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3 years ago

Who works alongside and assists the engineers?

Engineering

1 answer:

nika2105 [10]3 years ago

5 0

Answer:

Assistants works alongside and assists the engineers.

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What engineers call moment, scientists call

Svet_ta [14]

Answer:

yes

Explanation:

7 0

3 years ago

A heat pump operates on a Carnot heat pump cycle with a COP of 12.5. It keeps a space at 24°C by consuming 2.15 kW of power. Det

Vinil7 [7]

Answer:

a) $T_{L} = 273.378\,K\,(0.228\,^{\textdegree}C)$ , b) $\dot Q_{H} = 26.875\,kW$

Explanation:

a) The Coefficient of Performance of the Carnot Heat Pump is:

$COP_{HP} = \frac{T_{H}}{T_{H}-T_{L}}$

After some algebraic handling, the temperature of the cold reservoir is determined:

$T_{H}-T_{L} = \frac{T_{H}}{COP_{HP}}$

$T_{L} = T_{H}\cdot \left(1-\frac{1}{COP_{HP}} \right)$

$T_{L} = (297.15\,K)\cdot \left(1-\frac{1}{12.5}\right)$

$T_{L} = 273.378\,K\,(0.228\,^{\textdegree}C)$

b) The heating load provided by the heat pump is:

$\dot Q_{H} = COP_{HP}\cdot \dot W$

$\dot Q_{H} = (12.5)\cdot (2.15\,kW)$

$\dot Q_{H} = 26.875\,kW$

4 0

3 years ago

You are evaluating the lifetime of a turbine blade. The blade is 4 cm long and there is a gap of 0.16 cm between the tip of the

Tcecarenko [31]

Answer:

Explanation:

Given conditions

1)The stress on the blade is 100 MPa

2)The yield strength of the blade is 175 MPa

3)The Young’s modulus for the blade is 50 GPa

4)The strain contributed by the primary creep regime (not including the initial elastic strain) was 0.25 % or 0.0025 strain, and this strain was realized in the first 4 hours.

5)The temperature of the blade is 800°C.

6)The formula for the creep rate in the steady-state regime is dε /dt = 1 x 10-5 σ4 exp (-2 eV/kT)

where: dε /dt is in cm/cm-hr σ is in MPa T is in Kelvink = 8.62 x 10-5 eV/K

Young Modulus, E = Stress, $\sigma$ /Strain, ∈

initial Strain, $\epsilon_i = \frac{\sigma}{E}$

$\epsilon_i = \frac{100\times 10^{6} Pa}{50\times 10^{9} Pa}$

$\epsilon_i = 0.002$

creep rate in the steady state

$\frac{\delta \epsilon}{\delta t} = (1 \times {10}^{-5})\sigma^4 exp^(\frac{-2eV}{kT} )$

$\frac{\epsilon_{initial} - \epsilon _{primary}}{t_{initial}-t_{final}} = 1 \times 10^{-5}(100)^{4}exp(\frac{-2eV}{8.62\times10^{-5}(\frac{eV}{K} )(800+273)K} )$