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KatRina [158]
3 years ago
11

Refrigerant-134a enters a compressor at 180 kPa as a saturated vapor with a flow rate of 0.35 m3/min and leaves at 900 kPa. The

power supplied to the refrigerant during the compression process is 2.35 kW. What is the temperature of R-134a at the exit of the compressor

Physics
1 answer:
Alexus [3.1K]3 years ago
7 0

Answer:

52.5°C

Explanation:

The final enthalpy is determined from energy balance where initial enthalpy and specific volume are obtained from A-12 for the given pressure and state

mh1 + W = mh2

h2 = h1 + W/m

h1 + Wα1/V1

242.9 kJ/kg + 2.35.0.11049kJ/ 0.35/60kg

=287.4 kJ/kg

From the final enthalpy and pressure the final temperature is obtained A-13 using interpolation

i.e T2 = T1 + T2 -T1/h2 -h1(h2 - h1)

= 50°C + 60 - 50/295.15 - 284.79

(287.4 - 284.79)°C

= 52.5°C

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Now assume that the frictional force f is not at its maximum value. What is the relation between the torque Ï„ applied to each w
leva [86]

Answer:

a)The direction the frictional force will acts is in the positive x direction.

Explanation:

a)The direction the frictional force will acts is in the positive x direction

b)in the horizontal direction, the total force F(total) is equal to 4times the frictional force in the wheel.

F(total)=4f

''f'' is taken as the frictional force.

c)4times the normal force on each wheel minus the acceleration equals zero i.e 4N(wheel)-a=0

=4N(wheel)-mg=0

d) torque is the force that tends to bend rotation

ζ=rf

but acceleration=4×frictional force

cross multiply

f=ζ/r

f=ma/4

ma/4=ζ/r

a=4ζ/r

5 0
4 years ago
Read 2 more answers
Determine an appropriate size for a square cross-section solid steel shaft to transmit 260 hp at a speed of 550 rev/min if the m
Phantasy [73]

Answer:46.05 mm

Explanation:

Given

Power=260\ hp\approx 260\times 746=193.96\ KW

speed N=550\ rev/min

allowable shear stress (\tau )_{max}=15\ kpsi\approx 103.421\ MPa

Power is given by

P=\frac{2\pi NT}{60}

193.96=\frac{2\pi 550\times T}{60}

T=3367.6\ N-m

From Torsion Formula

\frac{T}{J}=\frac{\tau }{r}-----1

where J=Polar section modulus

T=Torque

\tau=shear stress

For square cross section

r=\frac{a}{2}

where a=side of square

J=\frac{a^4}{6}

Substituting the values in equation 1

\frac{3376.6}{\frac{a^4}{6}}=\frac{103.421\times 10^6}{\frac{a}{2}}

a=0.04605\ m

a=46.05\ mm

7 0
3 years ago
A point charge gives rise to an electric field with magnitude 2 N/C at a distance of 4 m. If the distance is increased to 20 m,
worty [1.4K]

Answer:

0.08 N/C

Explanation:

Electric Field: This is defined as the force per unit charge exerted at a point. The expression for electric field is given as,

E = Kq/r².............................. Equation 1

Where E = Electric Field, q = Charge, k = proportionality constant, r = distance.

making q the subject of the equation,

q = Er²/k............................... Equation 2

Given: E = 2 N/C, r = 4 m,

Substitute into equation 2

q = 2(4)²/k

q = 32/k C.

When r is increased to 20 m,

E = k(32/k)/20²

E = 32/400

E = 0.08 N/C.

Hence the electric Field = 0.08 N/C

5 0
4 years ago
A 6kg block starts from rest against a spring with a spring constant of 500N/m that is compressed by a distance of 2m. The groun
cestrela7 [59]

The velocity of the block leaving the spring is determined as 18.26 m/s.

<h3>Velocity of the block</h3>

The velocity of the block leaving the spring is calculated from the principle of conservation of energy.

K.E = Ux

¹/₂mv² = ¹/₂kx²

v² = kx²/m

v² = (500 x 2²)/6

v² = 333.33

v = √333.33

v = 18.26 m/s

Thus, the velocity of the block leaving the spring is determined as 18.26 m/s.

Learn more about velocity here: brainly.com/question/4931057

#SPJ1

7 0
2 years ago
A car moving at 11.6 m/s begins to accelerate at 5.22 m/s/s and reaches a final velocity of 31.5 m/s. How far did it travel duri
mr_godi [17]

Use the kinematic equation,

Vf^2 = Vi^2 + 2aX

31.5^2 = 11.6^2 + 2(5.22)(x)

Solve for x

Answer: 82.2m

8 0
3 years ago
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