Question

1: A hot-water stream at 95°C enters a mixing chamber where it is mixed with a stream of cold water at 25°C. Assume all the streams are at a pressure of 350 kPa. It is desired that the mixture with a mass flow rate of 0.7 kg/s leaves the chamber at 39°C. Each enthalpy is given by hi hf@95°c = 350 kJ/kg h2 z hr@25°C = 83 kJ/kg h = hf@ 39°C = 175 kJ/kg a. Write down the mass balance. b. Write down the energy balance. c. Determine the mass flow rate of the hot-water stream. Ti = 95°C H20 (P = 350 kPa) T3 = 39°C T2 = 25°C

Answer 1

Common assumptions for mixing chamber:

1. The mixing chamber is well-insulated so that heat loss is negligible. Q° = 0

2. W° = 0

3. There are negligible changes in the kinetic and potential energies of fluid streams. ΔP.E. = 0 = ΔK.E.

(a). Based on mass principle, the sum of incoming flow equals to sum of leaving fluid.

Mass balance: $\dot{m}$_in - $\dot{m}$_out = 0

$\dot{m}$₁ +  $\dot{m}$₂ =  $\dot{m}$₃$\Rightarrow$   $\dot{m}$₁ =  $\dot{m}$₃ -   $\dot{m}$₂ (1)

(b). Energy balance:   $\dot{E}$_in -  $\dot{E}$_out = 0 $\Rightarrow$  $\dot{E}$_in =  $\dot{E}$_out

Since, $\dot{Q}$ = $\dot{W}$ = $\Delta K.E.$ = $\Delta P.E.$ = 0

So,  $\dot{m}$₁h₁ +  $\dot{m}$₂h₂ =  $\dot{m}$₃h₃ (2)

(c)

Substituting (1) in (2) :

( $\dot{m}$₃ -   $\dot{m}$₂) h₁ =  $\dot{m}$₃h₃ - $\dot{m}$₂h₂

$\dot{m}$₃h₁ -  $\dot{m}$₂h₁ =  $\dot{m}$₃h₃ - $\dot{m}$₂h₂$\Rightarrow$    $\dot{m}$₂h₂-  $\dot{m}$₂h₁ = $\dot{m}$₃h₃ -  $\dot{m}$₃h₁

$\dot{m}$₂(h₂- h₁) =  $\dot{m}$₃(h₃ - h₁) $\Rightarrow$   $\dot{m}$₂ =  [$\dot{m}$₃(h₃ - h₁)] / (h₂ - h₁) (3)

Given, $\dot{m}$₃ = 0.7 kg/s

h₁ = 350 kJ/kg, h₂ = 83 kJ/kg, h₃ = 175 kJ/kg

Substituting in (3) :

$\dot{m}$₂ = [0.7 kg/s (175 kJ/kg - 350 kJ/kg)] / (83 kJ/kg - 350 kJ/kg)

$\dot{m}$₂ = [0.7 X (-175)] / (-267) = 122.5 / 267

$\dot{m}$₂ = 0.4588 kg/s

Substituting in (1) :

$\dot{m}$₁ = 0.7 kg/s - 0.4588 kg/s = 0.2412 kg/s

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