Question

A rigid, well-insulated tank of volume 0.9 m is initially evacuated. At time t = 0, air from the surroundings at 1 bar, 27°C begins to flow into the tank. An electric resistor transfers energy to the air in the tank at a constant rate for 5 minutes, after which time the pressure in the tank is 1 bar and the temperature is 457°C. Modeling air as an ideal gas, determine the power input to the tank, in kW.

Answer 1

`\dot{w}`= -0.303 KW

Step-by-step explanation:

This is the case of unsteady flow process because properties are changing with time.

From first law of thermodynamics for unsteady flow process

`(dU)/(dt)=\dot{m_i}h_i+\dot{Q}-\dot{m_e}h_i+\dot{w}`

Given that tank is insulated so
`\dot{Q}=0` and no mass is leaving so

`\dot{m_e}=0`

`\int dU=\int \dot{m_i}h_i\ dt-\int \dot{w}\ dt`

`m_2u_2-m_1u_1=(m_2-m_1)h_i- \dot{w}\Delta t`

Mass conservation
`m_2-m_1=m_e-m_i`

`m_1,m_2` is the initial and final mass in the system respectively.

Initially tank is evacuated so
`m_1=0`

We know that for air
`u=C_vT ,h=C_p T`,
`P_2v_2=m_2RT_2`

`m_2=0.42 kg`

So now putting values

`0.42 * 0.71 * 730=0.42* 1.005* 300- \dot{w} * 300`

`\dot{w}`= -0.303 KW