Crank OA rotates with uniform angular velocity 0  4 rad/s along counterclockwise. Take OA= r= 0.5

Given: A graphite-moderated nuclear reactor. Heat is generated uniformly in uranium rods of 0.05m diameter at the rate of 7.5 x

sineoko [7]

Answer:

The maximum temperature at the center of the rod is found to be 517.24 °C

Explanation:

Assumptions:

1- Heat transfer is steady.

2- Heat transfer is in one dimension, due to axial symmetry.

3- Heat generation is uniform.

Now, we consider an inner imaginary cylinder of radius R inside the actual uranium rod of radius Ro. So, from steady state conditions, we know that, heat generated within the rod will be equal to the heat conducted at any point of the rod. So, from Fourier's Law, we write:

Heat Conduction Through Rod = Heat Generation

-kAdT/dr = qV

where,

k = thermal conductivity = 29.5 W/m.K

q = heat generation per unit volume = 7.5 x 10^7 W/m³

V = volume of rod = π r² l

A = area of rod = 2π r l

using these values, we get:

dT = - (q/2k)(r dr)

integrating from r = 0, where T(0) = To = Maximum center temperature, to r = Ro, where, T(Ro) = Ts = surface temperature = 120°C.

To -Ts = qr²/4k

To = Ts + qr²/4k

To = 120°C + (7.5 x 10^7 W/m³)(0.025 m)²/(4)(29.5 W/m.°C)

To = 120° C + 397.24° C

5 0

4 years ago

The rectangular frame is composed of four perimeter two-force members and two cables AC and BD which are incapable of supporting

Rama09 [41]

Answer:

Your question is lacking some information attached is the missing part and the solution

A) AB = AD = BD = 0, BC = LC

AC = $\frac{5L}{3}T, CD = \frac{4L}{3} C$

B) AB = AD = BC = BD = 0

AC = $\frac{5L}{3} T, CD = \frac{4L}{3} C$

Explanation:

A) Forces in all members due to the load L in position A

assuming that BD goes slack from an inspection of Joint B

AB = 0 and BC = LC from Joint D, AD = 0 and CD = 4L/3 C

B) steps to arrive to the answer is attached below

AB = AD = BC = BD = 0

AC = $\frac{5L}{3} T, CD = \frac{4L}{3}C$

7 0

4 years ago

Select the correct answer.

Ulleksa [173]

The answer is A. Immediately inform her colleague

4 0

3 years ago

A liquid drug, with the viscosity and density of water, is to be administered through a hypodermic needle. The inside diameter o

vredina [299]

Answer:

(a) The maximum volume flow rate for which the flow will be laminar is 0.0190 cubic meter per second

(b) The pressure drop required to deliver the maximum flow rate is 148962.96 Pascal

(c) The corresponding wall shear stress is 7600 Pascal

Explanation:

Reynolds number = 2299, density of water = 1000kg/m^3, diameter of needle = 0.27mm = 0.00027m, Length of needle = 50mm = 0.05m, viscosity of water = 0.00089kg/ms, area = 0.05m × 0.05m = 0.0025m^2, coefficient of friction = 64 ÷ Reynolds number = 64 ÷ 2299 = 0.028

Velocity = (Reynolds number × viscosity) ÷ (density × diameter) = (2299 × 0.00089) ÷ (1000 × 0.00027) = 2.046 ÷ 0.27 = 7.58m/s

(a) Maximum volume flow rate = velocity × area of needle = 7.58 × 0.0025 = 0.0190 cubic meter per second

(b) Pressure drop = ( coefficient of friction × length × density × velocity^2) ÷ (2 × diameter) = (0.028 × 0.05 × 1000 × 7.58^2) ÷ (2 × 0.00027) = 80.44 ÷ 0.00054 = 148962.96 Pascal

(c) Wall shear stress = (density × volume flow rate) ÷ area = (1000 × 0.0190) ÷ 0.0025 = 7600 Pascal

7 0

3 years ago

7.13 An intersection approach has a saturation flow rate of 1500 veh/h, and vehicles arrive at the approach at the rate of 800 v

Naddik [55]

Answer:

23.34 seconds

Explanation:

Flow rate = 1500

Arrival = 800 vehicle per hour

Cycle c = 60 seconds

Dissipation time = 10 seconds

Arrival time = 800/3600 = 0.2222

Rate of departure = 1500/3600 = 0.4167

Traffic density p = 0.2222/0.4167 = 0.5332

Real time = r

r + to + 10 = c

to = c-r-10 ----1

t0 = p*r/1-p ----2

Equate both 1 and 2

C-r-10 = p*r/1-p

60-r-10 = 0.5332r/1-0.5332

50-r = 0.5332r/0.4668

50-r = 1.1422r

50 = 1.1422r + r

50 = 2.1422r

r = 50/2.1422

r = 23.34 seconds

7 0

3 years ago