A (an) __________________________ is a series of refinements to an original idea or product

What is this spray pattern defect most likely caused by:

DiKsa [7]

Answer:

fluid nozzle that is too large

6 0

2 years ago

A household refrigerator that has a power input of 450 W and a COP of 1.5 is to cool five large watermelons, 10 kg each, to 8°C.

Andreyy89

Answer:

6222.22 sec

Explanation:

Given data the power input to the refrigerator is 450 W

The COP of refrigerator is 1.5

Temperature $T_1=8^{\circ}C$

$T_2=28^{\circ}C$

mass of watermelon =10 kg

specific heat =4.2 KJ/kg°C

The amount of heat removed from 5 watermelon

$Q=mc_pdt=5\times 10\times 4.2\times (28-8)=4200 KJ$

We know that $COP=\frac{Q_1}{W}$

$1.5=\frac{Q_1}{450}$

$Q_1=675$ W=0.675 KW

so time required to cool the watermelon is

$t=\frac{Q_1}{Q_2}=\frac{4200}{0.675}=6222.22 sec$

4 0

3 years ago

g Consider a thin opaque, horizontal plate with an electrical heater on its backside. The front end is exposed to ambient air th

xxTIMURxx [149]

Answer:

The electrical power is 96.5 W/m^2

Explanation:

The energy balance is:

Ein-Eout=0

$qe+\alpha sGs+\alpha skyGsky-EEb(Ts)-qc=0$

if:

Gsky=oTsky^4

Eb=oTs^4

qc=h(Ts-Tα)

$\alpha s=\frac{\int\limits^\alpha _0 {\alpha l Gl} \, dl }{\int\limits^\alpha _0 {Gl} \, dl }$

$\alpha s=\frac{\int\limits^\alpha _0 {\alpha lEl(l,5800 } \, dl }{\int\limits^\alpha _0 {El(l,5800)} \, dl }$

if Gl≈El(l,5800)

$\alpha s=(1-0.2)F(0-2)+(1-0.7)(1-F(0-2))$

lt= 2*5800=11600 um-K, at this value, F=0.941

$\alpha s=(0.8*0.941)+0.3(1-0.941)=0.77$

The hemispherical emissivity is equal to:

$E=(1-0.2)F(0-2)+(1-0.7)(1-F(0-2))$

lt=2*333=666 K, at this value, F=0

$E=0+(1-0.7)(1)=0.3$

The hemispherical absorptivity is equal to:

$qe=EoTs^{4}+h(Ts-T\alpha )-\alpha sGs-\alpha oTsky^{4}=(0.3*5.67x10^{-8}*333^{4})+10(60-20)-(0.77-600)-(0.3*5.67x10^{-8}*233^{4})=96.5 W/m^{2}$

3 0

3 years ago

Some aircraft component is fabricated from an aluminum alloy that has a plane strain fracture toughness of 35 MPa m . It has bee

GaryK [48]

Answer: Fracture will not occur since Kc (32.2 MPa√m) ∠ KIc (35 MPa√m).

Explanation:

in this question we are asked to determine if an aircraft will fracture for a given fracture toughness.

let us begin,

from the question we have that;

stress = 325 MPa

fracture toughness (KIc) = 35 MPa√m

the max internal crack length = 1.0 m

using the formula;

Y = KIc/σ√(πα) ---------------(1)

solving for Y we have;

Y = 35 (MPa√m) / 250 (MPa) √(π × 2×10⁻3/2m)

Y = 2.50

so to calculate the fracture roughness;

Kc = Y × σ√(πα) = 2.5 × 3.25√(π × 1×10⁻³/2) = 32.2 MPa√m

Kc = 32.2 MPa√m

From our results we can say that fracture will not occur since Kc (32.2 MPa√m) is less than KIc (35 MPa√m) of the material.

cheers i hope this helps!!!!

8 0

3 years ago

The explosion of a hydrogen bomb can be approximated by a fireball with a temperature of 7200 K, according to a report published

Iteru [2.4K]

Answer:

a

The rate of radiation of the energy is $E_r = 1.523747635*10^9 W/m^2$

b

The irradiation is $G =46.177\ kW/m^2$

c

The amount of energy absorbed is $E_B = 461.772 KJ$

d

The oak Tree would catch fire because the temperature of the blast(7200 K) is higher than the flammability limit (650 K) of the oak tree and secondly the thickness is very small

Explanation:

From the question we are told that

The temperature is $T = 7200K$

The diameter of the ball is $d = 1.5 km = 1.5 *1000 = 1500m$

Hence the radius = $= \frac{1500}{2} = 750m$

The total energy radiated can be mathematically represented as

$E = \sigma A T^4$

Where $\sigma$ is the Stefan-Boltzmann constant $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = 5.67*10^{-8} W \ \cdot m^{-2} K^{-4}$

A is the area of a sphere = $\pi d^2 = 3.142 * 1500^2 = 7.069500 *10 ^6\ m^2$

Substituting values we have

$E = 5,67*10^{-8} * 7.069500*10^6 * 7200^4$

$=1.077*10^{15} W$

Now the state of the energy is mathematically represented as

$Rate \ of \ energy \ radiation (E_r)= \frac{E}{A} = \sigma T^4$

$= 5.67*10^{-8} * 7200^2$

$= 1.523747635*10^9 W/m^2$

A sketch illustrating the b part of the question is shown on the first uploaded image

looking at the height at which the blast occurs(16km) as compared to the height of the wall we notice that the height of the wall is negligibly small

from the diagram x can be calculated as follows

$x = \sqrt{40^2 + 16^2}$

$= 43.0813 Km$

This value of x represents the radius of the blast(assuming it is spherical ) when it is at that wall

Now the irradiation G is mathematically represented as

$G = \frac{E}{4 \pi r^2}$

Here r = 43.0813 Km = 43.0813 × 1000 = 43081.3 m

$G= \frac{1.077*10^15}{4 \pi (431081.3^2)}$

$G =46.177\ kW/m^2$

Generally the amount of energy absorbed can be mathematically represented as

$Amount \ of \ energy \ absorbed \ (E_B ) = G * t$

Where t is the time taken

Therefore $E_B = 46.177 *10 = 461.77 KJ$

6 0

3 years ago