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jenyasd209 [6]
3 years ago
14

g Warm water in a geothermal heating system enters the pipe of a radiator at 20 psia and 119oF with a flow rate of 235 cfm (ft3/

min). The diameter of the pipe decreases from 3.2" to 2.3" inside the radiator, and heat is transferred from the water to the room at a rate of 1,100 BTU/s. Determine the temperature of the water when it leaves the radiator. Assume any changes in density of the water are negligible.

Physics
1 answer:
Vika [28.1K]3 years ago
4 0

Answer:

See explanation

Explanation:

Notice that the condenser section includes both the hot water and space heater and station (3) is specified as being in the Quality region. Assume that 50°C is a reasonable maximum hot water temperature for home usage, thus at a high pressure of 1.6 MPa, the maximum power available for hot water heating will occur when the refrigerant at station (3) reaches the saturated liquid state. (Quick Quiz: justify this statement). Assume also that the refrigerant at station (4) reaches a subcooled liquid temperature of 20°C while heating the air.

Using the conditions shown on the diagram and assuming that station (3) is at the saturated liquid state

a) On the P-h diagram provided below carefully plot the five processes of the heat pump together with the following constant temperature lines: 50°C (hot water), 13°C (ground loop), and -10°C (outside air temperature)

b) Using the R134a property tables determine the enthalpies at all five stations and verify and indicate their values on the P-h diagram.

c) Determine the mass flow rate of the refrigerant R134a. [0.0127 kg/s]

d) Determine the power absorbed by the hot water heater [2.0 kW] and that absorbed by the space heater [0.72 kW].

e) Determine the time taken for 100 liters of water at an initial temperature of 20°C to reach the required hot water temperature of 50°C [105 minutes].

f) Determine the Coefficient of Performance of the hot water heater [COPHW = 4.0] (defined as the heat absorbed by the hot water divided by the work done on the compressor)

g) Determine the Coefficient of Performance of the heat pump [COPHP = 5.4] (defined as the total heat rejected by the refrigerant in the hot water and space heaters divided by the work done on the compressor)

h) What changes would be required of the system parameters if no geothermal water loop was used, and the evaporator was required to absorb its heat from the outside air at -10°C. Discuss the advantages of the geothermal heat pump system over other means of space and water heating

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alukav5142 [94]

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If he happens to be walking north, then it takes him

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If he's walking in any other direction, it takes him longer than that.

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What is R2 in the circuit?<br> WILL GIVE BRAINLIEST !!!!
Nataly_w [17]

Answer:

1. Rₑq = 4 Ω

2. R₂ = 6 Ω

3. Vₜ = 12 V, V₁ = 12 V, V₂ = 12 V

4. Iₜ = 3 A, I₁ = 1 A, I₂ = 2 A

Explanation:

1. Determination of the equivalent resistance

Voltage (V) = 12 V

Current (I) = 3 A

Resistance (Rₑq) =?

V= IRₑq

12 = 3 × Rₑq

Divide both side by 3

Rₑq = 12 / 3

Rₑq = 4 Ω

Thus, the equivalent resistance (Rₑq) = 4 Ω

2. Determination of R₂.

Equivalent resistance (Rₑq) = 4 Ω

Resistance 1 (R₁) = 12 Ω

Resistance 2 (R₂)

Since the resistor are in parallel arrangement, the value of R₂ can be obtained as follow:

Rₑq = R₁ × R₂ / R₁ + R₂

4 = 12 × R₂ / 12 + R₂

Cross multiply

4(12 + R₂) = 12R₂

48 + 4R₂ = 12R₂

Collect like terms

48 = 12R₂ – 4R₂

48 = 8R₂

Divide both side by 8

R₂ = 48 / 8

R₂ = 6 Ω

3. Determination of the total voltage (Vₜ), V₁ and V₂.

From the question given above, the total voltage is 12 V

Since the resistors are arranged in parallel connection, the same voltage will go through them.

Thus,

Vₜ = V₁ = V₂ = 12 V

4. Determination of the total current (Iₜ), I₁ and I₂

From the question given above, the total current (Iₜ) is 3 A

Next, we shall determine I₁. Since the resistors are arranged in parallel connection, different current will pass through each resistor respective.

Vₜ = V₁ = 12 V

R₁ = 12 Ω

I₁ =?

V₁ = I₁R₁

12 = I₁ ×12

Divide both side by 12

I₁ = 12 / 12

I₁ = 1 A

Next, we shall determine I₂. This can be obtained as follow:

Iₜ = 3 A

I₁ = 1 A

I₂ =?

Iₜ = I₁ + I₂

3 = 1 + I₂

Collect like terms

I₂ = 3 – 1

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The initial surface area is:

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