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Stels [109]
3 years ago
11

Define the terms (a) thermal conductivity, (b) heat capacity and (c) thermal diffusivity

Engineering
1 answer:
IceJOKER [234]3 years ago
6 0

Explanation:

<u>(a)</u>

<u>The measure of material's ability to conduct thermal energy (heat) is known as thermal conductivity.</u> For examples, metals have high thermal conductivity, it means that they are very efficient at conducting heat.<u> The SI unit of heat capacity is W/m.K.</u>

The expression for thermal conductivity is:

q=-\kappa \bigtriangledown T

Where,

q is the heat flux

\kappa is the thermal conductivity

\bigtriangledown T is the temperature gradient.

<u>(b)</u>

<u>Heat capacity for a substance is defined as the ratio of the amount of energy required to change the temperature of the substance and the magnitude of temperature change. The SI unit of heat capacity is J/K.</u>

The expression for Heat capacity is:

C=\frac{E}{\Delta T}

Where,

C is the Heat capacity

E is the energy absorbed/released

\Delta T is the change in temperature

<u>(c)</u>

<u>Thermal diffusivity is defined as the thermal conductivity divided by specific heat capacity at constant pressure and its density. The Si unit of thermal diffusivity is m²/s.</u>

The expression for thermal diffusivity is:

\alpha=\frac{\kappa}{C_p \times \rho}

Where,

\alpha is thermal diffusivity

\kappa is the thermal conductivity

C_p is specific heat capacity at constant pressure

\rho is density

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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
The hot combustion gases of a furnace are separated from the ambient air and its surroundings, which are at 25 oC, by a brick wa
yanalaym [24]

Answer:

T1 = 625.54 K

Explanation:

We are given;

T_α = Tsur = 25°C = 298K

h = 20 W/m².K,

L = 0.15 m

K = 1.2 W/m.K

ε = 0.8

Ts = T2 = 100°C = 373K

T1 = ?

Assumption:

-Steady- state condition

-One- dimensional conduction

-No uniform heat generation

-Constant properties

From Energy balance equation;

E°in - E°out = 0

Thus,

q"cond – q"conv – q"rad = 0

K[(T1 - T2)/L] - h(Ts-T_α) - εσ (Ts⁴ – Tsur⁴)

Where σ is Stephan-Boltzmann constant and has a value of 5.67 x 10^(-8)

Thus;

K[(T1 - T2)/L] - h(Ts-T_α) - εσ (Ts⁴ – Tsur⁴) = 1.2[(T1 - 373)/0.15] - 20(373 - 298] - 0.8x5.67x10^(-8)[373⁴ - 298⁴] = 0

This gives;

(8T1 - 2984) - (1500) - 520.31 = 0

8T1 = 2984 + 1500 + 520.31

8T1 = 5004.31

T1 = 5004.31/8

T1 = 625.54 K

7 0
3 years ago
Consider a cylindrical nickel wire 2.1 mm in diameter and 3.2 × 104 mm long. Calculate its elongation when a load of 280 N is ap
pshichka [43]

Answer:

Total elongation will be 0.012 m

Explanation:

We have given diameter of the cylinder = 2.1 mm

Length of wire L=3.2\times 10^4mm

So radius r=\frac{d}{2}=\frac{2.1}{2}=1.05mm=1.05\times 10^{-3}m

Load F = 280 N

Elastic modulus = 207 Gpa

Area of cross section A=\pi r^2=3.14\times (1.05\times 10^{-3})^2=3.461\times 10^{-6}m^2

We know that elongation in wire is given by \delta =\frac{FL}{AE}, here F is load, L is length, A is area and E is elastic modulus

So \delta =\frac{FL}{AE}=\frac{280\times 32}{3.461\times 10^{-6}\times 207\times 10^9}=0.012m

4 0
3 years ago
A 75-hp (shaft output) motor that has an efficiency of 91.0 percent is worn out and is replaced by a high-efficiency 75-hp motor
Naya [18.7K]

Answer:

4.536hp

Explanation:

The decrease in the heat gain of the room is determined from difference in electrical inputs:

Q = W_{shaft} (\frac{1}{n_{1} } - \frac{1}{n_{2} })\\Q = (75hp)*(\frac{1}{0.91 } - \frac{1}{0.963 })\\\\Q = 4.536 hp

8 0
3 years ago
Describe experimental factors that could be modified, and unalterable properties of materials used.
Sphinxa [80]

Answer:

a. mechanical properties

b. thermal properties

c. chemical properties

d. electical properties

e. magnetic properties

Explanation:

a. The mechanical properties of a material are those properties that involve a reaction to an applied load.The most common properties considered are strength, ductility, hardness, impact resistance, and fracture toughness, elasticity, malleability, youngs' modulus etc.

b. Thermal properties such as boiling point , coefficient of thermal expansion , critical temperature  , flammability  , heat of vaporization , melting point ,thermal conductivity , thermal expansion ,triple point , specific heat capacity

c. Chemical properties such as corrosion resistance , hygroscopy , pH , reactivity , specific internal surface area , surface energy , surface tension

d. electrical properties such as capacitance , dielectric constant , dielectric strength , electrical resistivity and conductivity , electric susceptibility , nernst coefficient (thermoelectric effect) , permittivity  etc.

e. magnetic properties such as diamagnetism,  hysteresis,  magnetostriction , magnetocaloric coefficient , magnetoresistance , permeability , piezomagnetism , pyromagnetic coefficient

3 0
3 years ago
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