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# The change in specific enthalpy the steam(Gas) turbine

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A steam turbine takes in steam at the rate of 6000 kg hour^-1 and its power output is 800 kW. Neglect any heat loss from the turbine. Find the change in the specific enthalpy of the steam as it passes through the turbine if (a) the entrance and exit are at the same elevation and the entrance and exit velocities are negligible (b) the entrance velocity is 50 ms^-1 and the exit velocity is 200 ms^-1, with the outlet pipe being 2 m above the inlet. Please see the attached file.

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## SOLUTION This solution is FREE courtesy of BrainMass!

As the steam turbine is steady state system, we can use steady flow energy equation as,

m[h1 + (C1^2/2) + z1*g] - Q =m[h1 + (C1^2/2) + z1*g] + W ----------------(1)

Where, m is flow rate (kg/s), h1 and h2 are enthalpy of steam at inlet and outlet respectively, C1 and C2 are velocities of steam at inlet and outlet, Z1 and Z2 are height of inlet and outlet, Q is heat loss (negative sign because heat is released), W is work done by system (output power per kg per s) and g is gravitational acceleration.

Now from eqn (1), the simplified form is

-Q - W = m[(h2 - h1) + (C2^2-C1^2/2) + (Z2-Z1)*g] ----------------(2)

Solution Part (a):

In this case we have,
Flow rate m = 6000 kg/hour = 6000/3600 kg/s = 1.67 kg/s,
Work done = 800 kW,
Velocities, C1=C2=0,
Heat loss Q = 0
Height of inlet and outlet are same i.e. Z1-Z2 = 0

Now from eqn (2),

-Q - W = m[(h2 - h1) + (C2^2-C1^2/2) + (Z2-Z1)*g]

Therefore, 0 - 800 = 1.67 [(h2 - h1) + 0 + 0]
Therefore, h2 - h1 = - 800/1.67 = 479.04 kJ/kg

Hence, the change in specific enthalpy is 479.04 kJ/kg

Solution Part (b):

In this case we have,
Flow rate m = 6000 kg/hour = 6000/3600 kg/s = 1.67 kg/s,
Work done = 800 kW = 800000 W,
Velocities, C1 = 50 m/s & C2= 200 m/s,
Heat loss Q = 0
Difference in Height of outlet and inlet is Z2-Z1 = 2 m,

Put these values in eqn 2, we get

0 - 800 = 1.67 [(h2 - h1) + (200^2-50^ 2/2) + 2*9.8]

Therefore, -800000= 1.67 [(h2 - h1) + ((40000-2500)/2) + 19.6]
-800000 = 1.67 [(h2 - h1) + 18750 + 19.6]
18769.6 - 800000 = 1.67 (h2 - h1)
(h2 - h1) = -781230.4/1.67
(h2 - h1) = 467802.63 J/kg = 467.8 kJ/kg

Hence, the change in specific enthalpy is 467.8 kJ/kg.

Reference

Book: Engineering thermodynamics, 3rd edition, by R. K. Rajput, 2007, Laxmi publications (P) LTD, page no 150 to 170

Note:
1 .For more detailed understanding see examples in pg 150 to 170 of this book. Students can use any eng thermodynamics book for this problem.
2. Feel free to ask me for further explanation of this solution.

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