m )2-29
UNF - PHY - 4803
Chapter 2 Heat Conduction Equation2-64 A large plane wall is subjected to specified heat flux and temperature on the left surface and noconditions on the right surface. The mathematical formulation, the variation of temperature in the plate,and the rig
UNF - PHY - 4803
Chapter 2 Heat Conduction Equation2-65 A large plane wall is subjected to specified heat flux and temperature on the left surface and noconditions on the right surface. The mathematical formulation, the variation of temperature in the plate,and the rig
UNF - PHY - 4803
Chapter 2 Heat Conduction Equation2-66E A large plate is subjected to convection, radiation, and specified temperature on the top surface andno conditions on the bottom surface. The mathematical formulation, the variation of temperature in theplate, an
UNF - PHY - 4803
Chapter 2 Heat Conduction Equation2-67E A large plate is subjected to convection and specified temperature on the top surface and noconditions on the bottom surface. The mathematical formulation, the variation of temperature in the plate,and the bottom
UNF - PHY - 4803
Chapter 2 Heat Conduction Equation2-68 A compressed air pipe is subjected to uniform heat flux on the outer surface and convection on theinner surface. The mathematical formulation, the variation of temperature in the pipe, and the surfacetemperatures
UNF - PHY - 4803
Chapter 2 Heat Conduction Equation(c) The inner and outer surface temperatures are determined by direct substitution to berInner surface (r = r1): T (r1 ) = 10 + 0.483 ln 1 + 12.61 = 10 + 0.483(0 + 12.61) = 3.91Cr1r 0.04Outer surface (r = r2): T (
UNF - PHY - 4803
Chapter 2 Heat Conduction Equation2-69&quot;GIVEN&quot;L=6 &quot;[m]&quot;r_1=0.037 &quot;[m]&quot;r_2=0.04 &quot;[m]&quot;k=14 &quot;[W/m-C]&quot;Q_dot=300 &quot;[W]&quot;T_infinity=-10 &quot;[C]&quot;h=30 &quot;[W/m^2-C]&quot;f_loss=0.15&quot;ANALYSIS&quot;q_dot_s=(1-f_loss)*Q_dot)/AA=2*pi*r_2*LT=T_infinity+(ln(r/r_1)+k/(h*r_1)
UNF - PHY - 4803
Chapter 2 Heat Conduction EquationAssumptions 1 Heat conduction is steady and one-dimensional since there is no change with time andthere is thermal symmetry about the mid point. 2 Thermal conductivity is constant. 3 There is no heatgeneration in the c
UNF - PHY - 4803
Chapter 2 Heat Conduction Equation&amp;&amp;Q = mC p T m =&amp;Q0.450 kJ / s== 0.00134 kg / s = 4.84 kg / hC p T (4.185 kJ / kg C)(100 20) C2-38
UNF - PHY - 4803
Chapter 2 Heat Conduction Equation2-71&quot;GIVEN&quot;r_1=0.40 &quot;[m]&quot;r_2=0.41 &quot;[m]&quot;k=1.5 &quot;[W/m-C]&quot;T_1=100 &quot;[C]&quot;Q_dot=500 &quot;[W]&quot;f_loss=0.10&quot;ANALYSIS&quot;q_dot_s=(1-f_loss)*Q_dot)/AA=4*pi*r_2^2T=T_1+(1/r_1-1/r)*(q_dot_s*r_2^2)/k &quot;Variation of temperature&quot;&quot;r i
UNF - PHY - 4803
Chapter 2 Heat Conduction EquationHeat Generation in Solids2-72C No. Heat generation in a solid is simply the conversion of some form of energy into sensible heatenergy. For example resistance heating in wires is conversion of electrical energy to heat
UNF - PHY - 4803
Chapter 2 Heat Conduction Equation2-78 Heat is generated in a long solid cylinder with a specified surfacetemperature. The variation of temperature in the cylinder is given byT (r ) =2&amp;gr02 r 1 + Tsk r0 80C(a) Heat conduction is steady since t
UNF - PHY - 4803
Chapter 2 Heat Conduction Equation2-79&quot;GIVEN&quot;r_0=0.04 &quot;[m]&quot;k=25 &quot;[W/m-C]&quot;g_dot_0=35E+6 &quot;[W/m^3]&quot;T_s=80 &quot;[C]&quot;&quot;ANALYSIS&quot;T=(g_dot_0*r_0^2)/k*(1-(r/r_0)^2)+T_s &quot;Variation of temperature&quot;&quot;r is the parameter to be varied&quot;r [m]00.0044440.0088890.01
UNF - PHY - 4803
Chapter 2 Heat Conduction Equation2-80E A long homogeneous resistance heater wire with specified convection conditions at the surface isused to boil water. The mathematical formulation, the variation of temperature in the wire, and thetemperature at th
UNF - PHY - 4803
Chapter 2 Heat Conduction EquationT ( 0) = T +&amp;&amp;g 2 gr0r0 +4k2h2= 212F +(1800 Btu/h.in 3 )(0.25 in) 2 12 in (1800 Btu/h.in 3 )(0.25 in ) 12 in + = 290.8F4 (8.6 Btu/h.ft.F)2 (820 Btu/h ft 2 F) 1 ft 1 ft Thus the centerline temperature will
UNF - PHY - 4803
Chapter 2 Heat Conduction Equation2-81E&quot;GIVEN&quot;r_0=0.25/12 &quot;[ft]&quot;k=8.6 &quot;[Btu/h-ft-F]&quot;&quot;g_dot=1800 [Btu/h-in^3], parameter to be varied&quot;T_infinity=212 &quot;[F]&quot;h=820 &quot;[Btu/h-ft^2-F]&quot;&quot;ANALYSIS&quot;T_0=T_infinity+(g_dot/Convert(in^3, ft^3)/(4*k)*(r_0^2-r^2)+(
UNF - PHY - 4803
Chapter 2 Heat Conduction Equation2-82 A nuclear fuel rod with a specified surface temperature is used as the fuel in a nuclear reactor. Thecenter temperature of the rod is to be determined.175CAssumptions 1 Heat transfer is steady since there is no i
UNF - PHY - 4803
Chapter 2 Heat Conduction Equation2-83 Both sides of a large stainless steel plate in which heat is generated uniformly are exposed toconvection with the environment. The location and values of the highest and the lowest temperatures in theplate are to
UNF - PHY - 4803
Chapter 2 Heat Conduction Equation2-85&quot;GIVEN&quot;L=0.05 &quot;[m]&quot;k=111 &quot;[W/m-C]&quot;g_dot=2E5 &quot;[W/m^3]&quot;T_infinity=25 &quot;[C]&quot;&quot;h=44 [W/m^2-C], parameter to be varied&quot;&quot;ANALYSIS&quot;T_min=T_infinity+(g_dot*L)/hT_max=T_min+(g_dot*L^2)/(2*k)h [W/m2.C]20253035404
UNF - PHY - 4803
Chapter 2 Heat Conduction Equation550500450T m in [ C]400350300250200150100203040506070809010080901002h [W /m -C]550500450T m ax [ C]4003503002502001501002030405060702h [W /m -C]2-49
UNF - PHY - 4803
Chapter 2 Heat Conduction Equation2-86 A long resistance heater wire is subjected to convection at its outer surface. The surface temperatureof the wire is to be determined using the applicable relations directly and by solving the applicabledifferenti
UNF - PHY - 4803
Chapter 2 Heat Conduction EquationAssumptions 1 Heat transfer is steady since there is no changewith time. 2 Heat transfer is one-dimensional since there isthermal symmetry about the center line and no change in theaxial direction. 3 Thermal conductiv
UNF - PHY - 4803
Chapter 2 Heat Conduction Equation2-89 Heat is generated uniformly in a spherical radioactive material with specified surface temperature.The mathematical formulation, the variation of temperature in the sphere, and the center temperature are tobe dete
UNF - PHY - 4803
Chapter 2 Heat Conduction Equation2-90&quot;GIVEN&quot;r_0=0.04 &quot;[m]&quot;g_dot=4E7 &quot;[W/m^3]&quot;T_s=80 &quot;[C]&quot;k=15 &quot;[W/m-C], Parameter to be varied&quot;&quot;ANALYSIS&quot;T=T_s+g_dot/(6*k)*(r_0^2-r^2) &quot;Temperature distribution as a function of r&quot;&quot;r is the parameter to be varied&quot;
UNF - PHY - 4803
Chapter 2 Heat Conduction Equation800700600T [C]500400300200100000.0050.010.0150.020.0250.030.0353003500.04r [m ]12001000T 0 [ C]8006004002000050100150200250k [W /m -C]2-54400