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Vadim26 [7]
4 years ago
11

Airflow through a long, 0.15-m-square air conditioning duct maintains the outer duct surface temperature at 10°C. If the horizon

tal duct is uninsulated and exposed to air at 35°C in the crawlspace beneath a home, what is the heat gain per unit length of the duct? Evaluate the properties of air at 300 K. For the sides of the duct, use the more accurate Churchill and Chu correlations for laminar flow on vertical plates.

Engineering
1 answer:
Ulleksa [173]4 years ago
4 0

The complete Question is:

Airflow through a long, 0.15-m-square air conditioning duct maintains the outer duct surface temperature at 10°C. If the horizontal duct is uninsulated and exposed to air at 35°C in the crawlspace beneath a home, what is the heat gain per unit length of the duct? Evaluate the properties of air at 300 K. For the sides of the duct, use the more accurate Churchill and Chu correlations for laminar flow on vertical plates.

What is the Rayleigh number for free convection on the outer sides of the duct?

What is the free convection heat transfer coefficient on the outer sides of the duct, in W/m2·K?

What is the Rayleigh number for free convection on the top of the duct?  

What is the free convection heat transfer coefficient on the top of the duct, in W/m2·K?

What is the free convection heat transfer coefficient on the bottom of the duct, in W/m2·K?

What is the total heat gain to the duct per unit length, in W/m?

Answers:

- 7709251  or 7.709 ×10⁶

- 4.87

- 965073

- 5.931 W/m² K

- 2.868 W/m² K

- 69.498 W/m

Explanation:

Find the given attachments for complete explanation

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3 years ago
Two loads connected in parallel draw a total of 2.4 kW at 0.8 pf lagging from a 120-V rms, 60-Hz line. One load absorbs 1.5 kW a
stealth61 [152]

Answer: a) 0.948 b) 117.5µf

Explanation:

Given the load, a total of 2.4kw and 0.8pf

V= 120V, 60 Hz

P= 2.4 kw, cos θ= 80

P= S sin θ - (p/cos θ) sin θ

= P tan θ(cos^-1 (0.8)

=2.4 tan(36.87)= 1.8KVAR

S= 2.4 + j1. 8KVA

1 load absorbs 1.5 kW at 0.707 pf lagging

P= 1.5 kW, cos θ= 0.707 and θ=45 degree

Q= Ptan θ= tan 45°

Q=P=1.5kw

S1= 1.5 +1.5j KVA

S1 + S2= S

2.4+j1.8= 1.5+1.5j + S2

S2= 0.9 + 0.3j KVA

S2= 0.949= 18.43 °

Pf= cos(18.43°) = 0.948

b.) pf to 0.9, a capacitor is needed.

Pf = 0.9

Cos θ= 0.9

θ= 25.84 °

(WC) V^2= P (tan θ1 - tan θ2)

C= 2400 ( tan (36. 87°) - tan (25.84°)) /2 πf × 120^2

f=60, π=22/7

C= 117.5µf

7 0
3 years ago
Coils of various dimensions designed to introduce specified amounts of inductance into a circuit are called (a) semiconductors (
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Answer: d) Inductors

Explanation:  Inductors are the electric component commonly known by the name of coils as well. The work done by inductors is when the electric current flows through the coils of the wire then there is production of the magnetic field. They are the component that have the inductance in particular amount according to the circuit.So, coils of various dimensions are designed to introduce specified amount of inductance  into a circuit are called inductors.

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What historical event allowed both aerospace fields to make enormous strides<br> forward? *
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Answer:

The world wars. Most notably world War II

Explanation:

The demand for aircrafts during these events led to extensive research into the design of aircrafts. Aircraft advanced within these years from a simple design to a more complex design; capable of carrying fire power and even became bomb equipped. Also, the material of choice of production moved from wood to metal and the engine was improved on to gain more speed and maneuverability.

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4 years ago
Source 1 can supply energy at the rate of 11000 kJ/min at 310°C. A second Source 2 can supply energy at the rate of 110000 kJ/mi
VladimirAG [237]

Answer:

Source 2.

Explanation:

The efficiency of the ideal reversible heat engine is given by the Carnot's power cycle:

\eta_{th} = 1 - \frac{T_{L}}{T_{H}}

Where:

T_{L} - Temperature of the cold reservoir, in K.

T_{H} - Temperature of the hot reservoir, in K.

The thermal efficiencies are, respectively:

Source 1

\eta_{th} = 1 - \frac{311.15\,K}{583.15\,K}

\eta_{th} = 0.466 \,(46.6\,\%)

Source 2

\eta_{th} = 1 - \frac{311.15\,K}{338.15\,K}

\eta_{th} = 0.0798 \,(7.98\,\%)

The power produced by each device is presented below:

Source 1

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Source 2

\dot W = (0.0798)\cdot (110000\,\frac{kJ}{min})\cdot (\frac{1\,min}{60\,s} )

\dot W = 146.3\,kW

The source 2 produces the largest amount of power.

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