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EXAMPLE 3.3-1: Transient Response of a Tank Wall A metal wall (Figure 1) is used to separate two tanks of liquid at different temperatures, Thot = 500 K and Tcold = 400 K. The thickness of the wall is th = 0.8 cm and its area is Ac= 1.0 m2. The properties of the wall material are ρ = 8000 kg/m3, c = 400 J/kg-K, and k = 20 W/m-K. The average heat transfer coefficient between the wall and the liquid in either tank is hliq = 5000 W/m2-K. th = 0.8 cm liquid at liquid at Tcold = 400 K Thot = 500 K hliq = 5000 W/m -K hliq = 5000 W/m -K 2 2 x k = 20 W/m-K ρ = 8000 kg/m3 c = 400 J/kg-K Figure 1: Tank wall exposed to fluid. a.) Initially, the wall is at steady-state. That is, the wall has been exposed to the fluid in the tanks for a long time and therefore the temperature within the wall is not changing in time. What is the rate of heat transfer through the wall? What are the temperatures of the two surfaces of the wall (i.e., what is Tx=0 and Tx=th)? The known information is entered in EES: "EXAMPLE 3.3-1: Transient Response of a Tank Wall" $UnitSystem SI MASS RAD PA K J $Tabstops 0.2 0.4 0.6 0.8 3.5 "Inputs" k=20 [W/m-K] "thermal conductivity" c=400 [J/kg-K] "specific heat capacity" rho=8000 [kg/m^3] "density" T_cold=400 [K] "cold fluid temperature" T_hot=500 [K] "hot fluid temperature" h_bar_liq=5000 [W/m^2-K] "liquid-to-wall heat transfer coefficient" th=0.8 [cm]*convert(cm,m) "wall thickness" A_c=1 [m^2] "wall area" There are three thermal resistances governing this problem, convection from the surface on either side (Rconv,liq) and conduction through the wall (Rcond): 1 Rconv ,liq = hliq Ac th Rcond = k Ac "Steady-state solution, part (a)" R_conv_liq=1/(h_bar_liq*A_c) "convection resistance with liquid" R_cond=th/(k*A_c) "conduction resistance" The total heat transfer is ( q ) is: Thot − Tcold q= 2 Rconv ,liq + Rcond and the temperatures at x=0 and x=th are: Tx =0 = Tcold + q Rconv ,liq Tx =th = Thot − q Rconv ,liq q_dot=(T_hot-T_cold)/(2*R_conv_liq+R_cond) "heat transfer" T_0=T_cold+q_dot*R_conv_liq "temperature of cold side of wall" T_L=T_hot-q_dot*R_conv_liq "temperature of hot side of wall" The rate of heat transfer through the wall is q =125 kW and the edges of the wall are at Tx=0 = 425 K and Tx=th = 475 K. At time, t = 0, both tanks are drained and then both sides of the wall are exposed to gas at Tgas = 300 K (Figure 2). The average heat transfer coefficient between the walls and the gas is hgas = 100 W/m2-K. Assume that the process of draining the tanks and filling them with gas occurs instantaneously so that the wall has the linear temperature distribution from part (a) at time t = 0. th = 0.8 cm gas at gas at Tgas = 300 K Tgas = 300 K hgas = 100 W/m -K 2 hgas = 100 W/m -K 2 x k = 20 W/m-K ρ = 8000 kg/m3 c = 400 J/kg-K Figure 2: Tank wall exposed to gas. b.) On the axes in Figure 3, sketch the temperature distribution in the wall (i.e., the temperature as a function of position) at t = 0 s (i.e., immediately after the process starts) and also at t = 0.5 s, 5 s, 50 s, 500 s, and 5000 s. Clearly label these different sketches. Be sure that you have the qualitative features of the temperature distribution drawn correctly. There are two processes that occur after the tank is drained. The tank material is not in equilibrium with itself due to its internal temperature gradient. Therefore, there is an internal equilibration process that will cause the wall material to come to a uniform temperature. Also, there is an external equilibration process as the wall transfers heat with its environment. The internal equilibration process is governed by a diffusive time constant, τdiff, discussed in Section 3.3.2 and provided, approximately, by Eq. (3-62): th 2 τ diff = 4α where α is the thermal diffusivity of the wall material: k α= ρc "Internal equilibration process" alpha=k/(rho*c) "thermal diffusivity" tau_diff=th^2/(4*alpha) "diffusive time constant" The diffusive time constant is about 2.6 seconds; this is, approximately, how long it will take for a thermal wave to pass from one side of the wall to the other and therefore this is about the amount of time that is required for the wall to internally equilibrate. If the edges of the wall were adiabatic, then the wall will be at a nearly uniform temperature after a few seconds. The external equilibration process is governed by a lumped time constant, discussed previously in Section 3.1 and defined for this problem as: τ lumped = Rconv , gas C where Rconv,gas is the thermal resistance to convection between the wall and the gas: 1 Rconv , gas = 2 hgas Ac and C is the thermal capacitance of the wall: C = Ac th ρ c "External equilibration process" h_bar_gas=100 [W/m^2-K] "gas-to-wall heat transfer coefficient" R_conv_gas=1/(2*h_bar_gas*A_c) "convection resistance with gas" C_total=A_c*th*rho*c "capacity of wall" tau_lumped=R_conv_gas*C_total "lumped time constant" The lumped time constant is about 130 s. Because the diffusive time constant is two orders of magnitude smaller than the lumped time constant, the wall will initially internally equilibrate rapidly and subsequently externally equilibrate more slowly. The initial internal equilibration process will be completed after about 5 to 10 s. Subsequently, the wall will equilibrate externally with the surrounding gas; this external equilibration process will be completed after 200 to 400 s. The temperature distributions sketched in Figure 3 are consistent with these time constants; they do not represent an exact solution, but they are consistent with the physical intuition that was gained through knowledge of the two time constants. Temperature (K) 500 t=0s 475 t = 0.5 s 450 t=5s t = 50 s 425 400 t = 500 s t = 5000 s 300 Position (cm) 0 0.8 Figure 3: Temperature distributions in the wall as it equilibrates. A more exact solution to this problem can be obtained using the techniques discussed in subsequent sections (see EXAMPLE 3.5-2). However, the calculation diffusive and lumped time constants provide important physical intuition about the problem.

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EXAMPLE 33-1 Transient Response of a Tank Wall

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