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A single insulated duct flow experiment using air operating at steady-state is performed in a lab. One measurement location (Sta

weqwewe [10]

Answer:

a) -0.0934 kJ/kg. K

b) The direction of flow is from right to left.

Explanation:

A free flow diagram of the horizontal insulated duct is as shown below.

NOW,

Let assume that the direction of flow is from left to right and consider the following relation for the entropy rate balance equation for a control volume as:

$\frac{\sigma_{cv}}{m}= (s_2-s_1) \geq 0 \ \ \ -------> \ \ \ 1$

Now; if the value for this relation is greater than zero; then we conclude that our assumption is correct.

If the value is less than zero; then we conclude that the assumption is wrong.

Then, the flow is said to be in the opposite direction

Formula for the change in specific entropy can be calculated as:

$s_2-s_1 = s^0(T_2) - s^0(T_1)-R \ In ( \frac{P_2}{P-1}) \ \ \ -------> \ \ \ 2$

where;

$s_1, s_2 , s^0(T_2), s^0(T1)$ are specific entropies

R = universal gas constant

$P_1$ = pressure at location 1

$P_2$ = pressure at location 2

We obtain the specific properties of air at temperature at $T_1$ = (67°C + 273)K = 340 K from the table A-22 ( Ideal gas properties of air)

$s^0(T1)$ = 1.8279 kJ/kg.K

We also obtain the specific properties of air at temperature $T_2$ = 22°C + 273) K = 295 K

From the table A- 22

$s^0(T_2)$ = 1.68515 kJ/kg . K

R = $\frac{8.314 kJ}{28.97 kg.K}$

$P_1 =$ 0.95 bar

$P_2 =$ 0.8 bar

Now replacing our values into equation (2) from above; we have;

$s_2-s_1 = s^0(T_2) -s^0(T_1)-R \ In (\frac{P_2}{P_1} )$

$s_2-s_1 = 1.68515 -1.8279-\frac{8.314}{28.97} \ In (\frac{0.8}{0.95} )$

$s_2-s_1 = 1.68515 -1.8279+ 0.0493$

$s_2-s_1 =-0.0934 \ kJ/kg.K$

Equating our result to equation (1)

$s_2-s_1 \geq 0\\-0.0934 \leq 0$

Therefore , our assumption is wrong and the direction of flow is said to be from right to left.

We therefore conclude that the direction of flow is from right to left.

3 0

4 years ago

Hull (1943) had rats push a lever that required 21 grams of force to budge. After they had learned to push the lever in order to

Zina [86]

Years of research have demonstrated that rats are intelligent creatures who experience pain and pleasure, care about one another, are able to read the emotions of others, and would assist other rats, even at their own expense.

<h3>Experiments:</h3>

In trials carried out at Brown University in the 1950s, rats were trained to press a lever for food, but they stopped pressing the lever when they noticed that with each press, a rat in an adjacent cage would scream in pain (after experiencing an electric shock).

Rats were trained to press a lever to lower a block that was hanging from a hoist by electric shocks administered by experimenters. A rat was subsequently hoisted into a harness by the experimenters, and according to their notes, "This animal normally shrieked and wriggled sufficiently while dangling, and if it did not, it was jabbed with a sharp pencil until it exhibited indications of discomfort." Even if it wasn't in danger of receiving a shock, a rat watching the scenario from the floor would pull a lever to lower the hapless rodent to safety.

Learn more about experiments on rats here:

brainly.com/question/13625715

#SPJ4

6 0

2 years ago

A ball is released from the top of a hill. How fast is the ball going when it reaches the base of the hill? Approximate g as 10

irina [24]

First I’ll show you this standard derivation using conservation of energy:
Pi=Kf,
mgh = 1/2 m v^2,
V = sqrt(2gh)
P is initial potential energy, K is final kinetic, m is mass of object, h is height from stopping point, v is final velocity.
In this case the height difference for the hill is 2-0.5=1.5 m. Thus the ball is moving at sqrt(2(10)(1.5))=
5.477 m/s.

5 0

3 years ago

Read 2 more answers

Calculate the acceleration of a bus that goes

Svet_ta [14]

The acceleration of the bus is 1.11 meters per second square to the direction of motion

Explanation:

Acceleration is the rate of change of velocity

The formula of the acceleration is $a=\frac{v_{2}-v_{1}}{t}$ , where

$v_{1}$ is the initial velocity

$v_{2}$ is the final velocity
t is the time

A bus that goes from 10 km/h to a speed of 50 km/h in 10 seconds

→ $v_{1}$ = 10 km/h

→ $v_{2}$ = 50 km/h

→ t = 10 seconds

Change the unite of the time from seconds to hour

→ 1 hour = 60 × 60 = 3600 seconds

→ 10 seconds = $\frac{10}{3600}=\frac{1}{360}$ hour

Substitute these values in the formula of the acceleration above

→ $a=\frac{50-10}{\frac{1}{360}}$

→ a = 14400 km/h²

To change the unit of acceleration to meter per second change the

kilometer to meter and the hour to seconds

→ 1 km = 1000 m

→ 1 hour = 3600 seconds

→ $a=\frac{14400*1000}{(3600)^{2}}=\frac{10}{9}$

→ a = 1.11 m/sec².

The acceleration of the bus is 1.11 meters per second square to the direction of motion

Learn more:

You can learn more about the acceleration in brainly.com/question/6323625

#LearnwithBrainly

4 0

4 years ago

Briefly explain how a resonance tube works

Lera25 [3.4K]

Answer:

As the tines of the tuning fork vibrate at their own natural frequency, they created sound waves that impinge upon the opening of the resonance tube. These impinging sound waves produced by the tuning fork force air inside of the resonance tube to vibrate at the same frequency.

4 0

2 years ago

Read 2 more answers