An insulated air tank, initially at a pressure of 0 MPa (gage) and 20°C, is connected by a valve to an air supply that provides air at a constant pressure of 0.5 MPa (gage) and the same temperature. When the valve is opened, the initial mass ﬂowrate into the tank is found to be

Question

An insulated air tank, initially at a pressure of 0 MPa (gage) and 20°C, is connected by a valve to an air supply that provides air at a constant pressure of 0.5 MPa (gage) and the same temperature. When the valve is opened, the initial mass ﬂowrate into the tank is found to be [latex]\dot M_0[/latex] = 0.05 g/s. If the volume of the tank is [latex]\sout{V}[/latex] = 0.05 m³, and the cross-sectional area of the supply line is 0.5 cm², determine the initial rate of change of density and temperature in the tank.

Accepted Answer

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, [latex]\int_{CV}[/latex] (∂ρ/∂t)dV + [latex]\int_{CS}^{}{}[/latex] 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) [latex]\sout{V}[/latex] − ρ₁V₁(t)A1 = 0. Upon rearrangement we ﬁnd

[latex]\frac{\partial ho }{\partial t}=\frac{ρ_1A_1V_1(t) }{\sout{V}}=\frac{\dot M(t)}{\sout{V}}[/latex] (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 [latex]\dot M_0[/latex] = 0.05 g/s, the initial rate of change of density inside the tank is calculated as

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

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

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

[latex]=\dot W_{power} +\dot W_{shaft} +\dot Q_C +\dot S [/latex]

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

[latex]\int_{CV}\frac{\partial}{\partial t}( ho u)dV+\int_{CS} ho \left(u+\frac{p}{ ho } ight)(u • n)dS= 0[/latex]

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:

[latex]\frac{\partial ( ho u)}{\partial t}\sout{V}−ρ_1V_1(t)A_1\left(u_1+\frac{p_1}{ ho _1} ight)=0[/latex]

Expanding the time derivative, the above yields [latex][u(∂ρ/∂t)+ρ(∂u/∂t)] \sout{V} −ρ_1V_1(t)A_1( u_1 + p_1/ρ_1) =0[/latex]. Now, [latex]\dot M(t)=ρ_1V_1(t)A_1[/latex], and from (A) we have [latex]∂ρ/∂t = \dot M(t)/ \sout{V}[/latex].

Thus the preceding expression becomes

[latex]\dot M(t)(u−u_1)+ρ\left(\frac{∂u}{∂t} ight) \sout{V}-\dot M(t)\left(\frac{p_1}{ ho _1} ight)=0[/latex]

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 [latex]\dot M (t)c_V[/latex](T − T₁) + ρc_V(∂T/∂t) [latex]\sout{V} - \dot M[/latex](t)RT₁ = 0. Solving for the rate of change of temperature in the tank, we have

[latex]\frac{\partial T}{\partial t}=\frac{\dot M(t)RT_1}{ ho (t)c_V \sout{V}}-\frac{\dot M(t)(T(t)-T_1)}{ ho (t) \sout{V}} [/latex] (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 = [latex]\dot M_0[/latex]RT₁/ρ₀ c_V [latex]\sout{V} [/latex]. 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

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

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

[latex]\left[\frac{\partial T}{\partial t} ight]_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) ight\} ^2 }{(1.01×10^5 \ N/m^2)[717(N-m)/(kg-K)](5×10^{−2} \ m^3)} [/latex]

= 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

[latex]\frac{1}{2}\dot M_0V_1(0)^2=\frac{1}{2}\dot M_0\left(\frac{\dot M_0}{ ho _1A_1} ight)^2=\frac{1}{2}\dot M_0^3\left(\frac{RT_1}{ ho _1A_1} ight)^2[/latex]

[latex]=\frac{1}{2}(5 imes 10^{-5}\ kg/s)^3\left\{\frac{[287(N-m)/(kg-K)](293 K)}{[(0.5+0.101)×10^6\ N/m^2](5×10^{−5}\ m^2)} ight\}^2 [/latex]

= 4.9 × 10⁻⁷ (kg-m²)/s³

The corresponding initial ﬂux of pressure potential energy is

[latex]\dot M_0\frac{p_1}{ ho _1} =\dot M_0[/latex]RT₁ = (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

			ln SI Units: J/(kg-k)			In BG Units: (ft-lb_f )/(slug-◦R)
Gas	MW	[latex]\gamma [/latex]	R	[latex]c_p[/latex]	[latex]c_v[/latex]	R	[latex]c_p[/latex]	[latex]c_p[/latex]
Air	29	1.40	287	1,004	717	1,716	6,006	4,290
CO₂	44	1.30	189	819	630	1,130	4,897	3,767
He	4	1.66	2077	5,224	3,147	12,420	31,238	18,818
H₂	2	1.41	4124	14,180	10,060	24,660	84,806	60,146
Methane	16	1.31	518	2,190	1,672	3,099	13,096	9,997
N₂	28	1.40	297	1,039	742	1,775	6,213	4,438
O₂	32	1.40	260	909	650	1,130	5,439	3,885

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

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