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german
2 years ago
14

What is mechanical engineer​

Engineering
1 answer:
Charra [1.4K]2 years ago
4 0
A man or women who design other machines inside building, etc. pls give brainliest
You might be interested in
The steady-state data listed below are claimed for a power cycle operating between hot and cold reservoirs at 1200K and 400K, re
Anni [7]

Answer:

a) W_cycle = 200 KW , n_th = 33.33 %  , Irreversible

b) W_cycle = 600 KW , n_th = 100 %     , Impossible

c) W_cycle = 400 KW , n_th = 66.67 %  , Reversible

Explanation:

Given:

- The temperatures for hot and cold reservoirs are as follows:

  TL = 400 K

  TH = 1200 K

Find:

For each case W_cycle , n_th ( Thermal Efficiency ) :

(a) QH = 600 kW, QC = 400 kW

(b) QH = 600 kW, QC = 0 kW

(c) QH = 600 kW, QC = 200kW

- Determine whether the cycle operates reversibly, operates irreversibly, or is impossible.

Solution:

- The work done by the cycle is given by first law of thermodynamics:

                                 W_cycle = QH - QC

- For categorization of cycle is given by second law of thermodynamics which states that:

                                 n_th < n_max     ...... irreversible

                                 n_th = n_max     ...... reversible

                                 n_th > n_max     ...... impossible

- Where n_max is the maximum efficiency that could be achieved by a cycle with Hot and cold reservoirs as follows:

                                n_max = 1 - TL / TH = 1 - 400/1200 = 66.67 %

And,                         n_th = W_cycle / QH

a) QH = 600 kW, QC = 400 kW

   - The work done by cycle according to First Law is:

                                W_cycle = 600 - 400 = 200 KW

   - The thermal efficiency of the cycle is given by n_th:

                                n_th = W_cycle / QH

                                n_th = 200 / 600 = 33.33 %

   - The type of process according to second Law of thermodynamics:

               n_th = 33.333 %                n_max = 66.67 %

                                       n_th < n_max  

      Hence,                Irreversible Process  

b) QH = 600 kW, QC = 0 kW

   - The work done by cycle according to First Law is:

                                W_cycle = 600 - 0 = 600 KW

   - The thermal efficiency of the cycle is given by n_th:

                                n_th = W_cycle / QH

                                n_th = 600 / 600 = 100 %

   - The type of process according to second Law of thermodynamics:

                 n_th = 100 %                 n_max = 66.67 %

                                     n_th > n_max  

      Hence,               Impossible Process              

c) QH = 600 kW, QC = 200 kW

   - The work done by cycle according to First Law is:

                                W_cycle = 600 - 200 = 400 KW

   - The thermal efficiency of the cycle is given by n_th:

                                n_th = W_cycle / QH

                                n_th = 400 / 600 = 66.67 %

   - The type of process according to second Law of thermodynamics:

               n_th = 66.67 %                 n_max = 66.67 %

                                     n_th = n_max  

      Hence,                Reversible Process

7 0
3 years ago
Fully developed conditions are known to exist for water flowing through a 25-mm-diameter tube at 0.01 kg/s and 27 C. What is the
Irina18 [472]

Answer:

0.0406 m/s

Explanation:

Given:

Diameter of the tube, D = 25 mm = 0.025 m

cross-sectional area of the tube = (π/4)D² = (π/4)(0.025)² = 4.9 × 10⁻⁴ m²

Mass flow rate = 0.01 kg/s

Now,

the mass flow rate is given as:

mass flow rate = ρAV

where,

ρ is the density of the water = 1000 kg/m³

A is the area of cross-section of the pipe

V is the average velocity through the pipe

thus,

0.01 = 1000 × 4.9 × 10⁻⁴ × V

or

V = 0.0203 m/s

also,

Reynold's number, Re = \frac{VD}{\nu}

where,

ν is the kinematic viscosity of the water = 0.833 × 10⁻⁶ m²/s

thus,

Re = \frac{0.0203\times0.025}{0.833\times10^{-6}}

or

Re = 611.39 < 2000

thus,

the flow is laminar

hence,

the maximum velocity =  2 × average velocity = 2 × 0.0203 m/s

or

maximum velocity = 0.0406 m/s

5 0
3 years ago
A train starts from rest at station A and accelerates at 0.6 m/s^2 for 60 s. Afterwards it travels with a constant velocity for
Aleks [24]

Answer:

The distance between the station A and B will be:

x_{A-B}=55.620\: km  

Explanation:

Let's find the distance that the train traveled during 60 seconds.

x_{1}=x_{0}+v_{0}t+0.5at^{2}

We know that starts from rest (v(0)=0) and the acceleration is 0.6 m/s², so the distance will be:

x_{1}=\frac{1}{2}(0.6)(60)^{2}

x_{1}=1080\: m

Now, we need to find the distance after 25 min at a constant speed. To get it, we need to find the speed at the end of the first distance.

v_{1}=v_{0}+at

v_{1}=(0.6)(60)=36\: m/s

Then the second distance will be:

x_{2}=v_{1}*1500

x_{2}=(36)(1500)=54000\: m        

The final distance is calculated whit the decelerate value:

v_{f}^{2}=v_{1}^{2}-2ax_{3}

The final velocity is zero because it rests at station B. The initial velocity will be v(1).

0=36^{2}-2(1.2)x_{3}

x_{3}=\frac{36^{2}}{2(1.2)}  

x_{3}=540\: m

Therefore, the distance between the station A and B will be:

x_{A-B}=x_{1}+x_{2}+x_{3}  

x_{A-B}=1080+54000+540=55.620\: km  

I hope it helps you!

 

7 0
2 years ago
3. Ang mga kababaihan noon ay tumutulong sa pakikipaglaban sa mga Espanyol upang makamit ang Kalayaan. Ganito pa rin ba ang mga
likoan [24]

NYC dp my frndsssssssssss

5 0
2 years ago
A heat pump with refrigerant-134a as the working fluid is used to keep aspace at 25°C by absorbing heat from geothermal water th
Anni [7]

Answer:12

Explanation:

7 0
3 years ago
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