Three return steam lines in a chemical processing plan enter a collection tank operating at a steady state at 1 bar. Steam enters inlet 1 with flow rate of 0.8 kg/s and a quality of 0.9. Steam enters inlet 2 with flow rate of 2kg/s at 200 degrees C. Steam enters inlet 3 with flow rate 1.2 kg/s at 95 degrees 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)Mass flow rate in kg/s (b) the temperature in degrees C.

Question

01-11-2019
Engineering

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Three return steam lines in a chemical processing plan enter a collection tank operating at a steady state at 1 bar. Steam enters inlet 1 with flow rate of 0.8 kg/s and a quality of 0.9. Steam enters inlet 2 with flow rate of 2kg/s at 200 degrees C. Steam enters inlet 3 with flow rate 1.2 kg/s at 95 degrees 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)Mass flow rate in kg/s (b) the temperature in degrees C.

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sebassandin sebassandin · Answer 1 · 2019-11-02T23:29:30+01:00

Answer:

a. 4kg/s

b. 99.97 °C

Explanation:

Hello,

a. The resulting mass balance turns out into:

[tex]F_1+F_2+F_3=F_{out}\\F_{out}=0.8kg/s+2kg/s+1.2kg/s=4kg/s[/tex]

b. Now, the energy balance is:

[tex]F_1h_1+F_2h_2+F_3h_3-Q_{out}=F_{out}h_{out}[/tex]

In such a way, the first enthalpy is taken as a liquid-vapor mixture at 1 bar and 0.9 quality, it means:

[tex]h_1=hf(1bar)+xhfg(1bar)\\h_1=419.06kJ/kg+0.9*2256.5kJ/kg=2449.91kJ/kg[/tex]

Second enthalpy is taken by identifying that stream as an overheated vapor at 1 bar and 200 °C, thus, the resulting enthalpy is:

[tex]h_2=2875.5kJ/kg[/tex]

Then, the third enthalpy is taken by considering that at 95°C and 1 bar the water is a saturated liquid, thus:

[tex]h_3=hf(95^0C)=398.09kJ/kg[/tex].

Now, by solving for [tex]h_{out}[/tex], we've got:

[tex]h_{out}=\frac{0.8kg/s*2449.91kJ/kg+2kg/s*2875.5kJ/kg+1.2kg/s*398.09kJ/kg-40kW}{4kg/s} \\h_{out}=\frac{8148.612kJ/s}{4kg/s} \\h_{out}=2037.153kJ/kg[/tex]

Finally, by searching for that value of enthalpy, one sees that at 1 bar, the exiting stream is a liquid-vapor mixture that is at 99.97 °C and has a 72%- quality.

(NOTE: all the data was extracted from Cengel's book 7th edition).

Best regards.