Which of the following is NOT a task role within a team?

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 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 part of diameter d is loaded by an axial force P. This causes a stress of P/A, where A = πd2/4. If the load is kno

Afina-wow [57]

Answer:

1.505

Explanation:

cylindrical part of diameter d is loaded by an axial force P. This causes a stress of P/A, where A = πd2/4. If the load is known with an uncertainty of ±11 percent, the diameter is known within ±4 percent (tolerances), and the stress that causes failure (strength) is known within ±20 percent, determine the minimum design factor that will guarantee that the part will not fail.

stress is force per unit area

stress=P/A

A = πd^2/4.

uncertainty of axial force P= +/-.11

s=+/-.20, strength

d=+/-.04 diameter

fail load/max allowed

minimum design=fail load/max allowed

minimum design =s/(P/A)

sA/P

A=( $\pi$ .96d^2)/4, so Amin=

$0.96^{2}$ (because the diameter at minimum is (1-0.04=0.96)

minimum design=Pmax/(sminxAmin)

1.11/(.80*.96^2)=

1.505

8 0

2 years ago

The outer surface of an engine is situated in a place where oil leakage can occur. When leaked oil comes in contact with ahot su

VladimirAG [237]

Answer:

TBC thickness of 4 mm is insufficient to prevent fire hazard

Explanation:

Given:

- Temperature of hot-fluid inner surface T_i = 333°C

- The convection coefficient hot-fluid h_i = 7 W/m^2K

- The thermal conductivity of engine cover k_1 = 14 W/mK

- The thickness of engine cover L_1 = 0.01 m

- The thermal conductivity of TBC layer k_2 = 1.1 W/mK ... (Typing error)

- The thickness of TBC layer L_2 = 0.004 m

- Temperature of ambient air outer surface T_o = 69°C

- The convection coefficient ambient air h_o = 7 W/m^2K

Find:

Would a TBC layer of 4 mm thickness be sufficient to keep the engine cover surface below autoignition temperature of 200°C to prevent fire hazard?

Solution:

- We will use thermal circuit analogy for the 1-D problem and steady state conduction with no heat generation in the cover or TBC layer.

The temperature at each medium interface and the Thermal resistance for each medium is given in the attachment schematic and circuit analogy.

- We will calculate the total heat flux for the entire system q:

q = ( T_i - T_o ) / R_total

- R_total is the equivalent thermal resistance of the entire circuit. Since all resistances are in series we have:

R_total = 1 / h_i + L_1 / k_1 + L_2 / k_2 + 1 / h_o

- Plug in the values and compute:

R_total = 1 / 7 + 0.01 / 14 + 0.004 / 1.1 + 1 / 7

R_total = 0.2900649351 T-m^2 / W

- Calculate the Total heat flux q:

q = ( 333 - 69 ) / 0.2900649351

q = 910.141 W / m^2

- Just like the total current in a circuit remains same, the total heat flux remains same. We will use the total heat flux q to calculate the temperature of outer engine surface T_2 as follows:

q = ( T_i - T_2 ) / R_i2

Where,

R_i2 = 1 / h_i + L_1 / k_1

R_i2 = 1 / 7 + 0.01 / 14 = 0.14357 T-m^2 / W

Hence,

( T_i - T_2 ) = q*R_i2

T_2 = T_i - q*R_i2

Plug the values in:

T_2 = 333 - 910.141*0.14357

T_2 = 202.33°C

- The outer surface of the engine cover has a temperature above T_ignition = 200°C. Hence, the TBC thickness of 4 mm is insufficient to prevent fire hazard

3 0

2 years ago

In a planetary geartrain with a form factor of 8, the sun gear rotates clockwise at 5 rad⁄s and the ring gear rotates clockwise

lina2011 [118]

Answer:

D. N= 11. 22 rad/s (CW)

Explanation:

Given that

Form factor R = 8

Speed of sun gear = 5 rad/s (CW)

Speed of ring gear = 12 rad/s (CW)

Lets take speed of carrier gear is N

From Algebraic method ,the relationship between speed and form factor given as follows

$\dfrac{N_{sun}-N}{N_{ring}-N}=-R$

here negative sign means that ring and sun gear rotates in opposite direction

Lets take CW as positive and ACW as negative.

Now by putting the values

$\dfrac{N_{sun}-N}{N_{ring}-N}=-R$

$\dfrac{5-N}{12-N}=-8$

N= 11. 22 rad/s (CW)

So the speed of carrier gear is 11.22 rad/s clockwise.

8 0

2 years ago