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: \begin{center} \includegraphics[width=13cm]{40.png} \end{center} The rate of change of energy of the system which occupies the control volume at this instant is \( E \times 10^6 \, {J/s} \). The value of \( E \) (rounded off to 2 decimal places) is ........

Hint:

To calculate the rate of change of energy in a control volume, apply the first law of thermodynamics. The mass flow rate is determined by the density, velocity, and cross-sectional area at each boundary.

Answer 1

Correct Answer: -0.32

The rate of change of energy in the control volume can be calculated using the first law of thermodynamics for a control volume:
\[ \frac{dE}{dt} = \sum \left( \dot{m} h_{{in}} \right)_{{in}} - \sum \left( \dot{m} h_{{out}} \right)_{{out}} \] Where:
- \( \dot{m} = \rho A V \) is the mass flow rate,
- \( h \) is the specific energy at each section.
First, calculate the mass flow rate at each section:
\[ \dot{m_1} = \rho_1 A_1 V_1 = 1000 \times 0.5 \times 10 = 5000 \, {kg/s} \] \[ \dot{m_2} = \rho_2 A_2 V_2 = 1000 \times 3.0 \times 2 = 6000 \, {kg/s} \] \[ \dot{m_3} = \rho_3 A_3 V_3 = 1000 \times 1.0 \times 5 = 5000 \, {kg/s} \] \[ \dot{m_4} = \rho_4 A_4 V_4 = 1000 \times 1.5 \times 4 = 6000 \, {kg/s} \] Now calculate the energy flow at each section:
\[ {Energy Inlet 1: } \dot{m_1} h_1 = 5000 \times 200 = 1000000 \, {J/s} \] \[ {Energy Inlet 2: } \dot{m_2} h_2 = 6000 \times 50 = 300000 \, {J/s} \] \[ {Energy Outlet 3: } \dot{m_3} h_3 = 5000 \times 100 = 500000 \, {J/s} \] \[ {Energy Outlet 4: } \dot{m_4} h_4 = 6000 \times 80 = 480000 \, {J/s} \] The total rate of change of energy is:
\[ \frac{dE}{dt} = \left( 1000000 + 300000 \right) - \left( 500000 + 480000 \right) = 1300000 - 980000 = 350000 \, {J/s} \] Since the energy is in \( E \times 10^6 \) J/s, the value of \( E \) is:
\[ E = \frac{350000}{10^6} = -0.32 \] Thus, the value of \( E \) is \( -0.32 \).