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3 years ago

13

Energy that causes a transfer of heat between marterials

1 answer:

notsponge [240]3 years ago

4 0

Thermal radiation is a form of heat transfer because the electromagnetic radiation emitted from the source carries energy away from the source to surrounding (or distant) objects.

I tried, but I hope this helps :)

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In software engineering how do you apply design for change?

Pavlova-9 [17]

Answer:

it is reducely very iloretable chance for a software engineer to give an end to this question

7 0

3 years ago

At a certain location, wind is blowing steadily at 5 mph. Suppose that the mass density of air is 0.0796 lbm/ft3 and determine t

nlexa [21]

Answer:

The radius of a wind turbine is 691.1 ft

The power generation potential (PGP) scales with speed at the rate of 7.73 kW.s/m

Explanation:

Given;

power generation potential (PGP) = 1000 kW

Wind speed = 5 mph = 2.2352 m/s

Density of air = 0.0796 lbm/ft³ = 1.275 kg/m³

Radius of the wind turbine r = ?

Wind energy per unit mass of air, e = E/m = 0.5 v² = (0.5)(2.2352)²

Wind energy per unit mass of air = 2.517 J/kg

PGP = mass flow rate * energy per unit mass

PGP = ρ*A*V*e

$PGP = \rho *\frac{\pi r^2}{2} *V*e \\\\r^2 = \frac{2*PGP}{\rho*\pi *V*e} , r=\sqrt{ \frac{2*PGP}{\rho*\pi *V*e}} = \sqrt{ \frac{2*10^6}{1.275*\pi *2.235*2.517}}$

r = 210.64 m = 691.1 ft

Thus, the radius of a wind turbine is 691.1 ft

PGP = CVᵃ

For best design of wind turbine Betz limit (c) is taken between (0.35 - 0.45)

Let C = 0.4

PGP = Cvᵃ

take log of both sides

ln(PGP) = a*ln(CV)

a = ln(PGP)/ln(CV)

a = ln(1000)/ln(0.4 *2.2352) = 7.73

The power generation potential (PGP) scales with speed at the rate of 7.73 kW.s/m

5 0

3 years ago

What is the diffrence between a small block and a big block lets see if yall know

ivann1987 [24]

Answer:

The difference in weight and size?

Explanation:

It explains itself :P

5 0

2 years ago

The heat required to raise the temperature of m (kg) of a liquid from T1 to T2 at constant pressure is Z T2CpT dT (1) In high sc

a_sh-v [17]

Answer:

(a)

<em>d</em>Q = m<em>d</em>q

<em>d</em>q = $C_p$ <em>d</em>T

$q = \int\limits^{T_2}_{T_1} {C_p} \, dT$ = $C_p$ (T₂ - T₁)

From the above equations, the underlying assumption is that $C_p$ remains constant with change in temperature.

(b)

Given;

V = 2L

T₁ = 300 K

Q₁ = 16.73 KJ , Q₂ = 6.14 KJ

ΔT = 3.10 K , ΔT₂ = 3.10 K for calorimeter

Let $C_{cal}$ be heat constant of calorimeter

Q₂ = $C_{cal}$ ΔT

Heat absorbed by n-C₆H₁₄ = Q₁ - Q₂

Q₁ - Q₂ = m $C_p$ ΔT

number of moles of n-C₆H₁₄, n = m/M

ρ = 650 kg/m³ at 300 K

M = 86.178 g/mol

m = ρv = 650 (2x10⁻³) = 1.3 kg

n = m/M => 1.3 / 0.086178 = 15.085 moles

Q₁ - Q₂ = m $C_p$ ' ΔT

$C_p$ = (16.73 - 6.14) / (15.085 x 3.10)

$C_p$ = 0.22646 KJ mol⁻¹ k⁻¹

6 0

3 years ago

An office building is served by an air-cooled chiller currently operating at 115 tons (404.5 kW). The measured chilled water sup

Andrei [34K]

Answer:

B.197 gpm and 12.4 L/s

Explanation:

Given that

Load Q = 404.5 KW

Water inlet temperature= 6.1 °C

Water outlet temperature= 13.9°C

We know that specific heat for water

$C_p=4.187\ \frac{KJ}{kg.K}$

Now from energy balance

$Q=\dot{m}C_p\Delta T$

by putting the values

$Q=\dot{m}C_p\Delta T$

$404.5=\dot{m}\times 4.187(13.9-6.1)$

$\dot{m}=12.38\ \frac{kg}{s}$ (1 Kg/s = 15.85 gal/min)

We can say that

$\dot{m}=196.31\ \frac{gal}{min}$

We know that

$\dot{m}=\rho\times volume\ flow\ rate$

12.38=1000 x volume flow rate

$volume\ flow\ rate\ = 12.38\times 10^{-3}\ \frac{m^3}{s}$

So

volume flow rate = 12.38 L/s

So the option B is correct.

8 0

3 years ago