Complete Question:
A concrete building slab, on a basement floor is 5 m long, 3 m wide and 0.6 m thick. During the winter, the temperature is normally 18°C on the upper surface of the slab (the inside), and it is -7°C on the lower surface, with a linear temperature profile in between. If the concrete has a thermal conductivity of 1.4 W/m-K, what is the rate of heat loss through the slab? If the basement is heated by a gas furnace operated with an efficiency of 66%, how much energy must be utilized to maintain the basement temperature for 90 days assuming that nearly all heat losses occur through the slab?
Answer:
a) Rate of heat loss,
b) Energy that must be utilized to maintain the basement temperature,
Em = 10309 MJ
Explanation:
Length of the slab, l = 5 m
Width of the slab, w = 3 m
Thickness of the slab, t = 0.6 m
Cross Sectional Area of the slab, A = l x b
A = 5 x 3
A = 15 m²
Upper Surface Temperature, T₁ = 18°C = 18 + 273 = 291 K
Lower Surface Temperature, T₂ = -7°C = -7 + 273 = 266 K
a) The rate of heat loss is given by the formula:
b) Energy,
t = 90 days = 90 * 24 * 3600 = 7776000 s
E = 875 * 7776000
E = 6804000000J
E = 6804 MJ
If Efficiency = 66%, Energy that must be utilized to maintain the basement temperature.
6804 = 66% * Em
6804 = 0.66 * Em
Em = 6804/0.66
Em = 10309 MJ
Answer:
The final temperature of water is 381.39 °C.
Explanation:
Given that
Mass of water = 5 kg
Heat transfer at constant pressure Q = 2960 KJ
Initial temperature = 240 °C
We know that heat transfer at constant pressure given as follows
We know that for water
Lets take final temperature of water is T
So
T=381.39 °C
So the final temperature of water is 381.39 °C.
Answer:
equivalent stiffness is 136906.78 N/m
damping constant is 718.96 N.s/m
Explanation:
given data
mass = 120 kg
amplitude = 120 N
frequency = 320 r/min
displacement = 5 mm
to find out
equivalent stiffness and damping
solution
we will apply here frequency formula that is
frequency ω = ω(n) √(1-∈ ²) ......................1
here ω(n) is natural frequency i.e = √(k/m)
so from equation 1
320×2π/60 = √(k/120) × √(1-2∈²)
k × ( 1 - 2∈²) = 33.51² ×120
k × ( 1 - 2∈²) = 134752.99 .....................2
and here amplitude ( max ) of displacement is express as
displacement = force / k × ( )
put here value
0.005 = 120/k × ( )
k ×∈ × √(1-2∈²) = 1200 ......................3
so by equation 3 and 2
solve it and we get
∈ = 1.00396
and
∈ = 0.08869
here small value we will consider so
by equation 2 we get
k × ( 1 - 2(0.08869)²) = 134752.99
k = 136906.78 N/m
so equivalent stiffness is 136906.78 N/m
and
damping is express as
damping = 2∈ √(mk)
put here all value
damping = 2(0.08869) √(120×136906.78)
so damping constant is 718.96 N.s/m
Answer:
189.15cy
Explanation:
To understand this problem we need to understand as well the form.
It is clear that there is four wall, two short and two long.
The two long are
The two long are
The two shors are
The height and the thickness are 14ft and 0.83ft respectively.
So we only calculate the Quantity of concrete,
That in cubic yards is equal to
Hence, we need order 5% plus that represent with the quantity