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3 years ago

14

A simple ideal Rankine cycle with water as the working fluid operates between the pressure limits of 4 MPa in the boiler and 25

kPa in the condenser and a turbine inlet temperature of 700°C. The boiler is sized to provide a steam flow of 50 kg/s. Determine the power produced by the turbine and consumed by the pump. Use steam tables.

2 answers:

Maru [420]3 years ago

6 0

Answer:

power produced by turbine = 79656.15 KW

pump work =227.57 KW

Explanation:

here At T1 = 700 C and P1 = 4 Mpa, from steam tabel : h1 =3906.41 KJ/Kg and s1 = 7.62 KJ/Kg.K

now s1 = s2 = 7.62 KJ/Kg.K as 1-2 is isentropic

so at P2 = 20 Kpa and s2 = 7.62 KJ/Kg.K , fom steam table : h2 = 2513.33 KJ/Kg

power produced by turbine = m.( h1 - h2) = 57.18 x ( 3906.41 - 2513.33) = 79656.15 KW

at P3 = 20 Kpa, 3 = vf = 0.001 m^3/kg

so pump work = v3.dP = 57.18 x 0.001 x ( 4000 - 20) =227.57 KW

chubhunter [2.5K]3 years ago

3 0

Answer:

PT = 70 000kw

Pp = 202.4kw

Explanation:

The full details of the explanation is attached.

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sertanlavr [38]

Answer:

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Explanation:

Duration = 6 hours

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Answer: Burning it.

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What is the physical significance of the Reynolds number?. How is defined for external flow over a plate of length L.

yanalaym [24]

Answer:

$Re=\dfrac{\rho\ v\ l}{\mu }$

Explanation:

Reynolds number:

Reynolds number describe the type of flow.If Reynolds number is too high then flow is called turbulent flow and Reynolds is low then flow is called laminar flow .

Reynolds number is a dimensionless number.Reynolds number given is the ratio of inertia force to the viscous force.

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For plate can be given as

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1. Consider a city of 10 square kilometers. A macro cellular system design divides the city up into square cells of 1 square kil

kakasveta [241]

Answer:

a) $n = 1000\,users$ , b) $\Delta t_{min} = \frac{1}{30}\,h$ , $\Delta t_{max} = \frac{\sqrt{2} }{30}\,h$ , $\Delta t_{mean} = \frac{1 + \sqrt{2} }{60}\,h$ , c) $n = 10000000\,users$ , $\Delta t_{min} = \frac{1}{3000}\,h$ , $\Delta t_{max} = \frac{\sqrt{2} }{3000}\,h$ , $\Delta t_{mean} = \frac{1 + \sqrt{2} }{6000}\,h$

Explanation:

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$\Delta t_{min} = \frac{l}{v}$

$\Delta t_{min} = \frac{1\,km}{30\,\frac{km}{h} }$

$\Delta t_{min} = \frac{1}{30}\,h$

The maximum time for traversing a cell is:

$\Delta t_{max} = \frac{\sqrt{2}\cdot l }{v}$

$\Delta t_{max} = \frac{\sqrt{2} }{30}\,h$

The approximate time is giving by the average of minimum and maximum times:

$\Delta t_{mean} = \frac{1+\sqrt{2} }{2}\cdot\frac{l}{v}$

$\Delta t_{mean} = \frac{1 + \sqrt{2} }{60}\,h$

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$n = \frac{10\times 10^{6}\,m^{2}}{100\,m^{2}}\cdot (100\,\frac{users}{cell} )$

$n = 10000000\,users$

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$l = \sqrt{100\,m^{2}}$

$l = 10\,m$

Minimum time for traversing a cell is:

$\Delta t_{min} = \frac{l}{v}$

$\Delta t_{min} = \frac{0.01\,km}{30\,\frac{km}{h} }$

$\Delta t_{min} = \frac{1}{3000}\,h$

The maximum time for traversing a cell is:

$\Delta t_{max} = \frac{\sqrt{2}\cdot l }{v}$

$\Delta t_{max} = \frac{\sqrt{2} }{3000}\,h$

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$\Delta t_{mean} = \frac{1+\sqrt{2} }{2}\cdot\frac{l}{v}$

$\Delta t_{mean} = \frac{1 + \sqrt{2} }{6000}\,h$

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NikAS [45]

Attached is the solution to the above question.

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