One of the best ways to avoid weather-induced hazards is to ____

The 8-mm-thick bottom of a 220-mm-diameter pan may be made from aluminum (k = 240 W/m ⋅ K) or copper (k = 390 W/m ⋅ K). When use

Artemon [7]

Answer:

For aluminum 110.53 C

For copper 110.32 C

Explanation:

Heat transmission through a plate (considering it as an infinite plate, as in omitting the effects at the borders) follows this equation:

$q = \frac{k * A * (th - tc)}{d}$

Where

q: heat transferred

k: conduction coeficient

A: surface area

th: hot temperature

tc: cold temperature

d: thickness of the plate

Rearranging the terms:

d * q = k * A * (th - tc)

$\frac{d * q}{k * A} = th - tc$

$th = \frac{d * q}{k * A} + tc$

The surface area is:

$A = \frac{\pi * d^2}{4}$

$A = \frac{\pi * 0.22^2}{4} = 0.038 m^2$

If the pan is aluminum:

$th = \frac{0.008 * 600}{240 * 0.038} + 110 = 110.53 C$

If the pan is copper:

$th = \frac{0.008 * 600}{390 * 0.038} + 110 = 110.32 C$

7 0

2 years ago

What is the physical significance of the Reynolds number?. How is defined for external flow over a plate of length L.

yanalaym [24]

Answer:

$Re=\dfrac{\rho\ v\ l}{\mu }$

Explanation:

Reynolds number:

Reynolds number describe the type of flow.If Reynolds number is too high then flow is called turbulent flow and Reynolds is low then flow is called laminar flow .

Reynolds number is a dimensionless number.Reynolds number given is the ratio of inertia force to the viscous force.

$Re=\dfrac{F_i}{F_v}$

For plate can be given as

$Re=\dfrac{\rho\ v\ l}{\mu }$

Where ρ is the density of fluid , v is the average velocity of fluid and μ is the dynamic viscosity of fluid.

Flow on plate is a external flow .The values of Reynolds number for different flow given as

$Reynolds\ number\is \ >\ 5 \times 10 ^5\ then\ flow\ will\ be\ turbulent.$

$Reynolds\ number\is \$

7 0

2 years ago

Bananas are to be cooled from 28°C to 12°C at a rate of 1140 kg/h by a refrigerator that operates on a vapor-compression refrige

Lera25 [3.4K]

Answer:

A) $COP = \frac{16.97}{9.8} = 1.731$

B) $P_{IN} = 0.4763$

C) Second law efficiency 4.85%

exergy destruction for the cycle = 9.3237 kW

Explanation:

Given data:

$T_1 = 28$ degree celcius

$T_2 = 12$ degree celcius

$\dot m = 1140 kg/h$

Power to refrigerator = 9.8 kW

Cp = 3.35 kJ/kg degree C

$A) Q = \dot m Cp \Delta T$

$= 1140 \times 3.35\times (28-12) = 61,104 kJ/h$

$Q_{abs} = 61,104 kJ/h = 16.97 kJ/sec$

$COP = \frac{16.97}{9.8} = 1.731$

b)

$COP ∝ \frac{1}{P_{in}}$

$P_{in}$ wil be max when COP maximum

taking surrounding temperature T_H = 20 degree celcius

$COP_{max} = \frac{T_L}{T_H- T_L} = \frac{285}{293 - 285} = 35.625$

we know that

$COP = \frac{heat\ obsorbed}{P_{in}}$

$P_{IN} = \frac{16.97}{35.62} = 0.4763$

c) second law efficiency

$\eta_{11} = \frac{COP_R}{(COP)_max} = \frac{1.731}{35.625} = 4.85\%$

exergy destruction os given as $X = W_{IN} - X_{Q2}$

= 9.8 - 0.473 = 9.3237 kW

8 0

2 years ago

A generator operating at 50 Hz delivers 1 pu power to an infinite bus through a transmission circuit in which resistance is igno

olga55 [171]

Answer:

critical clearing angle = 70.3°

Explanation:

Generator operating at = 50 Hz

power delivered = 1 pu

power transferable when there is a fault = 0.5 pu

power transferable before there is a fault = 2.0 pu

power transferable after fault clearance = 1.5 pu

using equal area criterion to determine the critical clearing angle

Attached is the power angle curve diagram and the remaining part of the solution.

The power angle curve is given as

= Pmax sinβ

therefore : 2sinβo = Pm

2sinβo = 1

sinβo = 0.5 pu

βo = $sin^{-1} (0.5) = 30$ ⁰

also ; 1.5sinβ1 = 1

sinβ1 = 1/1.5

β1 = $sin^{-1} (\frac{1}{1.5} )$ = 41.81⁰

∴ βmax = 180 - 41.81 = 138.19⁰

attached is the remaining solution

The critical clearing angle = $cos^{-1} 0.3372$ ≈ 70.3⁰

3 0

2 years ago

A 75,000 ft3 clarifier is to be used to treat wastewater. The recycle ratio is 50%, the sludge volume index (SVI) is 125, and th

Inessa05 [86]

Answer:

11 hours approximately

Explanation:

We are to calculate mean cell residence time mcrt

= Mass of solid in reactor/mass of solid wasted in a day

Q = Qe + We

Q = 2.5

Qw = 0.5

Qe = 2.5 - 0.5

= 2 MGD

10⁶/svi

= 10⁶/125

= 8000

X = 3500

Xe = 20mg/

1MGD = 0.1337million

Mcrt = 75000x3500/[0.5*8000*10⁶+2*20*10⁶] x 0.1337

= 262500000/[4000000000+40000000} x 0.1337

= 262500000/574800000

= 0.45668 days

= 0.45668 x 24 hours

= 10.9603 hours

Approximately 11 hours

3 0

2 years ago

One of the best ways to avoid weather-induced hazards is to _____.