The pressure drop across a valve through which air flows is expected to be 10 kPa. If this differential were applied to the two

Answer:

The change in entropy of the steam is 2.673 kJ/K

Explanation:

The mass of the liquid-vapor mixture is 1.5 kg

The mass in the liquid phase is calculated as 3/4 × 1.5 kg = 1.125 kg

The mass in the vapor phase is calculated as 1.5 - 1.125 = 0.375 kg

According to the steam tables

At a pressure of 200 kPa (200/100 = 2 bar), the specific entropy of steam is found to be 7.127 kJ/kgK

The entropy of steam can be calculated as specific entropy multiplied by mass = 7.127 × 0.375 = 2.673 kJ/K

Fatec – SP) Let A be a point on line r, which is contained within plane α. It holds true that: a) there is exactly one line that is perpendicular to line r at point A. b) there exists one unique line, not lying in plane α, that is parallel to line r. c) there are infinitely many distinct planes that are parallel to plane α and contain line r. d) there are infinitely many distinct planes that are perpendicular to plane α and contain line r. e) there are countless distinct lines contained within plane α that are parallel to line r.

Answer:

t = 5.27 years

Explanation:

Firstly, the corrosion penetration rate is defined by the formula;

CPR = (KW)/(ρAt)

Where;

K = constant based on exposed area A.

W - mass lost over time

t- duration

ρ - density

A - area exposed

From the problem, we have;

W = 7.6kg or 7.6 x 10^(6) mg

CPR = 4 mm/yr

ρ = 4.5 g/cm³

Area = 800 cm²

K is a constant valued at 87.6cm

Rearranging the CPR formula to isolate t, we derive;

t = KW/(ρA(CPR))

t = (87.6 x 7.6 x 10^(6))/(4.5 x 800 x 4) = 46233.3 hours

The duration in question needs to be expressed in years.

Thus, converting hours to years;

There are 8760 hours in a year.

Therefore;

t = 46233.3/8760 = 5.27 years.

Answer:

1.015 ha.

Explanation:

To calculate the landfill area required for 12,000 people producing waste over 10 years, follow these steps:[STEP ONE: Calculate the average solid waste generated per person per year (kg p^-1 ^y(kg/py)).

According to the problem, the average solid waste produced is 2.78 kg per person daily (kg/pd), hence converting to kg/py involves:

2.78 × 365 days = 1014.7 kg/py.

STEP TWO: Determine yearly volume of refuse per person.

Thus, volume = 1014.7 kg/py ÷ 500 kg/m^3 = 2.03 m^3 per person per year.

STEP THREE: Calculate total solid waste volume over 10 years for 12,000 individuals.

Total waste volume over 10 years = 10 × 12,000 × 2.03 = 243,600 m^3.

STEP FOUR: Find the required area for the landfill.

Note: The total height for the landfill should be 20 + 4 = 24m.

Thus, the area for the landfill = 243,600 m^3 / 24m = 10,150 m^2.

If 10,000 m^2 equals 1 ha, then 10,150 m^2 ÷ 10,000 m^2 = 1.015 ha.

(f). Ensure to expand the landfill area for enhancements.

Answer:

T2 ( final temperature ) = 576.9 K

a) 853.4 kJ/kg

b) 1422.3 kJ / kg

Explanation:

given data:

pressure ( P1 ) = 90 kPa

Temperature ( T1 ) = 30°c + 273 = 303 k

P2 = 450 kPa

To determine final temperature in an Isentropic process

----------- ( 1 )

T2 = 303 = 576.9K

The work performed in a piston-cylinder device is calculated using the subsequent formula

   ------- ( 2 )

where: cv = 3.1156 kJ/kg.k for helium gas

             T2 = 576.9K,    T1 = 303 K

substituting values into equation 2

= 853.4 kJ/kg

the work done in a steady flow compressor is determined using this

where: cp ( constant pressure of helium gas ) = 5.1926 kJ/kg.K

             T2 = 576.9 k, T1 = 303 K

plugging values back into equation 3

= 1422.3 kJ / kg