1. Add:<br>(i) 5xy, -2xy, -11xy, 8xy<br>(iv) 3a

Annealing is a process by which steel is reheated and then cooled to make it less brittle. Consider the reheat stage for a 100-m

kodGreya [7K]

Complete question is;

Annealing is a process by which steel is reheated and then cooled to make it less brittle. Consider the reheat stage for a 100 mm thick plate (ρ = 7830 kg/m3, Cp = 550 J/kg K, k = 48 W/m K). The plate initially is at 200 °C and is to be heated to a minimum temperature of 550 °C. Heating is effected in a gas-fired furnace where the products of combustion at T∞ = 800 °C maintain a convection heat transfer coefficient of h = 250 W/m.K on both surfaces of the plate. How long should the plate be left in the furnace?

Answer:

860 seconds

Explanation:

We are given;

Initial Temperature; Ti = 200 °C

Minimum Temperature; T_i = 550 °C

T∞ = 800 °C

convection coefficient; h = 250 W/m².K

ρ = 7830 kg/m³

Cp = 550 J/kg K

k = 48 W/m K

Plate thickness = 100mm

Thus,L = 100/2 = 50mm = 0.05 m

Let's find the biot number from the formula;

Bi = hL/K

Bi = (250 × 0.05)/48

Bi = 0.2604

Now, lowest temperature in the slab is given as;

θ_o = (T_o - T∞)/(T_i - T∞)

θ_o = (550 - 800)/(200 - 800)

θ_o = 0.4167

Now, from online tables calculation, we can find the root of the biot number.

Thus, root of the biot number Bi = 0.2604 is;

ζ1 = 0.488 rad

Also, C1 is gotten as 1.0396

Now,formula for thermal diffusivity is;

α = k/ρc

α = 48/(7830 × 550)

α = 1.115 × 10^(-5) m²/s

Also, from online tables, f(ζ1) = 0.401

Thus, we can find the time the plate should the plate be left in the furnace from;

-(ζ1)²(αt/L²) = In 0.401

-(ζ1)²(αt/L²) = -0.9138

t = (-0.9138 × 0.05²)/-(0.488² × 1.115 × 10^(-5))

t ≈ 860 s

8 0

2 years ago

A 150-lbm astronaut took his bathroom scale (a spring scale) and a beam scale (compares masses) to the moon where the local grav

kozerog [31]

Answer:

a)Wt =25.68 lbf

b)Wt = 150 lbf

F= 899.59 N

Explanation:

Given that

$g = 5.48 ft/s^2.$

m= 150 lbm

a)

Weight on the spring scale(Wt) = m g

We know that

$1\ lbf=32.17 \ lmb.ft/s^2$

Wt = 150 x 5.48/32 lbf

Wt =25.68 lbf

b)

On the beam scale

This is scale which does not affects by gravitational acceleration.So the wight on the beam scale will be 150 lbf.

Wt = 150 lbf

If the plane is moving upward with acceleration 6 g's then the for F

F = m a

We know that

$1\ ft/s^2= 0.304\ m/s^2$

$5.48\ ft/s^2= 1.66\ m/s^2$

a=6 g's

$a=9.99\ m/s^2$

So

F = 90 x 9.99 N

F= 899.59 N

3 0

2 years ago

A double-acting duplex pump with 6.5-in. liners, 2.5-in. rods, and 18-in. strokes was operated at 3,000 psig and 20 cycles/min.

Stella [2.4K]

Answer:

Pump factor = Fp = 7.854 gal/cycle

Ev = 82.00 %

P_H = 183.29 hp

Explanation:

Given data:

Dimension of duplex pump

6.5 inch liner

2.5 inch rod

18 inch strokes

Pressure 3000 psig

Pit dimension

7 ft wide

20 ft long

Ls = 18 inch

Velocity = (18)/10

volumetric efficiency is given as E_v = (Actual flow rate)/(Theortical flow rate) * 100

we know that flow rate is given as = Area * velocity

Theoritical flow rate $= \frac{\pi}{2}\times Ls(2d_l^2 - d_r^2)\times N$

$Ev = \frac{7\times 12 \times 20\times 12\times 12 \times \frac{18}{10} inch^3/min}{\frac{\pi}{2} \times 18 (2\times 6.5^2 -2.5^2) \times 20}$

Ev = 82.00 %

Pump factor $Fp = = \frac{\pi}{2}\times Ls(2d_l^2 - d_r^2)\times Ev$

$Fp =\frac{\pi}{2} \times 18 (2\times 6.5^2 -2.5^2) \times 0.82$

Fp = 1814.22 in^3/cyl

Fp = 7.854 gal/cycle

Flow rate $q = NFp = 20 \times 7.854 = 157.08 gal/min$

Power $Ph = \frac{\DeltaP q}{1714} = \frac{3000 \times 157.08}{1714} = 274.93 hp$

6 0

2 years ago

A cylindrical specimen of a brass alloy having a length of 60 mm (2.36 in.) must elongate only 10.8 mm (0.425 in.) when a tensil

Gemiola [76]

The radius of the specimen is 60 mm

<u>Explanation:</u>

Given-

Length, L = 60 mm

Elongated length, l = 10.8 mm

Load, F = 50,000 N

radius, r = ?

We are supposed to calculate the radius of a cylindrical brass specimen in order to produce an elongation of 10.8 mm when a load of 50,000 N is applied. It is necessary to compute the strain corresponding to this

elongation using Equation:

ε = Δl / l₀

ε = 10.8 / 60

ε = 0.18

We know,

σ = F / A

Where A = πr²

According to the stress-strain curve of brass alloy,

σ = 440 MPa

Thus,

$sigma = 50,000 / \pi (r)^2\\\\440 X 10^6 = \frac{50,000}{3.14 X (r)^2}\\\\r = 0.06m\\r = 60mm\\\\\\$

Therefore, the radius of the specimen is 60 mm

3 0

2 years ago

A conical enlargement in a vertical pipeline is 5 ft long and enlarges the pipe diameter from 12 in. to 24 in. diameter. Calcula

makkiz [27]

Answer:

F_y = 151319.01N = 15.132 KN

Explanation:

From the linear momentum equation theory, since flow is steady, the y components would be;

-V1•ρ1•V1•A1 + V2•ρ2•V2•A2 = P1•A1 - P2•A2 - F_y

We are given;

Length; L = 5ft = 1.52.

Initial diameter;d1 = 12in = 0.3m

Exit diameter; d2 = 24 in = 0.6m

Volume flow rate of water; Q2 = 10 ft³/s = 0.28 m³/s

Initial pressure;p1 = 30 psi = 206843 pa

Thus,

initial Area;A1 = π•d1²/4 = π•0.3²/4 = 0.07 m²

Exit area;A2 = π•d2²/4 = π•0.6²/4 = 0.28m²

Now, we know that volume flow rate of water is given by; Q = A•V

Thus,

At exit, Q2 = A2•V2

So, 0.28 = 0.28•V2

So,V2 = 1 m/s

When flow is incompressible, we often say that ;

Initial mass flow rate = exit mass flow rate.

Thus,

ρ1 = ρ2 = 1000 kg/m³

Density of water is 1000 kg/m³

And A1•V1 = A2•V2

So, V1 = A2•V2/A1

So, V1 = 0.28 x 1/0.07

V1 = 4 m/s

So, from initial equation of y components;

-V1•ρ1•V1•A1 + V2•ρ2•V2•A2 = P1•A1 - P2•A2 - F_y

Where F_y is vertical force of enlargement pressure and P2 = 0

Thus, making F_y the subject;

F_y = P1•A1 + V1•ρ1•V1•A1 - V2•ρ2•V2•A2

Plugging in the relevant values to get;

F_y = (206843 x 0.07) + (1² x 1000 x 0.07) - (4² x 1000 x 0.28)

F_y = 151319.01N = 15.132 KN

6 0

2 years ago

1. Add:(i) 5xy, -2xy, -11xy, 8xy(iv) 3a - 2b + c, 5a + 8b -70​

1. Add:(i) 5xy, -2xy, -11xy, 8xy(iv) 3a - 2b + c, 5a + 8b -70