Answer:
15.24°C
Explanation:
The quality of any heat pump pumping heat from cold to hot place is determined by its coefficient of performance (COP) defined as

Where Q_{in} is heat delivered into the hot place, in this case, the house, and W is the work used to pump heat
You can think of this quantity as similar to heat engine's efficiency
In our case, the COP of our heater is

Where T_{house} = 24°C and T_{out} is temperature outside
To achieve maximum heating, we will have to use the most efficient heat pump, and, according to the second law of thermodynamics, nothing is more efficient that Carnot Heat Pump
Which has COP of:

So we equate the COP of our heater with COP of Carnot heater

Rearrange the equation

Solve this simple quadratic equation, and you should get that the lowest outdoor temperature that could still allow heat to be pumped into your house would be
15.24°C
Answer:
- <em><u> Land, labor, and capital </u></em>
Explanation:
The <em>factors of production </em>are the resources that are used to produce goods and services.
By definition resources are scarce.
<em>Land</em> includes everything that comes from the land, that can be used as raw material to produce other materials; for instance, water, minerals, wood.
<em>Labor</em> is the work done by anybody, not just at a factory but at any enterpise that produce a good or a service. For instance, the work done by a person in a bank or a restaurant.
<em>Capital</em> is the facilites (buildings), machinery, equipments, tools that the persons use to produce goods or services. For instance, a computer, a chemical reactor, or a pencil.
Nowadays, also entrepreneurship is included as a <em>factor of production</em>, since it is the innovative skill of the entrepeneurs to combine land, labor and capital what permit the production of good and services.
Answer:
True
Explanation:
Nikola Tesla defeated Thomas Edison in the AC/DC battle of electric current.
Answer:
The answer to the question is
The heat transferred in the process is -274.645 kJ
Explanation:
To solve the question, we list out the variables thus
R-134a = Tetrafluoroethane
Intitial Temperaturte t₁ = 100 °C
Initial pressure = 3.5 bar = 350 kPa
For closed system we have m₁ = m₂ = m
ΔU = m×(u₂ - u₁) = ₁Q₂ -₁W₂
For constant pressure process we have
Work done = W =
= P×ΔV = P × (V₂ - V₁) = P×m×(v₂ - v₁)
From the tables we have
State 1 we have h₁ = (490.48 +489.52)/2 = 490 kJ/kg
State 2 gives h₂ = 206.75 + 0.75 × 194.57= 352.6775 kJ/kg
Therefore Q₁₂ = m×(u₂ - u₁) + W₁₂ = m × (u₂ - u₁) + P×m×(v₂ - v₁)
= m×(h₂ - h₁) = 2.0 kg × (352.6775 kJ/kg - 490 kJ/kg) =-274.645 kJ
Answer:
a) 3581.15067 kw
b) 95.4%
Explanation:
<u>Given data:</u>
compressor efficiency = 85%
compressor pressure ratio = 10
Air enters at: flow rate of 5m^3/s , pressure = 100kPa, temperature = 300 K
At turbine inlet : pressure = 950 kPa, temperature = 1400k
Turbine efficiency = 88% , exit pressure of turbine = 100 kPa
A) Develop a full accounting of the exergy increase of the air passing through the gas turbine combustor in kW
attached below is a detailed solution to the given question