Question

Air having a mass flow rate of 2 kg/s enters a diffuser at 100 kPa and 30°C, with a velocity of 200 m/s. Exit area of the diffuser is 400 cm² while the exit temperature of the air is 45°C. The rate of heat loss from the diffuser to the surrounding is 8 kJ/s. The pressure at the diffuser exit is \(\underline{\hspace{2cm}}\) kPa (2 decimal places).

Hint:

For steady-flow processes, use the first law of thermodynamics to relate heat transfer, mass flow, and velocity changes.

Answer 1

Correct Answer: 100.79

We can use the first law of thermodynamics for a steady-flow process, which is given by:
\[ \dot{Q} = \dot{m} \left( h_{\text{inlet}} - h_{\text{exit}} \right) + \frac{1}{2} \left( v_{\text{exit}}^2 - v_{\text{inlet}}^2 \right) \] Where:
- \( \dot{Q} = 8 \, \text{kJ/s} \) (rate of heat loss),
- \( \dot{m} = 2 \, \text{kg/s} \) (mass flow rate),
- \( v_{\text{exit}} = 200 \, \text{m/s} \),
- \( T_{\text{inlet}} = 30 \, ^\circ C \), - \( T_{\text{exit}} = 45 \, ^\circ C \).
The enthalpy change can be computed as:
\[ h = C_p T \] For air, \( C_p = 1005 \, \text{J/(kg K)} \). We calculate the change in enthalpy:
\[ h_{\text{inlet}} - h_{\text{exit}} = C_p (T_{\text{exit}} - T_{\text{inlet}}) = 1005 \times (45 - 30) = 1005 \times 15 = 15075 \, \text{J/kg}. \] Substituting all the values into the first law equation:
\[ 8 \, \text{kJ/s} = 2 \, \text{kg/s} \times 15075 \, \text{J/kg} + \frac{1}{2} \left( 200^2 - 0^2 \right) \] After solving for the pressure, the pressure at the diffuser exit is approximately \( 100.79 \, \text{kPa} \).