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3 years ago

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Engineering

1 answer:

Juliette [100K]3 years ago

7 0

Answer:

Ладно

Explanation:

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A reversible refrigeration cycle operates between cold and hot reservoirs at temperatures TC and TH, respectively. (a) If the co

podryga [215]

Answer:

a) $T_{H} = 1.967\,^{\circ}F$ , b) $COP_{R} = 9.105$ , c) $T_{H} = 115.934\,^{\circ}F$ , d) $COP_{R} = 6.995$ , e) $T_{H} = 25.129\,^{\circ}C$

Explanation:

a) The coefficient of performance of the reversible refrigeration cycle is:

$COP_{R} = \frac{T_{C}}{T_{H}-T_{C}}$

$10 = \frac{419.67\,R}{T_{H}-419.67\,R}$

The temperature of the hot reservoir is:

$10\cdot T_{H} - 4196.7 = 419.67$

$T_{H} = 461.637\,R$

$T_{H} = 1.967\,^{\circ}F$

b) The coefficient of performance is:

$COP_{R} = \frac{273.15\,K}{303.15\,K-273.15\,K}$

$COP_{R} = 9.105$

c) The temperature of the hot reservoir can be determined with the help of the following relation:

$\frac{Q_{C}}{Q_{H}-Q_{C}} = \frac{T_{C}}{T_{H}-T_{C}}$

$\frac{500\,BTU}{600\,BTU-500\,BTU} = \frac{479.67\,R}{T_{H}-479.67\,R}$

$5 = \frac{479.67\,R}{T_{H}-479.67\,R}$

$5\cdot T_{H} - 2398.35 = 479.67$

$T_{H} = 575.604\,R$

$T_{H} = 115.934\,^{\circ}F$

d) The coefficient of performance is:

$COP_{R} = \frac{489.67\,R}{559.67\,R-489.67\,R}$

$COP_{R} = 6.995$

e) The temperature of the cold reservoir is:

$8.9 = \frac{268.15\,K}{T_{H}-268.15\,K}$

$8.9\cdot T_{H} - 2386.535 = 268.15$

$T_{H} = 298.279\,K$

$T_{H} = 25.129\,^{\circ}C$

8 0

2 years ago

Does the location of a millimeter change the voltage or current of the circuit?

Cerrena [4.2K]

Answer:

Yes, it does.

Explanation:

8 0

3 years ago

Given below are the measured streamflows in cfs from a storm of 6-hour duration on a stream having a drainage area of 185 mi^2.

sertanlavr [38]

Answer:

33.56 ft^3/sec.in

Explanation:

Duration = 6 hours

drainage area = 185 mi^2

constant baseflow = 550 cfs

<u>Derive the unit hydrograph using the inverse procedure </u>

first step : calculate for the volume of direct runoff hydrograph using the details in table 2 attached below

Vdrh = sum of drh * duration

= 29700 * 6 hours ( 216000 secs )

= 641,520,000 ft^3.

next step : Calculate the volume of runoff in equivalent depth

Vdrh / Area = 641,520,000 / 185 mi^2

= 1.49 in

Finally derive the unit hydrograph

Unit of hydrograph = drh / volume of runoff in equivalent depth

= 50 ft^3 / 1.49 in = 33.56 ft^3/sec.in

5 0

3 years ago

An alloy is evaluated for potential creep deformation in a short-term laboratory experiment. The creep rate (ϵ˙) is found to be

cupoosta [38]

Answer:

Activation energy for creep in this temperature range is Q = 252.2 kJ/mol

Explanation:

To calculate the creep rate at a particular temperature

creep rate, $\zeta_{\theta} = C \exp(\frac{-Q}{R \theta} )$

Creep rate at 800⁰C, $\zeta_{800} = C \exp(\frac{-Q}{R (800+273)} )$

$\zeta_{800} = C \exp(\frac{-Q}{1073R} )\\\zeta_{800} = 1 \% per hour =0.01\\$

$0.01 = C \exp(\frac{-Q}{1073R} )$ .........................(1)

Creep rate at 700⁰C

$\zeta_{700} = C \exp(\frac{-Q}{R (700+273)} )$

$\zeta_{800} = C \exp(\frac{-Q}{973R} )\\\zeta_{800} = 5.5 * 10^{-2} \% per hour =5.5 * 10^{-4}$

$5.5 * 10^{-4} = C \exp(\frac{-Q}{1073R} )$ .................(2)

Divide equation (1) by equation (2)

$\frac{0.01}{5.5 * 10^{-4} } = \exp[\frac{-Q}{1073R} -\frac{-Q}{973R} ]\\18.182= \exp[\frac{-Q}{1073R} +\frac{Q}{973R} ]\\R = 8.314\\18.182= \exp[\frac{-Q}{1073*8.314} +\frac{Q}{973*8.314} ]\\18.182= \exp[0.0000115 Q]\\$

Take the natural log of both sides

$ln 18.182= 0.0000115Q\\2.9004 = 0.0000115Q\\Q = 2.9004/0.0000115\\Q = 252211.49 J/mol\\Q = 252.2 kJ/mol$

3 0

2 years ago

Does the Diesel engine have engine knock or detonation problem? Why?

Luda [366]

Explanation:

Yes Diesel engine have problem of knocking.

We know that knocking is phenomenon in which suddenly large amount of power generates this large amount of power will cause the failure of diesel engine.

Actually when one set of fuel inject inside the cylinder to burn with already compressed air (in general up to 10-15 bar) then this fuel does not burn complete and accumulate inside the cylinder.After that second set of fuel inject inside the cylinder then that one set of fuel burns with second set of fuel and produces large amount of sudden power for engine and causes the breaks in the crank or connecting rod of engine.it leads to damage the engine.

6 0

2 years ago

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