Relationship to Thermodynamics. Chapter One Section 1.3


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1 Relationship to Thermodynamics Chapter One Section 1.3
2 Alternative Formulations Alternative Formulations Time Basis: CONSERVATION OF ENERGY (FIRST LAW OF THERMODYNAMICS) An important tool in heat transfer analysis, often providing the basis for determining the temperature of a system. At an instant or Over a time interval Type of System: Control volume Control surface
3 CV at an Instant and over a Time Interval APPLICATION TO A CONTROL VOLUME At an Instant of Time: Note representation of system by a control surface (dashed line) at the boundaries. Surface Phenomena, E E in out Volumetric Phenomena E g E st : rate of thermal and/or mechanical energy transfer across the control surface due to heat transfer, fluid flow and/or work interactions. : rate of thermal energy generation due to conversion from another energy form (e.g., electrical, nuclear, or chemical); energy conversion process occurs within the system. : rate of change of energy storage in the system. Conservation of Energy de + = st dt E E E E in out g st Each term has units of J/s or W. (1.12c) Over a Time Interval + = E E E E in out g st Each term has units of J. (1.12b)
4 Closed System Special Cases (Linkages to Thermodynamics) (i) Transient Process for a Closed System of Mass (M) Assuming Heat Transfer to the System (Inflow) and Work Done by the System (Outflow). Over a time interval Q W E st tot = (1.12a) For negligible changes in potential or kinetic energy Q W = U t Internal thermal energy At an instant du q t W = dt
5 Example 1.4 Example 1.4: Application to thermal response of a conductor with Ohmic heating (generation): Involves change in thermal energy and for an incompressible substance. dut dt dt = Mc dt Heat transfer is from the conductor (negative q ) Generation may be viewed as electrical work done on the system (negative ) W
6 Example 1.6 Example 1.6: Application to isothermal solidliquid phase change in a container: U = U = Mh t lat sf Latent Heat of Fusion
7 Open System (ii) Steady State for Flow through an Open System without Phase Change or Generation: At an Instant of Time: 2 2 m ut pv V gz q mut + pv + V + gz 0 2 W = in out ( pv) flow work ( ut + pv) i enthalpy For an ideal gas w ith constant specific heat: ( ) i i = c T T in out p in out For an incompressible liquid: = ( ) ( pv) ( pv) 0 u u c T T in out in out in out (1.12d) For systems with significant heat transfer: 2 2 ( V ) 2 ( V 2 ) in ( gz) ( gz) in 0 out out 0
8 Surface Energy Balance THE SURFACE ENERGY BALANCE A special case for which no volume or mass is encompassed by the control surface. Conservation of Energy (Instant in Time): E in Eout = 0 (1.13) Applies for steadystate and transient conditions. With no mass and volume, energy storage and generation are not pertinent to the energy balance, even if they occur in the medium bounded by the surface. Consider surface of wall with heat transfer by conduction, convection and radiation. q q q cond conv rad = 0 T1 T k 2 h T T T T = L 4 4 ( ) ( ) sur 0
9 Methodology METHODOLOGY OF FIRST LAW ANALYSIS On a schematic of the system, represent the control surface by dashed line(s). Choose the appropriate time basis. Identify relevant energy transport, generation and/or storage terms by labeled arrows on the schematic. Write the governing form of the Conservation of Energy requirement. Substitute appropriate expressions for terms of the energy equation. Solve for the unknown quantity.
10 Problem: Silicon Wafer Problem 1.57: Thermal processing of silicon wafers in a twozone furnace. Determine (a) the initial rate of change of the wafer temperature and (b) the steadystate temperature. KNOWN: Silicon wafer positioned in furnace with top and bottom surfaces exposed to hot and cool zones, respectively. FIND: (a) Initial rate of change of the wafer temperature from a value of T w,i = 300 K, and (b) steadystate temperature. Is convection significant? Sketch the variation of wafer temperature with vertical distance. SCHEMATIC:
11 Problem: Silicon Wafer (cont.) ASSUMPTIONS: (1) Wafer temperature is uniform, (2) Hot and cool zones have uniform temperatures, (3) Radiation exchange is between small surface (wafer) and large enclosure (chamber, hot or cold zone), and (4) Negligible heat losses from wafer to pin holder. ANALYSIS: The energy balance on the wafer includes convection to the upper (u) and lower (l) surfaces from the ambient gas, radiation exchange with the hot and coolzones and an energy storage term for the transient condition. Hence, from Eq. (1.12c), Ein Eout = Est or, per unit surface area dt qrad, w h + qrad, c qcv, u qcv, l = cd dt ( sur, h w ) ( ) sur, c w u ( w ) l ( w ) dt T T + T T h T T h T T = cd w dt (a) For the initial condition, the time rate of change of the wafer temperature is determined using the foregoing energy balance with T = T, = 300 K, w w i 4 ( ) ( ) W / m K K W / m K K 8W / m 2 K K 4 W / m 2 K K = kg / m ( ) 104 K/s ( ) ( ) 875J / kg K m( dtw / dt ) i dt w / dt = i <
12 Problem: Silicon Wafer (cont.) (b) For the steadystate condition, the energy storage term is zero, and the energy balance can be solved for the steadystate wafer temperature, T = T,. w w ss ( T 4 w,ss ) + ( Tw, ss ) K K 4 ( w,ss ) ( w,ss ) 8W / m 2 K T 700 K 4 W / m 2 K T 700 K = 0 T w,ss = 1251 K To assess the relative importance of convection, solve the energy balances assuming no dt / dt = 101 K/s and T = 1262 K, we conclude that the radiation convection. With ( w ), i exchange processes control the initial rate of change and the steadystate temperature. w ss If the wafer were elevated above the present operating position, its temperature would increase, since the lower surface would begin to experience radiant exchange with progressively more of the hot zone. Conversely, by lowering the wafer, the upper surface would experience less radiant exchange with the hot zone, and its temperature would decrease. The temperaturedistance relation might appear as shown in the sketch. <
13 Problem: Cooling of Spherical Canister Problem 1.64: Cooling of spherical canister used to store reacting chemicals. Determine (a) the initial rate of change of the canister temperature, (b) the steadystate temperature, and (c) the effect of convection on the steadystate temperature. 535 J/kg K KNOWN: Inner surface heating and new environmental conditions associated with a spherical shell of prescribed dimensions and material. FIND: (a) Governing equation for variation of wall temperature with time and the initial rate of change, (b) Steadystate wall temperature and, (c) Effect of convection coefficient on canister temperature.
14 Problem: Cooling of Spherical Canister (cont.) SCHEMATIC: ASSUMPTIONS: (1) Negligible temperature gradients in wall, (2) Constant properties, (3) Uniform, timeindependent heat flux at inner surface. PROPERTIES: Table A.1, Stainless Steel, AISI 302: = 8055 kg/m 3, c p = 535 J/kgK. ANALYSIS: (a) Performing an energy balance on the shell at an instant of time, Ein Eout = Est. Identifying relevant processes and solving for dt/dt, ( 4 2 ) ( 4 2 )( ) 4 ( 3 3 ) q i r i h r dt o T T = r 3 o r i c p dt dt 3 = q 2 2 ( ) 3 3 i ri hro T T. dt cp ro r i ( ) <
15 Problem: Cooling of Spherical Canister (cont.) Substituting numerical values for the initial condition, find 5 W 2 W ( 0.5m) 500 ( 0.6m) ( ) K 2 2 dt = m m K dt kg J i ( 0.6) ( 0.5) m m 3 kg K dt dt = K/s < i (b) Under steadystate conditions with ( ) ( 4 )( ) qi ri = h ro T T E st = 0, it follows that T qi ri 10 W/m 0.5m = T + = 300K + 439K h r 500W/m 2 = o K 0.6m < (c) Parametric calculations show a sharp increase in temperature with decreasing values of h < 1000 W/m 2 K. For T > 380 K, boiling will occur at the canister surface, and for T > 410 K a condition known as film boiling (Chapter 10) will occur. The condition corresponds to a precipitous reduction in h and increase in T.
16 Problem: Cooling of Spherical Canister (cont.) Although the canister remains well below the melting point of stainless steel for h = 100 W/m 2 K, boiling should be avoided, in which case the convection coefficient should be maintained at h > 1000 W/m 2 K. COMMENTS: The governing equation of part (a) is a first order, nonhomogenous differential equation = S/R 1 e Rt + Rt ie, where T T, with constant coefficients. Its solution is ( )( ) S 3 q 2 / 3 3 i ri cp ( ro ri ), R 3 hro 2 / cp ( ro 3 ri 3 ) =. Note results for t and for S = 0.
17 Second law SECOND LAW OF THERMODYNAMICS An important tool to determine how heat transfer affects the efficiency of energy conversion. For a reversible heat engine neglecting heat transfer effects between the heat engine and large reservoirs, the Carnot efficiency is W Qout T C = = 1 = 1 c Qin Qin Th where Tc and Th are the absolute temperatures of large cold and hot reservoirs, respectively. For an internally reversible heat engine with heat transfer to and from the large reservoirs properly accounted for, the modified Carnot efficiency is out out ci, m = W Q q Q = T in Q = = in q in Thi, (1.15, 1.16) (1.17) where Tc, i Tc and Th, i Th are the absolute temperatures seen by the internally reversible heat engine. Note that qout and qin are heat transfer rates (J/s or W).
18 Second law (cont.) Heat transfer resistances associated with, for example, walls separating the internally reversible heat engine from the hot and cold reservoirs relate the heat transfer rates to temperature differences: Th Th, i = qi nrth, Tc, i Tc = qout Rtc ( ) ( ), In reality, heat transfer resistances (K/W) must be nonzero since according to the rate equations, for any temperature difference only a finite amount of heat may be transferred. The modified Carnot efficiency may ultimately be expressed as (1.18 a,b) T 1 c m = where R tot = Rt, h + Rt, c Th qinrtot (1.21) From Eq. 1.21, m = n C only if R tot could be made infinitely small. For realistic situations ( Rtot 0), m nc. Good heat transfer engineering is a key to improve the efficiency of heat engines.
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