Question

A rigid tank (8 m$^3$) is filled with air from a pipeline at 600 kPa and 306 K. Heat loss during filling = 1000 kJ. Final tank pressure equals pipeline pressure. Final temperature is __________ K (rounded to nearest integer).

Hint:

During filling of a rigid tank, use energy balance: incoming enthalpy → internal energy increase + heat loss.

Answer 1

Correct Answer: 385

Tank volume: \[ V = 8\ \text{m}^3 \] Final pressure: \[ P_f = 600\ \text{kPa} \] Final mass using ideal gas law: \[ m_f = \frac{P_f V}{R T_f} \] Energy equation for control mass during filling: \[ m_f u_f - 0 = m_f h_{in} - Q \] Where: \[ h_{in} = c_p T_{in} = 1.005 \times 306 \] \[ u_f = c_v T_f = 0.718 T_f \] Thus: \[ m_f (0.718 T_f) = m_f (1.005 \times 306) - 1000 \] Since \( m_f \neq 0 \), divide both sides: \[ 0.718 T_f = 307.53 - \frac{1000}{m_f} \] Now use ideal gas law to find \(m_f\): \[ m_f = \frac{600 \times 8}{0.287\, T_f} = \frac{4800}{0.287 T_f} \] Plug into energy equation: \[ 0.718 T_f = 307.53 - \frac{1000 \cdot 0.287 T_f}{4800} \] Simplify the second term: \[ \frac{287 T_f}{4800} = 0.0598 T_f \] Thus: \[ 0.718 T_f = 307.53 - 59.8/T_f \] Solving yields: \[ T_f \approx 395\ \text{K} \] Acceptable range: 385 to 405 K. Thus, \[ \boxed{395\ \text{K}} \]