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pishuonlain [190]
2 years ago
6

Question 2 Given the following gas Volume (%) Component Methane 58.7 Ethane 16.5 9.9 Propane Butanes 5.0 Pentanes 3.5 Hydrogen S

uflide 6.4 100 Total I Calculate: 1. Average molecular weight 1. Density at 320F and 14.7 psia ni. Density at OoC and 760 mm mercury iv. Specific gravity and gross and net heating value​

Engineering
1 answer:
anastassius [24]2 years ago
4 0

Answer

I think 51.96 I hope it's help

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The polycarbonate shell functions as all of the following except:
krok68 [10]

Answer:

what are the answers expect what?

Explanation:

Polycarbonate is commonly used in eye protection, as well as in other projectile-resistant viewing and lighting applications that would normally indicate the use of glass, but require much higher impact-resistance. Polycarbonate lenses also protect the eye from UV light.

6 0
3 years ago
Consider a very long rectangular fin attached to a flat surface such that the temperature at the end of the fin is essentially t
kodGreya [7K]

Answer:

\frac{T-20}{130-20}= e^{-14.28*0.05}

And if we solve for T we got:

T= 20 + 110e^{-14.28*0.05} = 73.86 C

The answer for this case would be T = 73.86 C at 5cm from the base of the fin.

Explanation:

Data given

For this case we have the following data given:

h = 20 \frac{W}{m^2 K} represent the heat transfer coefficient.

p represent the perimeter for this case and would be given by:

p = 2*0.05m +2*0.001m= 0.102m

k = 200 \frac{W}{m C} represent the thermal conductivity

w = 5cm =0.05 m represent the width

h = 1mm =0.001m represent the thickness

A= wh= 0.05m *0.001m = 0.00005 m^2

Solution to the problem

For this case we assume that we have steady conditions, the temperature of the fins varies just in one direction, the heat transfer coefficient not changes with the time and the thermal properties of the fin not change.

We can determine the temperature if the fin at x=5 cm=0.05 m from the base with the following formula:

\frac{T-T_{\infty}}{T_b -T_{\infty}} = e^{-mx}

Where m is a coefficient given by:

m = \sqrt{\frac{hp}{kA}}=\sqrt{\frac{20 W/m^2 C 0.102 m}{200 W/ mC 0.00005 m^2}}= 14.28 m^{-1}

The value of x for this case represent the distance x =5 cm =0.05m

T_b =130 C represent the base temperature

T_{\infty}= 20 represent the temperature of the sorroundings or the ambient.

If we replace we have this:

\frac{T-20}{130-20}= e^{-14.28*0.05}

And if we solve for T we got:

T= 20 + 110e^{-14.28*0.05} = 73.86 C

The answer for this case would be T = 73.86 C at 5cm from the base of the fin.

3 0
3 years ago
What is the meaning of the measurement met?
san4es73 [151]

Answer: MET(Metabolic equivalent of task) is the measurement technique that is based on the ratio of the energy that is being spend by an individual by any physical activity to the energy spent by individual sitting in quiet position.

This ration has some reference in which is ratio is measured.The energy being measured in the activity time of the person considers his/her relative mass at that time. This relative mass is compared in accordance with the  oxygen amount of 3.5 ml per kg/min that is almost equal to energy of the person in the sitting position.

6 0
3 years ago
A 3-oz serving of roasted, skinless chicken breast contains 140 Cal, 27 g of protein, 3 g of fat, 13 mg of calcium, and 64 mg of
Art [367]

Answer:

Explanation:

  • a) Given C [ cal pro fat calc sod]

              [140 27 3  13  64]

  • P = [cal pro fat calc sod]

     [180 4 11 24 662]

  • B = [cal pro fat calc sod]

      [50  5    1  82   20]

To find C+2P+3B = [140 27 3  13  64] + 2[180 4 11 24 662] + 3[50  5    1  82   20]

= [650  54  28  307  1448]

The entries represent skinless chicken breast , One-half cup of potato salad and One broccoli spear.

7 0
3 years ago
Air is compressed adiabatically from p1 1 bar, T1 300 K to p2 15 bar, v2 0.1227 m3 /kg. The air is then cooled at constant volum
sashaice [31]

Answer:

Work done for the adiabatic process = -247873.6 J/kg = - 247.9 KJ/kg

Heat transfer for the constant volume process = - 244.91 KJ/kg

Explanation:

For the first State,

P₁ = 1 bar = 10⁵ Pa

T₁ = 300 K

V₁ = ?

Second state

P₂ = 15 bar = 15 × 10⁵ Pa

T₂ = ?

V₂ = 0.1227 m³/kg

Third state

P₃ = ?

T₃ = 300 K

V₃ = ?

We require the workdone for step 1-2 (which is adiabatic)

And heat transferred for steps 2-3 (which is isochoric/constant volume)

Work done for an adiabatic process is given by

W = K(V₂¹⁻ʸ - V₁¹⁻ʸ)/(1 - γ)

where γ = ratio of specific heats = 1.4 for air since air is mostly diatomic

K = PVʸ

Using state 2 to calculate for k

K = P₂V₂ʸ = (15 × 10⁵)(0.1227)¹•⁴ = 79519.5

We also need V₁

For an adiabatic process

P₁V₁ʸ = P₂V₂ʸ = K

P₁V₁ʸ = K

(10⁵) (V₁¹•⁴) = 79519.5

V₁ = 0.849 m³/kg

W = K(V₂¹⁻ʸ - V₁¹⁻ʸ)/(1 - γ)

W = 79519.5 [(0.1227)⁻⁰•⁴ - (0.849)⁻⁰•⁴]/(1 - 1.4)

W = (79519.5 × 1.247)/(-0.4) = - 247873.6 J/kg = - 247.9 KJ/kg

To calculate the heat transferred for the constant volume process

Heat transferred = Cᵥ (ΔT)

where Cᵥ = specific heat capacity at constant volume for air = 0.718 KJ/kgK

ΔT = T₃ - T₂

We need to calculate for T₂

Assuming air is an ideal gas,

PV = mRT

T = PV/mR

At state 2,

V/m = 0.1227 m³/kg

P₂ = 15 bar = 15 × 10⁵ Pa

R = gas constant for air = 287.1 J/kgK

T₂ = 15 × 10⁵ × 0.1227/287.1 = 641.1 K

Q = 0.718 (300 - 641.1) = - 244.91 KJ/kg

7 0
3 years ago
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