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3 years ago

Welding and cutting done in confined spaces must

Engineering

2 answers:

Papessa [141]3 years ago

7 0

The answer is true i’m 99% sure

ELEN [110]3 years ago

4 0

The answer is true. Have a good day

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1 A power transmission includes a belt drive, a chain drive and a gear drive. Which of the following is the best arrangement bet

Setler79 [48]

Answer:d

Explanation:

(d) chain drive belt drive gear drive

3 0

3 years ago

What is the difference Plastic vs elastic deformation.

Reika [66]

Answer:

What is the difference Plastic vs elastic deformation

Explanation:

The elastic deformation occurs when a low stress is apply over a metal or metal structure, in this process, the stress' deformation is temporary and it's recover after the stress is removed. In other words, this DOES NOT affects the atoms separation.

The plastic deformation occurs when the stress apply over the metal or metal structure is sufficient to deform the atomic structure making the atoms split, this is a crystal separation on a limited amount of atoms' bonds.

8 0

3 years ago

A large heat pump should upgrade 5 MW of heat at 85°C to be delivered as heat at 150°C. Suppose the actual heat pump has a COP o

AysviL [449]

Answer:

W=2 MW

Explanation:

Given that

COP= 2.5

Heat extracted from 85°C

Qa= 5 MW

Lets heat supplied at 150°C = Qr

The power input to heat pump = W

From first law of thermodynamics

Qr= Qa+ W

We know that COP of heat pump given as

$COP=\dfrac{Qr}{W}$

$2.5=\dfrac{5}{W}$

W=2 MW

For Carnot heat pump

$COP=\dfrac{T_2}{T_2-T_1}$

$2.5=\dfrac{T_2}{T_2-(273+85)}$

2.5 T₂ - 895= T₂

T₂=596.66 K

T₂=323.6 °C

7 0

3 years ago

A gas turbine operates with a regenerator and two stages of reheating and intercooling. Air enters this engine at 14 psia and 60

Rzqust [24]

Answer:

flow(m) = 7.941 lbm/s

Q_in = 90.5184 Btu/lbm

Q_out = 56.01856 Btu/lbm

Explanation:

Given:

- T_1 = 60 F = 520 R

- T_6 = 940 = 1400 R

- Heat ratio for air k = 1.4

- Compression ratio r = 3

- W_net,out = 1000 hp

Find:

mass flow rate of the air

rates of heat addition and rejection

Solution:

- Using ideal gas relation compute T_2, T_4, T_10:

T_2 = T_1 * r^(k-1/k)

T_2 = T_4 = T_10 = 520*3^(.4/1.4) = 711.744 R

- Using ideal gas relation compute T_7, T_5, T_9:

T_7 = T_6 * r^(-k-1/k)

T_7 = T_5 = T_9 = 1400*3^(-.4/1.4) = 1022.84 R

- The mass flow rate is obtained by:

flow(m) = W_net,out / 2*c_p*(1400-1022.84-711.744+520)

flow(m) = 1000*.7068 / 2*0.24*(1400-1022.84-711.744+520)

flow(m) = 7.941 lbm/s

- The heat input is as follows:

Q_in = c_p*(T_6 - T_5)

Q_in = 0.24*(1400 - 1022.84)

Q_in = 90.5184 Btu/lbm

- The heat output is as follows:

Q_out = c_p*(T_10 - T_1)

Q_out = 0.24*(711.744 - 520)

Q_out = 56.01856 Btu/lbm

5 0

3 years ago

When the compression process is non-quasi-equilibrium, the molecules before the piston face cannot escape fast enough, forming a

muminat

Answer:

a. true

Explanation:

Firstly, we need to understand what takes places during the compression process in a quasi-equilibrium process. A quasi-equilibrium process is a process in during which the system remains very close to a state of equilibrium at all times. When a compression process is quasi-equilibrium, the work done during the compression is returned to the surroundings during expansion, no exchange of heat, and then the system and the surroundings return to their initial states. Thus a reversible process.

While for a non-quasi equilibrium process, it takes more work to move the piston against this high-pressure region.

5 0

4 years ago

Read 2 more answers