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rosijanka [135]
3 years ago
10

Carbon dioxide used as a natural refrigerant flows through a cooler at 10 MPa, which is supercritical, so no condensation occurs

. The inlet is at 220°C and the exit is at 50°C. Find the specific heat transfer.

Engineering
2 answers:
kupik [55]3 years ago
5 0

Answer:

The answer which is a calculation can be found as an attached document

Explanation:

MrRa [10]3 years ago
4 0

Answer:

Answer is -286.78 kJ/kg

Refer below.

Explanation:

Refer to the pictures for brief explanation.

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8 0
3 years ago
A piston/cylinder contains 1.5 kg of water at 200 kPa, 150°C. It is now heated by a process in which pressure is linearly relate
Fofino [41]

Answer:

final volume V2 = 0.71136 m³

work done in process W = -291.24 kJ

heat transfer Q = 164 kJ

Explanation:

given data

mass = 1.5 kg

pressure p1 = 200 kPa

temperature t1 = 150°C

final pressure p2 = 600 kPa

final temperature t2 = 350°C

solution

we will use here superheated water table that is

for pressure 200 kPa and 150°C temperature

v1 = 0.95964 m³/kg

u1 = 2576.87 kJ/kg

and

for pressure 600 kPa and 350°C temperature

v2 = 0.47424 m³/kg

u2 = 2881.12 kJ/kg

so v1 is express as

V1 = v1 × m    ............................1

V1 = 0.95964 × 1.5

V1 = 1.43946 m³

and

V2 = v2 × m    ............................2

V2 = 0.47424 × 1.5

final volume V2 = 0.71136 m³

and

W = P(avg) × dV      .............................3

P(avg) = \frac{p1+p2}{2}    = \frac{200+600}{2} = 400 × 10³

put here value

W = 400 × 10³ × (0.71136 - 1.43946 )

work done in process W = -291.24 kJ

and

heat transfer is

Q = m × (u2 - u1)  + W       .............................4

Q = 1.5 × (2881.12 - 2576.87)  + 292.24

heat transfer Q = 164 kJ

7 0
3 years ago
Which of the following is not caused by alcohol?
Brrunno [24]

Answer:

Inhibited comprehension

4 0
3 years ago
Read 2 more answers
A water skier leaves the end of an 8 foot tall ski ramp with a speed of 20 mi/hr and at an angle of 250. He lets go of the tow r
klemol [59]

Answer:

At highest point:

y1 = 10.4 ft

v1 = (26.5*i + 0*j) ft/s

When he lands:

x2 = 31.5 ft (distance he travels)

t2 = 1.19 s

V2 = (26.5*i - 25.9*j) ft/s

a2 = -44.3°

Explanation:

Since he let go of the tow rope upon leaving the ramp he is in free fall from that moment on. In free fall he is affected only by the acceleration of gravity. Gravity has a vertical component only, so the movement will be at constant acceleration in the vertical component and at constant speed in the horizontal component.

20 mi / h = 29.3 ft/s

If the ramp has an angle of 25 degrees, the speed is

v0 = (29.3 * cos(25) * i + 29.3 * sin(25) * j) ft/s

v0 = (26.5*i + 12.4*j) ft/s

I set up the coordinate system with the origin at the base of the ramp under its end, so:

R0 = (0*i + 8*j) ft

The equation for the horizontal position is:

X(t) = X0 + Vx0 * t

The equation for horizontal speed is:

Vx(t) = Vx0

The equation for vertical position is:

Y(t) = Y0 + Vy0 * t + 1/2 * a * t^2

The equation for vertical speed is:

Vy(t) = Vy0 + a * t

In this frame of reference a is the acceleration of gravity and its values is -32.2 ft/s^2.

In the heighest point of the trajectory the vertical speed will be zero because that is the point where it transitions form going upwards (positive vertical speed) to going down (negative vertical speed), and it crosses zero.

0 = Vy0 + a * t1

a * t1 = -Vy0

t1 = -Vy0 / a

t1 = -12.4 / -32.2 = 0.38 s

y1 = y(0.38) = 8 + 12.4 * 0.38 + 1/2 * (-32.2) * (0.38)^2 = 10.4 ft

The velocity at that moment will be:

v1 = (26.5*i + 0*j) ft/s

When he lands in the water his height is zero.

0 = 8 + 12.4 * t2 + 1/2 * (-32.2) * t2^2

-16.1 * t2^2 + 12.4 * t2 + 8 = 0

Solving this equation electronically:

t2 = 1.19 s

Replacing this time on the position equation:

X(1.19) = 26.5 * 1.19 = 31.5 ft

The speed is:

Vx2 = 26.5 ft/s

Vy2 = 12.4 - 32.2 * 1.19 = -25.9 ft/s

V2 = (26.5*i - 25.9*j) ft/s

a2 = arctg(-25.9 / 26.5) = -44.3

3 0
3 years ago
2. When performing an alignment, what action should be taken immediately after putting a vehicle on the rack?
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