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3 years ago

Before you disconnect the service battery from the discharged battery, it is good practice to place a load across the

Engineering

2 answers:

Lilit [14]3 years ago

6 0

Answer:

it is true i just did this test

Explanation:

alexgriva [62]3 years ago

3 0

The answer should be true

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Air enters the compressor of a cold air-standard Brayton cycle with regeneration and reheat at 100 kPa, 300 K, with a mass flow

yanalaym [24]

Answer:

a. 47.48%

b. 35.58%

c. 2957.715 KW

Explanation:

$T_2 =T_1 + \dfrac{T_{2s} - T_1}{\eta _c}$

T₁ = 300 K

$\dfrac{T_{2s}}{T_1} = \left( \dfrac{P_{2}}{P_1} \right)^{\dfrac{k-1}{k} }$

$T_{2s} = 300 \times (10) ^{\dfrac{0.4}{1.4} }$

$T_{2s}$ = 579.21 K

T₂ = 300+ (579.21 - 300)/0.8 = 649.01 K

T₃ = T₂ + $\epsilon _{regen}$ (T₅ - T₂)

T₄ = 1400 K

Given that the pressure ratios across each turbine stage are equal, we have;

$\dfrac{T_{5s}}{T_4} = \left( \dfrac{P_{5}}{P_4} \right)^{\dfrac{k-1}{k} }$

$T_{5s}$ = 1400× $\left( 1/\sqrt{10} \right)^{\dfrac{0.4}{1.4} }$ = 1007.6 K

T₅ = T₄ + ( $T_{5s}$ - T₄)/ $\eta _t$ = 1400 + (1007.6- 1400)/0.8 = 909.5 K

T₃ = T₂ + $\epsilon _{regen}$ (T₅ - T₂)

T₃ = 649.01 + 0.8*(909.5 - 649.01 ) = 857.402 K

T₆ = 1400 K

$\dfrac{T_{7s}}{T_6} = \left( \dfrac{P_{7}}{P_6} \right)^{\dfrac{k-1}{k} }$

$T_{7s}$ = 1400× $\left( 1/\sqrt{10} \right)^{\dfrac{0.4}{1.4} }$ = 1007.6 K

T₇ = T₆ + ( $T_{7s}$ - T₆)/ $\eta _t$ = 1400 + (1007.6 - 1400)/0.8 = 909.5 K

a. $W_{net \ out}$ = cp(T₆ -T₇) = 1.005 * (1400 - 909.5) = 492.9525 KJ/kg

Heat supplied is given by the relation

cp(T₄ - T₃) + cp(T₆ - T₅) = 1.005*((1400 - 857.402) + (1400 - 909.5)) = 1038.26349 kJ/kg

Thermal efficiency of the cycle = (Net work output)/(Heat supplied)

Thermal efficiency of the cycle = (492.9525 )/(1038.26349 ) =0.4748 = 47.48%

b. $bwr = \dfrac{W_{c,in}}{W_{t,out}}$

bwr = (T₂ -T₁)/[(T₄ - T₅) +(T₆ -T₇)] = (649.01 - 300)/((1400 - 909.5) + (1400 - 909.5)) = 35.58%

c. Power = 6 kg *492.9525 KJ/kg = 2957.715 KW

3 0

3 years ago

A flame ionization detector, which is often used in gas chromatography, responds to a change in

Lena [83]

Answer:

Option A

Explanation:

We know that ions are present in hydrogen-air flame and when the burning of an organic compound takes place in this flame more ions are produced in the flame.

Thus when we apply a voltage across this flame, the ion collector plate attracts the all the ions in the flame.

The presence of organic compounds increases the voltage across the hydrogen ion flame produced at the ion collector increases and as the voltage increases, the detection of the organic compound can be made in turn.

Thus flame ionization detector clearly responds to the variation in the collection of ions or electrons in a flame.

3 0

3 years ago

The bulk density of a compacted soil specimen (Gs = 2.70) and its water content are 2060 kg/m^3 and 15.3%, respectively. If the

yaroslaw [1]

Answer:

the saturated density should be

Explanation:

4 0

3 years ago

Mnsdcbjksdhkjhvdskjbvfdfkjbcv hjb dfkjbkjfvvfebjkhbvefgjdf

Julli [10]

Answer:

ik true eedcggftggggfffff

8 0

3 years ago

Read 2 more answers

Consider a very long, slender rod. One end of the rod is attached to a base surface maintained at Tb, while the surface of the r

9966 [12]

Answer:

(a) Calculate the rod base temperature (°C). = 299.86°C

(b) Determine the rod length (mm) for the case where the ratio of the heat transfer from a finite length fin to the heat transfer from a very long fin under the same conditions is 99 percent. = 0.4325m

Explanation:

see attached file below

3 0

3 years ago