What are common names assigned to instruction addresses in a PLC program called

A thin aluminum sheet is placed between two very large parallel plates that are maintained at uniform temperatures T1 = 900 K, T

Maru [420]

The net radiation heat transfer between the two plates per unit surface area of the plates with shield and without shied are respectively; 2282.76 W/m² and 9766.75 W/m²

<h3>How to find the net radiation heat transfer?</h3>

We are given;

Temperature 1; T₁

Temperature 2; T₂

Temperature 3; T₃

Emissivity 1; ε₁ = 0.3

Emissivity 2; ε₂ = 0.7

Emissivity 3; ε₃ = 0.2

The net rate of radiation heat transfer with a thin aluminum shield per unit area of the plates with shield is;

Q'₁₂ = σ(T₁⁴ - T₂⁴)]/[((1/ε₁) + (1/ε₂) - 1) + ((1/ε₃,₁) + (1/ε₃,₂) - 1)]

Q'₁₂ = 5.67 * 10⁻⁸(900⁴ - 300⁴)/[((1/0.3) + (1/0.7) - 1) + ((1/0.15) + (1/0.15) - 1)]

Q'₁₂,shield = 2282.76 W/m²

The net rate of radiation heat transfer with a thin aluminum shield per unit area of the plates with no shield is;

Q'₁₂,no shield = σ(T₁⁴ - T₂⁴)]/((1/ε₁) + (1/ε₂) - 1))

Q'₁₂,no shield = 5.67 * 10⁻⁸(900⁴ - 300⁴)/[(1/0.3) + (1/0.7) - 1)]

Q'₁₂,no shield = 9766.75 W/m²

Then the ratio of radiation heat transfer for the two cases becomes;

Q'₁₂,shield/Q'₁₂,no shield = 2282.76/9766.75 = 0.2337 or 4/17

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8 0

2 years ago

A gas cylinder system is composed of a cylinder fitted with a piston on which small weights are placed. Given that the initial p

Reika [66]

Answer:

the answer is going to be a I think

4 0

4 years ago

There are 20 forging presses in the forge shop of a small company. The shop produces batches of forgings requiring a setup time

Aleksandr-060686 [28]

Answer:

Considering the guidelines of this exercise.

The pieces produced per month are 504 000

The productivity ratio is 75%

Explanation:

To understand this answer we need to analyze the problem. First of all, we can only produce 2 batches of production by the press because we require 3 hours to set it up. So if we rest those 6 hours from the 8 of the shift we get 6, leaving 2 for an incomplete bath. So multiplying 2 batches per day of production by press we obtain 40 batches per day. So, considering we work in this factory for 21 days per month well that makes 40 x 21 making 840 then we multiply the batches for the pieces 840 x 600 obtaining 504000 pieces produced per month. To obtain the productivity ratio we need to divide the standard labor hours meaning 6 by the amount of time worked meaning 8. Obtaining 75% efficiency.

4 0

3 years ago

The flow rate of liquid metal into the downsprue of a mold = 0.7 L/sec. The cross-sectional area at the top of the sprue = 750 m

katrin [286]

Answer:

367.43 mm²

Explanation:

Given:

Flow rate, Q = 0.7 L/s

1000 L = 1 m³ = 10⁹ mm³

thus,

1 L = 10⁶ mm³

Therefore,

Q = 0.7 × 10⁶ mm³/s

Cross-sectional area at the top of the sprue = 750 mm²

Length of the sprue = 185 mm

Now,

Velocity = $\sqrt{2gh}$

where, g is the acceleration due to gravity = 9.81 m/s²

h is the height through which flow is taking place = 185 mm = 0.185

thus,

Velocity = $\sqrt{2\times9.81\times0.185}$

or

velocity = 1.9051 m/s = 1905.1 mm/s

Also,

Q = Area × Velocity

thus,

0.7 × 10⁶ = Area × 1905.1

or

Area = 367.43 mm²

3 0

3 years ago

"Given a nodal delay of 84.1ms when there is no traffic on the network (i.e. usage = 0%), what is the effective delay when netwo

shtirl [24]

Answer:

Explanation:

effective delay = delay when no traffic x $\frac{100}{100- network\r usage}$

effective delay = $84.1 \times \frac{100}{100-39.3}=138.55024711697ms$

4 0

3 years ago