Tech A says that some battery corrosion cannot be seen beneath the insulation of the cables. Tech B says

For two different air velocities, the Nusselt number for two different diameter cylinders in cross flow is the same. The average

densk [106]

C smaller than that of the larger-diameter cylinder

3 0

3 years ago

Principals of Construction intro

Nadusha1986 [10]

Answer:

The answer is safety documents

Explanation:

Every company is responsible

For people documents

Is very important for every company

8 0

3 years ago

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

4 years ago

In details and step-by-step, show how you apply the Bubble Sort algorithm on the following list of values. Your answer should sh

astraxan [27]

( 12 17 18 19 25 )

Explanation:

First Pass:

( 19 18 25 17 12 ) –> ( 18 19 25 17 12 ), Here, algorithm compares the first two elements, and swaps since 19 > 18.

( 18 19 25 17 12 ) –> ( 18 19 25 17 12 ), Now, since these elements are already in order (25 > 19), algorithm does not swap them.

( 18 19 25 17 12 ) –> ( 18 19 17 25 12 ), Swap since 25 > 17

( 18 19 17 25 12 ) –> ( 18 19 17 12 25 ), Swap since 25 > 12

Second Pass:

( 18 19 17 12 25 ) –> ( 18 19 17 12 25 )

( 18 19 17 12 25 ) –> ( 18 17 19 12 25 ), Swap since 19 > 17

( 18 17 19 12 25 ) –> ( 18 17 12 19 25 ), Swap since 19 > 12

( 18 17 12 19 25 ) –> ( 18 17 12 19 25 )

Third Pass:

( 18 17 12 19 25 ) –> ( 17 18 12 19 25 ), Swap since 18 > 17

( 17 18 12 19 25 ) –> ( 17 12 18 19 25 ), Swap since 18 > 12

( 17 12 18 19 25 ) –> ( 17 12 18 19 25 )

Fourth Pass:

( 17 12 18 19 25 ) –> ( 12 17 18 19 25 ), Swap since 17 > 12

( 12 17 18 19 25 ) –> ( 12 17 18 19 25 ), Swap since 18 > 12

( 12 17 18 19 25 ) –> ( 12 17 18 19 25 )

Now, the array is already sorted, but our algorithm does not know if it is completed. The algorithm needs one whole pass without any swap to know it is sorted.

Fifth Pass:

( 12 17 18 19 25 ) –> ( 12 17 18 19 25 )

6 0

3 years ago

As resistors are added in series to a circuit, the total resistance will

olganol [36]

The equivalent of the resistance connected in the series will be Req=R₁+R₂+R₃.

<h3>
What is resistance?</h3>

Resistance is the obstruction offered whenever the current is flowing through the circuit.

So the equivalent resistance is when three resistances are connected in series. When the resistances are connected in series then the voltage is different and the current remain same in each resistance.

V eq = V₁ + V₂ + V₃

IReq = IR₁ + IR₂ + IR₃

Req = R₁ + R₂ + R₃

Therefore the equivalent of the resistance connected in the series will be Req=R₁+R₂+R₃.

To know more about resistance follow

brainly.com/question/24858512

#SPJ4

8 0

2 years ago