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3 years ago

Question # 3

Engineering

2 answers:

oksian1 [2.3K]3 years ago

7 0

False. Because large corporation hires the general contractor

tino4ka555 [31]3 years ago

4 0

Answer:

I think it's False.

Apologies if I am wrong.

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1. A copper block of volume 1 L is heat treated at 500ºC and now cooled in a 200-L oil bath initially at 20◦C. Assuming no heat

MaRussiya [10]

Answer:

final temperature T = 24.84ºC

Explanation:

given data

copper volume = 1 L

temperature t1 = 500ºC

oil volume = 200 L

temperature t2 = 20ºC

solution

Density of copper $\rho$ cu = 8940 Kg/m³

Density of light oil $\rho$ oil = 889 Kg/m³

Specific heat capacity of copper Cv = 0.384 KJ/Kg.K

Specific heat capacity of light oil Cv = 1.880 KJ/kg.K

so fist we get here mass of oil and copper that is

mass = density × volume ................1

mass of copper = 8940 × 1 × $10^{-3}$ = 8.94 kg

mass of oil = 889 × 200 × $10^{-3}$ = 177.8 kg

so we apply here now energy balance equation that is

$M(cu)\times Cv \times (T-T1)_{cu} + M(oil) \times Cv\times (T-T2)_{oil}$ = 0

put here value and we get T2

$8.94\times 0.384 \times (T-500) + 177.8 \times 1.890\times (T-20)$ = 0

solve it we get

T = 24.84ºC

7 0

3 years ago

A petrol engine produces 20 hp using 35 kW of heat transfer from burning fuel. What is its thermal efficiency, and how much powe

hoa [83]

Answer:

efficiency =42.62%

AMOUNT OF POWER REJECTED IS 20.080 kW

Explanation:

given data:

power 20 hp

heat energy = 35kW

power production = 20 hp = 20* 746 W = 14920 Watt [1 hp =746 watt]

$efficiency = \frac{power}{heat\ required}$

$efficiency = \frac{14920}{35*10^3}$

$= 0.4262*10^100$

=42.62%

b) $heat\ rejected = heat\ required - amount\ of\ power\ generated$

$= 35*10^3 - 14920$

= 20.080 kW

AMOUNT OF POWER REJECTED IS 20.080 kW

5 0

3 years ago

g A smooth pipeline with a diameter of 5 cm carries glycerin at 20 degrees Celsius. The flow rate in the pipe is 0.01 m3/s. What

earnstyle [38]

Answer:

The friction factor is 0.303.

Explanation:

The flow velocity ( $v$ ), measured in meters per second, is determined by the following expression:

$v = \frac{4\cdot \dot V}{\pi \cdot D^{2}}$ (1)

Where:

$\dot V$ - Flow rate, measured in cubic meters per second.

$D$ - Diameter, measured in meters.

If we know that $\dot V = 0.01\,\frac{m^{3}}{s}$ and $D = 0.05\,m$ , then the flow velocity is:

$v = \frac{4\cdot \left(0.01\,\frac{m^{3}}{s} \right)}{\pi\cdot (0.05\,m)^{2}}$

$v \approx 5.093\,\frac{m}{s}$

The density and dinamic viscosity of the glycerin at 20 ºC are $\rho = 1260\,\frac{kg}{m^{3}}$ and $\mu = 1.5\,\frac{kg}{m\cdot s}$ , then the Reynolds number ( $Re$ ), dimensionless, which is used to define the flow regime of the fluid, is used:

$Re = \frac{\rho\cdot v \cdot D}{\mu}$ (2)

If we know that $\rho = 1260\,\frac{kg}{m^{3}}$ , $\mu = 1.519\,\frac{kg}{m\cdot s}$ , $v \approx 5.093\,\frac{m}{s}$ and $D = 0.05\,m$ , then the Reynolds number is:

$Re = \frac{\left(1260\,\frac{kg}{m^{3}} \right)\cdot \left(5.093\,\frac{m}{s} \right)\cdot (0.05\,m)}{1.519 \frac{kg}{m\cdot s} }$

$Re = 211.230$

A pipeline is in turbulent flow when $Re > 4000$ , otherwise it is in laminar flow. Given that flow has a laminar regime, the friction factor ( $f$ ), dimensionless, is determined by the following expression: