Question

I kg of a fluid expands reversibly according to a linear law from 4.2 bar to 1.4 bar; the initia] and final volumes are 0.004 m) and 0.02 m). The fluid is then cooled reversibly at constant pressure, and finally compressed reversibly according to a law po = constant back to the initial conditions of 4.2 bar and 0.004 m '. Calculate the work done in each process and the net work of the cycle. Sketch the cycle on a. p-v diagram,

Answer 1

Answer on Question #78936, Physics / Molecular Physics | Thermodynamics

Question. $1 \, kg$ of a fluid expands reversibly according to a linear law from 4.2 bar to 1.4 bar; the initial and final volumes are $0.004 \, m^3$ and $0.02 \, m^3$ . The fluid is then cooled reversibly at constant pressure, and finally compressed reversibly according to a law $pV = \text{constant}$ back to the initial conditions of 4.2 bar and $0.004 \, m^3$ . Calculate the work done in each process and the net work of the cycle. Sketch the cycle on a $p - V$ diagram.

Given. $m = 1$ kg; $p_1 = 4.2$ bar $= 4.2 \cdot 10^5$ Pa; $p_2 = 1.4$ bar $= 1.4 \cdot 10^5$ Pa; $V_1 = 0.004$ m³; $V_2 = 0.02$ m³.

Find. $W_{12}, W_{23}, W_{31}, W - ?$

Solution

$W_{12}$ (area of the trapezoid) $= \frac{1}{2} (p_1 + p_2)(V_2 - V_1) = \frac{1}{2} (4.2\cdot 10^5 +1.4\cdot 10^5)(0.02 - 0.004) = 4480J$

The work done by the fluid.

$$W _ {2 3} (\text {a r e a o f t h e r e c t a n g l e}) = p _ {2} \left(V _ {3} - V _ {2}\right)$$ $$p _ {3} V _ {3} = p _ {1} V _ {1} \text {a n d} p _ {3} = p _ {2}$$ $$V _ {3} = \frac {p _ {1}}{p _ {2}} V _ {1} = \frac {4 . 2 \cdot 1 0 ^ {5}}{1 . 4 \cdot 1 0 ^ {5}} 0. 0 0 4 = 0. 0 1 2 m ^ {3}$$ $$W _ {2 3} = 1. 4 \cdot 1 0 ^ {5} (0. 0 1 2 - 0. 0 2) = - 1 1 2 0 J$$

The work done on the fluid.

$$\begin{array}{l} W _ {3 1} = \int_ {V _ {3}} ^ {V _ {1}} p d V = \int_ {V _ {3}} ^ {V _ {1}} \frac {m}{M} \frac {R T}{V} d V = \frac {m}{M} R T \ln \frac {V _ {1}}{V _ {3}} = p _ {3} V _ {3} \ln \frac {V _ {1}}{V _ {3}} = p _ {1} V _ {1} \ln \frac {V _ {1}}{V _ {3}} = 4. 2 \cdot 1 0 ^ {5} \cdot 0. 0 0 4 \cdot \ln \frac {0 . 0 0 4}{0 . 0 1 2} = \\ = - 1 8 4 5 J \\ \end{array}$$

The work done on the fluid.

$$W = W _ {1 2} + W _ {2 3} + W _ {3 1} = 4 4 8 0 - 1 1 2 0 - 1 8 4 5 = 1 5 1 5 J$$

The work done by the fluid.

Answer. $W_{12} = 4480J$ ; $W_{23} = -1120J$ ; $W_{31} = -1845J$ ; $W = 1515J$ .

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