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galina1969 [7]
2 years ago
8

How is the air delivery temperature controlled during A/C operation?

Engineering
1 answer:
shepuryov [24]2 years ago
3 0
A IS THE CORRECT ANSWER
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Answer: Yes

Explanation: -4 x 2 = 8

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1<br> Create an example of a positive correlation
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A hollow shaft of diameter ratio 3/8 (internal dia to outer dia) is to transmit 375 kW power at 100 rpm. The maximum torque bein
Schach [20]

Answer:

External diameter = 158.15 mm mm

Internal diameter = 59.31 mm

Explanation:

We are given;

Diameter ratio; d_i = ⅜d_o

Where d_i is internal diameter and d_o is external diameter

Power;P = 375 KW = 375000 W

Rotational speed;N = 100 rpm

Max torque is 20% greater than mean torque; T_max = 1.2T_avg

Shear stress;τ = 60 N/mm²

Length; L = 4m = 4000 mm

Angle of twist; θ = 2° = 2π/180 radians

Modulus of rigidity;G = 0.85 X 10^(5) N/mm²

Formula for the power transmitted by the shaft is;

P = 2πNT_avg/60

Plugging in the relevant values, we have ;

375000 = 2π × 100T_avg/60

T_avg = (375000 × 60)/(2π × 100) = 35809.862 N.m = 35809862 N.mm

Since T_max = 1.20T_avg

Thus, T_max = 1.20(35809862) = 42971834.4 N.mm

Checking for strength, we'll use;

τ = Tr/J_p

Or since r = d/2

It can be written as;

τ = T(d_o)/2J_p - - - (1)

Where T is T_max

But Polar moment of inertia of hollow shaft is;

J_p = [π(d_o)⁴ - π(d_i)⁴]/32

Now, we are told that d_i = ⅜d_o

Thus;

J_p = [π(d_o)⁴ - π(⅜d_o)⁴]/32

J_p = (π/32) × d_o⁴(1 - 3⁴/8⁴)

J_p = 0.0926 d_o⁴

Plugging this for J_p in eq 1,we have;

τ = T(d_o)/2(0.0926d_o⁴)

Making d_o the subject gives;

d_o³ = T/(2 × 0.0926τ)

Plugging in the relevant values to give;

d_o³ = 42971834.4/(2 × 0.0926 × 60)

d_o³ = 3867155.7235421166

d_o = ∛3867155.7235421166

d_o = 156.96 mm

Thus, d_i = ⅜ × 156.96 = 58.86 mm

Checking for stiffness, we'll use;

T/J_p = Gθ/L

Again T is T_max

Plugging in the relevant values, we have;

42971834.4/0.0926 d_o⁴ = (0.85 × 10^(5) × 2π/180)/4000

464058686.825054/d_o⁴ = 0.7417649321

d_o⁴ = 464058686.825054/0.7417649321

d_o⁴ = 625614216.5028806

d_o = ∜625614216.5028806

d_o = 158.15 mm

d_i = ⅜ × 158.15 = 59.31 mm

So we will pick the highest values.

Thus;

d_o = 158.15 mm

d_i = 59.31 mm

3 0
3 years ago
Nowadays power supply on board ship is determine low and high voltage. From what range low &amp; high voltage is being determine
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Any Voltage used on board a ship if less than 1kV (1000 V) then it is called as LV (Low Voltage) system and any voltage above 1kV is termed as High Voltage. Typical Marine HV systems operate usually at 3.3kV or 6.6kV.

Explanation:

trust

6 0
1 year ago
C&amp;A's potato chip filling process has a lower specification limit of 9.5 oz. and an upper specification limit of 10.5 oz. Th
Law Incorporation [45]

Answer:

Process capability index = (USL - LSL)/6Sigma = 1/6(0.3)= 0.56

Answer (a) 0.56

Explanation:

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