Question

6. Two air flows are combined to a single flow. One flow is 1 m3/s at 20 oC and the other is 2 m3/s at 200 oC, both at 100 kPa. They mix without any heat transfer to produce an exit flow at 100 kPa. Neglect kinetic energies and find the exit temperature and volume flow rate.

Answer 1

`T_(3) = 393.20\,K`,
`\dot V_(3) = 3.002\,(m^(3))/(s)`

Step-by-step explanation:

This is a case of a mix chamber, which is modelled after the First Law of the Thermodynamics:

`\dot m_(1) \cdot h_(1) + \dot m_(2) \cdot h_(2) - \dot m_(3) \cdot h_(3) = 0`

According to the Principle of Mass Conservation:

`\dot m_(1) + \dot m_(2) - \dot m_(3) = 0`

Let assume that air behaves as an ideal gas. The density has the following expression:

`P\cdot V = n\cdot R_(u)\cdot T`

`P \cdot V = (m)/(M)\cdot R_(u)\cdot T`

`\rho = (P\cdot M)/(R_(u)\cdot T)`

Densities at inlets are, respectively:

`\rho_(1) = ((100\,kPa)\cdot (28.02\,(kg)/(kmol) ))/((8.314\,(kPa\cdot m^(2))/(kmol\cdot K) )\cdot (293.15\,K))`

`\rho_(1) = 1.149\,(kg)/(m^(3))`

`\rho_(2) = ((100\,kPa)\cdot (28.02\,(kg)/(kmol) ))/((8.314\,(kPa\cdot m^(2))/(kmol\cdot K) )\cdot (473.15\,K))`

`\rho_(2) = 0.712\,(kg)/(m^(3))`

The mass flow at outlet is:

`\dot m_(3) = (1.149\,(kg)/(m^(3)) )\cdot (1\,(m^(3))/(s) ) + (0.712\,(kg)/(m^(3)) )\cdot (2\,(m^(3))/(s) )`

`\dot m_(3) = 2.573\,(kg)/(s)`

Specific enthalpies depends on temperature only. The required variable for inlet are obtained from property tables:

`h_(1) = 293.31\,(kJ)/(kg)`

`h_(2) = 475.46\,(kJ)/(kg)`

Specific enthalpy at outlet is:

`h_(3) = (\dot m_(1)\cdot h_(1) + \dot m_(2)\cdot h_(2))/(\dot m_(3))`

`h_(3) = ((1.149\,(kg)/(s) )\cdot (293.31\,(kJ)/(kg) )+(1.424\,(kg)/(s) )\cdot (475.46\,(kJ)/(kg) ))/(2.573\,(kg)/(s) )`

`h_(3) = 394.11\,(kJ)/(kg)`

The exit temperature is:

`T_(3) = 393.20\,K`

The density of air at outlet is:

`\rho_(3) = ((100\,kPa)\cdot (28.02\,(kg)/(kmol) ))/((8.314\,(kPa\cdot m^(2))/(kmol\cdot K) )\cdot (393.20\,K))`

`\rho_(3) = 0.857\,(kg)/(m^(3))`

The volume flow rate at outlet is:

`\dot V_(3) = (\dot m_(3))/(\rho_(3))`

`\dot V_(3) = (2.573\,(kg)/(s) )/(0.857\,(kg)/(m^(3)) )`

`\dot V_(3) = 3.002\,(m^(3))/(s)`

Answer 2

The exit temperature is 392.6 K and the volume flow rate is 3 m³/s

Step-by-step explanation:

Given:

V₁ = 1 m³/s

T₁ = 20°C = 293 K

V₂ = 2 m³/s

T₂ = 200°C = 473 K

P₁ = P₂ = 100 kPa

P₃ = 100 kPa

Applying ideal gas law:

`v_(1) =(RT_(1) )/(P_(1) ) =(0.287*293)/(100) =0.841m^(3) /kg`

`v_(2) =(RT_(2) )/(P_(2) ) =(0.287*473)/(100) =1.358m^(3) /kg`

The mass flow rates are:

`m_(1) =(V_(1) )/(v_(1) ) =(1)/(0.841) =1.189kg/s`

`m_(2) =(V_(2) )/(v_(2) ) =(2)/(1.358) =1.473kg/s`

`m_(3) =m_(1) +m_(2) =1.189+1.473=2.662kg/s`

The exit temperature is obtained from the energy expression:

`m_(1) CpT_(1)+m_(2) CpT_(2) =m_(3) CpT_(3)\\T_(3) =(m_(1)T_(1) )/(m_(3) ) +(m_(2)T_(2) )/(m_(3) )`

`T_(3) =(1.189*293)/(2.662) +(1.473*473)/(2.662) =392.6K`

The specific volume at exit is:

`v_(3) =(RT_(3) )/(P_(3) ) =(0.287*392.6)/(100) =1.127m^(3) /kg`

The volume rate is:

`V_(3) =m_(3) v_(3) =2.662*1.127=3m^(3) /s`