All categories

3 years ago

Match the following parts of a crane

Engineering

1 answer:

trasher [3.6K]3 years ago

5 0

It is auxillary sorry i couldn’t help it happens to the best of us

You might be interested in

Can you solve this question

Alecsey [184]

Answer:

eojcjksjsososisjsiisisiiaodbjspbcpjsphcpjajosjjs ahahhahahahahahahahahahahahahhhahahahaahahhahahahahaahahahahaha

6 0

3 years ago

Read 2 more answers

Need help with both giving out brainlest for the people to help me

sammy [17]

The first one is d or the 4th answer choice and the second one is false. Hope this helps!

4 0

3 years ago

John, a team member, has completed e0 - agile for beginners he wants to contribute to tcs agile vision. he wants to find out wha

Jet001 [13]

Answer:

John should detail his Scrum Master.

Explanation:

The Team Lead or Scrum Master coordinates the tasks of individual team members and supports the progress of the team. The Scrum Master usually receives instructions from the Product Owner and then ensures that the tasks are performed accordingly. She also coaches the Development Team and works with the Product Owner to carry out daily development activities. She also drives the Scrum Values and Principles, ensuring that the team members understand and practice them.

7 0

3 years ago

A stationary gas-turbine power plant operates on a simple ideal Brayton cycle with air as the working fluid. The air enters the

ololo11 [35]

Answer:

A) W' = 15680 KW

B) W' = 17113.87 KW

Explanation:

We are given;

Temperature at state 1; T1 = 290 K

Temperature at state 3; T3 = 1100 K

Rate of heat transfer; Q_in = 35000 kJ/s = 35000 Kw

Pressure of air into compressor; P_c = 95 kPa

Pressure of air into turbine; P_t = 760 kPa

A) The power assuming constant specific heats at room temperature is gotten from;

W' = [1 - ((T4 - T1)/(T3 - T2))] × Q_in

Now, we don't have T4 and T2 but they can be gotten from;

T4 = [T3 × (r_p)^((1 - k)/k)]

T2 = [T1 × (r_p)^((k - 1)/k)]

r_p = P_t/P_c

r_p = 760/95

r_p = 8

Also,k which is specific heat capacity of air has a constant value of 1.4

Thus;

Plugging in the relevant values, we have;

T4 = [(1100 × (8^((1 - 1.4)/1.4)]

T4 = 607.25 K

T2 = [290 × (8^((1.4 - 1)/1.4)]

T2 = 525.32 K

Thus;

W' = [1 - ((607.25 - 290)/(1100 - 525.32))] × 35000

W' = 0.448 × 35000

W' = 15680 KW

B) The power accounting for the variation of specific heats with temperature is given by;

W' = [1 - ((h4 - h1)/(h3 - h2))] × Q_in

From the table attached, we have the following;

At temperature of 607.25 K and by interpolation; h4 = 614.64 KJ/K

At T3 = 1100 K, h3 = 1161.07 KJ/K

At T1 = 290 K, h1 = 290.16 KJ/K

At T2 = 525.32 K, and by interpolation, h2 = 526.12 KJ/K

Thus;

W' = [1 - ((614.64 - 290.16)/(1161.07 - 526.12))] × 35000

W' = 17113.87 KW

4 0

2 years ago

A reservoir delivers water to a horizontal pipeline 39 long The first 15 m has a diameter of 50 mm, after which it suddenly beco

allsm [11]

Answer:

The difference of head in the level of reservoir is 0.23 m.

Explanation:

For pipe 1

$d_1=50 mm,f_1=0.0048$

For pipe 2

$d_2=75 mm,f_2=0.0058$

Q=2.8 l/s

$Q=2.8\times 10^{-3]$

We know that Q=AV

$Q=A_1V_1=A_2V_2$

$A_1=1.95\times 10^{-3}m^2$

$A_2=4.38\times 10^{-3} m^2$

$So V_2=0.63 m/s,V_1=1.43 m/s$

head loss (h)

$h=\dfrac{f_1L_1V_1^2}{2gd_1}+\dfrac{f_2L_2V_2^2}{2gd_2}+0.5\dfrac{V_1^2}{2g}$

Now putting the all values

$h=\dfrac{0.0048\times 15\times 1.43^2}{2\times 9.81\times 0.05}+\dfrac{0.0058\times 24\times 0.63^2}{2\times 9.81\times 0.075}+0.5\dfrac{1.43^2}{2\times 9.81}$

So h=0.23 m

So the difference of head in the level of reservoir is 0.23 m.

8 0

3 years ago