Thermodynamics 1 question Three return steam 1ines in a chem

Thermodynamics 1 question.

Three return steam 1ines in a chemical processing plant enter collection tank operating at steady state at 1 bar. Steam enters inlet 1 with flow rate of 0.8 kgs and quality of 1 0.9. Steam enters inlet 2 with flow rate of 2 kg/s at 200 C. Steam enters inlet 3 with flow rate of 1.2 kg/s at 95 C. Steam exits the tank at 1 bar. The rate of heat transfer from the collection tank is 40 kW. Neglecting kinetic and potential energy effects, determine for the steam exiting the tank: (a) the mass flow rate in kg/s), and (b) the temperature (in C). Collection Tank Heat transfer from the tank

Solution

a) it is a steady state, so mass flow is conserved:

m₁+m₂+m₃=m₄

m₄ = .8 + 2 + 1.2 = 4 kg/s

b) for the first inlet, at 1 bar (100kPa) from saturated water tables

h_f = 417.51, h_fg =2257.5 steam quality = 0.9

h₁ = 417.51 + .9x2257.5 = 2449.26 kJ/kg

so energy flowing in will be 2449.26 kJ/kg x .8 kg/s = 1959.4 kJ/s

for second inlet, at 1 bar, boiling temperature is around 100C, so look in superheated tables, for 200C

h₂ = 2875.5

energy flowing in through 2 is 2875.5 kJ/kg x 2 kg/s =5751 kJ/s

for third inlet

this is a compressed liquid as at 1 bar, saturation temperature is 99.6 C

so from compressed liquid table, h₃ = 398.1 kJ/kg

energy flow is 398.1 x 1.2 = 477.72 kJ/s

now, total energy flowing in is 1959.4+5751+477.72 = 8188.12 kJ/s

but 40 kW flows out as heat, that is 40 kJ/s, so total energy flowing out with steam will be

8188.12-40=8148.12kJ/s

mass flow rate out of the tank is 4kg/s, therefore, enthalpy of water going out willbe

8148.12 / 4 = 2037.03 kJ/kg

at 1 bar, enthalpy of saturated liquid water is 417.51 and of saturated vapour is 2675 kJ/kg

therefore the water going out will be a liquid vapuor mixture with temperature 99.61^oC which is the saturation temperature

and the steam quality will be 0.7