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Goryan [66]
3 years ago
13

Consider a cubical furnace with a side length of 3 m. The top surface is maintained at 700 K. The base surface has emissivity of

0.90 and is maintained at 950 K. The side surface is black and is maintained at 450 K. Heat is supplied from the base surface at a rate of 340 kW. Determine the emissivity of the top surface and the net rates of heat transfer between the top and bottom surfaces, and between the bottom and side surfaces. Answers: 0.44; 54.4 kW; 285.6 kW

Engineering
1 answer:
Ket [755]3 years ago
8 0

Answer:

Check the explanation

Explanation:

Assumptions.

1. The surfaces are diffuse, may and opaque

2. steady operating conditions exist

3. Heat transfer from and to the surfaces is only due to Radiation  

Consider the base surface to be surface 2 the top surface to be surface and the side surfaces to surface 3 1. cubical furnace can be considered to be three-surface enclosure. the areas and black body emissive powers of surfaces can be calculated as seen in the attached images below.

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Consider the cascade of the three LTI systems having impulse responses: h-1(t) = e^-tu(t + 3) h_2(t) = rect((1 -1)/2) h_3(t) = d
omeli [17]

Explanation:

There are two ways to find out the equivalent impulse response of the system.

1. Convolution in time domain

2. Simple multiplication in Laplace domain

2nd method is efficient, easy and is less time consuming.

Step 1: Take the Laplace transform of the given three impulse response functions to convert time domain signals into s-domain

Step 2: Once we get signals in s-domain, multiply them algebraically to get the equivalent s-domain response.

Step 3: Take inverse Laplace transform of the equivalent impulse response to convert from s-domain into time domain.

Solution using Matlab:

Step 1: Take Laplace Transform

Ys1 =  1/(s + 1)

Ys2 =  1/s - exp(-s/2)/s

Ys3 =  exp(-3*s)

Step 2: Multiplication in s-domain

Y =  (exp(-(7*s)/2)*(exp(s/2) - 1))/(s*(s + 1))

Step 3: Inverse Laplace Transform (Final Solution in Time Domain)

h =  heaviside(t - 7/2)*(exp(7/2 - t) - 1) - heaviside(t - 3)*(exp(3 - t) - 1)

8 0
3 years ago
A piston/cylinder contains 1.5 kg of water at 200 kPa, 150°C. It is now heated by a process in which pressure is linearly relate
Fofino [41]

Answer:

final volume V2 = 0.71136 m³

work done in process W = -291.24 kJ

heat transfer Q = 164 kJ

Explanation:

given data

mass = 1.5 kg

pressure p1 = 200 kPa

temperature t1 = 150°C

final pressure p2 = 600 kPa

final temperature t2 = 350°C

solution

we will use here superheated water table that is

for pressure 200 kPa and 150°C temperature

v1 = 0.95964 m³/kg

u1 = 2576.87 kJ/kg

and

for pressure 600 kPa and 350°C temperature

v2 = 0.47424 m³/kg

u2 = 2881.12 kJ/kg

so v1 is express as

V1 = v1 × m    ............................1

V1 = 0.95964 × 1.5

V1 = 1.43946 m³

and

V2 = v2 × m    ............................2

V2 = 0.47424 × 1.5

final volume V2 = 0.71136 m³

and

W = P(avg) × dV      .............................3

P(avg) = \frac{p1+p2}{2}    = \frac{200+600}{2} = 400 × 10³

put here value

W = 400 × 10³ × (0.71136 - 1.43946 )

work done in process W = -291.24 kJ

and

heat transfer is

Q = m × (u2 - u1)  + W       .............................4

Q = 1.5 × (2881.12 - 2576.87)  + 292.24

heat transfer Q = 164 kJ

7 0
3 years ago
. Consider the single-engine light plane described in Prob. 2. If the specific fuel consumption is 0.42 lb of fuel per horsepowe
Trava [24]

Answer:

Hence the Range and Endurance of single engine plane is given by

650.644 miles and 5.3528 hrs at standard sea level.

Explanation:

Given :

A single engine light plane with ,

Specific fuel consumption 0.42lb/hr/hp.

Fuel capacity =44 gal.

Gross weight =3400 lb.

To find :

Range and Endurance of the plane.

Solution:

Consider  all standard measures of standard single engine propeller plane

as

Wing span =35.8 fts.

Wing swing area=174 sq ft

parasite drag coefficient  =Cd.o.=0.025

Oswald's eff. factor= 0.8

ρ=0.002377= corresponds to standard sea level constant.

Now

Formula for Range is given by, Breguent formula.

R=(η/c)  *(Cl/Cd)*ln(W1/W0)

here η is Oswald's constant,

Now calculating lift(Cl) and drag coefficient (Cd)

Cl=W/(1/2*ρ*v^2*S)

W=Gross weight

ρ=0.002377

Assume v=200 ft/sec normally,

S=174 Sq .ft.

CI=3400/(1/2*0.002377*200*200*174)

=6800/16543.9

=0.4110

Now calculating drag constant,

AR=(wing span)^2/wing swing area

=(35.8)^2/174

=7.37

Now

Drag Coefficient

Cd=Cd.o.+ (Cl^2)/(pie*e*AR)

=0.025+(0.4110)^2/(3.142*0.8*7.36)

=0.0342

Given that 44 gal fuel capacity and in Aviation weight of fuel is 5.64 lb/gal

hence weight of fuel=W1=3400- (44*5.64)

=3151.84

Now

for specific fuel consumption=0.42  lb/hp/hr

=0.42  lb*(1/550 ft)*(1/3600)sec

=2.12 *10^-7 lb/ft/sec

Now further calculating range

R=(η/c)  *(Cl/Cd)*ln(W1/W0)

={0.8/(2.12*10^-7)}*(0.4110/0.0342)*ln(3151.84/3400)

=0.024908/0.072504

=0.34354*10^7

=3.4353 *10^6 fts.

1mi =5280 ft

=(3.4353/5280)*10^6

=650.644 miles

Now

For Endurance

E=(η/c)*{(Cl^3/2)/Cd}*(2*ρ*S)^1/2*[1/(W1)^1/2  -1/(W0)^1/2].

=(0.8/2.12*10^-7)*{(0.4110^3/2)/0.0342}*(2*0.002377*174)^1/2*[1/(3151.84)^1/2  -1/(3400)^1/2]

=3.7735*10^6*7.7043*0.8272*0.0006629

=0.01927*10^6

=1.927*10^4 sec

here 1hr =3600 sec

E=(1.927/3600)*10^4

=5.3528 hrs

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nata0808 [166]

Answer:

D. protruding steel rebars .. #answerwithquality #BAL

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