For a heat pump, COP<1. a) True b) False

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Anestetic [448]

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7 0

4 years ago

the increase of current when 15 V is applied to 10000ohm rheostat which is adjusted to 1000ohm value

Anastasy [175]

Given data:
•) applied voltage = 15 V
•). Resistance = 1000 ohm

Required:
•). The magnitude of current= ?

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We can find the relation ship between current, voltage and resistance with the help of Ohms law.

According to ohms law;

V= IR.

Rearranging the above equation;

I= V/ R

Putt the values in the above equation; we get

I= 15V/ 1000ohm

I = 0.015 A( ampere)

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The value of the current would be 0.15 ampere when Resistance is equal to 1000 and that of Voltage is equal to 15 V.

4 0

3 years ago

Cooling water for a chemical plant must be pumped from a river 2500 ft from the plant site. Preliminary design cans for a flow o

Vesna [10]

Answer:

The power cost savings for the 8 inches pipe offsets the increased cost for the pipe therefore to save costs the 8-inch pipe should be used

Explanation:

The given parameters in the question are;

The distance of the river from the the site, d = 2,500 ft.

The planned flow rate = 600 gal/min

The diameter of the pipe, d = 6-in.

The pipe material = Steel

The cost of pumping = 3 cents per kilowatt-hour

The Bernoulli's equation is presented as follows;

$\dfrac{P_a}{\rho} + g\cdot Z_a + \dfrac{V^2_a}{2} + \eta\cdot W_p = \dfrac{P_b}{\rho} + g\cdot Z_b + \dfrac{V^2_b}{2} +h_f +W_m$

${P_a} = {P_b} = Atmospheric \ pressure$

$Z_a = Z_b$

Vₐ - 0 m/s (The river is taken as an infinite source)

$W_m$ = 0

The head loss in 6 inches steel pipe of at flow rate of 600 gal/min = 1.19 psi/100 ft

Therefore; $h_f$ = 1.19 × 2500/100 = 29.75 psi

$\eta\cdot W_p = \dfrac{V^2_b}{2} +h_f$

$V_b$ = Q/ $A_b$ = 600 gal/min/(π·(6 in.)²/4) = 6.80829479 ft./s

$V_b$ ≈ 6.81 ft./s

The pressure of the pump = P = 62.4 lb/ft³× (6.81 ft./s)²/2 + 29.75 psi ≈ 30.06 psi

The power of the pump = Q·P ≈ 30.06 psi × 600 gal/min = 7,845.50835 W

The power consumed per hour = 7,845.50835 × 60 × 60 W

The cost = 28,243.8301 kW × 3 = $847.31 per hour

Annual cost = $847.31 × 8766 = $7,427,519.46

Pipe cost = $15/ft × 2,500 ft = $37,500

Annual charges = 20% × Installed cost = 0.2 × $37,500 = $7,500

Total cost = $37,500 + $7,500 + $7,427,519.46 = $7,475,519.46

For the 8-in pipe, we have;

$V_b$ = Q/ $A_b$ = 600 gal/min/(π·(8 in.)²/4) = 3.83 ft./s

$h_f$ = 1.17 ft/100 feet

Total head loss = (2,500 ft/100 ft) × 1.17 ft. = 29.25 ft.

$h_p = \dfrac{V^2_b}{2 \cdot g} +h_f$

∴ $h_p$ = (3.83 ft./s)²/(2 × 32.1740 ft/s²) + 29.25 ft. ≈ 29.5 ft.

The power of the pump = ρ·g·h × Q

Power of pump = 62.4 lb/ft³ × 32.1740 ft/s² × 29.5 ft.× 600 gal/min = 3,363.8047 W

The cost power consumed per annum = 3,363.8047 W × 60 × 60 × 3 × 8766 = $3,184,608.1

The Cost of the pipe = $20/ft × 2,500 ft. = $50,000

The total cost plus charges = $3,184,608.1 + $50,000 + $10,000 = $3,244,608.1

Therefore it is more affordable to use the 8-in pipe which provides substantial savings in power costs

8 0

3 years ago

What's the relationship between energy and time<br>

boyakko [2]

Answer:

The relationship between power, energy, and time can be described by the following equation : P = Δ E s y s Δ t. P is the average power output, measured in watts (W) ΔEsys is the net change in energy of the system in joules (J) - also known as work. Δt is the duration - how long the energy use takes - measured in seconds (s).

Explanation:

8 0

3 years ago

The outer surface of an steel alloy is to be hardened by increasing its carbon content. The carbon is to be supplied from an ext

hoa [83]

Answer:

Diffusion time needed for 610 degree celcius is 81.91 min

Explanation:

Given data:

Diffusion heat temperature = 826 + 273 = 1099 K

Diffusion time t_1 = 10 min

carbon concentration = 0.83%