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Dahasolnce [82]
3 years ago
7

A two-phase mixture of water and steam with a quality of 0.63 and T = 300F expands isothermally until only saturated vapor rema

ins. The initial volume is 0.114 ft3 . During the process, 16.2 Btu of heat are added. Find the work done. Also draw this process in the P-v space indicating the initial and final states with respect to the liquid-vapor dome.

Engineering
1 answer:
VMariaS [17]3 years ago
5 0

Answer:

Explanation:

Hello!

To solve this problem you must follow the following steps, which are fully registered in the attached image.

1. Draw the complete outline of the problem.

2. Through laboratory tests, thermodynamic tables were developed, these allow to know all the thermodynamic properties of a substance (entropy, enthalpy, pressure, specific volume, internal energy etc ..)

through prior knowledge of two other properties.

3. Use temodynamic tables to find the density of water in state 1, by means of temperature and quality, with this value and volume we can find the mass.

3. Use thermodynamic tables to find the internal energy in state 1 and two using temperature and quality.

4. uses the first law of thermodynamics that states that the energy in a system is always conserved, replaces the previously found values ​​and finds the work done.

5. draw the pV diagram using the 300F isothermal line

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An aggregate blend is composed of 55% aggregate A (Sp. Gr. 2.631), 25% aggregate B (Sp. Gr. 2.331) and 20% sand (Sp. Gr. 2.609).
Phoenix [80]

Answer:

The right choice would be Option b (2.545).

Explanation:

The given values are:

The aggregate blend will be:

W_A = 55%

G_ A = 2.631

W_ B = 25%

G_B = 2.331

W_C = 20%

G_C = 2.609

Now,

On applying the formula, we get

⇒ G_{BA}=\frac{W_A+W_B+W_C}{\frac{W_A}{G_A} +\frac{W_B}{G_B} +\frac{W_C}{G_C}}

On substituting the values, we get

⇒          =\frac{55+25+20}{\frac{55}{2.631} +\frac{25}{2.331} +\frac{20}{2.609}}

⇒          = \frac{100}{\frac{55}{2.631} +\frac{25}{2.331} +\frac{20}{2.609}}

⇒          =2.545

8 0
3 years ago
At the instant under consideration, the hydraulic cylinder AB has a length L = 0.75 m, and this length is momentarily increasing
Inessa [10]

Answer:

vB = - 0.176 m/s   (↓-)

Explanation:

Given

(AB) = 0.75 m

(AB)' = 0.2 m/s

vA = 0.6 m/s

θ = 35°

vB = ?

We use the formulas

Sin θ = Sin 35° = (OA)/(AB) ⇒  (OA) = Sin 35°*(AB)

⇒   (OA) = Sin 35°*(0.75 m) = 0.43 m

Cos θ = Cos 35° = (OB)/(AB) ⇒  (OB) = Cos 35°*(AB)

⇒   (OB) = Cos 35°*(0.75 m) = 0.614 m

We apply Pythagoras' theorem as follows

(AB)² = (OA)² + (OB)²

We derive the equation

2*(AB)*(AB)' = 2*(OA)*vA + 2*(OB)*vB

⇒  (AB)*(AB)' = (OA)*vA + (OB)*vB

⇒  vB = ((AB)*(AB)' - (OA)*vA) / (OB)

then we have

⇒  vB = ((0.75 m)*(0.2 m/s) - (0.43 m)*(0.6 m/s) / (0.614 m)

⇒  vB = - 0.176 m/s   (↓-)

The pic can show the question.

7 0
3 years ago
Read 2 more answers
The domain of discourse is the members of a chess club. The predicate B(x, y) means that person x has beaten person y at some po
satela [25.4K]

Answer:A. No one has ever beat Nancy.

Explanation:

The dormain of discourse in a simple language is the set of entities upon which our discussions are based when discussing about something.

The dormain of discourse is also known simply as universe, can also be said to be a set of entities o

upon which certain variables of interest in some formal treatment may range.

The dormain of discourse is generally attributed to Augustus De Morgan, it was also extensively used by George Boole in his Laws of Thought.

THE LOGICAL UNDERSTANDING OF THE THE QUESTION IS THAT NO ONE HAS EVER BEAT NANCY.

8 0
3 years ago
A safety interlock module operates by monitoring the voltage from the
In-s [12.5K]

Answer: its an Ignition coil

8 0
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A square silicon chip (k = 152 W/m·K) is of width 7 mm on a side and of thickness 3 mm. The chip is mounted in a substrate such
Harrizon [31]

Answer:

The steady-state temperature difference is 2.42 K

Explanation:

Rate of heat transfer = kA∆T/t

Rate of heat transfer = 6 W

k is the heat transfer coefficient = 152 W/m.K

A is the area of the square silicon = width^2 = (7/1000)^2 = 4.9×10^-5 m^2

t is the thickness of the silicon = 3 mm = 3/1000 = 0.003 m

6 = 152×4.9×10^-5×∆T/0.003

∆T = 6×0.003/152×4.9×10^-5 = 2.42 K

7 0
3 years ago
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