Question

Steam enters a turbine operating at steady state with a mass flow of 10 kg/s, a specific enthalpy of 3100 kJ/kg, and a velocity of 30 m/s. At the exit, the specific enthalpy is 2300 kJ/kg and the velocity is 45 m/s. The elevation of the inlet is 3 m higher than at the exit. Heat transfer from turbine to its surroundings occur at rate of 1.1 kJ per kg of steam flowing. Let g 9.81 m/s. Determine the power developed by the turbine, in kW.

Answer 1

Answer:7989.86KW

Step-by-step explanation:

Given data

`h_1`=3100 KJ/kg

`\dot{m}`=10kg/s

`V_1`=30m/s

`Z_1`=3m

`h_2`=2300KJ/kg

`V_2`=45m/s

`Z_2`=0

using steady flow energy equation which is

`\dot{m}`
`\left (h_1+gZ_1+(V_1^2)/(g)\right )`+
`\dot{Q}`=
`\dot{m}`
`\left (h_2+gZ_2+(V_2^2)/(g)\right )\+`
`\dot{W}`

`10\left (3100+(9.81* 3)/(1000)+(30^2)/(9.81* 1000)\right )`+
`\dot{Q}`=
`10\left (2300+(9.81* 0)/(1000)+(45^2)/(9.81* 1000)\right )`+
`\dot{W}`

`10\left ( \left ( 3100-2300\right )+(30^(2)-45^(2))/(2* 9.81* 1000)+(9.82* 3)/(1000)\right )`-
`1.1* 10`=
`\dot{W}`

`\dot{W}`=7989.8673 KW