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2 years ago

What are the ordinary and absolute temperature scales in the SI and the English system?

Engineering

2 answers:

Romashka-Z-Leto [24]2 years ago

8 0

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Aleks [24]2 years ago

7 0

Answer:

The answer is below

Explanation:

Absolute temperature is the temperature of a body with reference to 0 as the absolute zero.

English system of measurement are measurements used in united states while the S.I units are units that are universally accepted.

The temperature scale in S.I are Celsius (°C) for ordinary temperature and kelvin(K) for absolute temperature while the temperature scale in English system is Fahrenheit (°F) for ordinary temperature and Rankine (R) for absolute temperature.

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A 1-kW electric resistance heater submerged in 10-kg water is turned on and kept on for 15 min. During the process, 400 kJ of he

hichkok12 [17]

Answer:

ΔT= 11.94 °C

Explanation:

Given that

mass of water = 10 kh

Time t= 15 min

Heat lot from water = 400 KJ

Heat input to the water = 1 KW

Heat input the water= 1 x 15 x 60

=900 KJ

By heat balancing

Heat supply - heat rejected = Heat gain by water

As we know that heat capacity of water

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

$Q=mC_p\Delta T$

Now by putting the values

900 - 400 = 10 x 4.187 x ΔT

So rise in temperature of water ΔT= 11.94 °C

6 0

2 years ago

An air standard cycle with constant specific heats is executed in a closed system with 0.003 kg of air and consists of the follo

Vsevolod [243]

Answer:

a) Please see attached copy below

b) 0.39KJ

c) 20.9‰

Explanation:

The three process of an air-standard cycle are described.

Assumptions

1. The air-standard assumptions are applicable.

2. Kinetic and potential energy negligible.

3. Air in an ideal gas with a constant specific heats.

Properties:

The properties of air are gotten from the steam table.

b) T₁=290K ⇒ u₁=206.91 kj/kg, h₁=290.16 kj/kg.

P₂V₂/T₂=P₁V₁/T₁⇒ T₂=P₂T₁/P₁ = 380/95(290K)= 1160K

T₃=T₂(P₃/P₂)⁽k₋1⁾/k =(1160K)(95/380)⁽⁰°⁴/₁.₄⁾ =780.6K

Qin=m(u₂₋u₁)=mCv(T₂-T₁)

=0.003kg×(0.718kj/kg.k)(1160-290)K= 1.87KJ

Qout=m(h₃₋h₁)=mCp(T₃₋T₁)

=0.003KG×(1.005kj/kg.k(780.6-290)K= 1.48KJ

Wnet, out= Qin-Qout = (1.87-1.48)KJ =0.39KJ

c)ηth= Wnet/W₍in₎ =0.39KJ/1.87KJ = 20.9‰

7 0

2 years ago

In order to fill a tank of 1000 liter volume to a pressure of 10 atm at 298K, an 11.5Kg of the gas is required. How many moles o

lesya [120]

Answer:

The molecular weight will be "28.12 g/mol".

Explanation:

The given values are:

Pressure,

P = 10 atm

= $10\times 101325 \ Pa$

= $1013250 \ Pa$

Temperature,

T = 298 K

Mass,

m = 11.5 Kg

Volume,

V = 1000 r

= $1 \ m^3$

R = 8.3145 J/mol K

Now,

By using the ideal gas law, we get

⇒ $PV=nRT$

⇒ $n=\frac{PV}{RT}$

By substituting the values, we get

$=\frac{1013250\times 1}{8.3145\times 298}$

$=408.94 \ moles$

As we know,

⇒ $Moles(n)=\frac{Mass(m)}{Molecular \ weight(MW)}$

or,

⇒ $MW=\frac{m}{n}$

$=\frac{11.5}{408.94}$

$=0.02812 \ Kg/mol$

$=28.12 \ g/mol$

3 0

2 years ago

Các đặc điểm chính của đường dây dài siêu cao áp .

rodikova [14]

Answer:

Đường dây siêu cao áp 500kV: Những chuyện giờ mới kể ... Ngày 27/5/1994, hệ thống đường dây điện siêu cao áp 500kV Bắc - Nam chính thức đưa ... Tại thời điểm đó, các nước như Pháp, Úc, Mỹ khi xây dựng đường dây dài nhất ... và chế ra các máy kéo dây theo đặc thù công việc của từng đơn vị.

Explanation:

8 0

2 years ago

Timken rates its bearings for 3000 hours at 500 rev/min. Determine the catalog rating for a ball bearing running for 10000 hours

svet-max [94.6K]

Answer:

C₁₀ = 6.3 KN

Explanation:

The catalog rating of a bearing can be found by using the following formula:

C₁₀ = F [Ln/L₀n₀]^1/3

where,

C₁₀ = Catalog Rating = ?

F = Design Load = 2.75 KN

L = Design Life = 1800 rev/min

n = No. of Hours Desired = 10000 h

L₀ = Rating Life = 500 rev/min

n₀ = No. of Hours Rated = 3000 h

Therefore,

C₁₀ = [2.75 KN][(1800 rev/min)(10000 h)/(500 rev/min)(3000 h)]^1/3

C₁₀ = (2.75 KN)(2.289)

3 0

2 years ago