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Mice21 [21]
1 year ago
10

Why is the back-work ratio much higher in the brayton cycle than in the rankine cycle?

Engineering
1 answer:
zloy xaker [14]1 year ago
4 0

The back-work ratio much higher in the Brayton cycle than in the Rankine cycle because a gas cycle is the Brayton cycle, while a steam cycle is the Rankine cycle. Particularly, the creation of water droplets will be a constraint on the steam turbine's efficiency. Since gas has a bigger specific volume than steam, the compressor will have to work harder while using gas.

<h3>What are modern Brayton engines?</h3>

Even originally Brayton exclusively produced piston engines, modern Brayton engines are virtually invariably of the turbine variety. Brayton engines are also gas turbines.

<h3>What is the ranking cycle?</h3>

A gas cycle is the Brayton cycle, while the Ranking cycle is a steam cycle. The production of water droplets will especially decrease the steam turbine's performance. Gas-powered compressors will have to do more work since gas's specific volume is greater than steam's.

Th

To know more about Rankine cycle, visit: brainly.com/question/13040242

#SPJ4

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The wave-particle duality theory is the first adequate explanation of which one of the following observations about the hydrogen
Bingel [31]

Answer:

None of these is correct.

Explanation:

The wave particle duality has to do with Louis de Broglie's proposition that matter could exist as waves or particles.

According to him, matter poseses an associated wavelength. Hence, a certain wavelength is traceable to the hydrogen atom.

This wavelength is the ratio of Plank's constant to the momentum of the hydrogen atom

3 0
3 years ago
Two loads connected in parallel draw a total of 2.4 kW at 0.8 pf lagging from a 120-V rms, 60-Hz line. One load absorbs 1.5 kW a
stealth61 [152]

Answer: a) 0.948 b) 117.5µf

Explanation:

Given the load, a total of 2.4kw and 0.8pf

V= 120V, 60 Hz

P= 2.4 kw, cos θ= 80

P= S sin θ - (p/cos θ) sin θ

= P tan θ(cos^-1 (0.8)

=2.4 tan(36.87)= 1.8KVAR

S= 2.4 + j1. 8KVA

1 load absorbs 1.5 kW at 0.707 pf lagging

P= 1.5 kW, cos θ= 0.707 and θ=45 degree

Q= Ptan θ= tan 45°

Q=P=1.5kw

S1= 1.5 +1.5j KVA

S1 + S2= S

2.4+j1.8= 1.5+1.5j + S2

S2= 0.9 + 0.3j KVA

S2= 0.949= 18.43 °

Pf= cos(18.43°) = 0.948

b.) pf to 0.9, a capacitor is needed.

Pf = 0.9

Cos θ= 0.9

θ= 25.84 °

(WC) V^2= P (tan θ1 - tan θ2)

C= 2400 ( tan (36. 87°) - tan (25.84°)) /2 πf × 120^2

f=60, π=22/7

C= 117.5µf

7 0
3 years ago
Can you solve this question​
Alecsey [184]

Answer:

eojcjksjsososisjsiisisiiaodbjspbcpjsphcpjajosjjs ahahhahahahahahahahahahahahahhhahahahaahahhahahahahaahahahahaha

6 0
3 years ago
Read 2 more answers
Line.
Veronika [31]

Air supplied to a pneumatic system is supplied through the C. Actuator

Explanation

Pneumatic systems are like hydraulic systems, it is just that these systems uses compressed air rather than hydraulic fluid.  Pneumatic systems are used widely across the industries. these pneumatic systems needs a constant supply of compressed air to operate. This is provided by an air compressor. The compressor sucks in air at a very high rate from the environment and stores it in a pressurized tank. the Air is supplied thereafter with the help of a actuator valve that is a more sophisticated form of a valve.

From the above statement it is clear that Air supplied to a pneumatic system is supplied through the  Actuator

7 0
3 years ago
A reservoir is 1 km wide and 10 km long and has an average depth of 100m. Every hour, 0.1% of the reservoir's volume drops throu
Ksju [112]

Answer:

250.7mw

Explanation:

Volume of the reservoir = lwh

Length of reservoir = 10km

Width of reservoir = 1km

Height = 100m

Volume = 10x10³x10³x100

= 10⁹m³

Next we find the volume flow rate

= 0.1/100x10⁹x1/3600

= 277.78m³/s

To get the electrical power output developed by the turbine with 92 percent efficiency

= 0.92x1000x9.81x277.78x100

= 250.7MW

7 0
3 years ago
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