Question

In a heating system, cold outdoor air at \(7^{\circ} \mathrm{C}\) flowing at a rate of \(4 \mathrm{kg} / \mathrm{min}\) is mixed adiabatically with heated air at \(70^{\circ} \mathrm{C}\) flowing at a rate of \(3 \mathrm{kg} / \mathrm{min} .\) The exit temperature of the mixture is \((a) 34^{\circ} \mathrm{C}\) (b) \(39^{\circ} \mathrm{C}\) \((c) 45^{\circ} \mathrm{C}\) \((d) 63^{\circ} \mathrm{C}\) \((e) 77^{\circ} \mathrm{C}\)

Short answer

Answer: (a) 34°C

Step-by-step solution

Conservation of Energy

For an adiabatic process, the total energy of the incoming streams should be equal to the total energy of the outgoing stream. Therefore, the energy equation for this process can be expressed as: \(m_1 c_p(T_1) + m_2 c_p(T_2) = (m_1 + m_2) c_p(T_x)\) where - \(m_1\) and \(m_2\) are the mass flow rates of the two streams (in kg/min), - \(c_p\) is the specific heat capacity of air (in J/kg K), - \(T_1\) and \(T_2\) are the temperatures of the two incoming streams (in °C), - \(T_x\) is the exit temperature we want to find (in °C).

Solve for the exit temperature \(T_x\)

Now we can plug in the known values and solve the equation for \(T_x\): \(4 \, kg/min \cdot c_p \cdot (7 ^{\circ} C) + 3 \, kg/min \cdot c_p \cdot (70^{\circ} C) = (4 \, kg/min + 3 \, kg/min) \cdot c_p \cdot (T_x)\). Since \(c_p\) is the same for both flow rates and temperatures, we can cancel it out from the equation: \((4)(7^{\circ} C) + (3)(70^{\circ} C) = (4 + 3)(T_x)\).

Calculate the exit temperature

Solve the equation for \(T_x\): \(28^{\circ} C + 210^{\circ} C = 7(T_x)\). \(238^{\circ} C = 7(T_x)\). Now, divide both sides of the equation by 7 to find the exit temperature: \(T_x = 34^{\circ} C\). So, the exit temperature of the mixture is \((a) \, 34^{\circ} C\).