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lions [1.4K]
3 years ago
13

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.
Engineering
1 answer:
zubka84 [21]3 years ago
3 0

Answer:

Betbtybrbytntrnyrnrynunjhjhnthnnhtnnthnhtnnhnhrnntnthhnhnhtnthn

Explanation:

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I am just finishing up my work

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Water flows through a horizontal plastic pipe with a diameter of 0.15 m at a velocity of 15 cm/s. Determine the pressure drop pe
Sonja [21]

Answer:0.1898 Pa/m

Explanation:

Given data

Diameter of Pipe\left ( D\right )=0.15m

Velocity of water in pipe\left ( V\right )=15cm/s

We know viscosity of water is\left (\mu\right )=8.90\times10^{-4}pa-s

Pressure drop is given by hagen poiseuille equation

\Delta P=\frac{128\mu \L Q}{\pi D^4}

We have asked pressure Drop per unit length i.e.

\frac{\Delta P}{L} =\frac{128\mu \ Q}{\pi D^4}

Substituting Values

\frac{\Delta P}{L}=\frac{128\times8.90\times10^{-4}\times\pi \times\left ( 0.15^{3}\right )}{\pi\times 4 \times\left ( 0.15^{2}\right )}

\frac{\Delta P}{L}=0.1898 Pa/m

4 0
3 years ago
A family member who hasn’t worked with computers before has decided to change jobs. You’ve been asked to explain some of the bas
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That due to the specific tasks that needs to be accomplished by each program to make an all encompassing program would be inefficient and full of bugs
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A large heat pump should upgrade 5 MW of heat at 85°C to be delivered as heat at 150°C. Suppose the actual heat pump has a COP o
AysviL [449]

Answer:

W=2 MW

Explanation:

Given that

COP= 2.5

Heat extracted from 85°C  

Qa= 5 MW

Lets heat supplied at 150°C   = Qr

The power input to heat pump = W

From first law of thermodynamics

Qr= Qa+ W

We know that COP of heat pump given as

COP=\dfrac{Qr}{W}

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2.5=\dfrac{5}{W}

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For Carnot heat pump

COP=\dfrac{T_2}{T_2-T_1}

2.5=\dfrac{T_2}{T_2-(273+85)}

2.5 T₂ -  895= T₂

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T₂=323.6 °C

7 0
3 years ago
In using the drag coefficient care needs to be taken to use the correct area when determining the drag force. What is a typical
stealth61 [152]

Answer:

Explanation:

We know that Drag forceF_D

  F_D=\dfrac{1}{2}C_D\rho AV^2

Where

             C_D is the drag force constant.

                 A is the projected area.

                V is the velocity.

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Form the above expression of drag force we can say that drag force depends on the area .So We should need to take care of correct are before finding drag force on body.

Example:

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6 0
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