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DIA [1.3K]
3 years ago
11

A heat pump with a COP of 3.15 is used to heat an air-tight house. When running, the heat pump consumes 5 kW of power. If the te

mperature in the house is 7°C when the heat pump is turned on, how long will it take for the heat pump to raise the temperature of the house to 22°C? Assume the entire mass within the house (air, furniture, etc.) is equivalent to 1500 kg of air. The constant volume specific heat of air at room temperature is cv = 0.718 kJ/kg·°C.
Physics
2 answers:
cricket20 [7]3 years ago
6 0

Answer:

1026s, 17.1m suitability aye

Explanation:

Jet001 [13]3 years ago
5 0

Answer: 1026s, 17.1m

Explanation:

Given

COP of heat pump = 3.15

Mass of air, m = 1500kg

Initial temperature, T1 = 7°C

Final temperature, T2 = 22°C

Power of the heat pump, W = 5kW

The amount of heat needed to increase temperature in the house,

Q = mcΔT

Q = 1500 * 0.718 * (22 - 7)

Q = 1077 * 15

Q = 16155

Rate at which heat is supplied to the house is

Q' = COP * W

Q' = 3.15 * 5

Q' = 15.75

Time required to raise the temperature is

Δt = Q/Q'

Δt = 16155 / 15.75

Δt = 1025.7 s

Δt ~ 1026 s

Δt ~ 17.1 min

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This question involves the concepts of dynamic pressure, volume flow rate, and flow speed.

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First, we will use the formula for the dynamic pressure to find out the flow speed of water:

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Therefore,

v=\sqrt{\frac{2(379212\ Pa)}{1000\ kg/m^3}}

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Now, we will use the formula for volume flow rate of water coming from the hose to find out the time taken by the pool to be filled:

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where,

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A = Area of the mouth of hose = \frac{\pi (0.015875\ m)^2}{4} = 1.98 x 10⁻⁴ m²

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V = [\frac{\pi (9.144\ m)^2}{4}][1.524\ m] = 100.1 m³

Therefore,

t = \frac{(100.1\ m^3)}{(1.98\ x\ 10^{-4}\ m^2)(27.54\ m/s)}\\\\

<u>t = 18353.5 s = 305.9 min = 5.1 hours</u>

Learn more about dynamic pressure here:

brainly.com/question/13155610?referrer=searchResults

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