Question Details

An air-conditioning system provides a continuous flow of air to a room using an intake duct and an exit duct, as shown in the figure. To maintain the quality of the indoor air, the intake duct supplies a mixture of fresh air with a cold air stream. The two streams are mixed in an insulated mixing chamber located upstream of the intake duct. Cold air enters the mixing chamber at 5 °C, 105 kPa with a volume flow rate of 1.25 m3 /s during steady state operation. Fresh air enters the mixing chamber at 34 °C and 105 kPa. The mass flow rate of the fresh air is 1.6 times of the cold air stream. Air leaves the room through the exit duct at 24 °C.

Assuming the air behaves as an ideal gas with 𝒄𝒑 = 1.005 kJ/kg.K and 𝑹 = 0.287 kJ/kg.K, the rate of heat gain by the air from the room is ____________ kW(round off to two decimal places).

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Correct Answer :

Correct answer is : 4.96

Solution :

The correct answer is 4.96.

Step 1: Analyze the Cold Air Stream (State 1)
We are given the following properties for the cold air stream entering the mixing chamber:
Temperature: Tc=5°C=278 K
Pressure: Pc=105 kPa
Volume flow rate: V˙c=1.25 m3/s
Characteristic gas constant: R=0.287 kJ/(kg·K)

Using the ideal gas equation of state, we can find the mass flow rate of the cold air stream (m˙c):

Pc V˙c = m˙c R Tc

105 × 1.25 = m˙c × 0.287 × 278

m˙c = 131.25 79.786 1.645 kg/s

Step 2: Analyze the Fresh Air Stream (State 2)
We are given that the temperature of the fresh air is:
Tf=34°C=307 K
The mass flow rate of the fresh air is 1.6 times that of the cold air stream:

m˙f = 1.6 × m˙c = 1.6 × 1.645025 2.632 kg/s

Step 3: Energy Balance for the Insulated Mixing Chamber (State 3)
Assuming steady-state operation and an insulated mixing chamber, the energy balance equation is:

m˙c hc + m˙f hf = ( m˙c + m˙f ) h3

Using h=cpT with a constant specific heat cp=1.005 kJ/(kg·K), the equation simplifies to:

m˙c Tc + m˙f Tf = ( m˙c + m˙f ) T3

Substituting the values into the equation:

1.645 × 278 + 2.632 × 307 = ( 1.645 + 2.632 ) × T3

457.31 + 808.024 = 4.277 × T3

T3 = 1265.334 4.277 295.846 K

Step 4: Rate of Heat Gain by the Air from the Room (State 4)
The air enters the room at temperature T3 and exits the room through the exit duct at a temperature of:
T4=24°C=297 K
The rate of heat gain by the air (Q˙) is given by the energy balance of the room:

Q˙ = m˙total cp ( T4 - T3 )

Substituting the values:

Q˙ = 4.277 × 1.005 × ( 297 - 295.846 )

Q˙ = 4.277 × 1.005 × 1.154 4.96 kW

Thus, the rate of heat gain by the air from the room is 4.96 kW.

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