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olchik [2.2K]
3 years ago
5

A commercial refrigerator with refrigerant-134a as the working fluid is used to keep the refrigerated space at 2308C by rejectin

g its waste heat to cooling water that enters the condenser at 188C at a rate of 0.25 kg/s and leaves at 268C. The refrigerant enters the condenser at 1.2 MPa and 658C and leaves at 428C. The inlet state of the compressor is 60 kPa and 2348C and the compressor is estimated to gain a net heat of 450 W from the surroundings. Determine (a) the quality of the refrigerant at the evaporator inlet, (b) the refrigeration load, (c) the COP of the refrigerator, and (d) the theoretical maximum refrigeration load for the same power input to the compressor.

Engineering
1 answer:
lord [1]3 years ago
5 0

Answer:

hello your question is incomplete attached below is the missing part and also attached is the solution

answer: a) 0.4801

              b) 5.398 kw

              c) 2.14

              d) 12.72

Explanation:

The quality of the refrigerant at the evaporator inlet

h4 = hf4 + x4(hfx4)

Refrigeration load

Ql = m(h1-h4)

COP of the refrigerator

Ql / m(h2-h1) - Qm

Theoretical maximum refrigeration load

( Ql )max = COPr.rev * [m(h2-h1) - Qin]

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The 1000-lb elevator is hoisted by the pulley system and motor M. The motor exerts a constant force of 500 lb on the cable. The
klemol [59]

The power that must be supplied to the motor is 136 hp

<u>Explanation:</u>

Given-

weight of the elevator, m = 1000 lb

Force on the table, F = 500 lb

Distance, s = 27 ft

Efficiency, ε = 0.65

Power  = ?

According to the equation of motion:

F = ma

3(500) - 1000 = \frac{1000}{32.2} * a

a = 16.1 ft/s²

We know,

v^2 - u^2 = 2a (S - So)\\\\v^2 - (0)^2 = 2 * 16.1 (27-0)\\\\v = 29.48m/s

To calculate the output power:

Pout = F. v

Pout = 3 (500) * 29.48

Pout = 44220 lb.ft/s

As efficiency is given and output power is known, we can calculate the input power.

ε = Pout / Pin

0.65 = 44220 / Pin

Pin = 68030.8 lb.ft/s

Pin = 68030.8 / 500 hp

     = 136 hp

Therefore, the power that must be supplied to the motor is 136 hp

5 0
3 years ago
A race car is travelling around a circular track. The velocity of the car is 20 m/s, the radius of the track is 300 m, and the m
Zielflug [23.3K]

Answer:

μ = 0.136

Explanation:

given,

velocity of the car = 20 m/s

radius of the track = 300 m

mass of the car = 2000 kg

centrifugal force

F_c = \dfrac{mv^2}{r}

F_c = \dfrac{2000\times 20^2}{300}

F c = 2666. 67 N

F f= μ N

F f = μ m g

2666.67  =  μ × 2000 × 9.8

μ = 0.136

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A rectangular steel bar, with 8" x 0.75" cross-sectional dimensions, has equal and opposite moments applied to its ends.
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Answer:

Part a: The yield moment is 400 k.in.

Part b: The strain is 8.621 \times 10^{-4} in/in

Part c: The plastic moment is 600 ksi.

Explanation:

Part a:

As per bending equation

\frac{M}{I}=\frac{F}{y}

Here

  • M is the moment which is to be calculated
  • I is the moment of inertia given as

                         I=\frac{bd^3}{12}

Here

  • b is the breath given as 0.75"
  • d is the depth which is given as 8"

                     I=\frac{bd^3}{12}\\I=\frac{0.75\times 8^3}{12}\\I=32 in^4

  • y is given as

                     y=\frac{d}{2}\\y=\frac{8}{2}\\y=4"\\

  • Force is 50 ksi

\frac{M_y}{I}=\frac{F_y}{y}\\M_y=\frac{F_y}{y}{I}\\M_y=\frac{50}{4}{32}\\M_y=400 k. in

The yield moment is 400 k.in.

Part b:

The strain is given as

Strain=\frac{Stress}{Elastic Modulus}

The stress at the station 2" down from the top is estimated by ratio of triangles as

                        F_{2"}=\frac{F_y}{y}\times 2"\\F_{2"}=\frac{50 ksi}{4"}\times 2"\\F_{2"}=25 ksi

Now the steel has the elastic modulus of E=29000 ksi

Strain=\frac{Stress}{Elastic Modulus}\\Strain=\frac{F_{2"}}{E}\\Strain=\frac{25}{29000}\\Strain=8.621 \times 10^{-4} in/in

So the strain is 8.621 \times 10^{-4} in/in

Part c:

For a rectangular shape the shape factor is given as 1.5.

Now the plastic moment is given as

shape\, factor=\frac{Plastic\, Moment}{Yield\, Moment}\\{Plastic\, Moment}=shape\, factor\times {Yield\, Moment}\\{Plastic\, Moment}=1.5\times400 ksi\\{Plastic\, Moment}=600 ksi

The plastic moment is 600 ksi.

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