Question

For these problems, assume air behaves as an ideal gas with R = 0.287 k J k g K. A compressor operates at steady state and takes in air from ambient 0 kPa, gage and 300 K. The outlet pressure is 50 kPa, gage and 400 K. Determine: the mass flow rate if the inlet area is 10 cm2 and the inlet pressure is -2 kPa, gage. the minimum outlet temperature that is possible for this compressor. the isentropic efficiency of the compressor, assuming no heat loss. if there is a heat loss of 30 kJ/kg, the work required to run the compressor, and the new isentropic efficiency. A turbine receives air at 50 kPa, gage and 800 K. It discharges to 0 kPa, gage, and the outlet temperature is measured as 500 K. The mass flow rate is the same as in the previous problem. Determine: the maximum work the turbine can do under these conditions. the heat loss, if the turbine works isentropically.

Answer 1

Part 1

1) 0.252 kg/s

2) 457.06 K

3) 63.45%

4) 17.96 kJ

5) 44.85%

Part 2

1) 65.92 kJ

2) 57.62 kJ/kg

Step-by-step explanation:

1) The mass flow rate

The flow velocity is given by the Bernoulli relation;

`U =\sqrt{ (\Delta P)/(\rho ) }`

Where:

ΔP = The difference in pressure = 50 - (-2) = 52 kPa

ρ = Density of air = 1.225 kg/m³

`U =\sqrt{ (52,000)/(1.225 ) } = 206.03 m/s`

The volume flow rate, V = U × A

Where:

A = Cross sectional area of the of the inlet = 10 cm² = 0.001 m²

Therefore, V = 0.001 × 206.03 = 0.206 m³/s

The mass flow rate = ρ × V = 1.225 × 0.206 = 0.252 kg/s

2) The minimum outlet temperature

P₁v₁/T₁ = P₂v₂/T₂

v₁ = v₂

∴ P₁/T₁ = P₂/T₂

T₂ = P₂T₁/P₁ = 151.325*300/99.325 = 457.06 K

3) The isentropic efficiency no heat loss

h₁ = 300.4 kJ/kg

`h_((out \ actual))` = 401.3 kJ/kg

`h_((out \ isentropic))` = 441.9 + (457.06 - 440)/(460 - 440)*(462.3 - 441.9) = 459.30 kJ/kg

The isentropic efficiency,
`\eta _(S)`, is given by the expression;

`\eta _(S) = (h_(in) - h_((out \ actual)))/(h_(in) -h_((out \ isentropic)) ) = (300.4 - 401.3)/(300.4 - 459.3) = 0.6345`

Therefore, the isentropic efficiency,
`\eta _(S)` in percentage = 63.45%

4) Where there is an heat loss of 30 kJ/kg, we have;

`h_((out \ actual \ new))` =
`h_((out \ actual))` - Heat loss = 401.3- 30 = 371.3 kJ/kg

The work done = (371.3 - 300.04)*0.252= 17.96 kJ/s

The new isentropic efficiency is given by the relation;

`\eta _(S, new) =(300.4 - 371.3)/(300.4 - 459.3) = 0.4485`

Therefore, the isentropic efficiency,
`\eta _(S, new)`, in percentage = 44.85%

Part 2

1) Turbine mass flow rate = 0.252 kg/s

From

T₂ = P₂T₁/P₁ = 101.325*800/151.325= 535.67 K

h₁ = 822.2 kJ/kg

`h_((out \ actual))` = 503.3 kJ/kg

`h_((out \ isentropic))` = 544.7 + (535.67 - 520)/(540 - 520)*(544.7 - 524.0) = 560.92 kJ/kg

The maximum work,
`W_(max)`, is given by the expression;

`W_(max)` = Mass flow rate×(h₁ -
`h_((out \ actual))`)

`W_(max)` = (822 - 503.3)*0.252 = 65.92 kJ/s

2) The heat lost,
`h_(loss)`, is given by the relation;

`h_(loss)` =
`h_((out \ isentropic))` -
`h_((out \ actual))` = 560.92 - 503.3 = 57.62 kJ/kg.