Question

Steam flows steadily into a turbine with a mass flow rate of 26 Kg/s and a negligible velocity at 6 MPa and 600°C. The steam leaves the turbine at 0.5 MPa and 200°C with a velocity of 180 m/s. The rate of work done by the steam in the turbine is measured to be 20 MW. If the elevation change between the turbine inlet and exit is negligible, Determine: (a) The rate of heat transfer associated with this process.

Answer 1

Answer:

heat transfer = -454.2 kW

and -ve sign show heat transfer system to surrounding here

Explanation:

given data

flow rate = 26 Kg/s

velocity v1 = 6 MPa

temperature t1 = 600°C

work done W = 20 MW = 20000 kW

steam leaves = 0.5 MPa

velocity v2 = 180 m/s

temperature t2 = 200°C

to find out

rate of heat transfer

solution

we know here energy flow equation that is

h1  + v1²/2 + z1g + Q = h2 + v2²/2 + g2 + W    

here we know z is datum height and Q is heat transfer

and h1 specific enthalpy in inlet = 3658.7 kJ/kg

and h2 specif enthalpy in outlet = 2855.8 kJ/kg

so put all these value in equation of rate of heat transfer  that is Q

heat transfer = m × ( h2-h1) + mv2² / 2000  + W

heat transfer = 26 × ( 2855.8-3658.7) + 26(180)² /2000  + 20000

heat transfer = -454.2 kW

so heat transfer = -454.2 kW

and -ve sign show heat transfer system to surrounding here

Answer 2

Answer:

$Q_{out}=456.8kW$

Explanation:

Hello,

To solve this problem, an energy balance is proposed as follows:

$mh_{in}=m[h_{out}+\frac{V_{out}^2}{2}]+W_{out}+Q_{out}$

The inlet conditions correspond to an oversaturated steam and the outlet conditions to an oversaturated steam as well, so the inlet and outlet enthalpies are extracted from the tables of oversaturated steam and have the following values:

$h_{in}=3658.8kJ/kg;h_{out}=2855.8kJ/kg$

Now, solving for the outgoing heat, we've got:

$Q_{out}=m[h_{in}-h_{out}-\frac{V_{out}^2}{2}]-W_{out} \\Q_{out}=26\frac{kg}{s}(3658.8\frac{kJ}{kg}-2855.8\frac{kJ}{kg}-\frac{(180\frac{m}{s})^2}{2} ) -20MW*\frac{1000kW}{1MW} \\Q_{out}=26\frac{kg}{s}(803\frac{kJ}{kg}-16200\frac{J}{kg}*\frac{1kJ}{1000J} ) -20000kW\\Q_{out}=456.8kW$

This heat is positive due to the balance, but if we have used the classic notation it'll be negative indicating that is a lost heat.

Best regards.