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Phoenix [80]
3 years ago
6

A turbine operates at steady state, and experiences a heat loss. 1.1 kg/s of water flows through the system. The inlet is mainta

ined at 100 bar, 520 Celsius. The outlet is maintained at 10 bar, 280 Celsius. A rate of heat loss of 60 kW is measured. Determine the rate of work output from the turbine, in kW.
Engineering
1 answer:
strojnjashka [21]3 years ago
5 0

Answer:

\dot W_{out} = 399.47\,kW

Explanation:

The turbine is modelled after the First Law of Thermodynamics:

-\dot Q_{out} -\dot W_{out} + \dot m\cdot (h_{in}-h_{out}) = 0

The work done by the turbine is:

\dot W_{out} = \dot m \cdot (h_{in}-h_{out})-\dot Q_{out}

The properties of the water are obtained from property tables:

Inlet (Superheated Steam)

P = 10\,MPa

T = 520\,^{\textdegree}C

h = 3425.9\,\frac{kJ}{kg}

Outlet (Superheated Steam)

P = 1\,MPa

T = 280\,^{\textdegree}C

h = 3008.2\,\frac{kJ}{kg}

The work output is:

\dot W_{out} = \left(1.1\,\frac{kg}{s}\right)\cdot \left(3425.9\,\frac{kJ}{kg} -3008.2\,\frac{kJ}{kg}\right) - 60\,kW

\dot W_{out} = 399.47\,kW

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GenaCL600 [577]

Answer:

6.6 kilo volts = 6.6 k volts

Explanation:

A prefix is a word, number or a letter that is added before another word. In physics we have different prefixes for the exponential powers of 10, that are placed before units in place of those powers. Some examples are:

deci (d)   ------  10⁻¹

centi (c)   ------  10⁻²

milli (m)   ------   10⁻³

kilo (k)     ------   10³

mega (M) -----   10⁶

giga (G)   ------   10⁹

We have:

6600 volts

converting to exponential form:

=> 6.6 x 10³ volts

Thus, we know that the prefix of kilo (k) is used for 10³.

Hence,

=> <u>6.6 kilo volts = 6.6 k volts</u>

7 0
3 years ago
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Bas_tet [7]

Answer:

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Explanation:

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3 years ago
A square silicon chip (k = 152 W/m·K) is of width 7 mm on a side and of thickness 3 mm. The chip is mounted in a substrate such
Harrizon [31]

Answer:

The steady-state temperature difference is 2.42 K

Explanation:

Rate of heat transfer = kA∆T/t

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k is the heat transfer coefficient = 152 W/m.K

A is the area of the square silicon = width^2 = (7/1000)^2 = 4.9×10^-5 m^2

t is the thickness of the silicon = 3 mm = 3/1000 = 0.003 m

6 = 152×4.9×10^-5×∆T/0.003

∆T = 6×0.003/152×4.9×10^-5 = 2.42 K

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