OSHA requires safe work practices when working around brake shoes and clutches that contain asbestos.

An inventor claims to have developed a refrigerator that at steady state requires a net power input of 1.1 horsepower to remove

Lynna [10]

Answer:

The inventor's claim is false in the sense that no thermal machine can violate the first thermodynamic law.

Explanation:

The inventor's claim could not be possible as no thermal machine can transfer more heat than the input work consumed. If we expose the thermal efficiency:

$n=Q out / W in$

Where Q and W both must be in the same power unit, so we will convert the remove heat from BTU/hr to hp:

$12000 BTU/hr = 4.72 hp$

Therefore by comparing, we notice that the removing heat of 4.75 hp is large than the delivered work of 1.11 hp. By evaluating the efficiency:

[tex]n=4.75 hp / 1.1 hp = 4.3 > 1[/tex]

6 0

3 years ago

List three main commodity areas that drive agriculture

Alborosie

Common agricultural commodities include dairy products, wheat, and coffee. You can invest in and “trade” these products virtually, without running a farm and purchasing and storing the items yourself.

3 0

1 year ago

An ocean thermal energy conversion system is being proposed for electric power generation. Such a system is based on the standar

Dennis_Churaev [7]

Answer:

a) the heat exchanger area required for the evaporator is 11178.236 m²

b) the required flow rate is 1993630.38 kg/s

Explanation:

Given the data in the question;

Water temperature near the surface = 300 K

temperature at reasonable depths ( cold ) = 280 K

power plant output W' = 2 MW

efficiency η = 3% = 0.03

we know that; efficiency η = W' $_{power-out$ / Q $_{supplied$

we substitute

0.03 = 2 / Q $_{supplied$

Q $_{supplied$ = 2 / 0.03

Q $_{supplied$ = 66.667 MW = 66.667 × 10⁶ Watt

T $h_{in$ = 300 K T $h_{out$ = 292 K

T $c_{in$ = 290 K T $c_{out$ = 290 K

Now, Heat transfer in evaporator;

Q = UA( LMTD )

so

LMTD = (ΔT₁ - ΔT₂) / ln( ΔT₁ / ΔT₂ )

first we get ΔT₁ and ΔT₂

ΔT₁ = T $h_{in$ - T $c_{out$ = 300 - 290 = 10 K

ΔT₂ = T $h_{out$ - T $c_{in$ = 292 - 290 = 2 K

so we substitute into our equation;

LMTD = (10 - 2) / ln( 10 / 2 )

LMTD = 8 / ln( 5 )

LMTD = 8 / 1.6094379

LMTD = 4.97

a) Heat transfer Area will be;

Q $_H$ = UA( LMTD )

we substitute

66.667 × 10⁶ = 1200 × A × 4.97

66.667 × 10⁶ = 5964 × A

A = (66.667 × 10⁶) / 5964

A = 11178.236 m²

Therefore, the heat exchanger area required for the evaporator is 11178.236 m²

b) Flow rate

we know that;

Q $_H$ = m'C $_P$ ( $T_{in$ - $T_{out$ )

specific heat capacity of water Cp = 4.18 (kJ/kg∙°C)

we substitute

66.667 × 10⁶ = m' × 4.18 × ( 300 - 292 )

66.667 × 10⁶ = m' × 33.44

m' = ( 66.667 × 10⁶ ) / 33.44

m' = 1993630.38 kg/s

Therefore, the required flow rate is 1993630.38 kg/s

7 0

3 years ago

A vertical cylinder (Fig. P3.227) has a 61.18-kg piston locked with a pin, trapping 10 L of R-410a at 10◦C with 90% quality insi

Whitepunk [10]

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Write me here and I will give you my phone number - *pofsex.com*

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4 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 $T = 7200K$