What type of intersection is this?

Give two causes that can result in surface cracking on extruded products.

Andreas93 [3]

Answer:

1. High friction

2. High extrusion temperature

Explanation:

Surface cracking on extruded products are defects or breakage on the surface of the extruded parts. Such cracks are inter granular.

Surface cracking defects arises from very high work piece temperature that develops cracks on the surface of the work piece. Surface cracking appears when the extrusion speed is very high, that results in high strain rates and generates heat.

Other factors include very high friction that contributes to surface cracking an d chilling of the surface of high temperature billets.

6 0

3 years ago

A large heat pump should upgrade 5 MW of heat at 85°C to be delivered as heat at 150°C. Suppose the actual heat pump has a COP o

AysviL [449]

Answer:

W=2 MW

Explanation:

Given that

COP= 2.5

Heat extracted from 85°C

Qa= 5 MW

Lets heat supplied at 150°C = Qr

The power input to heat pump = W

From first law of thermodynamics

Qr= Qa+ W

We know that COP of heat pump given as

$COP=\dfrac{Qr}{W}$

$2.5=\dfrac{5}{W}$

W=2 MW

For Carnot heat pump

$COP=\dfrac{T_2}{T_2-T_1}$

$2.5=\dfrac{T_2}{T_2-(273+85)}$

2.5 T₂ - 895= T₂

T₂=596.66 K

T₂=323.6 °C

7 0

3 years ago

Write down the types and tasks of the pressure control valves ?

Yuki888 [10]

Answer:

There are 6 types of pressure control valves and their function is to regulate the pressure below a threshold level within safe limits and to maintain and control pressure of a particular circuit.

Explanation:

The six type of Pressure valve with their functions are given below:

a. Unloading Valve:

These type of pressure valve are used to pour fluid into the container at very low or no pressure.

b. Safety valve:

These are used when the pressure within the vessel is in excess as inside temperature is near about preset [point point then these valves are open to release the extra pressure and are closed once normal conditions are regained.

c. Pressure Reducing Valve:

These are basically used for the control of the pressure in downstream not exceeding the design limits.

d. Pressure Relief Valves:

These are basically used to limit and regulate the pressure of any system.

e. Counter Balance Valve:

These are used to develop pressure in the reverse direction at the actuator's return line in order to keep the load under control.

f. Sequence Valve:

These are used to maintain sequence or order in the operations of two parts or branches.

8 0

3 years ago

PLZ ASAP WILL GIVE BRAINLIST

gavmur [86]

Answer: A, B, C & F (interacting w computers, making decisions & solving problems, evaluating information & getting information).

Explanation: Those are the correct & verified answers.

7 0

3 years ago

Read 2 more answers

(3) Calculate the heat flux through a sheet of brass 7.5 mm (0.30 in.) thick if the temperatures at the two faces are 150°Cand 5

bezimeni [28]

Answer:

a.) 1.453MW/m2, b.) 2,477,933.33 BTU/hr c.) 22,733.33 BTU/hr d.) 1,238,966.67 BTU/hr

Explanation:

Heat flux is the rate at which thermal (heat) energy is transferred per unit surface area. It is measured in W/m2

Heat transfer(loss or gain) is unit of energy per unit time. It is measured in W or BTU/hr

1W = 3.41 BTU/hr

Given parameters:

thickness, t = 7.5mm = 7.5/1000 = 0.0075m

Temperatures 150 C = 150 + 273 = 423 K

50 C = 50 + 273 = 323 K

Temperature difference, T = 423 - 323 = 100 K

We are assuming steady heat flow;

a.) Heat flux, Q" = kT/t

K= thermal conductivity of the material

The thermal conductivity of brass, k = 109.0 W/m.K

Heat flux, $Q" = \frac{109 * 100}{0.0075} = 1,453,333.33 W/m^{2} \\ Heat flux, Q" = 1.453MW/m^{2} \\$

b.) Area of sheet, A = 0.5m2

Heat loss, Q = kAT/t

Heat loss, $Q = \frac{109*0.5*100}{0.0075} = 726,666.667W$

Heat loss, Q = 726,666.667 * 3.41 = 2,477,933.33 BTU/hr

c.) Material is now given as soda lime glass.

Thermal conductivity of soda lime glass, k is approximately 1W/m.K

Heat loss, $Q=\frac{1*0.5*100}{0.0075} = 6,666.67W$

Heat loss, Q = 6,666.67 * 3.41 = 22,733.33 BTU/hr

d.) Thickness, t is given as 15mm = 15/1000 = 0.015m

Heat loss, $Q=\frac{109*0.5*100}{0.015} =363,333.33W$

Heat loss, Q = 363,333.33 * 3.41 = 1,238,966.67 BTU/hr

5 0

3 years ago