Question 8.10: A CONTINUATION OF EXAMPLE 5.6, WITH THE ADDITION SHOWN IN IT...
A CONTINUATION OF EXAMPLE 5.6, WITH THE ADDITION SHOWN IN ITALIC TYPE
Suppose 0.100 \text{lbm } of Refrigerant-134a initially at 180.°\text{F } and 100. \text{psia} in a cylinder with a movable piston undergoes the following two-part process. First, the refrigerant is expanded adiabatically to 30.0 \text{psia} and 120.°\text{F }, then it is isobarically compressed to half its initial volume. Determine
a. The work transport of energy during the adiabatic expansion.
b. The heat transport of energy during the isobaric compression.
c. The final temperature at the end of the isobaric compression.
d. The total entropy production for both processes if the heat transport of energy and the boundary temperature are related by the formula dQ = KdT_b, where K = 5.00 \text{Btu/R} and T_b is in R.
First, draw a sketch of the system (Figure 8.10).
The unknowns here are (a) _1W_2, (b) _2Q_3, (c) T_3, and (d) _1(S_P)_3. The system is just the R-134a, and the state variables are as follows:
| \underline{\text{State 1}} | \underrightarrow{\text{Adiabatic} } \text{pansion} |
\underline{\text{State 2}} | \underrightarrow{\text{Isobaric} } \text{compression} |
\underline{\text{State 3}} |
| p_1 = 100 \text{psia} | p_2 = 30.0 \text{psia} | p_3 = p_2 = 30.0 \text{psia} | ||
| T_1 = 180.°\text{F} | T_2 = 120.°\text{F} | v_3 = v_1/2 | ||
| v_1 = 0.6210 \text{ft³/lbm} | v_2 = 1.9662 \text{ft³/lbm} | v_3 = 0.3105 \text{ft³/lbm} | ||
| h_1 = 137.49 \text{Btu/lbm} | h_2 = 126.39 \text{Btu/lbm} | x_3 = 0.1952 | ||
| s_1 = 0.2595 \text{Btu/(lbm.R)} | s_2 = 0.2635 \text{Btu/(lbm.R)} | s_3 = 0.07241 \text{Btu/(lbm.R)} |
The solutions to the first three parts of this problem are given in Example 5.6 as
a. _1W_2 = 1.05 \text{Btu}.
b. _2Q_3 = –9.31 \text{Btu}.
c. T_3 = 15.4°\text{F}.
The solution to part d can be determined by the indirect method (entropy balance) as follows. From Eq. (8.1), we have
\begin{matrix}_1(S_P)_2 = m(s_2-s_1)-\underbrace{\int_{Σ}(\frac{\overline{d} Q}{T_b} )}_{0} \\ \quad \quad \quad \quad \quad \quad \quad\quad \quad \quad \quad\text{(adiabatic process) }\end{matrix}
From Table C.7e, we find that
s_1 = 0.2595 \text{Btu/(lbm.R)}
and
s_2 = 0.2635 \text{Btu/(lbm.R)}
Since the process from 1 to 2 is adiabatic, \overline{d} Q = 0. and
_1(S_P)_2 =(0.100 \text{lbm})[0.2635 − 0.2595 \text{Btu/(lbm.R)}] − 0 = 4.00 × 10^{-4} \text{Btu/R}
Similarly,
_2(S_P)_3 = m(s_3-s_2)- \int_{Σ}(\frac{\overline{d} Q}{T_b} )
In this process, we are given that \overline{d} Q = KdT_b , where K = 5.00 \text{Btu/R} . Therefore, _2Q_3 = K(T_{b3} – T_{b2}) + C = –9.31 \text{Btu}, where C is an integration constant. Then,
\int_{Σ}(\frac{\overline{d} Q}{T_b}) = \int_{T_{b2} }^{T_{b3}}K (\frac{dT_b}{T_b} )= K \text{ln} \frac{T_{b3}}{T_{b2}}
Therefore,
_2(S_P)_3 = m(s_3-s_2) – K \text{ln} \frac{T_{b3}}{T_{b2}}
where T_{b2} = T_2 and T_{b3} = T_3.
Now, s_3 = s_{f 3} + x_3 s_{f g3}, and from Table C.7b in Thermodynamic Tables to accompany Modern Engineering Thermodynamics, at 30.0 \text{psia} , we get
s_{f 3} = 0.0364 \text{Btu/(lbm.R)}
s_{f g3} = 0.2209 \text{Btu/(lbm.R)}
Then,
s_3 = 0.0364 + (0.1952) (0.2209 − 0.0364) = 0.07241 \text{Btu/(lbm.R)}
and
_2(S_P)_3 = (0.100 \text{lbm})[0.07241−0.2635 \text{Btu/(lbm.R)}] − (5.00 \text{Btu/R}) \text{ln} \frac{15.38 + 459.67}{120. + 459.67}
= 0.976 \text{Btu/R}
Finally, the entropy production for the entire process is given by
_1(S_P)_3 = _1(S_P)_2 + _1(S_P)_2 = 4.00× 10^{-4} +0.976 = 0.976 \text{Btu/R} \text{(to 3 significant figures)}
In this example, a special Q = Q(T_b) relation is introduced, which is similar to that used to describe convection heat transfer processes. These relations, often called thermal constitutive equations, are mathematical models developed to describe specific heat transfer mechanisms. The following exercises illustrate this concept.
