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defon
3 years ago
12

The density of oxygen contained in a tank is 2.0 kg/m3 when the temperature is 25 °C. Determine the gage pressure of the gas if

the atmospheric pressure is 97 kPa.
Engineering
1 answer:
Verdich [7]3 years ago
8 0

Answer:

57.8408 kPa

Explanation:

Data provided in the question:

Density of oxygen contained, ρ = 2.0 kg/m³

Temperature, T = 25 °C = 25 +273 = 298 K

Atmospheric pressure = 97 kPa.

Now,

Pressure due to gas = ρRT

Here,

R is the gas constant = 259.8 J/kg.K

Thus,

Absolute Pressure due to oxygen = 2 × 259.8 × 298

= 154840.8 Pa

= 154840.8 × 10⁻³ kPa

= 154.84 kPa

Thus,

Gage pressure = Absolute Pressure - Atmospheric pressure

=  154.84 kPa - 97 kPa

= 57.8408 kPa

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A hypothetical metal alloy has a grain diameter of 1.7 102 mm. After a heat treatment at 450C for 250 min, the grain diameter ha
Romashka-Z-Leto [24]

Answer:

the required time for the specimen is  1109.4 min

Explanation:

Given that;

diameter of metal alloy d₀ = 1.7 × 10² mm

Temperature of heat treatment T = 450°C = 450 + 273 = 723 K

Time period of heat treatment t = 250 min

Increased grain diameter 4.5 × 10² mm

grain diameter exponent n = 2.1

First we calculate the time independent constant K

dⁿ - d₀ⁿ = Kt

K = (dⁿ - d₀ⁿ) / t

we substitute

K = (( 4.5 × 10² )²'¹ - ( 1.7 × 10² )²'¹) / 250

K = (373032.163378 - 48299.511117) / 250

K = 1298.9306 mm²/min

Now, we calculate the time required for the specimen to achieve the given grain diameter ( 8.7 × 10² mm )

dⁿ - d₀ⁿ = Kt

t = (dⁿ - d₀ⁿ) / K

t = (( 8.7 × 10² )²'¹ - ( 1.7 × 10² )²'¹) / 1298.9306

t = ( 1489328.26061158 - 48299.511117) / 1298.9306

t = 1441028.74949458 / 1298.9306

t = 1109.4 min

Therefore, the required time for the specimen is  1109.4 min

4 0
3 years ago
In the formula shown, k is a correction factor for nonideal mixing. In the worst case, k is usually estimated to be:_______. Cpp
LuckyWell [14K]

Answer:

d. 1.0

Explanation:

Correlation identifies the relationship between two variables. In the given scenario there is strong relation between non ideal mixing. The correction factor can be between -1 to 1 depending on the intensity of the relationship and dependency. The non ideal mixing efficiency is highly dependent on the factors that govern it this means there is high intensity relation so the k is estimated to be nearly 1.

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If you get a flat in the front of your car, your car will:
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Answer:

stop and might even crash

Explanation:

6 0
3 years ago
A 15 Watt desk-type fluorescent lamp has an effective resistance of 200 ohms when operating (note: the 15 Watts is only associat
Aleks04 [339]

The question is incomplete! Complete question along with answers and explanation is provided below.

Question:

A 15 Watt desk-type fluorescent lamp has an effective resistance of 200 ohms when operating (note: the 15 Watts is only associated with the lamp). It is in series with a ballast that has a resistance of 80 ohms and an inductance of .9H. The lamp and ballast are operated at 120V, 60Hz.

a) Draw the circuit

b) Calculate the power drawn by the lamp

c) Calculate the apparent power

d) Calculate the power factor

e) Calculate the reactive power

f) Calculate the size of the capacitor necessary to provide unity power factor correction

Explanation:

a) draw the circuit

Refer to the attached image.

As you can see in the attached drawing, it is a series circuit containing  two resistors and one inductor.

In a series circuit, current remains same throughout the circuit

The circuit is powered by an AC voltage source having voltage of 120 V and frequency 60 Hz.

The current flowing in the circuit can be found by ohm's law

 I = V/Z

where V is the voltage and Z is the total impedance of the circuit

 Z = R + XL

where  XL is the inductive reactance

XL = j2 π f L

XL = j2*π*60*0.9

XL = j339.29Ω

Total resistance is

R =200 + 80 = 280 Ω

Total impedance is

Z = 280 + j339.29 Ω

b) Calculate the power drawn by the lamp

First calculate the current

I = V/Z

I = 120/(280 + j339.29)

I = 0.272<-50.46° A  (complex notation)

P = I²R

P = (0.272)²200

P ≈ 15 W

Power drawn by the circuit

P=V*I*cos(50.46°)

P=20.77 W

c) Calculate the apparent power

A = VI*

A = 120*0.272<50.46°

A = 32.64<50.46° VA

d) Calculate the power factor

PF = cos(50.46)

PF = 0.63

e) Calculate the reactive power

Q = VIsin(50.46)

Q = 120*0.272<-50.46*sin(50.46)

Q = 25.13<-50.46  VAR

f) Calculate the size of the capacitor necessary to provide unity power factor correction

The required reactive compensation power is

Qc = P (tan(old) - tan(new))

Qc = 20.77 (tan(50.46) - tan(0))

Qc = 25.16 VAR

C = Qc/2πfV²

C = 25.16/2*π*60*120²

C = 4.63 uF

Hence adding a capacitor of 4.63 uF parallel to the load will improve the PF from 0.63 to 1.

5 0
3 years ago
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