Question 6.11: A standard propane tank, like those typically used with a ba...

In the NIST WebBook, we navigate to Thermophysical Properties of Fluid Systems, select propane, select saturation properties, select the desired units, and choose 25°C as the desired temperature. We find P^{sat} = 9.522 bar, the

V^l=2.031 \mathrm{~L} \cdot \mathrm{kg}^{-1}, V^v=48.51 \mathrm{~L} \cdot \mathrm{kg}^{-1}, U^l=263.4 \mathrm{~kJ} \cdot \mathrm{kg}^{-1}, U^v=555.0 \mathrm{~kJ} \cdot \mathrm{kg}^{-1} \text {, }

H^l=265.3 \mathrm{~kJ} \cdot \mathrm{kg}^{-1} \text {, and } H^v=601.2 \mathrm{~kJ} \cdot \mathrm{kg}^{-1}

(a) The volume of the liquid and vapor phases must sum to the total volume of the tank, and their masses must sum to 20  lb_m  or  9.072  kg. Thus,  m^l + m^v = 9.072  kg  and  m^l V^l+m^v V^v=21.5 \mathrm{~L} Combining these and using the values of V^l  and  V^v gives:

2.031(9.072  –  m^v) + 48.51 m^v = 21.5

Solving yields m^v = 0.066  kg,  from  which  m^l = 9.072  –  0.066 = 9.006  kg. The total volume occupied by the liquid is 9.006 · 2.031 = 18.3 L, and the volume of vapor is 3.2 L.

(b)   If the temperature remains constant as propane is removed, then the pressure also stays constant at the vapor pressure, as long as two phases are present. Thus, the properties of the vapor and liquid phase remain the same. As a quick estimate, we would expect that the heat added will be approximately equal to the heat of vaporization of the propane that leaves the tank, although slightly more must vaporize to fill the additional space left in the tank. From the NIST WebBook data, ΔH^v = 601.2 – 265.3 = 5  lb_m = 335.9  kJ·kg^−1. Thus,  if  5  lb_m = 2.268  kg of propane is vaporized, the total heat of vaporization is 335.9 · 2.268 = 762 kJ. For a more rigorous analysis, we must write an energy balance on the propane in the tank, as:

\frac{d(m U)}{d t}+\dot{m} H^v=\dot{Q}

This is Eq. (2.27) with a single outflow stream, no shaft work, and no changes in kinetic or potential energy. Here, mU is the total internal energy of the tank contents (both phases) while the relevant H is that of the vapor because propane leaves the tank as vapor. Formally integrating over the whole process of removing 5 lb_m of propane yields:

\frac{d(m U)_{\mathrm{cv}}}{d t}+\Delta\left[\left(H+\frac{1}{2} u^2+z g\right) \dot{m}\right]_{\mathrm{fs}}=\dot{Q}+\dot{W}      (2.27)

Q = Δ(mU) + H^v  · –  Δm

Note that the total outflow from the tank is minus Δm, if m refers to the mass remaining in the tank. To compute the first term on the right-hand side, we require the total internal energy in the tank at the beginning and end of the process. At the start, we have:

m U_{\text {before }}=m^I U^l+m^v U^v=9.006 \cdot 263.4+0.066 \cdot 555.0=2409 \mathrm{~kJ}

Repeating the analysis of part (a) for 15  l_bm (6.804 kg) propane remaining in the tank gives m^l = 6.639  kg, m^v = 0.165  kg . Thus, the total internal energy in the tank after removal of 5  lb_m (2.268 kg) of vapor is:

mU_{after}= 6.639 · 263.4 + 0.165 · 555.0 = 1840  kJ

Using this value, we find Q = (1840 – 2409) + 601.2 . 2.268 = 795 kJ. This is about 4% greater than our rough estimate using only the heat of vaporization at 25°C.