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3 years ago

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8.28 Water is the working fluid in an ideal Rankine cycle with superheat and reheat. Steam enters the first-stage turbine at 140

0 lbf/in.2 and 10008F, expands to a pressure of 350 lbf/in.2, and is reheated to 9008F before entering the secondstage turbine. The condenser pressure is 2 lbf/in.2 The net power output of the cycle is 1 3 109 Btu/h. Determine for the cycle (a) the mass flow rate of steam, in lb/h. (b) the rate of heat transfer, in Btu/h, to the working fluid passing through the steam generator. (c) the rate of heat transfer, in Btu/h, to the working fluid passing through the reheater. (d) the thermal efficiency.

1 answer:

zubka84 [21]3 years ago

3 0

Answer:

Betbtybrbytntrnyrnrynunjhjhnthnnhtnnthnhtnnhnhrnntnthhnhnhtnthn

Explanation:

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(20 points) (Assessment of Outcome 1): A plant has two identical standby generator units for emergency use. In the area of the g

erastovalidia [21]

Answer:

It wouldn't get any louder then maybe 3db more

Explanation:

There's even a equation if you wanted to check this out but, if they are the same generator same model and all and made the same precise noise it wouldn't increase more then 3db.

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4 years ago

A solid shaft and a hollow shaft of the same material have same length and outer radius R. The inner radius of the hollow shaft

alexandr402 [8]

Answer with Explanation:

By the equation or Torque we have

$\frac{T}{I_{p}}=\frac{\tau }{r}=\frac{G\theta }{L}$

where

T is the torque applied on the shaft

$I_{p}$ is the polar moment of inertia of the shaft

$\tau$ is the shear stress developed at a distance 'r' from the center of the shaft

$\theta$ is the angle of twist of the shaft

'G' is the modulus of rigidity of the shaft

We know that for solid shaft $I_{p}=\frac{\pi R^4}{2}$

For a hollow shaft $I_{p}=\frac{\pi (R_o^4-R_i^4)}{2}$

Since the two shafts are subjected to same torque from the relation of Torque we have

1) For solid shaft

$\frac{2T}{\pi R^4}\times r=\tau _{solid}$

2) For hollow shaft we have

$\tau _{hollow}=\frac{2T}{\pi (R^4-0.7R^4)}\times r=\frac{2T}{\pi 0.76R^4}$

Comparing the above 2 relations we see

$\frac{\tau _{solid}}{\tau _{hollow}}=0.76$

Similarly for angle of twist we can see

$\frac{\theta _{solid}}{\theta _{hollow}}=\frac{\frac{LT}{I_{solid}}}{\frac{LT}{I_{hollow}}}=\frac{I_{hollow}}{I_{solid}}=1.316$

Part b)

Strength of solid shaft = $\tau _{max}=\frac{T\times R}{I_{solid}}$

Weight of solid shaft = $\rho \times \pi R^2\times L$

Strength per unit weight of solid shaft = $\frac{\tau _{max}}{W}=\frac{T\times R}{I_{solid}}\times \frac{1}{\rho \times \pi R^2\times L}=\frac{2T}{\rho \pi ^2R^5L}$

Strength of hollow shaft = $\tau '_{max}=\frac{T\times R}{I_{hollow}}$

Weight of hollow shaft = $\rho \times \pi (R^2-0.7R^2)\times L$

Strength per unit weight of hollow shaft = $\frac{\tau _{max}}{W}=\frac{T\times R}{I_{hollow}}\times \frac{1}{\rho \times \pi (R^2-0.7^2)\times L}=\frac{5.16T}{\rho \pi ^2R^5L}$

Thus $\frac{Strength/Weight _{hollow}}{Strength/Weight _{Solid}}=5.16$

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4 years ago

Which of the following is an example of a hardwood? A maple B spruce C pine D fir

bearhunter [10]

Answer:

A. Maple

Explanation:

Maple is a hardwood.

Hope that helps!

7 0

2 years ago

A steady, incompressible, two-dimensional velocity field is given by the following components in the x-y plane: u=1.85+2.05x+0.6

AnnZ [28]

Answer:

$\frac{du}{dt} = 1.515$

$\frac{dv}{dt} = 5.511$

Explanation:

The acceleration field is obtained by deriving the components in function of the time. That is to say:

$\frac{du}{dt}=2.05\cdot \frac{dx}{dt}+0.656\cdot \frac{dy}{dt}\\\frac{dv}{dt} = -2.18\cdot \frac{dx}{dt} -2.05\cdot \frac{dy}{dt}$

Where $\frac{dx}{dt} = u$ and $\frac{dy}{dt} = v$ .

The velocity components at given point are, respectively:

$u = 2.424$

$v=-5.266$

Lastly, the acceleration components are found:

$\frac{du}{dt} = 1.515$

$\frac{dv}{dt} = 5.511$

7 0

3 years ago

Carbon dioxide (CO2) is used in industrial and research application, and is sometimes stored at very high pressure in rigid meta

12345 [234]

Answer:

a)m = 247.43 kg

b) m = 123.71 kg

Explanation:

a)

Given data:

volume =0.8 m^3

P = 18,000 kPa

T =35 degree C = 308 K

By ideal gas equation we have following relation

PV = mRT

where R is gas constant

$R = \frac{8.314}{44} = 0.18895 kJ/ kg K$

$m =\frac{PV}{RT}$

$m = \frac{18000\times 0.8}{0.18895 \times 308}$

m = 247.43 kg

b)

when pressure = 9000 kPa

from ideal gas equation

PV = mRT

where R is gas constant

$R = \frac{8.314}{44} = 0.18895 kJ/ kg K$

$m =\frac{PV}{RT}$

$m = \frac{9000\times 0.8}{0.18895 \times 308}$

m = 123.71 kg

3 0

3 years ago