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A rigid tank whose volume is 2 m3, initially containing air at 1 bar, 295 K, is connected by a valve to a large vessel holding air at 6 bar, 295 K. The valve is opened only as long as required to fill the tank with air to a pressure of 6 bar and a temperature of 350 K. Assuming the ideal gas model for the air, determine the heat transfer between the tank contents and the surroundings, in kJ

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Answer:


Q_(cv)=-339.347kJ

Step-by-step explanation:

First we calculate the mass of the aire inside the rigid tank in the initial and end moments.


P_iV_i=m_iRT_i (i could be 1 for initial and 2 for the end)

State1


1bar*|(100kPa)/(1)|*2=m_1*0.287*295


m_1=232kg

State2


8bar*|(100kPa)/(1bar)|*2=m_2*0.287*350


m_2=11.946

So, the total mass of the aire entered is


m_v=m_2-m_1\\m_v=11.946-2.362\\m_v=9.584kg

At this point we need to obtain the properties through the tables, so

For Specific Internal energy,


u_1=210.49kJ/kg

For Specific enthalpy


h_1=295.17kJ/kg

For the second state the Specific internal Energy (6bar, 350K)


u_2=250.02kJ/kg

At the end we make a Energy balance, so


U_(cv)(t)-U_(cv)(t)=Q_(cv)-W{cv}+\sum_i m_ih_i - \sum_e m_eh_e

No work done there is here, so clearing the equation for Q


Q_(cv) = m_2u_2-m_1u_1-h_1(m_v)


Q_(cv) = (11.946*250.02)-(2.362*210.49)-(295.17*9.584)


Q_(cv)=-339.347kJ

The sign indicates that the tank transferred heat to the surroundings.

User Johans
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