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Slav-nsk [51]
3 years ago
13

A heat pump is used to warm a building in the winter. This system implements an ideal vapor-compression cycle with R-134a as the

working fluid. The cycle operates between 200 kPa and 800 kPa, with a mass flow rate of 0.02 kg/s. Determine: (a) the heating load that can be met kW (b) the power input required for this system kW (c) the coefficient of performance (d) the warmest outside temperature at which this particular cycle is unable to operate ∘C
Engineering
1 answer:
Marrrta [24]3 years ago
5 0

Answer:

See attachment for detailed answer.

Explanation:

Download pdf
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What are the three elementary parts of a vibrating system?
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Answer:

the three part are mass, spring, damping

Explanation:

vibrating system consist of three elementary system namely

1) Mass - it is a rigid body due to which system experience vibration and kinetic energy due to vibration is directly proportional to velocity of the body.

2) Spring -  the part that has elasticity and help to hold mass

3) Damping - this part considered to have zero mass and  zero elasticity.

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Those in this career install, maintain, and repair electrical wiring,
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What might cause a shotgun to explode?
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When you pull the trigger to shoot a shotshell from a shotgun or a cartridge from a rifle or handgun, the firing pin strikes the primer in the base of the cartridge or shotshell. This causes the primer to explode. The spark from the primer ignites the gunpowder, which burns rapidly and converts to a gas.

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Determine whether or not each of the following four transaction execution histories is serializable. If a history is serializabl
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Option D. w1[x] w2[u] w2[y] w1[y] w3[x] w3[u] w1[z]

Explanation:

The execution in the option D is correct. This is because there is more than one reasonable criterion.

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3 years ago
The dam cross section is an equilateral triangle, with a side length, L, of 50 m. Its width into the paper, b, is 100 m. The dam
lisabon 2012 [21]

Answer:

Explanation:

In an equilateral trinagle the center of mass is at 1/3 of the height and horizontally centered.

We can consider that the weigth applies a torque of T = W*b/2 on the right corner, being W the weight and b the base of the triangle.

The weigth depends on the size and specific gravity.

W = 1/2 * b * h * L * SG

Then

Teq = 1/2 * b * h * L * SG * b / 2

Teq = 1/4 * b^2 * h * L * SG

The water would apply a torque of elements of pressure integrated over the area and multiplied by the height at which they are apllied:

T1 = \int\limits^h_0 {p(y) * sin(30) * L * (h-y)} \, dy

The term sin(30) is because of the slope of the wall

The pressure of water is:

p(y) = SGw * (h - y)

Then:

T1 = \int\limits^h_0 {SGw * (h-y) * sin(30) * L * (h-y)} \, dy

T1 = \int\limits^h_0 {SGw * sin(30) * L * (h-y)^2} \, dy

T1 = SGw * sin(30) * L * \int\limits^h_0 {(h-y)^2} \, dy

T1 = SGw * sin(30) * L * \int\limits^h_0 {(h-y)^2} \, dy

T1 = SGw * sin(30) * L * \int\limits^h_0 {h^2 - 2*h*y + y^2} \, dy

T1 = SGw * sin(30) * L * (h^2*y - h*y^2 + 1/3*y^3)(evaluated between 0 and h)

T1 = SGw * sin(30) * L * (h^2*h - h*h^2 + 1/3*h^3)

T1 = SGw * sin(30) * L * (h^3 - h^3 + 1/3*h^3)

T1 = 1/3 * SGw * sin(30) * L * h^3

To remain stable the equilibrant torque (Teq) must be of larger magnitude than the water pressure torque (T1)

1/4 * b^2 * h * L * SG > 1/3 * SGw * sin(30) * L * h^3

In an equilateral triangle h = b * cos(30)

1/4 * b^3 * cos(30) * L * SG  > 1/3 * SGw * sin(30) * L * b^3 * (cos(30))^3

SG > SGw * 4/3* sin(30) * (cos(30))^2

SG > 1/2 * SGw

For the dam to hold, it should have a specific gravity of at leas half the specific gravity of water.

This is avergae specific gravity, including holes.

6 0
3 years ago
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