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neonofarm [45]
3 years ago
14

With thermodynamics, one cannot determine ________.

Engineering
1 answer:
zloy xaker [14]3 years ago
4 0

Answer:

Option B is correct.

Explanation:

With thermodynamics, one cannot determine the speed of the reaction.

Speed of the reaction is defined as the rate at which one component is changes to another component. With factor cannot be calculated from the thermodynamics branch of engineering.

With thermodynamics we can determine the value of the equilibrium constant,

the extent of a reaction, the temperature at which a reaction will be spontaneous & the direction of a spontaneous reaction. But we cannot determine  the speed of a reaction.

Therefore option B is correct.

You might be interested in
simple Brayton cycle using air as the working fluid has a pressure ratio of 10. The minimum and maximum temperatures in the cycl
Irina18 [472]

Answer:

a) 764.45K

b) 210.48 kJ/kg

c) 30.14%

Explanation:

pressure ratio = 10

minimum temperature = 295 k

maximum temperature = 1240 k

isentropic efficiency for compressor = 83%

Isentropic efficiency for turbine = 87%

<u>a) Air temperature at turbine exit </u>

we can achieve this by interpolating for enthalpy

h4 = 783.05 kJ/kg ( calculated in the background ) at state 4 using Table A-17  for  Ideal gas properties of air

T4 ( temperature at Turbine exit ) = 760 + ( 780 - 760 ) (\frac{783.05-778.18}{800.13-778.18} ) = 764.45K

<u>b) The net work output </u>

first we determine the actual work input to compressor

Wc = h2 - h1  ( calculated values )

     = 626.57 - 295.17 =  331.4 kJ/kg

next determine the actual work done by Turbine

Wt = h3 - h4  ( calculated values )

     = 1324.93 - 783.05 = 541.88 kJ/kg

finally determine the network output of the cycle

Wnet = Wt - Wc

         = 541.88 - 331.4  = 210.48 kJ/kg

<u>c) determine thermal efficiency </u>

лth = Wnet / qin  ------ ( 1 )

where ; qin = h3 - h2

<em>equation 1 becomes </em>

лth = Wnet / ( h3 - h2 )

      = 210.48 / ( 1324.93 - 626.57 )

      = 0.3014  =  30.14%

6 0
3 years ago
An air conditioning performance check requires the use of a
Lostsunrise [7]

Answer:

aA

Explanation:

air conditioning performance check requires the use of aA.

Hope this helps!

4 0
3 years ago
write a paragraph on how iron and copper(separately) is manufactured, what it consists of and processes that it went through to
trasher [3.6K]

Answer:

Iron is manufactured in a blast furnace. First, iron ore is mixed with coke and heated to form an iron-rich clinker called 'sinter'. Sintering is an important part of the overall process as it reduces waste and provides an efficient raw material for iron making. Coke is produced from carefully selected grades of coal.

Copper is typically extracted from oxide and sulfide ores that contain between 0.5 and 2.0% copper. ... Regardless of the ore type, mined copper ore must first be concentrated to remove gangue or unwanted materials embedded in the ore. The first step in this process is crushing and powdering ore in a ball or rod mill.

5 0
3 years ago
Water enters the pump of a steam power plant as saturated liquid at 20 kPa at a rate of 45 kg/s and exits at 6 MPa. Neglecting t
Natasha2012 [34]

Answer:

\dot W_{in} = 273.69\,kW

Explanation:

The pump is modelled after the First Law of Thermodynamics. A reversible process means that fluid does not report any positive change in entropy:

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

The properties of the fluid at entrance and exit are, respectively:

Inlet (Saturated Liquid)

P = 20\,kPa

T = 60.06\,^{\textdegree}C

h = 251.42\,\frac{kJ}{kg}

s = 0.8320\,\frac{kJ}{kg\cdot K}

Outlet (Subcooled Liquid)

P = 6000\,kPa

T = 60.06\,^{\textdegree}C

h = 257.502\,\frac{kJ}{kg}

s = 0.8320\,\frac{kJ}{kg\cdot K}

The power input to the pump is computed hereafter:

\dot W_{in} = \left(45\,\frac{kg}{s} \right)\cdot \left(257.502\,\frac{kJ}{kg} -251.42\,\frac{kJ}{kg} \right)

\dot W_{in} = 273.69\,kW

8 0
4 years ago
Propane is to be compressed from 0.4 MPa and 360 K to 4 MPa using a two-stage compressor. An interstage cooler returns the tempe
kow [346]

Answer:

a. 81 kj/kg

b. 420.625K

c.  101.24kj/kg

Explanation:

\frac{t2}{t1} =[\frac{p2}{p1} ]^{\frac{y-1}{y} }

t1 = 360

p1 = 0.4mpa

p2 = 1.20

y = 1.13

substitute these values into the equation

\frac{t2}{360} =[\frac{1.20}{0.4} ]^{\frac{1.13-1}{1.13} }

\frac{t2}{360} =[\frac{1.2}{0.4} ]^{0.1150}\\\frac{t2}{360} =1.1347

when we cross multiply

t2 = 360 * 1.1347

= 408.5

a. the work required in the firs compressor

w=c(t2-t1)

c=1.67x10³

t1 = 360

t2 = 408.5

w = 1670(408.5-360)

= 1670*48.5

= 80995 J

= 81KJ/kg

b. n=\frac{t2-t1}{t'2-t1}

n = 80%

t2 = 408.5

t1 = 360

0.80 = 408.5-360 ÷ t'2-360

0.80 =\frac{48.5}{t'2-360}

cross multiply to get the value of t'2

0.80(t'2-360) = 48.5

0.80t'2 - 288 = 48.5

0.8t'2 = 48.5+288

0.8t'2 = 336.5

t'2 = 336.5/0.8

= 420.625

this is the temperature at the exit of the first compressor

c. cooling requirement

w = c(t2-t1)

= 1.67x10³(420.625-360)

= 1670*60.625

= 101243.75

= 101.24kj/kg

8 0
3 years ago
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