# 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 steel plate ( 7830 kg/m3 , c 550 J/kg K, k 48 W/m K), which is initially at a uniform temperature of Ti 200 C and is to be heated to a minimum temperature of 550 C. Heating is effected in a gas-fired furnace, where products of combustion at T 800 C maintain a convection coefficient of h 250 W/m2 K on both surfaces of the plate. How long should the plate be left in the furnace

T = 858.25 s

Explanation:

Given data:

Reheat stage for a 100-mm-thick steel plate ( 7830 kg/m3,  c 550 J/kg K, k 48 W/m K),

initial uniform temperature ( Ti ) = 200 c

Final temperature = 550 c

convection coefficient  = 250 w/m^2 k

products combustion temp = 800 c

calculate how long the plate should be left in the furnace ( to attain 550 c )

first calculate/determine the Fourier series Number ( Fo )

= 0.4167 =

therefore Fo =  3.8264

Now determine how long the plate should be left in the furnace

Fo =

k = 48

p = 7830

L = 0.1

Input the values into the relation and make t subject of the formula

hence t = 858.25 s

## Related Questions

The compressibility factor provides a quick way to assess when the ideal gas law is valid. Use a solver to find the minimum temperature where the fluid has a vapor phase compressibility factor greater than 0.95 at 3 MPa. Report the value in oC, without units.

Explanation:

The compressibility factor

A site is compacted in the field, and the dry unit weight of the compacted soil (in the field) is determined to be 18 kN/m3. Determine the relative compaction if the maximum dry unit weight was measured to be 17 kN/m3. Express your answer as a percentage (but do not write the percentage sign in the answer box).

the relative compaction is 105.88 %

Explanation:

Given;

dry unit weight of field compaction, = 18 kN/m³

maximum dry unit weight measured, = 17 kN/m³

Relative compaction (RC) of the site is given as the ratio of dry unit weight of field compaction and maximum dry unit weight measured

Relative compaction (RC) = dry unit weight of field compaction / maximum dry unit weight measured

substitute the given values;

RC (%) = 105.88 %

Therefore, the relative compaction is 105.88 %

How much heat is lost through a 3’× 5' single-pane window with a storm that is exposed to a 60°F temperature differential?A. 450 Btu/hB. 900 Btu/hC. 1350 Btu/hD. 1800 Btu/h

A. 450 btu/h

Explanation:

We solve this problem by using this formula:

Q = U x TD x area

U = U value of used material

TD = Temperature difference = 60°

Q = heat loss

Area = 3x5 = 15

We first find U

R = 1/u

2 = 1/U

U = 1/2 = 0.5

Then when we put these values into the formula above, we would have:

Q = 0.5 x 15 x 60

Q = 450Btu/h

Under the right conditions, it is possible, due to surface tension,to have metal objects float on water. Consider placing a shortlength of a small diameter steel ( γ = 490 lb/ft3)rod on a surface of water. What is the maximum diameter that therod can have before it will sink? Assume that the surface tensionforces act vertically upward. Note: A standard paper cliphas a diameter of 0.036 in. Partially unfold a paper clip and seeif you can get it to float on water. Do the results of thisexperiment support your analysis?

A) 0.0614 inches

b) The standard steel paper clip should float on water

Explanation:

The maximum diameter that the rod can have before it will sink

we can calculate this using this formula :

D = ----- 1

∝ = value of surface tension of water at 60⁰f  = 5.03×10^−3  lb/ft

y = 490 Ib/ft^3

input the given values into equation 1 above

D =

= 5.11 * 10^-3 ft   convert to inches

= 5.11 *10^-3 ( 12 in/ 1 ft ) = 0.0614 inches

B) The diameter of a standard paper Cliphas = 0.036 inches

and the diameter of the rod = 0.0614. Hence the standard steel paper clip should float on water

Why or why not the following materials will make good candidates for the construction of a) Turbine blade for a jet turbine and
b) Thermal barrier coating. Cement, aluminum, engineering ceramic, super alloy, steel and glass

Explanation:

3.] a] A turbine blade is the individual component which makes up the turbine section of a gas turbine. The blades are responsible for extracting energy from the high temperature, high pressure gas produced by the combustor.

The turbine blades are often the limiting component of gas turbines. To survive in this difficult environment, turbine blades often use exotic materials like superalloys and many different methods of cooling, such as internal air channels, boundary layer cooling, and thermal barrier coatings. The blade fatigue failure is one of the major source of outages in any steam turbines and gas turbines which is due to high dynamic stresses caused by blade vibration and resonance within the operating range of machinery.

To protect blades from these high dynamic stresses, friction dampers are used.

b] Thermal barrier coatings (TBC) are highly advanced materials systems usually applied to metallic surfaces, such as on gas turbine or aero-engine parts, operating at elevated temperatures, as a form ofexhaust heat management.

These 100μm to 2mm coatings serve to insulate components from large and prolonged heat loads by utilizing thermally insulating materials which can sustain an appreciable temperature difference between the load-bearing alloys and the coating surface.

In doing so, these coatings can allow for higher operating temperatures while limiting the thermal exposure of structural components, extending part life by reducing oxidation and thermal fatigue.

In conjunction with active film cooling, TBCs permit working fluid temperatures higher than the melting point of the metal airfoil in some turbine applications.

Due to increasing demand for higher engine operation (efficiency increases at higher temperatures), better durability/lifetime, and thinner coatings to reduce parasitic weight for rotating/moving components, there is great motivation to develop new and advanced TBCs.

An aquifer has three different formations. Formation A has a thickness of 8.0 m and hydraulic conductivity of 25.0 m/d. Formation B has a thickness of 2.0 m and a conductivity of 142 m/d. Formation C has a thickness of 34 m and a conductivity of 40 m/d. Assume that each formation is isotropic and homogeneous. Compute both the overall horizontal and vertical conductivities.

The horizontal conductivity is 41.9 m/d.

The vertical conductivity is 37.2 m/d.

Explanation:

Given that,

Thickness of A = 8.0 m

Conductivity = 25.0 m/d

Thickness of B = 2.0 m

Conductivity = 142 m/d

Thickness of C = 34 m

Conductivity = 40 m/d

We need to calculate the horizontal conductivity

Using formula of horizontal conductivity

Put the value into the formula

We need to calculate the vertical conductivity

Using formula of vertical conductivity

Put the value into the formula

Hence, The horizontal conductivity is 41.9 m/d.

The vertical conductivity is 37.2 m/d.