Question

A hot-water stream at 80 ℃ enters a mixing chamber with a mass flow rate of 0.5 kg/s where it is mixed with a stream of cold water at 20 ℃. If it is desired that the mixture leave the chamber at 42 ℃, determine the mass flow rate of the cold-water stream. Assume all the streams are at a pressure of 250 kPa

Answer 1

Answer:

$\dot{m_{2}}=0.865 kg/s$

Explanation:

$\dot{m_1}= 0.5kg/s$

from steam tables , at 250 kPa, and at

T₁ = 80⁰C ⇒ h₁ = 335.02 kJ/kg

T₂ = 20⁰C⇒ h₂ = 83.915 kJ/kg

T₃ = 42⁰C ⇒ h₃ = 175.90 kJ/kg

we know

$\dot{m_{in}}=\dot{m_{out}}$

$\dot{m_{1}}+\dot{m_{2}}=\dot{m_{3}}$

according to energy balance equation

$\dot{m_{in}}h_{in}=\dot{m_{out}}h_{out}$

$\dot{m_{1}}h_{1}+\dot{m_{2}}h_{2}=\dot{m_{3}}h_{3}$

$\dot{m_{1}}h_{1}+\dot{m_{2}}h_{2}=(\dot{m_{1}}+\dot{m_{2}})h_{3}\\(0.5\times 335.02)+(\dot{m_{2}}\times 83.915)=(0.5+\dot{m_{2}})175.90\\\dot{m_{2}}=0.865 kg/s$

Answer 2

The mass flow rate of the cold-water stream ( m₂ ) = 0.865 kg/s

Given data :

m₁ = 0.5 kg/s

m₂ = ?

From steam tables

At 250 kPa

at T1 = 80℃ , h₁ = 335.02 kJ/kg

at T2 = 20℃, h₂ = 83.915 kJ/kg

at T3 = 42℃, h₃ = 175.90 kJ/kg

Determine the mass flow rate of the cold water stream

Given that :

Min = Mout    also m₁ + m₂ = m₃

applying the principle of energy balance

M₁h₁ + M₂ h₂  = ( m₁ + m₂ ) h₃ ---- ( 1 )

insert values into equation ( 1 )

m₂ = 0.865 kg/s

Hence In conclusion The mass flow rate of the cold-water stream ( m₂ ) = 0.865 kg/s

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