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Paladinen [302]
3 years ago
7

What is a ton of refrigeration?

Engineering
1 answer:
AURORKA [14]3 years ago
6 0

Explanation:

The unit refrigeration is generally is given in terms of tons.In refrigeration compressor consume some amount of work to produce the cooling effect  with the help of evaporator and condenser.

In the simple words ton is the cooling load of refrigeration system.

So  

1 ton = 3.5 KW

1 ton = 12,000 BTU/hr

 

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A four-cylinder, four-stroke internal combustion engine has a bore of 3.7 in. and a stroke of 3.4 in. The clearance volume is 16
abruzzese [7]

Answer:

1) The three possible assumptions are

a) All processes are reversible internally

b) Air, which is the working fluid circulates continuously in a closed loop

cycle

c) The process of combustion is depicted as a heat addition process

2) The diagrams are attached

5) The net work per cycle is 845.88 kJ/kg

The power developed in horsepower ≈ 45374 hP

Explanation:

1) The three possible assumptions are

a) All processes are reversible internally

b) Air, which is the working fluid circulates continuously in a closed loop

cycle

c) The process of combustion is depicted as a heat addition process

2) The diagrams are attached

5) The dimension of the cylinder bore diameter = 3.7 in. = 0.09398 m

Stroke length = 3.4 in. = 0.08636 m.

The volume of the cylinder v₁= 0.08636 ×(0.09398²)/4 = 5.99×10⁻⁴ m³

The clearance volume = 16% of cylinder volume = 0.16×5.99×10⁻⁴ m³

The clearance volume, v₂  = 9.59 × 10⁻⁵ m³

p₁ = 14.5 lbf/in.² = 99973.981 Pa

T₁ = 60 F = 288.706 K

\dfrac{T_{2}}{T_{1}} = \left (\dfrac{v_{1}}{v_{2}}  \right )^{K-1}

Otto cycle T-S diagram

T₂ = 288.706*6.25^{0.393} = 592.984 K

The maximum temperature = T₃ = 5200 R = 2888.89 K

\dfrac{T_{3}}{T_{4}} = \left (\dfrac{v_{4}}{v_{3}}  \right )^{K-1}

T₄ = 2888.89 / 6.25^{0.393} = 1406.5 K

Work done, W = c_v×(T₃ - T₂) - c_v×(T₄ - T₁)

0.718×(2888.89  - 592.984) - 0.718×(1406.5 - 288.706) = 845.88 kJ/kg

The power developed in an Otto cycle = W×Cycle per second

= 845.88 × 2400 / 60  = 33,835.377 kW = 45373.99 ≈ 45374 hP.

8 0
3 years ago
In using the drag coefficient care needs to be taken to use the correct area when determining the drag force. What is a typical
stealth61 [152]

Answer:

Explanation:

We know that Drag forceF_D

  F_D=\dfrac{1}{2}C_D\rho AV^2

Where

             C_D is the drag force constant.

                 A is the projected area.

                V is the velocity.

                ρ is the density of fluid.

Form the above expression of drag force we can say that drag force depends on the area .So We should need to take care of correct are before finding drag force on body.

Example:

 When we place our hand out of the window in a moving car ,we feel a force in the opposite direction and feel like some one trying to pull our hand .This pulling force is nothing but it is drag force.

6 0
3 years ago
Can you prove that the two bleu areas are the same without numbers please?
Svet_ta [14]

Answer:

\small{\boxed{\tt{\colorbox{green}{✓Verified\:answer}}}}\:

Just draw a line from point D join to point E

The triangle formed DME will be congruent to AMC

6 0
3 years ago
A good rule of thumb in hazardous conditions is to _____.
Aloiza [94]

Answer:

C. Have your hazard lights on

Explanation:

Speeding up will cause an accident

Counter steering is not easy to do

Slowing down my result in you being rear ended

5 0
3 years ago
Read 2 more answers
Find the mechanical average of a wheel axle System of the wheel has a radius of 1.5 feet in the accident has a radius of 6 inche
kvasek [131]

Answer:

Mechanical average of a wheel = 3

Explanation:

Given:

Radius of wheel = 1.5 ft = 1.5 x 12 = 18 inches

Radius of axle = 6 inches

Find:

Mechanical average of a wheel

Computation:

Mechanical average of a wheel = Radius of wheel / Radius of axle

Mechanical average of a wheel = 18 / 6

Mechanical average of a wheel = 3

4 0
3 years ago
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