Question

R 134a enters a air to fluid heat exchanger at 700 kPa and 50 oC. Air is circulated into the heat exchanger to cool the R134a to saturated liquid at 700 kPa. Ambient air at 20 oC and 100 kPa is circulated into the heat exchanger to cool the R134a. If The mass flow rate of R134a is 0.099 kg/sec and the circulated air increases in temperature by 5 oC, what is the volumetric flow rate of air through the heat exchanger

Answer 1

The volumetric flow rate of air through the heat exchanger is 3.36 m³/s.

Step-by-step explanation:

Here we have,

`-\dot {m_1} c_1dt_1 = \dot {m_2} c_2dt_2`

The properties of R 134a at 700 kPa Saturated temperature = 26.7 °C and enthalpy = 88.8 kJ/kg

While at super heated temperature and pressure of 70 kPa and 50 °C the enthalpy is 288.53 kJ/kg

Therefore, we have, the heat lost per kg = 288.53 kJ/kg - 88.8 kJ/kg = 199.73 kJ/kg

For air we have at 20 ° and 100 kPa, enthalpy = 294 kJ/kg

At 25 °
`c_p` = 1.012 J·g⁻¹·K⁻¹

20 ° C
`c_p` = 1.006 kJ/(kg·K)

Therefore,
`c_p` air = 1.006 kJ/(kg·K)

Plugging in the values into the above equation, we have

`-\dot {m_1} c_1dt_1 = \dot {m_2} c_2dt_2` = 0.099 kg/sec × 199.73 kJ/kg =
`\dot {m_2} * 1.006 * 5 \textdegree`

`\dot {m_2}` = 19.77/5.03 = 3.93 kg/s

At 100 kPa and 25° C the density of air is 1.16843 kg/m³

Therefore the volume flow rate = 3.93/1.16843 =3.36 m³/s.