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zhenek [66]
3 years ago
15

10. In CCS, Northing and Easting are

Engineering
1 answer:
NeTakaya3 years ago
4 0
C. Positive or negative depending on the zone
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A mass of 5 kg of saturated liquid-vapor mixture of water is contained in a piston-cylinder device at 125 kPa. Initially, 2 kg o
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Answer:

See attached pictures.

Explanation:

See attached pictures for detailed explanation.

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4 years ago
) A brick of clay is being dried in a batch dryer using constant drying conditions. You have been asked to determine the amount
dimaraw [331]

Answer:

The drying time is calculated as shown

Explanation:

Data:

Let the moisture content be = 0.6

the free moisture content be = 0.08

total moisture of the clay  = 0.64

total drying time for the period = 8 hrs

then if the final dry and wet masses are calculated, it follows that

t = (X0+ Xc)/Rc) + (Xc/Rc)* ln (Xc/X)

 = 31.3 min.

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3 years ago
A CUSTOMER BRINGS HER CAR INTO THE
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a bc if the bulbs are in a bad conditio. than u know that u dont have to remove it but only repair it.

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Air is contained in a cylinder device fitted with a piston-cylinder. The piston initially rests on a set of stops, and a pressur
Veseljchak [2.6K]

Answer:

The amount of heat transferred to the air is 340.24 kJ

Explanation:

From P-V diagram,

Initial temperature T1 = 27°C

Initial pressure P1 = 100 kPa

final pressure P3 = P2 = 300 kPa

volume at point 2, V2 = V1 = 0.4 m³

final temperature T2 = T3 = 1200 K

To determine the final pressure V3, use ideal gas equation

PV = mRT

Where R is the specific gas constant = 0.2870 KPa m³ kg K

But,

from initial condition, mass m = PV/RT

m = (P1*V1)/R*T1

T1 = 27+273 = 300K

m = (100*0.4)/(0.2870*300) = 0.4646 kg

Then;

Final volume V3 = mRT3/P3

V3 = (0.4646*0.2870*1200)/300

V3 = 0.5334 m³

Total work done W is determined where there is volume change which from point 2 to 3.

W = P3*(V3-V2)

W = 300*(0.5334-0.4) = 40.02 kJ

To get the internal energy, the heat capacity at room temperature Cv is 0.718 kJ/kg K

∆U = m*Cv*(T2-T1)

∆U = 0.4646*0.718(1200-300)

∆U = 300.22 kJ

The heat transfer Q = W + ∆U

Q = 40.02 + 300.22 = 340.24 kJ

Determine the amount of heat transferred to the air, in kJ, while increasing the temperature to 1200 K is 340.24 kJ

The attached file shows the Pressure - Volume relationship (P -V graph)

7 0
4 years ago
Thomas Edison modification of the filament of the light bulb is an example of the "identify the problem" step of the engineering
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