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ozzi
4 years ago
9

Consider a 2.4-kW hooded electric open burner in an area where the unit costs of electricity and natural gas are $0.10/kWh and $

1.20/therm (1 therm = 105,500 kJ), respectively. The efficiency of open burners can be taken to be 73 percent for electric burners and 38 percent for gas burners. Determine the rate of energy consumption and the unit cost of utilized energy for both electric and gas burners.
Engineering
1 answer:
goldfiish [28.3K]4 years ago
4 0

Answer:

1.75 kW

$0.137 kWh

4.61 kW

$3.16 therm

Explanation:

Utilized power input of the burner is

P(ui) = total power input * efficiency

P(ui) = 2400 W * 0.73

P(ui) = 1752 W or 1.75 kW

Unit cost of utilized energy is

C(ui) = Unit cost of electricity/efficiency

C(ui) = $0.1 / 0.73 kWh

C(ui) = $0.137 kWh

Power input to the gas burner is

P(gi) = Utilized power input of the burner / efficiency of the burner

P(gi) = 1.75 / 0.38

P(gi) = 4.61 kW

Unit cost of utilized energy is

C(gi) = Unit cost of gas /efficiency

C(gi) = $1.2 / 0.38 kWh

C(gi) = $3.16 therm

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In an analog addressable system, the determination of the alarm state is performed by the _____?
Nostrana [21]

Answer:

Control panel

Explanation:

The Control Panel is a component of Microsoft Windows which provides the ability to view and change system settings. It consists of a set simple computer programs (applets) that include adding or removing hardware and software, controlling user accounts, changing accessibility options, and accessing networking settings.

6 0
3 years ago
An uninsulated, thin-walled pipe of 100-mm diameter is used to transport water to equipment that operates outdoors and uses the
Viefleur [7K]

Answer:

4.6 mm

Explanation:

Given data includes:

thin-walled pipe diameter = 100-mm =0.1 m

Temperature of pipe T_p = -15° C = (-15 +273)K =258 K

Temperature of water T_w = 3° C = (3 + 273)K = 276 K

Temperature of ice T_i = 0° C = (0 +273)K =273 K

Thermal conductivity (k) from the ice table = 1.94 W/m.K  ;  R = 0.05

convection coefficient Lh_l =2000 W/m².K

The energy balance can be expressed as:

q_{conduction} =q_{convention}

where;

q_{conduction} = \frac{2\pi LK(T_i-T_p)}{In(R/r)}       -------------   equation (1)

q_{convention} = \pi DLh_l(T_w-T_i)  ------------ equation(2)

Equating both equation (1) and (2); we have;

\frac{2\pi LK(T_i-T_p)}{In(R/r)} = \pi DLh_l(T_w-T_i)

Replacing the given data; we have:

\frac{2\pi (1)(1.94)(273-258)}{In(0.05/r)} = \pi (0.1)*2000(276-273)

\frac{182.84}{In(\frac{0.05}{r}) } = 1884.96

In(\frac{0.05}{r})*1884.96 = 182.84

In(\frac{0.05}{r}) = \frac{182.84}{1884.96}

In(\frac{0.05}{r}) =0.0970

\frac {0.05}{r} =e^{0.0970}

\frac {0.05}{r} =1.102

r=\frac{0.05}{1.102}

r = 0.0454

The thickness (t) of the ice layer can now be calculated as:

t = (R - r)

t = (0.05 - 0.0454)

t = 0.0046 m

t = 4.6 mm

6 0
4 years ago
Two steel plates of 15 mm thickness each are clamped together with an M14 x 2 hexagonal head bolt, a nut, and one 14R metric pla
Blababa [14]

Answer:

a) 50 mm

b) 808.24 MN/m

Explanation:

Given:

Thickness of each steel plate = 15mm

a) To find suitable length of bolt, we'll use:

Length of bolt = grip length + height of nut

To find the grip length since there is a washer, we'll use:

Grip length = plate thickness + washer thickness

Since we have 2 plates of 15mm thickness,

Plate thickness = 15 + 15 = 30mm

Using the table, a metric plain washer has a thickness of 3.5mm

Grip length = 30 + 3.5 =33.5 mm

Height of nut: Using table A-31, height of hexagonal nut is 12.8 mm

Therefore,

Length of bolt = 33.5 + 12.8 = 46.3

Rounde up to the nearest 5mm, we'll get 50mm

Length of bolt = 50mm

b) Bolt stiffness:

Threaded length for L ≤ 125mm

LT = 2*d + 6

Where d = 14mm, from table(8-7)

= 2*14 + 6

= 34 mm

Area of unthreaded portion:

Ad= \frac{\pi}{4} d^2 = \frac{pi}{4} * 14^2 = 153.94 mm^2

Length of unthreaded portion in grip:

Ld = 50 - 34 = 16mm

Length of threaded portion in grip:

Lt = 33.5 - 16 = 17.5mm

Bolt stiffness = \frac{A_d A_t E}{A_d l_t + A_t l_d}

= \frac{153.94 * 115 * 207}{(153.94*17.5)+(115*16)} = 808.24

Bolt stiffness = 808.24 MN/m

7 0
3 years ago
What are the five types of civil engineering
KatRina [158]

Answer: The five types of the civil engineering projects is construction and management, geotechnical, structural, transport, water, and architecture

Explanation: Hope this helps

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3 years ago
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Suppose we have two negative charges, one located at the origin and carrying charge −9e, and the other located on the positive x
Elanso [62]

Answer:

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q = 36e x^2/d^2

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q/(d-x)^2 = 9e/d^2

36 x^2/d^2 = 9/d^2 (d-x)^2

36 x^2 = 9 (d-x)^2

x=d/3

q = 36e (d/3)^2/d^2 = 36e/3^2 = 4 e

7 0
4 years ago
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