Question

A fixed control volume has four one-dimensional boundary sections (1, 2, 3, and 4). For a steady flow inside the control volume, the flow properties at each section are tabulated below:

The rate of change of energy of the system which occupies the control volume at this instant is $E \times 10^6$ J/s. Find the value of $E$ (rounded off to 2 decimal places).

Hint:

Always calculate mass flow = $\rho V A$ and then multiply with specific energy for energy transport.

Answer 1

Correct Answer: 0.32

Step 1: Energy balance for control volume.
For steady flow: \[ \dot{E}_{cv} = \sum \dot{m}_{in} e_{in} - \sum \dot{m}_{out} e_{out}. \]

Step 2: Mass flow rates.
\[ \dot{m} = \rho V_n A. \] - Section 1: $\dot{m}_1 = 1000 \times 10 \times 0.5 = 5000$ kg/s. - Section 2: $\dot{m}_2 = 1000 \times 2 \times 3 = 6000$ kg/s. - Section 3: $\dot{m}_3 = 1000 \times 5 \times 1 = 5000$ kg/s. - Section 4: $\dot{m}_4 = 1000 \times 4 \times 1.5 = 6000$ kg/s.

Step 3: Energy flow.
- Inlets: \[ \dot{E}_{in} = \dot{m}_1 e_1 + \dot{m}_2 e_2 = 5000(200) + 6000(50) = 1,000,000 + 300,000 = 1.3 \times 10^6 \, \text{J/s}. \] - Outlets: \[ \dot{E}_{out} = \dot{m}_3 e_3 + \dot{m}_4 e_4 = 5000(100) + 6000(80) = 500,000 + 480,000 = 0.98 \times 10^6 \, \text{J/s}. \]

Step 4: Net rate of energy change.
\[ \dot{E}_{cv} = 1.3 \times 10^6 - 0.98 \times 10^6 = 0.32 \times 10^6 \,\text{J/s}. \] So, \[ E = 0.32. \] Final Answer:
\[ \boxed{0.32} \]