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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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sergey [27]

Answer:

See attached picture.

Explanation:

See attached picture for explanation.

4 0
3 years ago
A flywheel made of Grade 30 cast iron (UTS = 217 MPa, UCS = 763 MPa, E = 100 GPa, density = 7100 Kg/m, Poisson's ratio = 0.26) h
hram777 [196]

Answer:

N = 38546.82 rpm

Explanation:

D_{1} = 150 mm

A_{1}= \frac{\pi }{4}\times 150^{2}

              = 17671.45 mm^{2}

D_{2} = 250 mm

A_{2}= \frac{\pi }{4}\times 250^{2}

              = 49087.78 mm^{2}

The centrifugal force acting on the flywheel is fiven by

F = M ( R_{2} - R_{1} ) x w^{2} ------------(1)

Here F = ( -UTS x A_{1} + UCS x A_{2} )

Since density, \rho = \frac{M}{V}

                        \rho = \frac{M}{A\times t}

                        M = \rho \times A\times tM = 7100 \times \frac{\pi }{4}\left ( D_{2}^{2}-D_{1}^{2} \right )\times t

                        M = 7100 \times \frac{\pi }{4}\left ( 250^{2}-150^{2} \right )\times 37

                        M = 8252963901

∴ R_{2} - R_{1} = 50 mm

∴ F = 763\times \frac{\pi }{4}\times 250^{2}-217\times \frac{\pi }{4}\times 150^{2}

  F = 33618968.38 N --------(2)

Now comparing (1) and (2)

33618968.38 = 8252963901\times 50\times \omega ^{2}

∴ ω = 4036.61

We know

\omega = \frac{2\pi N}{60}

4036.61 = \frac{2\pi N}{60}

∴ N = 38546.82 rpm

7 0
3 years ago
A gas turbine receives a mixture having the following molar analysis: 10% CO2, 19% H2O, 71% N2 at 720 K, 0.35 MPa and a volumetr
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Answer:

2074.2 KW

Explanation:

<u>Determine power developed at steady state </u>

First step : Determine mass flow rate  ( m )

m / Mmax = ( AV )₁ P₁ / RT₁   -------------------- ( 1 )

<em> where : ( AV )₁ = 8.2 kg/s,  P₁ = 0.35 * 10^6 N/m^2,   R = 8.314 N.M / kmol , </em>

<em>  T₁  = 720 K . </em>

insert values into equation 1

m  = 0.1871  kmol/s  ( mix )

Next : calculate power developed at steady state ( using ideal gas tables to get the h values of the gases )

W( power developed at steady state )

W = m [ Yco2 ( h1 - h2 )co2

Attached below is the remaining  part of the detailed solution

4 0
3 years ago
Q1: The first option should always be to get out safely (RUN)
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Answer:

Q1 true

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And other I am confuse

6 0
3 years ago
You find an unnamed fluid in the lab we will call Fluid A. Fluid A has a specific gravity of 1.65 and a dynamic viscosity of 210
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Answer:

1.2727 stokes

Explanation:

specific gravity of fluid A = 1.65

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Pa = Pw * Specific gravity =  1000 * 1.65 = 1650 kg/m^3

next : convert dynamic viscosity to kg/m-s

210 centipoise = 0.21 kg/m-s

Kinetic viscosity of Fluid A = dynamic viscosity / density of fluid A

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4 0
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