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mario62 [17]
3 years ago
15

An old refrigerator consumes 247 W of power. Assuming that the refrigerator operates for 19 hours everyday, what is the annual o

perating cost of the refrigerator, if the cost of electricity is $0.06 per kWh
Engineering
2 answers:
german3 years ago
6 0

Answer:

The annual operating cost of the refrigerator is $102.78.

Explanation:

Power consumed by the refrigerator = 247 W = 247/1000 = 0.247 kW

Daily operation of the refrigerator = 19 hours

Annual operation of the refrigerator = 365 × 19 = 6,935 hours

Annual energy consumed = 0.247 kW × 6,935 hours = 1712.945 kWh

1 kWh of electricity cost $0.06

1712.945 kWh will cost 1712.945 × $0.06 = $102.78

Annual operating cost = $102.78

BabaBlast [244]3 years ago
3 0
<h2>Answer:</h2>

$102.7767

<h2>Explanation:</h2>

Energy, E, consumed by a device is the product of the power, P, consumed by the device and the duration (time), t, of consumption. i.e;

E = p x t         ----------------(i)

The refrigerator consumes 247W of power and operates 19 hours per day.

This implies that the refrigerator consumes 0.247kW of power and operates 19 hours per day.

This means that the energy used per day is given by substituting p = 0.247kW and t = 19h into equation (i) as follows;

E = 0.247kW x 19h

E = 4.693kWh

From the question, it is given that the cost of electricity is $0.06 per kWh. i.e

$0.06 = 1 kWh

Then;

4.693kWh = 4.693kWh x $0.06 / 1kWh = $0.28158

Therefore, the cost of electricity per day due to the refrigerator is $0.28158.

Now, to get the annual (365 days) cost of electricity, we multiply the result by 365 as follows;

Annual cost = 365 x $0.28158

Annual cost = $102.7767

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Methane gas is 304 C with 4.5 tons of mass flow per hour to an uninsulated horizontal pipe with a diameter of 25 cm. It enters a
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Answer:

a) h_c = 0.1599 W/m^2-K

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d) \dot{Q} = 25.125 W

Explanation:

Non horizontal pipe diameter, d = 25 cm = 0.25 m

Radius, r = 0.25/2 = 0.125 m

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Exit temperature, T₂ = 284 + 273 = 557 K

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A = 2π * 0.125 * 10

A = 7.855 m²

Mass flow rate,

\dot{ m} = 4.5 tons/hr\\\dot{m} = \frac{4.5*1000}{3600}  = 1.25 kg/sec

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\dot{Q} = \dot{m} c_p ( T_1 - T_2)\\\dot{Q} = 1.25 * 1.005 * (577 - 557)\\\dot{Q} = 25.125 W

a) To calculate the convection coefficient relationship for heat transfer by convection:

\dot{Q} = h_c A (T_1 - T_2)\\25.125 = h_c * 7.855 * (577 - 557)\\h_c = 0.1599 W/m^2 - K

Note that we cannot calculate the heat loss by the pipe to the environment without first calculating the surface temperature of the pipe.

c) The surface temperature of the pipe:

Smear coefficient of the pipe, k_c = 0.8

\dot{Q} = k_c A (T_s - T_a)\\25.125 = 0.8 * 7.855 * (T_s - 298)\\T_s = 302 K

b) Heat loss from the pipe to the environment:

H_{loss} = h_c A(T_s - T_a)\\H_{loss} = 0.1599 * 7.855( 302 - 298)\\H_{loss} = 5.02 W

d) The required fan control power is 25.125 W as calculated earlier above

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Answer:

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Explanation:

Given data:

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From above information, the distance between  given two charges d = 2 cm

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Therefore work done by electric field  is 0.06 J

8 0
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