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3 years ago

A cold air standard gas turbine engine with a thermal efficiency of 56.9 % has a minimum pressure of 100 kP

Engineering

1 answer:

Aleks04 [339]3 years ago

4 0

Answer:

a) 5.2 kPa

b) 49.3%

Explanation:

Given data:

Thermal efficiency ( л ) = 56.9% = 0.569

minimum pressure ( P1 ) = 100 kpa

<u>a) Determine the pressure at inlet to expansion process</u>

P2 = ?

r = 1.4

efficiency = 1 - [ 1 / (rp) $\frac{r-1}{r}$ ]

0.569 = 1 - [ 1 / (rp)^0.4/1.4

1 - 0.569 = 1 / (rp)^0.285

∴ (rp)^0.285 = 0.431

rp = 0.0522

note : rp = P2 / P1

therefore P2 = rp * P1 = 0.0522 * 100 kpa

= 5.2 kPa

b) Thermal efficiency

Л = 1 - [ 1 / ( 10.9 )^0.285 ]

= 0.493 = 49.3%

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A circular ceramic plate that can be modeled as a blackbody is being heated by an electrical heater. The plate is 30 cm in diame

denis23 [38]

Answer:

Q = 125.538 W

Explanation:

Given data:

D = 30 cm

Temperature $T_\infity = 15$ degree celcius

$T_S = 220 + 273 = 473 K$

Heat coefficient = 12 W/m^2 K

Efficiency 80% = 0.8

$Q = hA(T_S - T_{\infty}) \eta$

$= 12(\frac{\pi}{4} 0.3^2) (473 - 288) 0.8$

Q = 125.538 W

5 0

3 years ago

"Water is flowing in a metal pipe. The pipe OD (outside diameter) is 61 cm. The pipe length is 120 m. The pipe wall thickness is

Katarina [22]

Answer:

Total wight =640.7927 KN

Explanation:

Given that

do= 61 cm

L =120

t= 0.9 cm

That is why inner diameter of the pipe

di= 61 - 2 x 0.9 cm

di=59.2 cm

Water density ,ρ = 1 kg/L = 1000 kg/m³

Weight of the pipe ,wt = 2500 N/m

wt = 2500 x 120 N = 300,000 N

The wight of the water

wt ' = ρ V g

$wt'=1000\times \dfrac{\pi}{4}\times (0.61^2-0.0592^2)\times 9.81\times 120 N$

wt'=340792.47 N

That is why total wight

Total wight = wt + wt'

Total wight =300,000+ 340792.47 N

Total wight =640,792.47 N

Total wight =640.7927 KN

7 0

4 years ago

A wing component on an aircraft is fabricated from an aluminum alloy that has a plane strain fracture toughness of 27 . It has b

Mamont248 [21]

Answer:

The stress level at which fracture will occur for a critical internal crack length of 6.2mm is 135.78MPa

Explanation:

Given data;

Let,

critical stress required for initiating crack propagation Cc = 112MPa

plain strain fracture toughness = 27.0MPa

surface length of the crack = a

dimensionless parameter = Y.

Half length of the internal crack, a = length of surface crack/2 = 8.8/2 = 4.4mm = 4.4*10-³m

Also for 6.2mm length of surface crack;

Half length of the internal crack = length of surface crack/2 = 6.2/2 = 3.1mm = 3.1*10-³m

The dimensionless parameter

Cc = Kic/(Y*√pia*a)

Y = Kic/(Cc*√pia*a)

Y = 27/(112*√pia*4.4*10-³)

Y = 2.05

Now,

Cc = Kic/(Y*√pia*a)

Cc = 27/(2.05*√pia*3.1*10-³)

Cc = 135.78MPa

The stress level at which fracture will occur for a critical internal crack length of 6.2mm is 135.78MPa

For more understanding, I have provided an attachment to the solution.

4 0

4 years ago

The cult of personality that surrounded Joseph Stalin in the Soviet Union led soviet citizens to believe that there was undisput

evablogger [386]

Answer:

The cult of personality that surrounded Joseph Stalin in the Soviet Union led soviet citizens to believe that there was undisputed support for Stalin both among the government and the common people. In turn, this fueled self-censorship and made political change harder. This cult of personality was achieved through propaganda and censorship, as the Communist Party had control of all mass media. This desire to make himself a "god-like" figure was also an attempt to increase acceptance of communism among the people and to boost morale.

Explanation:

7 0

3 years ago

A pool of contaminated water is lined with a 40 cm thick containment barrier. The contaminant in the pit has a concentration of

konstantin123 [22]

This question is incomplete, the complete question is;

A pool of contaminated water is lined with a 40 cm thick containment barrier. The contaminant in the pit has a concentration of 1.5 mol/L, while the groundwater circulating around the pit flows fast enough that the contaminate concentration remains 0. There is initially no contaminant in the barrier material at the time of installation. The governing second order, partial differential equation for diffusion of the contaminant through the barrier is:

dC/dt = D( d²C / dz²)

where c(z,t) represent the concentration of containment of any depth into the barrier at anytime and D is the diffusion coefficient (a constant) for the containment in the barrier material.

a) write all boundary and initial conditions needed to solve this equation for C(z, t)

b) Find the steady state solution (infinite time) for C(z)

Answer:

a) At t = 0, z= 0, c = 1.5 mol/L

at t =0, z = 0.4m, c = 0 mol/L

b) C(z) = z² - 4.15z + 1.5

Explanation:

The boundary and initial conditions are as follows

At t = 0, z= 0, c = 1.5 mol/L

at t =0, z = 0.4m, c = 0 mol/L

The governing second order, partial differential equation for diffusion of the contaminant through the barrier is :

(dC/dt) = D*(d²C/dz²) ..............equ(1)

For steady state, above equation becomes,

(d²C/dz²) =0

Integrating above equation,

(dC/dz) = Z + C1 { where C1 is integration constant) }

again integrating above equation,

C = z² + C1*z + C2 ...................equ(2)

applying boundary condition : at t =0, z= 0, c = 1.5 mol/L, to above equation

C = z² + C1*z + C2

1.5 = 0 + 0*0 + c2

C2 = 1.5

applying boundary condition : at t =0, z= 0.4m, c = 0 mol/L, to equation (2) ,

0 = 0.4² + C1*0.4 + 1.5

0 = 0.16 + 0.4C1 + 1.5

0.4C1 = - 1.66

C1 = -1.66/0.4

C1 = -4.15

So, the steady state solution for C(z) is:

C(z) = z² - 4.15z + 1.5

6 0

3 years ago