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2 years ago

What quantity measures the effect of change?

Engineering

2 answers:

alexandr1967 [171]2 years ago

8 0

Answer:

The Dependent Variable is the Effect, Its value depends on changes from the Independent Variable

Hope this Helps

Fed [463]2 years ago

4 0

Answer:

independent variable

Explanation:

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The volume of microbial culture is observed to increase according to the formula

saw5 [17]

The expression of V(m³)=e^(t(s)) to make V in in³ and t in minutes is;

V(in³) = (¹/₆₁₀₂₄)a $e^{\frac{1}{60}bt(h)$

We are given that;

Volume of microbial culture is observed to increase according to the formula;

V = e^(t)

where;

t is in seconds

V is in m³

We want to now express V in in³ and t in minutes.

Now, from conversions;

1 m³ = 61024 in³

Also; 1 second = 1/60 minutes

according to formula for exponential decay, we know that;

V = ae^(bt)

Thus, we have;

61024V = ae^(¹/₆₀b(t(h))

V(in³) = (¹/₆₁₀₂₄)a $e^{\frac{1}{60}bt(h)$

Read more about subject of formula at; brainly.com/question/790938

3 0

2 years ago

Steam at 4 MPa and 350°C is expanded in an adiabatic turbine to 125kPa. What is the isentropic efficiency (percent) of this turb

guajiro [1.7K]

Answer:

$\eta_{turbine} = 0.603 = 60.3\%$

Explanation:

First, we will find actual properties at given inlet and outlet states by the use of steam tables:

AT INLET:

At 4MPa and 350°C, from the superheated table:

h₁ = 3093.3 KJ/kg

s₁ = 6.5843 KJ/kg.K

AT OUTLET:

At P₂ = 125 KPa and steam is saturated in vapor state:

h₂ = $h_{g\ at\ 125KPa}$ = 2684.9 KJ/kg

Now, for the isentropic enthalpy, we have:

P₂ = 125 KPa and s₂ = s₁ = 6.5843 KJ/kg.K

Since s₂ is less than $s_g$ and greater than $s_f$ at 125 KPa. Therefore, the steam is in a saturated mixture state. So:

$x = \frac{s_2-s_f}{s_{fg}} \\\\x = \frac{6.5843\ KJ/kg.K - 1.3741\ KJ/kg.K}{5.91\ KJ/kg.K}\\\\x = 0.88$

Now, we will find $h_{2s}$ (enthalpy at the outlet for the isentropic process):

$h_{2s} = h_{f\ at\ 125KPa}+xh_{fg\ at\ 125KPa}\\\\h_{2s} = 444.36\ KJ/kg + (0.88)(2240.6\ KJ/kg)\\h_{2s} = 2416.088\ KJ/kg$

Now, the isentropic efficiency of the turbine can be given as follows:

$\eta_{turbine} = \frac{h_1-h_2}{h_1-h_{2s}}\\\\\eta_{turbine} = \frac{3093.3\ KJ/kg-2684.9\ KJ/kg}{3093.3\ KJ/kg-2416.088\ KJ/kg}\\\\\eta_{turbine} = \frac{408.4\ KJ/kg}{677.212\ KJ/kg}\\\\\eta_{turbine} = 0.603 = 60.3\%$

3 0

2 years ago

Steam at 40 bar and 500o C enters the first-stage turbine with a volumetric flow rate of 90 m3 /min. Steam exits the turbine at

a_sh-v [17]

Answer:

(a) 62460 kg/hr

(b) 17,572.95 kW

Explanation:

Volumetric flow rate, G = 30 m³ / 1 min => 90 / 60 => 1.5

Calculate for h₁ , h₂ , h₃

h₁ is h at P = 40 bar, 500°C => 3445.84 KJ/Kg

Specific volume steam, ц = 0.086441 m³kg⁻¹

h₂ is h at P = 20 bar, 400°C => 3248.23 KJ/Kg

h₃ is h at P = 20 bar, 500°C => 3468.09 KJ/Kg

h₄ is hg at P = 0.6 bar from saturated water table => 2652.85 KJ/Kg

Mass flow rate of the steam, m = G / ц

m = 1.5 / 0.086441

m = 17.35 kg/s

mass per hour is m = 62460 kg/hr

Total Power produced by two stages

= m (h₁ - h₂) + m (h₃ - h₁)

= m [(3445.84 - 3248.23) + (3468.09 - 2652.85)]

= m [ 197.61 + 815.24 ]

= 17.35 [1012.85]

= 17,572.95 kW

Rate of heat transfer to the steam through reheater

= m (h₃ - h₂)

= 17.35 x (3468.09 - 3248.23)

= 17.35 x 219.86

= 3,814.57 kW

8 0

3 years ago

To find the quotient of 8 divided by 1/3, multiply 8 by?

iVinArrow [24]

Answer: 8.33333333 or 6.1989778

Explanation:

8 0

2 years ago

Affect the amount and rate the alcohol reaches the<br> bloodstream.

just olya [345]

Answer:

Answer to the following question is as follows;

Explanation:

The amount of alcohol consumption can be influenced by a variety of things, including food.

The proportion and pace at which alcohol reaches the circulation is affected by drinking rate, body mass, and the size of the beverage. Alcohol enters your system as soon as it reaches that first sip, as per the National Institute on Drug Abuse and Alcoholism. After 10 minutes, the results are noticeable.

6 0

3 years ago