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EXAMPLE 33-1 Transient Response of a Tank Wall

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EXAMPLE 33-1 Transient Response of a Tank Wall Powered By Docstoc
					               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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Description: EXAMPLE 33-1 Transient Response of a Tank Wall