Ask question

Ask question

All categories

3 years ago

9

A 2.0-in-thick slab is 10.0 in wide and 12.0 ft long. Thickness is to be reduced in three steps in a hot rolling operation. Each

step will reduce the slab to 75% of its previous thickness. It is expected that for this metal and reduction, the slab will widen by 3% in each step. If the entry speed of the slab in the first step is 40 ft/min, and roll speed is the same for the three steps, determine: (a) length and (b) exit velocity of the slab after the final reduction

1 answer:

DerKrebs [107]3 years ago

5 0

Answer:

26.02 ft

86.7690 ft/min

Explanation:

After 3 steps

0.75³(2.0 thickness)

T = 0.84375

W = (1+0.03)³10

= 10.92727 inches

A To get length

2.0 x 10 x 12 x12 = 0.84375 x 10.92727x lf

= 2880 = 9.21988Lf

Lf = 2880/9.21988

= 312.368 inches

Convert to feet

322.368 x 0.0833

= 26.02 ft

B.

= 2 x 10 x 40 = 0.84375 x 10.92727 x vf

800 = 9.21988vf

Vf = 800/9.21988

Vf = 86.7690 ft/min

You might be interested in

A Michelson interferometer operating at a 500 nm wavelength has a 3.73-cm-long glass cell in one arm. To begin, the air is pumpe

LekaFEV [45]

Answer:

The number of bright-dark fringe is 42

Solution:

As per the question:

Wavelength of light, $\lambda = 500\ nm = 500\times 10^{- 9}\ m$

Length of the glass cell, x = 3.73 cm = 0.0373 m

Refractive index, $\mu = 1.00028$

Now,

To calculate the bright-dark fringe shifts, we use the formula given below:

$d_{m} = \frac{2x}{\lambda }\times (\mu - 1)$

Now, substituting the appropriate values in the above formula:

$d_{m} = \frac{2\times 0.0373}{500\times 10^{- 9}}\times (1.00028 - 1)$

$d_{m} = 41.77$ ≈ 42

6 0

3 years ago

Are engineers needed in today’s society ? Why or why not ? I need a short three paragraph essay !!! Please help me !!!

masha68 [24]

Of course they are needed because without them the society wouldn’t be as nice as it is right now and plus there would be no more buildings ! :)

8 0

3 years ago

Define a separate subroutine for each of the following tasks respectively.

Valentin [98]

Answer:

I HAVE NO CLUEEEE

Explanation:

???????????????????/

5 0

3 years ago

Read 2 more answers

Technician A says independent shops are not affiliated with vehicle manufacturers, but it is easy for technicians who work in th

KatRina [158]

Answer:

b

Explanation:

i did it yeater dayajsbs

8 0

3 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

4 years ago