Three return steam lines in a chemical processing plant enter a collection tank operating at 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 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 engery effects, determine for the steam exiting the tank
(a) the mass flow rate, in kg/s
(b) the temperature, in degrees C
(a) the mass flow rate, in kg/s
(b) the temperature, in degrees C
1 answer:
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Answer:
Temperature T = 394.38 K
Explanation:
The full solution and detailed explanation regarding the above question and its specified conditions can be found below in the accompanying document. I trust my explanation will assist you in grasping this particular topic.
Answer:
Volume change percentage is 2.60%
Water level increase is 4.138 mm
Explanation:
Provided data
Water volume V = 500 L
Initial temperature T1 = 20°C
Final temperature T2 = 80°C
Diameter of the vat = 2 m
Objective
We aim to determine percentage change in volume and the rise in water level.
Solution
We will apply the bulk modulus equation, which relates the change in pressure to the change in volume.
It can similarly relate to density changes.
Thus,
E = ................1
And ............2
Here, ρ denotes density. The density at 20°C = 998 kg/m³.
The density at 80°C = 972 kg/m³.
Plugging in these values into equation 2 gives
dV = 0.0130 m³
Therefore, the percentage change in volume will be
dV % = × 100
dV % = × 100
dV % = 2.60 %
Hence, the percentage change in volume is 2.60%
Initial volume v1 = ................3
Final volume v2 = ................4
From equations 3 and 4, subtract v1 from v2.
v2 - v1 =
dV =
Substituting all values yields
0.0130 =
Thus, dl = 0.004138 m.
Consequently, the water level rises by 4.138 mm.