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Svetlanka [38]
3 years ago
15

In a vapor-compression refrigeration cycle, ammonia exits the evaporator as saturated vapor at -10°C. The refrigerant enters the

condenser at 16 bar and 160°C, and saturated liquid exits at 16 bar. There is no significant heat transfer between the compressor and its surroundings, and the refrigerant passes through the evaporator with a negligible change in pressure. If the refrigerating capacity is 150 kW, determine (a) the mass flow rate of refrigerant, in kg/s (b) the power input to the compressor, in kw (c) the coefficient of performance (d) the isentropic compressor efficiency Round answers to 3 significant digits. (a) 0,1444 (b) 52.99 (c) 2.83 (d) 88.26
Engineering
1 answer:
sergij07 [2.7K]3 years ago
8 0

Answer:  (a) 0,142 (b) 52.99 (c) 2.83 (d) 88.26

Explanation:

If the refrigarating capacity is 150kw

(a) the mass flow rate of refrigerant, in kilograms per second  is 0.142

(b) the power input to the compressor, in kilowatts is 52.99

(c) the coefficient of performance is 2.83

(d) the isentropic compressor efficiency is 68.6 per cent

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What is the mechanical advantage of a simple block and tackle with one rope?
NISA [10]

Answer:

A block and tackle is characterized by the use of a single continuous rope to transmit a tension force around one or more pulleys to lift or move a load. Its mechanical advantage is the number of parts of the rope that act on the load. The mechanical advantage of a tackle dictates how much easier it is to haul or lift the load.

Explanation:

Hope this helps- have a good day ^w

6 0
3 years ago
Read 2 more answers
Pendulum impacting an inclined surface of a block attached to a spring-Dependent multi-part problem assign all parts NOTE: This
Art [367]

Answer:

vA = -2.55 m/s

vB = 0.947 m/s

Explanation:

Given:-

- The initial angle of rope, α = 30°

- The angle of rope just before impact or wedge angle, θ = 20°

- The weight of sphere, Ws = 1-lb

- The initial position velocity, vi = 4 ft/s

- The coefficient of restitution, e = 0.7

- The weight of the wedge, Ww = 2-lb

- The spring constant, k = 1.8 lb/in

- The length of rope, L = 2.6 ft

Find:-

 Determine the velocities of A and B immediately after the impact.

Solution:-

- We can first consider the ball ( acting as a pendulum ) to be isolated for study.

- There are no unbalanced fictitious forces acting on the sphere ball. Hence, we can reasonably assume that the energy is conserved.

- According to the principle of conservation for the initial point and the point just before impact.

Let,

              vA : The speed of sphere ball before impact

               

                  Change in kinetic energy = Change in potential energy

                  ΔK.E = ΔE.P

                  0.5*ms* ( uA^2 - vi^2 ) = ms*g*L*( cos ( θ ) - cos ( α ) )

                  uA^2 = 2*g*L*( cos ( θ ) - cos ( α ) ) + vi^2

                  uA = √ [ 2*32*2.6*( cos ( 20 ) - cos ( 30 ) ) + 4^2 ] = √28.25822

                  uA = 5.316 ft/s

- The coefficient of restitution (e) can be thought of as a measure of the extent to which mechanical energy is conserved when an object bounces off a surface:

                 e^2 = ( K.E_after impact / K.E_before impact )

- The respective Kinetic energies are:

               

                K.E_after impact = K.E_sphere + K.E_block

                                             = 0.5*ms*vA^2 + 0.5*mb*vB^2

                K.E_before impact = K.E = Ws*L*( cos ( θ ) - cos ( α ) )

                                                         = 1*2.6*( cos ( 20 ) - cos ( 30 ) )

                                                         = 0.1915 J

                32*2*0.1915*0.7^2 = Ws*vA^2 + Wb*vB^2  

                6.00544 = vA^2 + 2*vB^2  ... Eq1

- From conservation of linear momentum we have:

                vB = e*( uA - uB )*cos ( 20 ) + vA

                vB = 0.7*( 5.316 - 0 )*cos ( 20)   + vA

                vB = 3.49678 + vA  .... Eq 2

- Solve two equation simultaneously:

               

               6.00544 = vA^2 + 2*(3.49678 + vA)^2

               6.00544 = 3vA^2 + 13.98*vA + 24.455

               3vA^2 + 14.8848*vA + 18.4495 = 0

               vA = -2.55 m/s

               vB = 0.947 m/s

                                 

5 0
4 years ago
A 1 phase load operates at 600 V and consumes 75 kW with a 0.85 lagging power factor. Compute the complex power consumed by the
Free_Kalibri [48]

Answer:

Complex power=84 W

Explanation:

using equation

s=vs^2/2z

s=600^2/2z

s=84+29.4j

using s=P+jq

complex power=P=84 W

8 0
4 years ago
Read 2 more answers
Uses of P-N junction
Lubov Fominskaja [6]

Answer:

Explanation:

Two that come to mind:

  1. a semiconductor diode is essentially a PN junction
  2. a transistor is made of two pn junctions.
4 0
3 years ago
A thick aluminum block initially at 26.5°C is subjected to constant heat flux of 4000 W/m2 by an electric resistance heater whos
Yanka [14]

Given Information:

Initial temperature of aluminum block = 26.5°C

Heat flux = 4000 w/m²

Time = 2112 seconds

Time = 30 minutes = 30*60 = 1800 seconds

Required Information:

Rise in surface temperature = ?

Answer:

Rise in surface temperature = 8.6 °C after 2112 seconds

Rise in surface temperature = 8 °C after 30 minutes

Explanation:

The surface temperature of the aluminum block is given by

T_{surface} = T_{initial} + \frac{q}{k} \sqrt{\frac{4\alpha t}{\pi} }

Where q is the heat flux supplied to aluminum block, k is the conductivity of pure aluminum and α is the diffusivity of pure aluminum.

After t = 2112 sec:

T_{surface} = 26.5 + \frac{4000}{237} \sqrt{\frac{4(9.71\times 10^{-5}) (2112)}{\pi} }\\\\T_{surface} = 26.5 + \frac{4000}{237} (0.51098)\\\\T_{surface} = 26.5 + 8.6\\\\T_{surface} = 35.1\\\\

The rise in the surface temperature is

Rise = 35.1 - 26.5 = 8.6 °C

Therefore, the surface temperature of the block will rise by 8.6 °C after 2112 seconds.

After t = 30 mins:

T_{surface} = 26.5 + \frac{4000}{237} \sqrt{\frac{4(9.71\times 10^{-5}) (1800)}{\pi} }\\\\T_{surface} = 26.5 + \frac{4000}{237} (0.4717)\\\\T_{surface} = 26.5 + 7.96\\\\T_{surface} = 34.5\\\\

The rise in the surface temperature is

Rise = 34.5 - 26.5 = 8 °C

Therefore, the surface temperature of the block will rise by 8 °C after 30 minutes.

5 0
3 years ago
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