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Citrus2011 [14]
3 years ago
9

Required information NOTE: This is a multi-part question. Once an answer is submitted, you will be unable to return to this part

. The gas-turbine cycle 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 rate of total heat input.
Engineering
1 answer:
ra1l [238]3 years ago
4 0

Answer:

h1 = 290.16kj/kg

P = 1.2311

Prandil expression at 8

P=p1/p7×pr

=8(1.2311)

=9.85

Enthalpy state at 8 corresponding to 9.85

h1 = 526.13kj/kg

Now prandtl state at 9 that correspond to 1400k.

h9 = 1515.42kj/kg

Pr = 450.5

Prandtl expression at state 10

P= p10/p9×pr

=1/8(450.5)

=56.31

Enthalpy at state 10 corresponding to prandtl 56.31

h10 = 860.39kj/kg

At 520k

h11 = 523.63kj/kg

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Answer:

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8 0
3 years ago
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A double-pane insulated window consists of two 1 cm thick pieces of glass separated by a 1.8 cm layer of air. The window measure
Elanso [62]

Answer:

(b). T = 22.55 ⁰C

(c). q = 557.8 W

Explanation:

we take follow a step by step process to solving this problem.

from the question, we have that

The two glass pieces is separated by a 1.8 cm distance layer of air.

the thickness of glass piece is 1 cm

width = 4 m

the height = 3 m

(a). the sketch of the thermal circuit is uploaded in the picture below.

(b).  the thermal resistance due to the conduction in the first glass plane is given thus;

R₁ = Lg / Kg A ................(1)

given that Kg rep. the thermal conductivity of the glass plane

A = conduction surface area

Lg = Thickness of glass plane4

taking the thermal conductivity of glass plane as Kg = 0.78 w/mk

inputting values into equation (1) we have,

R₁ = [1 (cm) ˣ 1 (m)/100 (cm)] / [(0.78 w/mk)(4m ˣ 3m)]

R₁ = 1.068 ˣ 10 ⁻³ k/w

Being that we have same thermal resistance in the first and second plane,

therefore R₁ = R₃ = 1.068 ˣ 10 ⁻³ k/w

⇒ Also the thermal resistance between air and glass as a result of the conduction by the layer is given thus

R₂ = La/KaA .....................(2)

given Ka = thermal conductivity of air

A = surface area

La = thickness of air

substituting values into the equation we have

R₂ = [1.8 (cm) ˣ 1 (m)/100 (cm)] / [(0.0262 w/mk)(4m ˣ 3m)]

R₂ = 5.73 ˣ 10⁻² k/w

Given the thermal resistance on the outer surface due to convection, we have

R₄ = 1/hA

inputting value gives R₄ = 1 / (12 w/m² ˣ 12m) = 6.94 ˣ 10⁻³k/w

R₄ = 6.94 ˣ 10⁻³k/w

Finally the sum total of thermal resistance = R₁ + R₂ + R₃ + R₄

R-total = 0.0663 kw

From this we can calculate the rate of heat loss

using  q = Ti - To / R-total ..............(3)

given Ti and To is the inside and outside temperature i.e. 27⁰C and -10⁰C

from equation (3),

q = 27- (-10) / 0.0063 = 557.8 W

q = 557.8 W  

⇒ Applying the heat transfer formula for inside surface glass temperature gives;

q = Ti - T₂ / R₃ + R₄

T₂ = Ti - q (R₃ + R₄)

T₂ = 27 - 557.8 (1.068ˣ10⁻³ + 6.94ˣ10⁻³ ) = 22.55°C

T₂ = 22.55°C

cheers i hope this helps

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