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ser-zykov [4K]
4 years ago
13

Air is compressed in the compressor of a turbojet engine. Air enters the compressor at 270 K and 58 kPa and exits the compressor

at 465 K and 350 kPa. Take the average specific heat at constant pressure (cp,avg) of air as 1.015 kJ/ kg.K and Rair = 0.287 kJ/ kg.K. Find the mass specific entropy change associated with the compression process
Engineering
1 answer:
tia_tia [17]4 years ago
8 0

Answer:

Recall the Entropy  Balance (Second Law of Thermodynamics) equations for the system below

And for ideal gas, we know that

Change in Entropy of the air kJ/Kg.K, ΔS = S₁ - S ₂ = Cp Ln (T₂/T₁) - R Ln(P₂/P₁)

where R is gas constant =0.287kJ/kg.K and given

Cp=1.015 kJ/ kg.K

T₂ = 465k T₁=270k, P1=58kPa, P2=350kPa

substituting these values into the eqn above, we have

S = (1.015x 0.544) - (0.287x1.798)

S = 0.0363kJ/Kg.K

Hence, the mass specific entropy change associated with the compression process = 0.0363kJ/Kg.K

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A 15-ft beam weighing 570 lb is lowered by means of two cables unwinding from overhead cranes. As the beam approaches the ground
7nadin3 [17]

Answer:

I. Tension (cable A) ≈ 6939 lbf

II. Tension (cable B) ≈ 17199 lbf

Explanation:

Let's begin by listing out the data that we were given:

mass of beam (m) = 570 lb, deceleration (cable A) = -20 ft/s², deceleration (cable B) = -2 ft/s²,

g = 32.17405 ft/s²

The tension on an object is given by the product of mass of the object by gravitational force plus/minus the product of mass by acceleration.

Mathematically represented thus:

T = mg + ma

where:

T = tension, m = mass, g = gravitational force,

a = acceleration

I. For Cable A, we have:

T = mg + ma = (570 * 32.17405) + [570 * (-20)]

T = 18339.2085 - 11400 = 6939.2085

T ≈ 6939 lbf

II. For Cable B, we have:

T = mg + ma = (570 * 32.17405) + [570 * (-2)]

T = 18339.2085 - 1140 = 17199.2085

T ≈ 17199 lbf

4 0
4 years ago
Metal wireways are sheet metal troughs with _____________ for housing and protecting electrical conductors and cable.
levacccp [35]

Answer:

Metal wireways are sheet metal "U"s with removable housing for protecting electrical equipment, wires, and cables.

Explanation:

These are especially used to run wire in manufacturing environments.

5 0
2 years ago
Convert 250 lb·ft to N.m. Express your answer using three significant figures.
vfiekz [6]

Answer:

It will be equivalent to 338.95 N-m

Explanation:

We have to convert 250 lb-ft to N-m

We know that 1 lb = 4.45 N

So foe converting from lb to N we have to multiply with 4.45

So 250 lb = 250×4.45 =125 N

And we know that 1 feet = 0.3048 meter

Now we have to convert 250 lb-ft to N-m

So 250lb-ft=250\times 4.45N\times 0.348M=338.95N-m

So 250 lb-ft = 338.95 N-m

6 0
3 years ago
A parallel plates capacitor is filled with a dielectric of relative permittivity ε = 12 and a conductivity σ = 10^-10 S/m. The c
monitta

Answer:

t = 1.06 sec

Explanation:

Once disconnected from the battery, the capacitor discharges through the internal resistance of the dielectric, which can be expressed as follows:

R = (1/σ)*d/A, where d is is the separation between plates, and A is the area of one of  the plates.

The capacitance C , for a parallel plates capacitor filled with a dielectric of a relative permittivity ε, can be expressed in this way:

C = ε₀*ε*A/d = 8.85*10⁻¹² *12*A/d

The voltage in the capacitor (which is proportional to the residual charge as it discharges through the resistance of the dielectric) follows an exponential decay, as follows:

V = V₀*e(-t/RC)

The product RC (which is called the time constant of the circuit) can be calculated as follows:

R*C = (1/10⁻¹⁰)*d/A*8.85*10⁻¹² *12*A/d

Simplifying common terms, we finally have:

R*C = 8.85*10⁻¹² *12 / (1/10⁻¹⁰) sec = 1.06 sec

If we want to know the time at which the voltage will decay to 3.67 V, we can write the following expression:

V= V₀*e(-t/RC) ⇒ e(-t/RC) = 3.67/10 ⇒ -t/RC = ln(3.67/10)= -1

⇒ t = RC = 1.06 sec.

3 0
3 years ago
A pumping test was made in pervious gravels and sands extending to a depth of 50 ft. ,where a bed of clay was encountered. The n
Vikki [24]

Answer:per minute from the pumping well, a steady state was attained in about 24 hr. The draw-down at a distance of 10 ft. was 5.5 ft. and at 25 ft. was 1.21 ft.

Explanation:

6 0
3 years ago
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