# 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.