Pressurized Line and Rigid Tank Question

Using a pressurized line, a fluid is charged in a rigid tank with volume
$2m^3$. Derive the expression and calculate:


1. The internal energy $u\_t$ of the fluid in that tank supposing the initial mass is
, and specific enthalpy $h\_i=1670kJ/kg
2. Assuming a fluid is an ideal gas, derive and calculate the tank temperature using the specific heat ratio of k = 1.67 and the initial temperature of $T\_i=400 ^{\circ}C$
3. Assuming a fluid obeys ideal gas law and a flow is choked derive and calculate the tank pressure as a function of time using $\dot{m}=0.5kg/s,t=2s,k=1.67, T\_i=400^{\circ}C$.


Assume that charging occurs over a short period of time.


Starting with the control volume energy rate balance,
$$\frac{dE\_{cv}}{dt}=\dot{Q}\_{cv}-\dot{W}\_{cv}+\sum\_{i}\dot{m}\_i(h\_i+\frac{V\_i^2}{2}+gz\_i)-\sum\_{e}\dot{m}\_e(h\_e+\frac{V\_e^2}{2}+gz\_e)$$


we have, after ignoring kinetic energy and potential energy effects, taking into account that no mass is exiting the tank, and also assuming the tank is rigid and insulated, we have
$$\frac{dU\_{cv}}{dt}=\dot{Q}\_{cv}-\dot{W}\_{cv}+\sum\_{i}\dot{m}\_ih\_i-\sum\_{e}\dot{m}\_eh\_e$$
Upon integration and solving numerically, we have
$$\int\_{0}^{t}\frac{dU\_{cv}}{dt}dt=h\_i\int\_{0}^{t}\frac{dm}{dt}dt\Rightarrow U\_{cv}=mu=mh\_i\Rightarrow h\_i=u=1670\frac{kJ}{kg}$$
For part 2, using the ideal gas model we have
$$u(T)-u(T\_{i})=c\_v(T-T\_i)=RT\_i$$
$$h(T)-h(T\_i)=c\_p(T-T\_i)=RT\_i+R(T-T\_i)$$


I'm not sure where to go from here.


