b) Due to hydraulic shock, the pipe ruptured resulting in release of liquid ammonia into the atmosphere. Liquid ammonia immediately flashes in the surrounding at ambient temperature, Tamb = 298 K and atmospheric pressure, Patm = 101325 Nm2. Accordingly, i) starting from first principles, and stating any assumptions made, show that the mass flow rate of liquid ammonia escaping through the ruptured pipe upon failure is given by: 2(Ps Patm) Im = PAC. u? + (1) P Additional Information: Density of liquid ammonia at 240 K and 104800 Nm2 = 681.75 kgm n-3 Discharge coefficient for release = 0.61 Ruptured pipe release area = 0.003 m Molecular weight of ammonia = 17 gmol! Specific heat ratio of ammonia = 1.35 Gas constant = 8.314 Jmol K-1 Ambient temperature = 298 K Pipe length = 10 m Pipe diameter = 0.304 m ii) using the Eq. (1), determine the mass flow rate of liquid ammonia, Qm through the ruptured pipe assuming a constant feed flow rate of 50 kgs in the pipe at 240 K and 104800 Nm-2 iii) starting with the chocked mass flow rate equation for isothermal ideal gas flow discharging through a small puncture in a vessel given by m = = C,APF Y+1 My 2 y-1 RT\y+1 (2 ** (2) where A is the puncture area, y the ratio of specific heat capacities, M molecular weight, C. discharge coefficient, with subscript, f denoting fluid, show that, the variation of the upstream pressure, P(t) in the vessel as a function of time (t), is given by: Pf,t=0 P(t) CitRT e MV (3) where V is the volume of the gas and y+1 My 2 Y-1 C1 = CoA RT, \y + 1) (4)