An engineer designs a new bus that can drive 30 miles per gallon of fuel. Which of the following was likely one of the client’s
2 answers:
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Answer:
a) the coefficient of performance of the air conditioner is 3.5729
b)
- the power input required for a reversible air conditioner is 0.645 hp
- the coefficient of performance for the reversible air conditioner is 18.2759
Explanation:
Given the data in the question;
Lower Temperature = 70°F = ( 70 + 460 )R = 530 R
Higher Temperature = 99° F = ( 99 + 460 )R = 559 R
Cooling Load = 30000 Btu/h
we know that 1 hp = 2544.43 Btu/h
Net power input P = 3.3 hp = ( 3.3 × 2544.43 )Btu/h = 8396.619 Btu/h
a)
Coefficient of performance of the air conditioner;
= Cooling Load
/ power P
we substitute
= 30000 Btu/h / 8396.619 Btu/h
= 3.5729
Therefore, the coefficient of performance of the air conditioner is 3.5729
b)
- Power input required ( in hp )
/
=
/ (
-
)
we substitute
30000 Btu/h / = 530 R / ( 559 R - 530 R )
30000 Btu/h / = 530 R / 29 R
we solve for
= ( 30000 Btu/h × 29 R ) / 530 R
= ( 870000 Btu/h / 530 )
= 1641.5094 Btu/h
we know that; 1 hp = 2544.43 Btu/h
so;
= ( 1641.5094 / 2544.43 ) hp
= 0.645 hp
Hence, the power input required for a reversible air conditioner is 0.645 hp
- the coefficient of performance for the reversible air conditioner;
=
/ (
-
)
we substitute
= 530 R / ( 559 R - 530 R )
= 530 R / 29 R
= 18.2759
Hence, the coefficient of performance for the reversible air conditioner is 18.2759
Answer:
a) Δh = 2 cm, b) Δh = 0.4 cm
Explanation:
Let's start by using Bernoulli's equation for the Pitot tube, we define two points 1 for the small entry point and point 2 for the larger diameter entry point.
P₁ + ½ ρ v₁² + ρ g y₁ = P₂ + ½ ρ v₂² + ρ g y₂
Point 1 is called the stagnation point where the fluid velocity is reduced to zero (v₁ = 0), in general pitot tubes are used in such a way that the height of point 2 of is the same of point 1
y₁ = y₂
subtitute
P₁ = P₂ + ½ ρ v₂²
P₁ -P₂ = ½ ρ v²
where ρ is the density of fluid
now we measure the pressure on the included beforehand as a pair of communicating tubes filled with mercury, we set our reference system at the point of the mercury bottom surface
ΔP =ρ_{Hg} g h - ρ g h
ΔP = (ρ_{Hg} - ρ) g h
as the static pressure we can equalize the equations
ΔP = P₁ - P₂
(ρ_{Hg} - ρ) g h = ½ ρ v²
v =
in this expression the densities are constant
v = A √h
A =
They indicate the density of mercury rhohg = 13600 kg / m³, the density of dry air at 20ºC is rho air = 1.29 kg/m³
we look for the constant
A =
A = 454.55
we substitute
v = 454.55 √h
to calculate the uncertainty or error of the velocity
h =
Δh = Δv
Suppose we have a height reading of h = 20 cm = 0.20 m
a) uncertainty 2.5 m / s ( 0.05)
= 2 0.05
Δh = 0.1 h
Δh = 0.1 20 cm
Δh = 2 cm
b) uncertainty 0.5 m / s ( Δv/v= 0.01)
= 2 0.01
Δh = 0.02 h
Δh = 0.02 20
Δh = 0.1 20 cm
Δh = 0.4 cm = 4 mm