a circular pile, 19 m long is driven into a homogeneous sand layer. The piles width is 0.5 m. The standard penetration resistanc
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The rate of heat transfer by the air conditioner using constant specific heat of 1.004kj/kg.K is 15.06 kW.
<h3>What is the rate of heat transfer?</h3>Rate of heat transfer is the power rating of the machine.
Work done and changes in potential and kinetic energy are neglected since it is a steady state process.
The specific heat in terms of specific heat capacity and temperature change is given as:
The rate of heat transfer, is then determined as follows:
- Qout = flow rate × specific heat
Qout = 0.75 × 20.08 = 15.06 kW
Therefore, the rate of heat transfer by the air conditioner is 15.06 kW.
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Answer:
F = 9.11 x 10³ N = 9.11 KN
Explanation:
The areas, lengths, young's modulus, and coefficient of linear thermal expansion are given in the diagram. First we find the equivalent change in length due to temperature change:
ΔL = (ΔL)steel + (ΔL)brass + (ΔL)Copper
ΔL = (∝s)(Ls)(ΔT) + (∝b)(Lb)(ΔT) + (∝c)(Lc)(ΔT)
where,
ΔL = Equivalent Change in Length = ?
ΔT = Change in Temperature = 25°C - 12°C = 13°C
Ls = Length of Steel Segment = 300 mm = 0.3 m
Lb = Length of Brass Segment = 200 mm = 0.2 m
Lc = Length of Copper Segment = 100 mm = 0.1 m
Therefore,
ΔL = (12 x 10⁻⁶ °C⁻¹)(0.3 m)(13 °C) + (21 x 10⁻⁶ °C⁻¹)(0.2 m)(13 °C) + (17 x 10⁻⁶ °C⁻¹)(0.1 m)(13 °C)
ΔL = 46.8 x 10⁻⁶ m + 54.6 x 10⁻⁶ m + 22.1 x 10⁻⁶ m
ΔL = 123.5 x 10⁻⁶ m ----------------------- equation (1)
Now, we calculate this deflection in terms of an applied force (F):
ΔL = (F)(Ls)/(Es)(As) + (F)(Lb)/(Eb)(Ab) + (F)(Lc)/(Ec)(Ac)
ΔL = (F)(0.3 m)/(200 x 10⁹ Pa)(200 x 10⁻⁶ m²) + (F)(0.2 m)/(100 x 10⁹ Pa)(450 x 10⁻⁶ m²) + (F)(0.1 m)/(120 x 10⁹ Pa)(515 x 10⁻⁶ m²)
ΔL = F(7.5 x 10⁻⁹ m/N + 4.44 x 10⁻⁹ m/N + 1.61 x 10⁻⁹ m/N)
ΔL = F(13.55 x 10⁻⁹ m/N) --------------------- equation (1)
Comparing equation (1) and equation (2):
123.5 x 10⁻⁶ m = F(13.55 x 10⁻⁹ m/N)
F = (123.5 x 10⁻⁶ m)/(13.55 x 10⁻⁹ m/N)
<u>F = 9.11 x 10³ N = 9.11 KN</u>
Answer:
The rate at which this vacuum cleaner supply energy to the room when running is 285.71 Watts.
Explanation:
power efficiency of electric motor = 70% = 0.70
The power output of the vacuum cleaner == 200 W
The power output of the vacuum cleaner =
The rate at which this vacuum cleaner supply energy to the room when running is 285.71 Watts.