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Likurg_2 [28]
3 years ago
6

A steam turbine receives steam at 1.5MPa and 220oC, and exhausts at 50kPa, 0.75 dry. Neglecting heat losses and changes in kinet

ic and potential energy, estimate the work output per kg steam.
If when allowance is made for friction, radiation and leakages losses, the actual work of that estimated in (a), calculate the power output of the turbine when consuming 600kg of steam per minute.
Engineering
1 answer:
SIZIF [17.4K]3 years ago
4 0

Answer:

Can you make friend with me ?

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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:

a) The total number of users that can be accomodated in the system is:

n = \frac{10\,km^{2}}{1\,\frac{km^{2}}{cell} }\cdot (100\,\frac{users}{cell} )

n = 1000\,users

b) The length of the side of each cell is:

l = \sqrt{1\,km^{2}}

l = 1\,km

Minimum time for traversing a cell is:

\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

c) The total number of users that can be accomodated in the system is:

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

n = 10000000\,users

The length of each side of the cell is:

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

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} }{6000}\,h

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