Question

1. A feedwater heater operating at steady state has two inlets and one exit. At inlet 1, water vapor enters at p1 = 7 bar, T1 = 200^oC with a mass flow rate of 40 kg/s. At inlet 2, liquid water at p2 = 7 bar, T2 = 40^oC, density = 992.260 kg/m³ enters through an area A2 = 25 cm². Liquid water at 7 bar with a density of 902.527 kg/m³ exits at exit 3 with a volumetric flow rate of 0.06 m³/s. Draw the diagram and identify the inputs, output, and the system. Write the general form of Conservation of Mass equation, and simplify it for this case.

1. Determine the mass flow rate at inlet 2, in kg/s.
2. Determine the mass flow rate at the exit, in kg/s.
3. Determine the velocity at inlet 2, in m/s.

Answer 1

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Answer 2

Given:

• Inlet 1 (Vapor):
  ${\dot{m}}_1=40\text{kg/s}$

• Inlet 2 (Liquid):
  ${\rho }_2=992.260{\text{kg/m}}^3$, $A_2=25{\text{cm}}^2=0.0025{\text{m}}^2$

• Exit 3 (Liquid):
  ${\rho }_3=902.527{\text{kg/m}}^3$, ${\dot{V}}_3=0.06{\text{m}}^3\mathrm{/}\text{s}$

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a. Mass Flow Rate at Inlet 2 (${\dot{m}}_2$):

$${\dot{m}}_3={\rho }_3\times {\dot{V}}_3=902.527\times 0.06=54.1516\text{kg/s}$$$${\dot{m}}_2={\dot{m}}_3-{\dot{m}}_1=54.1516-40=\boxed{{\displaystyle 14.15\text{kg/s}}}$$

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b. Mass Flow Rate at Exit (${\dot{m}}_3$):

$${\dot{m}}_3={\rho }_3\times {\dot{V}}_3=902.527\times 0.06=\boxed{{\displaystyle 54.15\text{kg/s}}}$$

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c. Velocity at Inlet 2 ($v_2$):

$$v_2=\frac{{\dot{m}}_2}{{\rho }_2\times A_2}=\frac{14.1516}{992.260\times 0.0025}=\boxed{{\displaystyle 5.70\text{m/s}}}$$

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Final Answers:

Part	Answer
(a)	${\dot{m}}_2=14.15\text{kg/s}$
(b)	${\dot{m}}_3=54.15\text{kg/s}$
(c)	$v_2=5.70\text{m/s}$