Question

A 200 liter tank is connected to a line flowing nitrogen (N2) at 30°C and 10 MPa pressure. The tank initially contains nitrogen (N2) at 30°C, 101 kPa. The valve is opened, allowing nitrogen to flow into the tank until the pressure inside is 9 MPa. At this point the valve is closed. This filling process occurs rapidly so that there is no heat transfer. For simplicity, you can treat nitrogen as an ideal gas with constant specific heats. a. What is the temperature of the nitrogen in the tank when the valve is closed? This is very similar to an example from class, except the tank isn't initially empty in this problem. b. The tank is then placed in storage where it eventually returns to the environment's temperature, 20 °C. What is the final pressure in MPa?

Answer 1

Answer:

a. $T_2=30°C$

b. $P_{new}=8.703MPa$

Explanation:

Hello,

a. In this case, the following energy balance must be stated:

$E_{in}-E_{out}=$Δ$E$

$m_{N_2,in}h_{N_2,in}=m_2h_2-m_1h_1$,

Now, at 30°C which is both the initial and inlet temperature of nitrogen, the referred enthalpy has a value of 8,815 J/mol (Cengel's book), thus, one can say that:

$h_{N_2,in}=h_1$

In such a way, we could factor as follows:

$(m_{N_2,in}+m_1)h_{N_2,in}=m_2h_2$

Now, since $m_2$ is defined as the addition between the initial mass and the inlet mass of nitrogen, one sum up that:

$h_{N_2,in}=h_2$

Thus, since the initial enthalpy, computed at 30°C remains constant, it means that the final temperature remains constant, so:

$T_2=30°C$

This is evident due to the fact that the inlet nitrogen has the same initial nitrogen and no heat is transferred.

b. Considering the final temperature, one proceeds to compute the final mass as follows:

$m_2=\frac{0.014kg/mol*9x10^6Pa*2m^3}{8.314\frac{Pa*m^3}{mol*K}*303.15K}= 99.98kgN_2$

Now, as long as the temperature changes to 20°C, the new pressure is computed as:

$P=\frac{99.98kg*8.314\frac{Pa*m^3}{mol*K}*293.15K}{2m^3*0.014kg/mol}= 8.703MPa$