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kondaur [170]
3 years ago
11

A 500-m^3 rigid tank is filled with saturated liquid-vapor mixture of water at 200 kPa. If 20% of the mass is liquid and the 80%

of the mass is vapor, the total mass in the tank is: (a) 635 kg (b) 2809 kg (c) 258 kg (d) 500 kg (e) 705 kg
Engineering
1 answer:
guapka [62]3 years ago
3 0

Answer:

(a) 705 kg

Explanation:

it is given that volume =500m^3

pressure = 200 kPa

it is given that 80% mass is vapor so dryness fraction =x=0.8

from the standard table at 200 kPa

v_f=0.001061 \frac{m^3}{kg}

v_g=0.88578\frac{m^3}{kg}

specific volume v=v_f+x\left ( v_g-v_f \right )

v=0.00106+0.8\left ( 0.88578-0.00106 \right )=0.7088\frac{m^3}{kg}

we know that specific\ volume =\frac{volume}{mass}

0.7088=\frac{500}{mass}

mass=\frac{500}{0.7088}=705.456 \ kg

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Answer:

Considering the guidelines of this exercise.

The pieces produced per month are 504 000

The productivity ratio is 75%

Explanation:

To understand this answer we need to analyze the problem. First of all, we can only produce 2 batches of production by the press because we require 3 hours to set it up. So if we rest those 6 hours from the 8 of the shift we get 6, leaving 2 for an incomplete bath. So multiplying 2 batches per day of production by press we obtain 40 batches per day. So, considering we work in this factory for 21 days per month well that makes 40 x 21  making 840 then we multiply the batches for the pieces 840 x 600 obtaining 504000 pieces produced per month. To obtain the productivity ratio we need to divide the standard labor hours meaning 6 by the amount of time worked meaning 8. Obtaining 75% efficiency.

4 0
3 years ago
A cylindrical specimen of a hypothetical metal alloy is stressed in compression. If its original and final diameters are 30.00 a
IrinaVladis [17]

Answer:

The original length of the specimen l_{o} = 104.7 mm

Explanation:

Original diameter d_{o} = 30 mm

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Final length l_{1} = 105.20 mm

Elastic modulus E = 65.5 G pa = 65.5 × 10^{3} M pa

Shear modulus G = 25.4 G pa = 25.4 × 10^{3} M pa

We know that the relation between the shear modulus & elastic modulus is given by

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25.5 = \frac{65.5}{2 (1 + \mu)}

\mu = 0.28

This is the value of possion's ratio.

We know that the possion's ratio is given by

\mu = \frac{\frac{0.04}{30} }{\frac{change \ in \ length}{l_{o} } }

{\frac{change \ in \ length}{l_{o} } = \frac{\frac{0.04}{30} }{0.28}

{\frac{change \ in \ length}{l_{o} } = 0.00476

\frac{l_{1} - l_{o}  }{l_{o}  } = 0.00476

\frac{l_{1} }{l_{o} } = 1.00476

Final length l_{o} = 105.2 m

Original length

l_{o} = \frac{105.2}{1.00476}

l_{o} = 104.7 mm

This is the original length of the specimen.

5 0
3 years ago
Technician A says amperage cannot exist without both voltage and resistance. Technician B says if amperage is high, then you kno
Ivan

Answer:

Technician A

Explanation:

Ohms law:  I= E/R so rest resistance must be present along with E/potential difference.  Even if just wire shorted together there is resistance but very little.

Tech B: Again ohms law.  Current flow is directly proportional to the voltage and inversely  proportional to R (resistance or impedance).

8 0
3 years ago
A motor driven water pump operates with an inlet pressure of 96 kPa (absolute) and mass flow rate of 120 kg/min. The motor consu
NeX [460]

Answer:

The maximum water pressure at the discharge of the pump (exit) = 496 kPa

Explanation:

The equation expressing the relationship of the power input of a pump can be computed as:

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where;

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the pressure at the inlet P_1 = 96 kPa

the pressure at the exit P_2 = ???

the pressure \rho = 1000 kg/m³

∴

0.8 \times 10^{3} \ W = \dfrac{(120 \ kg/min * 1min/60 s)(P_2-96000)}{1000}

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Shkiper50 [21]

Answer:

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Q = 54.78 Watt per foot.

This is the value of heat transferred watt per foot length.

4 0
3 years ago
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