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Sergio039 [100]
3 years ago
13

Refrigerant 134a enters a horizontal pipe operating at steady state at 40oC, 300 kPaand a velocity of 40 m/s. At the exit, the t

emperature is 50oC and the pressure is 240 kPa. The pipe diameter is 0.04 m. Determine (a) the mass flow rate of the refrigerant, in kg/s, (b) the velocity at the exit, in m/s, and (c) the rate of heat transfer between the pipe and its surroundings, in kW
Engineering
1 answer:
zubka84 [21]3 years ago
6 0

Answer:

1. Mass flow rate = 0.621Kg/s

2. Velocity = 52.2m/s

3. Rate = +6.82kW

Explanation:

Given

Refrigerant = R134a

Diameter, d = 0.04m

T1 = 40°C

P1 = 300kPa

V1 = 40m/s

T2 = 50°C

P2 = 240kPa

a.

The specific volume of refrigerant R134a at T1 and P1 from Refrigerant table (A12) = v1 = 0.08089 m³/kg

The specific volume of refrigerant R134a at T2 and P2 from Refrigerant table (A12) = v2 = 0.10562 m³/kg.

Mass flow rate is calculated as follows...

= AV/v

Where A = Area =πd²/4

V = velocity = ,40

v = v1 = 0.08089 m³/kg.

So. Mass flow rate= (π * 0.04²/4 * 40 ) / 0.0809

Mass flow rate = 0.621Kg/s

b.

Calculating Velocity at the exit.

This is given as;

m * v2 / A

0.621 * 0.10562 / (π * 0.04²/4)

= 52.2m/s

c. Calculating the rate of heat transfer between the pipe and its surroundings

From energy equation, we have.

Q -> m * [(h2 - h1) + ((,V2² - V1²)/2)]

Where h2 is the specific enthalpy for refrigerant at T1 and P1

h2 = 234.05 Kj/Kg

And h1 is the specific enthalpy for refrigerant at T2 and P2

h2 = 244.47 Kj/Kg

Rate = the specific enthalpy for refrigerant at T1 and P1

h2 = 234.05 Kj/Kg

Rate = 0.621[(234.05 - 244.47) + (52.2² - 40²)/2]

Rate = 0.621[(234.05 - 244.47) +

562.42] --- Convert the kW

Rate = 0.621((244.47 - 234.05) +

562.42/1000)

Rate = +6.82kW

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The current entering the positive terminal of a device is i(t)= 6e^-2t mA and the voltage across the device is v(t)= 10di/dtV.
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Answer:

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b) P(t) = -720*e^(-4t) uW

c) -180 uJ

Explanation:

Given:

                           i (t) = 6*e^(-2*t)

                           v (t) = 10*di / dt

Find:

( a) Find the charge delivered to the device between t=0 and t=2 s.

( b) Calculate the power absorbed.

( c) Determine the energy absorbed in 3 s.

Solution:

-  The amount of charge Q delivered can be determined by:                      

                                       dQ = i(t) . dt

                  Q = \int\limits^2_0 {i(t)} \, dt = \int\limits^2_0 {6*e^(-2t)} \, dt = 6*\int\limits^2_0 {e^(-2t)} \, dt

- Integrate and evaluate the on the interval:

                   = 6 * (-0.5)*e^-2t = - 3*( 1 / e^4 - 1) = 2.945 C

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                 v(t) = 10* di / dt = 10*d(6*e^(-2*t)) /dt

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                 P(t) = -720*e^(-4t) uW

- The amount of energy W absorbed can be evaluated using P(t) as follows:

                 W = \int\limits^3_0 {P(t)} \, dt = \int\limits^2_0 {-720*e^(-4t)} \, dt = -720*\int\limits^2_0 {e^(-4t)} \, dt

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An aluminum alloy tube with an outside diameter of 3.50 in. and a wall thickness of 0.30 in. is used as a 14 ft long column. Ass
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Answer:

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wall thickness 0.30 inc

length of column is 14 ft

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