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hodyreva [135]
2 years ago
5

Give me top 5 British snacks

Engineering
1 answer:
tia_tia [17]2 years ago
6 0

Answer:

1. Mini Cheddars

2. Sausage Roll

3. Monster Munch

4. Cheese Twists

5. Flapjacks

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Regeneration can only increase the efficiency of a Brayton cycle when working fluid leaving the turbine is hotter than the worki
storchak [24]

Answer:

True, <em>Regeneration is the only process where increases the efficiency of a Brayton cycle when working fluid leaving the turbine is hotter than working fluid leaving the compressor</em>.

Option: A

<u>Explanation: </u>

To increase the efficiency of brayton cycle there are three ways which includes inter-cooling, reheating and regeneration. <em>Regeneration</em> technique <em>is used when a turbine exhaust fluids have higher temperature than the working fluid leaving the compressor of the turbine. </em>

<em>Thermal efficiency</em> of a turbine is increased as <em>the exhaust fluid having higher temperatures are used in heat exchanger where the fluids from the compressor enters and increases the temperature of the fluids leaving the compressor. </em>

6 0
3 years ago
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Army people are good people right
Phantasy [73]

Answer: Yes army people are good people but it also depends on how you fraze that some have been in trouble before but it doesnt mean there bad people we all make mistakes

5 0
3 years ago
Who plays blox burg???
xeze [42]

Answer:

I don't have robux

Explanation:

but i love adopt me

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3 years ago
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A heat pump with an ideal compressor operates between 0.2 MPa and 1 MPa. Refrigerant R134a flows through the system at a rate of
solmaris [256]

Answer:

The mass flow rate of refrigerant is 0.352 kg/s

Explanation:

Considering the cycle of an ideal heat pump, provided in the attachment, we first find enthalpy at state B and D. For that purpose, we use property tables of refrigerant R134a:

<u>At State A</u>:

From table, we see the enthalpy and entropy value of saturated vapor at 0.2 MPa. Therefore:

ha = 244.5 KJ/kg

Sa = 0.93788 KJ/kg.k

<u>At State B</u>:

Since, the process from state A to B is isentropic. Therefore,

Sb = Sa = 0.93788 KJ/Kg

From table, we see the enthalpy value of super heated vapor at 1 MPa and Sb. Therefore:

hb = 256.85 KJ/kg                          (By interpolation)

<u>At State C</u>:

From table, we see the enthalpy and entropy value of saturated liquid at 1 MPa. Therefore:

hc = 107.34 KJ/kg

Now, from the diagram it is very clear that:

Heat Loss = m(hb = hc)

m = (Heat Loss)/(hb - hc)

where,

m = mass flow rate = ?

Heat Loss = (180,000 Btu/hr)(1.05506 KJ/1 Btu)(1 hr/3600 sec)

Heat Loss = 52.753 KW

Therefore,

m = (52.753 KJ/s)/(256.85 KJ/kg - 107.34 KJ/kg)

<u>m = 0.352 kg/s</u>

5 0
3 years ago
A transformer winding contains 900 turns of wire which creates 400 ohms of primary importance how many terms of the same size wi
Eduardwww [97]

Answer: 255

255 turns are required to create 25 ohms of  secondary impedance.

Explanation:

Given that,

Number of turns in primary wire N₁ = 900

impedance on Primary wire Z₁ = 400 ohms

Number of turns in Secondary wire N₂ = ?

impedance on Secondary wire Z₂ = 25 ohms

we know that, the relationship between turn and impedance is

Zp / Zs = ( Np / Ns )²

(Primary impedance / secondary impedance) = Number of turns in primary wire / Number of turns in secondary wire)²

there fore

Z₁ / Z₂ = ( N₁ / N₂ )²

Now we substitute

( 400 / 25 ) = ( 900 / N₂ )²

400 / 25 = 900² / N₂²

we cross multiple to get our N₂

400 × N₂² =  900² × 25

N₂² = ( 900² × 25 ) / 400

N₂² = ( 810000 × 25 ) / 400

N₂² = 20250000 / 400

N₂² = 50625

N₂ = √50625

N₂ = 225

Therefore 255 turns are required to create 25 ohms of  secondary impedance.

4 0
3 years ago
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