Liquid water at 300 kPa and 20°C is heated in a chamber by mixing it with superheated steam at 300 kPa and 300°C. Cold water enters the chamber at a rate of 2.6 kg/s. If the mixture leaves the mixing chamber at 60°C.

Required:
Determine the mass flow rate of the superheated steam required.

Answer:

0.154kg/s

Explanation:

From this question we have the following information:

P1 = 300kpa

T1 = 20⁰c

M1 = 2.6kg/s

For superheated system

P2 = 300kpa

T2 = 300⁰c

M2 = ??

T2 = 60⁰c

From saturated water table

h1 = 83.91kj/kg

h3 = 251.18kj/kg

From superheated water,

h2 = 3069.6kj/kg

The equation of energy balance

m1h1 + m2h2 = m3h3

When we input all the corresponding values:

We get

m2 = -434.902/-2818.42

m2 = 0.15430

m2 = 0.154kg/s

This is the mass flow rate of the superheated steam

Please check attachment for more detailed explanation.

thank you!

This question involves the concepts of energy balance and mass flow rate.

The mass flow rate of the superheated steam required is "0.15 kg/s".

Applying the energy balance in this situation, we get:

[tex]m_1h_1+m_2h_2=m_3h_3[/tex]

where,

m₁ = mass flow rate of liquid water at 300 KPa and 200°C = 2.6 kg/s

m₂ = mass flow rate of superheated at 300 KPa and 300°C = ?

h₁ = enthalpy of liquid water at 300 KPa and 200°C = 83.91 KJ/kg (from saturated steam table)

h₂ = enthalpy of superheated at 300 KPa and 300°C = 3069.6 KJ/kg (from superheated steam table)

h₃ = enthalpy of exiting fluid at 60°C = 251.18 KJ/kg (from saturated steam table)

m₃ = mass flow rate of exiting fluid = 2.6 kg/s + m₂

Therefore,

[tex](2.6\ kg/s)(83.91\ KJ/kg)+(m_2)(3069.6\ KJ/kg)=(2.6\ kg/s+m_2)(251.18\ KJ/kg)\\m_2(3069.6\ KJ/kg-251.18\ KJ/kg)=(2.6\ kg/s)(251.18\ KJ/kg-83.91\ KJ/kg)\\\\m_2=\frac{434.902\ KW}{2818.42\ KJ/kg}[/tex]

m₂ = 0.15 kg/s

Learn more about energy balance here:

https://brainly.com/question/9839609?referrer=searchResults

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