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EleoNora [17]
3 years ago
14

Only respond if your the person im talkin to

Engineering
1 answer:
bagirrra123 [75]3 years ago
5 0

Answer: um wuh anyways thxs for the points!

Explanation: ....:/

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Give me uses of a grinding machine in agriculture.
Tju [1.3M]

Answer:

The grinding machine is used for roughing and finishing flat, cylindrical, and conical surfaces; finishing internal cylinders or bores; forming and sharpening cutting tools; snagging or removing rough projections from castings and stampings; and cleaning, polishing, and buffing surfaces.

4 0
2 years ago
Water vapor at 6 MPa, 600 degrees C enters a turbine operating at steady state and expands to 10kPa. The mass flow rate is 2 kg/
kirill115 [55]

Answer:

Explanation:

Obtain the following properties at 6MPa and 600°C from the table "Superheated water".

h_1=3658.8KL/Kg\\s_1=7.1693kJ/kg.k

Obtain the following properties at 10kPa from the table "saturated water"

h_{f2}=191.81KJ/Kg.K\\h_{fg2}=2392.1KJ/Kg\\s_{f2}=0.6492KJ/Kg.K\\s_{fg2}=7.4996KJ/Kg.K

Calculate the enthalpy at exit of the turbine using the energy balance equation.

\frac{dE}{dt}=Q-W+m(h_1-h_2)

Since, the process is isentropic process Q=0

0=0-W+m(h_1-h_2)\\h_2=h_1-\frac{W}{m}\\\\h_2=3658.8-\frac{2626}{2}\\\\=2345.8kJ/kg

Use the isentropic relations:

s_1=s_{2s}\\s_1=s_{f2}+x_{2s}s_{fg2}\\7.1693=6492+x_{2s}(7.4996)\\x_{2s}=87

Calculate the enthalpy at isentropic state 2s.

h_{2s}=h_{f2}+x_{2s}.h_{fg2}\\=191.81+0.87(2392.1)\\=2272.937kJ/kg

a.)

Calculate the isentropic turbine efficiency.

\eta_{turbine}=\frac{h_1-h_2}{h_1-h_{2s}}\\\\=\frac{3658.8-2345.8}{3658.8-2272.937}=0.947=94.7%

b.)

Find the quality of the water at state 2

since h_f at 10KPa <h_2<h_g at 10KPa

Therefore, state 2 is in two-phase region.

h_2=h_{f2}+x_2(h_{fg2})\\2345.8=191.81+x_2(2392.1)\\x_2=0.9

Calculate the entropy at state 2.

s_2=s_{f2}+x_2.s_{fg2}\\=0.6492+0.9(7.4996)\\=7.398kJ/Kg.K

Calculate the rate of entropy production.

S=\frac{Q}{T}+m(s_2-s_1)

since, Q = 0

S=m(s_2-s_1)\\=2\frac{kg}{s}(7.398-7.1693)kJ/kg\\=0.4574kW/k

6 0
3 years ago
determine the position d of the 6- kn load so that the average normal stress in each rod is the same.
Zinaida [17]

The load is placed at distance 0.4 L from the end of $$12 \mathrm{~mm}^{2} $ area.

<h3>What is meant by torque?</h3>

The force that can cause an object to rotate along an axis is measured as torque. Similar to how force accelerates an item in linear kinematics, torque accelerates an object in an angular direction. A vector quantity is torque.

Let the beam is of length L

Now the stress on both the end is the same now we can say that torque on the beam due to two forces must be zero

$N_1 * x=N_2 *(L-x)$

also, we know that stress at both ends are same

$\frac{N_1}{12}=\frac{N_2}{8}$

$2 * N_1=3 * N_2$

Now from two equations we have

$\frac{3}{2} N_2 * x=N_2 *(L-x)

solving the above equation we have

$x=\frac{2}{5} L

so the load is placed at distance 0.4 L from the end of $$12 \mathrm{~mm}^{2} $ area.

The complete question is:

47. the beam is supported by two rods ab and cd that have cross-sectional areas of $$12mm^2 and $$8mm^2, respectively. determine the position d of the 6-kn load so that the average normal stress in each rod is the same.

To learn more about torque refer to:

brainly.com/question/20691242

#SPJ4

7 0
2 years ago
What causes the moon to be illuminated? PLEASE HELP MEH ILL GIVE YOU BRAINLEIST!!
Likurg_2 [28]

Answer:

D Reflection

Explanation:

It reflects the light from the sun. Then when the earth gets in the way, it casts a shadow causing crescent moons.

7 0
2 years ago
A race car is travelling around a circular track. The velocity of the car is 20 m/s, the radius of the track is 300 m, and the m
Zielflug [23.3K]

Answer:

μ = 0.136

Explanation:

given,

velocity of the car = 20 m/s

radius of the track = 300 m

mass of the car = 2000 kg

centrifugal force

F_c = \dfrac{mv^2}{r}

F_c = \dfrac{2000\times 20^2}{300}

F c = 2666. 67 N

F f= μ N

F f = μ m g

2666.67  =  μ × 2000 × 9.8

μ = 0.136

so, the minimum coefficient of friction between road surface and car tyre is equal to μ = 0.136

5 0
3 years ago
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