Question 7.13: An insulated air tank, initially at a pressure of 0 MPa (gag...

We are asked to determine the initial rate of change of density and temperature during the filling of an air tank. Figure 7.25 sketches the tank and connection to the air supply as well as an appropriate fixed control volume for use in the solution of this problem. We will assume that the air obeys the perfect gas law and that the values of fluid properties on the section of the control surface defining the port are uniform. We will also assume that the temperature, density, and other fluid properties inside the tank are spatially uniform, that air velocities are small, and that the effects of gravity are negligible.

To find ∂ρ/∂t, the rate of change of density in the tank, recall that the mass balance Eq. 7.11, \int_{CV} (∂ρ/∂t)dV + \int_{CS}^{}{} ρ(u • n)dS = 0, contains this term. Using our assumptions of a spatially uniform density inside the tank and uniform properties at the port, and applying this equation to the selected control volume, we find (∂ρ/∂t) \sout{V} − ρ₁V₁(t)A1 = 0. Upon rearrangement we find

\frac{\partial \rho }{\partial t}=\frac{ρ_1A_1V_1(t) }{\sout{V}}=\frac{\dot M(t)}{\sout{V}}                                    (A)

which agrees with our mass balance analysis in Example 7.2. We see that the rate of change of density in the tank at any time is given by the instantaneous mass flowrate into the tank divided by the volume of the tank. A check shows that (A) has the expected dimensions of density per unit time. Since the initial mass flowrate is known to be \dot M_0 = 0.05 g/s, the initial rate of change of density inside the tank is calculated as

\left[\frac{\partial \rho }{\partial t}\right]_0=\frac{\dot M_0}{\sout{V}}=\frac{5\times 10^{-5}\ kg/s}{5\times 10^{-2} \ m^3}=1\times 10^{-3} \ kg/(m^3-s)

To find an expression for the rate of change of temperature in the tank, we apply an energy balance to the CV, recalling that for a perfect gas internal energy and temperature are related. According to Eq. 7.33 we have

\int_{CV}\frac{\partial}{\partial t} \left[\rho \left(u+\frac{1}{2}u  •  u+gz\right) \right]dV+\int_{CS}\rho \left(u+\frac{p}{\rho } +\frac{1}{2}u  •  u+gz\right)(u  •  n)dS

=\dot W_{power} +\dot W_{shaft} +\dot Q_C +\dot S

Since there is no fluid power, shaft power, heat, or other energy input to this CV, the right-hand side of this equation is zero. In addition, we will assume velocities are low enough to permit us to neglect the kinetic energy terms. Since we have also decided to neglect the effects of gravity the energy balance reduces to

\int_{CV}\frac{\partial}{\partial t}(\rho u)dV+\int_{CS}\rho \left(u+\frac{p}{\rho } \right)(u  •  n)dS= 0

With the assumption of uniform properties inside the tank and uniform properties on the control surface defining the port which are the same as in the air supply, we evaluate the integrals to find:

Thus the preceding expression becomes

\dot M(t)(u−u_1)+ρ\left(\frac{∂u}{∂t} \right) \sout{V}-\dot M(t)\left(\frac{p_1}{\rho _1}\right)=0

For a perfect gas, the internal energy change in going from state 1 to state 2 is related to the temperature change by Eq. 2.21a as u₂ − u₁ = c_V(T₂ − T₁). We can use this relationship to write u − u₁ = c_V(T − T₁), and ∂u/∂t = c_V(∂T/∂t). We will also make use of the fact that p = ρRT, and write the energy balance as \dot M (t)c_V(T − T₁) + ρc_V(∂T/∂t) \sout{V} – \dot M(t)RT₁ = 0. Solving for the rate of change of temperature in the tank, we have

\frac{\partial T}{\partial t}=\frac{\dot M(t)RT_1}{\rho (t)c_V \sout{V}}-\frac{\dot M(t)(T(t)-T_1)}{\rho (t) \sout{V}}                                         (B)

This result is valid at any instant of time. To find the initial rate of change of temperature in the tank, we make use of the fact that at the initial instant the temperature inside the tank is the same as that in the supply line. Thus T(0) = T₁, the second term in (B) is zero, and [∂T/∂t]₀ = \dot M_0RT₁/ρ₀ c_V \sout{V} . We are not given the density in the tank at the initial instant, but we can make use of the perfect gas law to write the density inside the tank as ρ₀ = p₀/RT(0) = p₀/RT₁, once again making use of the fact that T(0) = T₁. We now have

\left[\frac{\partial T}{\partial t}\right]_0=\frac{\dot M_0(RT_1)^2 }{p_0c_V \sout{V}}                                        (C)

Inserting the given data and values for R and c_V from Table 2.4 into (C) we find

\left[\frac{\partial T}{\partial t}\right]_0=\frac{\dot M_0(RT_1)^2 }{p_0c_V \sout{V}}=\frac{(5×10^{−5}\ kg/s)\left\{[287(N-m)/(kg-m)](293\ K)\right\} ^2 }{(1.01×10^5 \ N/m^2)[717(N-m)/(kg-K)](5×10^{−2} \ m^3)}

= 9.8 × 10⁻² K/s

Note that in using the perfect gas law we are required to use the absolute pressure and temperature.

It is instructive to calculate the magnitude of the neglected kinetic energy flux into the tank at the initial instant. This flux is seen to be

The corresponding initial flux of pressure potential energy is

\dot M_0\frac{p_1}{\rho _1} =\dot M_0RT₁ = (5 × 10⁻⁵ kg/s)[287(N-m)/(kg-K)](293 K) = 4.2 (kg-m²)/s³

which can be seen to be many orders of magnitude larger. This is typical for gas flows and justifies our neglect of the kinetic energy term in this flow.
TABLE 2.4 Perfect Gas Parameters for Several Common Gases