Please solve part two

compressors, the gas is often cooled while being compressed to reduce the power consumed by the compressor. explain how cooling

ASHA 777 [7]

The amount of work done by steady flow devices varies with the particular gas volume. The kinetic energy of gas particles decreases during cooling.

When the gas is subjected to intermediate cooling during compression, the gas specific volume is reduced, which lowers the compressor's power consumption. Compression is less adiabatic and more isothermal because the compressed gas must be cooled between stages since compression produces heat. The system's thermodynamic cycle's cold sink temperature is lowered by cooling the compressor coils. By increasing the temperature difference between the heat source and the cold sink, this improves efficiency.

Learn more about thermodynamics here-

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8 0

1 year ago

Technician A says mismatching tires of the same size on a heavy vehicle will generally not affect ABS operation. Technician B sa

marysya [2.9K]

Technician A is correct

3 0

3 years ago

The electron concentration in silicon at T = 300 K is given by

puteri [66]

Answer:

$E=1.44*10^-7-2.6exp(\frac{-x}{18} )v/m$

Explanation:

From the question we are told that:

Temperature of silicon $T=300k$

Electron concentration $n(x)=10^{16}\exp (\frac{-x}{18})$

$\frac{dn}{dx}=(10^{16} *(\frac{-1}{16})\exp\frac{-x}{16})$

Electron diffusion coefficient is $Dn = 25cm^2/s \approx 2.5*10^{-3}$

Electron mobility is $\mu n = 960 cm^2/V-s \approx0.096m/V$

Electron current density $Jn = -40 A/cm^2 \approx -40*10^{4}A/m^2$

Generally the equation for the semiconductor is mathematically given by

$Jn=qb_n\frac{dn}{dx}+nq \mu E$

Therefore

$-40*10^{4}=1.6*10^{-19} *(2.5*10^{-3})*(10^{16} *(\frac{-1}{16})\exp\frac{-x}{16})+(10^{16}\exp (\frac{-x}{18}))*1.6*10^{-19}*0.096* E$

$E=\frac{-2.5*10^-^7 exp(\frac{-x}{18})+40*10^{4}}{1.536*10^-4exp(\frac{-x}{18} )}$

$E=1.44*10^-7-2.6exp(\frac{-x}{18} )v/m$

7 0

2 years ago

Limited time only for christmas give yourself free 100 points YES YES muhahahahhaha

Setler79 [48]

Answer:

Thank you so much and may god bless you.

6 0

2 years ago

Read 2 more answers

Determine the design moment strength for a W21x73 steel beam with a simple span of 18 ft when lateral bracing for the compressio

SVETLANKA909090 [29]

This question is incomplete, the complete question is;

Determine the design moment strength (ϕMn) for a W21x73 steel beam with a simple span of 18 ft when lateral bracing for the compression flange is provided at the ends only (i.e., Lb = 18 ft). Report the result in kip-ft.

Use Fy=50 ksi and assume Cb=1.0 (if needed).

Answer: the design moment strength for the W21x73 steel beam is 566.25 f-ft

Explanation:

Given that;

section W 21 x 73 steel beam;

now from the steel table table for this section;

Zx = Sx = 151 in³

also given that; fy = 50 ksi and Cb = 1.0

QMn = 0.9 × Fy × Zx

so we substitute

QMn = 0.9 × 50 × 151

QMn = 6795 k-inch

we know that;

12inch equals 1 foot

so

QMn = 6795 k-inch / 12

QMn = 566.25 f-ft

Therefore the design moment strength for the W21x73 steel beam is 566.25 f-ft

7 0

3 years ago