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KIM [24]
3 years ago
9

Similarity of surfaceand groundwaler​

Engineering
1 answer:
almond37 [142]3 years ago
3 0
Nice to knowwwwwwwwww
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Suppose there is a mobile application that can run in two modes: Lazy or Eager. In Lazy Mode, the execution time is 3.333 second
lara [203]
N didn’t do it for you toroeriot everyone wwas wowowowoww
3 0
2 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
3 years ago
Describe two characteristics that bridges and skyscrapers have in common.
Elan Coil [88]

Answer:

skyscrapers and bridges both have structural builds accounting for the influence of wind and they both cost big money

Explanation:

7 0
3 years ago
Explain what the engineering team should advise in the following scenario.
Evgesh-ka [11]

Answer: its c

Explanation:

7 0
3 years ago
A piston-cylinder device contains an ideal gas mixture of 3 kmol of He gas and 7 kmol of Ar gas (both gases are monatomic) at 27
lidiya [134]

Answer:

Q = 62    ( since we are instructed not to include the units in the answer)

Explanation:

Given that:

n_{HCl} = 3 \ kmol\\n_{Ar} = 7 \ k mol

T_1 = 27^0 \ C = ( 27+273)K =  300 K

P_1 = 200 \ kPa

Q = ???

Now the gas expands at constant pressure until its volume doubles

i.e if V_1 = x\\V_2 = 2V_1

Using Charles Law; since pressure is constant

V \alpha T

\frac{V_2}{V_1}  =\frac{T_2}{T_1}

\frac{2V_1}{V_1}  =\frac{T_2}{300}

T_2 = 300*2\\T_2 = 600

mass of He =number of moles of He × molecular weight of He

mass of He = 3 kg  × 4

mass of He = 12 kg

mass of Ar =number of moles of Ar × molecular weight of Ar

mass of He = 7 kg  × 40

mass of He = 280 kg

Now; the amount of  Heat  Q transferred = m_{He}Cp_{He} \delta T  + m_{Ar}Cp_{Ar} \delta T

From gas table

Cp_{He} = 5.9 \ kJ/Kg/K\\Cp_{Ar}  = 0.5203 \  kJ/Kg/K

∴ Q = 12*5.19*10^3(600-300)+280*0.5203*10^3(600-300)

Q = 62.389 *10^6

Q = 62 MJ

Q = 62    ( since we are instructed not to include the units in the answer)

5 0
3 years ago
Read 2 more answers
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