Question

Air enters a cmpressor at 20 deg C and 80 kPa and exits at 800 kPa and 200 deg C. The power input is 400 kW. Find the heat transfer rate. The air exits the compressor at 20 m/s through a tube of 10 cm diameter.

Answer 1

The heat is transferred is at the rate of 752.33 kW

Solution:

As per the question:

Temperature at inlet,
`T_(i) = 20^(\circ)C` = 273 + 20 = 293 K

Temperature at the outlet,
`T_(o) = 200{\circ}C` = 273 + 200 = 473 K

Pressure at inlet,
`P_(i) = 80 kPa = 80* 10^(3) Pa`

Pressure at outlet,
`P_(o) = 800 kPa = 800* 10^(3) Pa`

Speed at the outlet,
`v_(o) = 20 m/s`

Diameter of the tube,
`D = 10 cm = 10* 10^(- 2) m = 0.1 m`

Input power,
`P_(i) = 400 kW = 400* 10^(3) W`

Now,

To calculate the heat transfer,
`Q`, we make use of the steady flow eqn:

`h_(i) + (v_(i)^(2))/(2) + gH  + Q = h_(o) + (v_(o)^(2))/(2) + gH' + p_(s)`

where

`h_(i)` = specific enthalpy at inlet

`h_(o)` = specific enthalpy at outlet

`v_(i)` = air speed at inlet

`p_(s)` = specific power input

H and H' = Elevation of inlet and outlet

Now, if

`v_(i) = 0` and H = H'

Then the above eqn reduces to:

`h_(i) + gH + Q = h_(o) + (v_(o)^(2))/(2) + gH + p_(s)`

`Q = h_(o) - h_(i) + (v_(o)^(2))/(2) + p_(s)` (1)

Also,

`p_(s) = (P_(i))/( mass, m)`

Area of cross-section, A =
`(\pi D^(2))/(4) =(\pi 0.1^(2))/(4) = 7.85* 10^(- 3) m^(2)`

Specific Volume at outlet,
`V_(o) = A* v_(o) = 7.85* 10^(- 3)* 20 = 0.157 m^(3)/s`

From the eqn:

`P_(o)V_(o) = mRT_(o)`

`m = (800* 10^(3)* 0.157)/(287* 473) = 0.925 kg/s`

Now,

`p_(s) = (400* 10^(3))/(0.925) = 432.432 kJ/kg`

Also,

`\Delta h = h_(o) - h_(i) = c_(p)\Delta T =c_(p)(T_(o) - T_(i)) = 1.005(200 - 20) = 180.9 kJ/kg`

Now, using these values in eqn (1):

`Q = 180.9 + (20^(2))/(2) + 432.432 = 813.33 kW`

Now, rate of heat transfer, q:

q = mQ =
`0.925* 813.33 = 752.33 kW`