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iogann1982 [59]
3 years ago
12

NOTE: This is a multi-part question. Once an answer is submitted, you will be unable to return to this part. The gas-turbine cyc

le of a combined gas–steam power plant has a pressure ratio of 8. Air enters the compressor at 290 K and the turbine at 1400 K. The combustion gases leaving the gas turbine are used to heat the steam at 15 MPa to 450°C in a heat exchanger. The combustion gases leave the heat exchanger at 247°C. Steam expands in a high-pressure turbine to a pressure of 3 MPa and is reheated in the combustion chamber to 500°C before it expands in a low-pressure turbine to 10 kPa. The mass flow rate of steam is 17 kg/s. Assume isentropic efficiencies of 100 percent for the pump, 85 percent for the compressor, and 90 percent for the gas and steam turbines. Determine the thermal efficiency of the combined cycle. The thermal efficiency of the combined cycle is %.、
Engineering
1 answer:
Nastasia [14]3 years ago
8 0

Answer:

the thermal efficiency of the combined cycle is =91.7%

Explanation:

to determine the combine thermal efficiency of the systems, we add all efficiency of the system since they are give and we hake the average,

mathematically we say\frac{100+85+90}{3} =91.7%

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A solid circular rod that is 600 mm long and 20 mm in diameter is subjected to an axial force of P = 50 kN The elongation of the
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Answer:

a) V = 0.354

b)  G = 25.34 GPA

Explanation:

Solution:

We first determine Modulus of Elasticity and Modulus of rigidity

Elongation of rod ΔL = 1.4 mm

Normal stress, δ = P/A

Where P = Force acting on the cross-section

A = Area of the cross-section

Using Area, A = π/4 · d²

= π/4 · (0.0020)²  = 3.14 × 10⁻⁴m²

δ = 50/3.14 × 10⁻⁴    = 159.155 MPA

E(long) = Δl/l  = 1.4/600 = 2.33 × 10⁻³mm/mm

Modulus of Elasticity Е = δ/ε

= 159.155 × 10⁶/2.33 × 10⁻³    = 68.306 GPA

Also final diameter d(f) = 19.9837 mm

Initial diameter d(i) = 20 mm

Poisson said that V = Е(elasticity)/Е(long)

= -  <u>( 19.9837 - 20 /20)</u>

        2.33 × 10⁻³                  

= 0.354,

∴ v = 0.354

Also G = Е/2. (1+V)

=  68.306 × 10⁹/ 2.(1+ 0.354)

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Weight of water required = 4.6% × 396,542.7 = 18241 kN

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Volume of water required = 1859 kL

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