Question

Steam enters a turbine steadily at 7 MPa and 600°C with a velocity of 60 m/s and leaves at 25 kPa with a quality of 95 percent. A heat loss of 20 kJ/kg occurs during the process. The inlet area of the turbine is 150 cm2, and the exit area is 1400 cm2. Determine (a) the mass flow rate of the steam, (b) the exit velocity, and (c) the power output.

Answer 1

a)
`\dot m = 16.168\,(kg)/(s)`, b)
`v_(out) = 680.590\,(m)/(s)`, c)
`\dot W_(out) = 18276.307\,kW`

Step-by-step explanation:

A turbine is a steady-state devices which transforms fluid energy into mechanical energy and is modelled after the Principle of Mass Conservation and First Law of Thermodynamics, whose expressions are described hereafter:

Mass Balance

`(v_(in)\cdot A_(in))/(\\u_(in)) - (v_(out)\cdot A_(out))/(\\u_(out)) = 0`

Energy Balance

`-q_(loss) - w_(out) + h_(in) - h_(out) = 0`

Specific volumes and enthalpies are obtained from property tables for steam:

Inlet (Superheated Steam)

`\\u_(in) = 0.055665\,(m^(3))/(kg)`

`h_(in) = 3650.6\,(kJ)/(kg)`

Outlet (Liquid-Vapor Mix)

`\\u_(out) = 5.89328\,(m^(3))/(kg)`

`h_(out) = 2500.2\,(kJ)/(kg)`

a) The mass flow rate of the steam is:

`\dot m = (v_(in)\cdot A_(in))/(\\u_(in))`

`\dot m = (\left(60\,(m)/(s) \right)\cdot (0.015\,m^(2)))/(0.055665\,(m^(3))/(kg) )`

`\dot m = 16.168\,(kg)/(s)`

b) The exit velocity of steam is:

`\dot m = (v_(out)\cdot A_(out))/(\\u_(out))`

`v_(out) = (\dot m \cdot \\u_(out))/(A_(out))`

`v_(out) = (\left(16.168\,(kg)/(s) \right)\cdot \left(5.89328\,(m^(3))/(kg) \right))/(0.14\,m^(2))`

`v_(out) = 680.590\,(m)/(s)`

c) The power output of the steam turbine is:

`\dot W_(out) = \dot m \cdot (-q_(loss) + h_(in)-h_(out))`

`\dot W_(out) = \left(16.168\,(kg)/(s) \right)\cdot \left(-20\,(kJ)/(kg) + 3650.6\,(kJ)/(kg) - 2500.2\,(kJ)/(kg)\right)`

`\dot W_(out) = 18276.307\,kW`

Answer 2

The mass flow rate = 16.17kg/sec

The exit velocity = 680.84 m/sec

`W_o_u_t=14562.37kw`

Step-by-step explanation:

Attached is a copy of the diagram to this question

Let's write out the given data

T= 600⁰C

`v_1=0.055665m^3/kg\\h_1=3650.60kJ/kg\\at p = 25kpa\\v_f=0.00102m^3/kg\\v_g=6.2034m^3/kg\\hfg=2345.5kJ/kg\\hf=271.96kJ/kg`

The enthalpy at exit is

`h_2=271.96+(95)/(100) *2345.50=2500.185 kJ/kg\\`

The specific volume is

`v_2=v_f+x_2vfg\\\\v_2=0.001020+(95)/(100)(6.2034-0.00102)=5.893m^3/kg`

a) To calculate the mass flow rate

`m=(V_1m)/(v_1)=(60*0.015)/(0.055665)\\m=16.17kg/s`

The mass flow rate is 16.17kg/s

b) Exit velocity

`v_2=(mV_2)/(A_2)=(16.17*5.893)/(0.14)=680.64m/sec`

The exit velocity is 680.64m/s

c) The power output

`W_o_u_t=-mQ_o_u_t+m(h_1-h_2+(v_1^2-v_2^2)/(2))\\ W_o_u_t=-(16.17*20)+16.17(3650.60-2500.185+(60^2-680.64^2)/(2)*(1)/(1000) \\W_o_u_t=14562.37kw`

The power output of the machine is 14562.37kw