An air duct heater consists of an aligned array of electrical heating elements in which the longitudinal and transverse pitches

Answer:

The change in entropy of the steam is 2.673 kJ/K

Explanation:

The mass of the liquid-vapor mixture is 1.5 kg

The mass in the liquid phase is calculated as 3/4 × 1.5 kg = 1.125 kg

The mass in the vapor phase is calculated as 1.5 - 1.125 = 0.375 kg

According to the steam tables

At a pressure of 200 kPa (200/100 = 2 bar), the specific entropy of steam is found to be 7.127 kJ/kgK

The entropy of steam can be calculated as specific entropy multiplied by mass = 7.127 × 0.375 = 2.673 kJ/K

An item of protective gear that shields individuals passing by from stray sparks or metal during the welding process performed by another worker is known as: E. Welding Screens.

An operator is a person tasked with joining two or more metals using a technique called wielding.

In the course of wielding, both sparks and tiny metallic fragments are released, which pose a danger to the operator and others working nearby.

As a result, the equipment outlined below should be worn or utilized directly by a worker actuating the wielding process:

• Visors.

• Goggles.

• Protective clothing.

• Dark walls.

Nonetheless, a type of protective gear that defends other workers nearby from stray sparks or metallic fragments while the operator (worker) is in the act of welding is called welding screens.

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Answer:

The velocity at exit U_2 is 578.359 m/s

The exit diameter d_e is 1.4924 cm

Explanation:

Provided data includes:

Length of the nozzle L = 25 cm

Inlet diameter d_i = 5 cm

At the nozzle entrance (state 1): Temperature T_1 = 325 °C, Pressure P_1 = 700 kPa, Velocity U_1 = 30 m/s, Enthalpy H_1 = 3112.5 kJ/kg, Volume V_1 = 388.61 cm³/gAt the nozzle exit (state 2): Temperature T_2 = 250 °C, Pressure P_2 = 350 kPa, Velocity U_2, Enthalpy H_2 = 2945.7 kJ/kg, Volume V_2 = 667.75 cm³/g To determine:a. Exit Velocity U_2b. Exit Diameter d_e

a.The Energy Equation can be represented by:ΔH + ΔU² / 2 + gΔz = Q + WAssuming Q = W = Δz = 0Substituting the values yields:

(H_2 - H_1) + (U²_2 - U²_1)  / 2 = 0From which we can derive U_2 = sqrt((2* (H_1 - H_2 )) + U²_1) with the calculations leading to U_2 = sqrt ( 2 * 10^3 * (3112.5 -2945.7) + 900) yielding U_2 = 578.359 m/s

b.

Using mass balance approach, we have U_1 * A_1 / V_1 = U_2 * A_2 / V_2

This leads to U_1 * d_i² / V_1 = U_2 * d_e² / V_2, thus d_e = d_i * sqrt((U_1 / U_2) * (V_2 / V_1)). Hence, d_e = 5 * sqrt((30 / 578.359) * (667.75 / 388.61)) computes to d_e = 1.4924 cm

Response:

The cutting speed is calculated at 365.71 m/min

Clarification:

Given parameters include

diameter D = 250 mm

length L = 625 mm

Feed f = 0.30 mm/rev

cut depth = 2.5 mm

n = 0.25

C = 700

To find

the cutting speed that ensures the tool life coincides with the cutting time for the three parts

The formula for cutting time is given as

Tc =

where D refers to diameter, L refers to length and f refers to feed while V represents speed

Tc =

Tc =

Given the tool life is expressed as

T = 3 × Tc............................2

where T denotes tool life and Tc is the cutting duration

Calculating tool life by substituting values into equation 2 yields

T = 3 ×

This yields V = 365.71

Thus, the cutting speed calculates to 365.71 m/min