1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Zepler [3.9K]
3 years ago
6

4. An aluminum alloy fin of 12 mm thick, 10 mm width and 50 mm long protrudes from a wall, which is maintained at 120C. The amb

ient air temperature is 22C. The heat transfer coefficient and conductivity of the fin material are 140 W/m2K and 55 W/mk respectively. Determine a. Temperature at the end of the fin b. Temperature at the middle of the fin. c. Calculate the heat dissipation energy of the fin
Engineering
1 answer:
Minchanka [31]3 years ago
7 0

Answer:

a) 84.034°C

b) 92.56°C

c) ≈ 88 watts

Explanation:

Thickness of aluminum alloy fin = 12 mm

width = 10 mm

length = 50 mm

Ambient air temperature = 22°C

Temperature of aluminum alloy is maintained at 120°C

<u>a) Determine temperature at end of fin</u>

m = √ hp/Ka

   = √( 140*2 ) / ( 12 * 10^-3 * 55 )

   = √ 280 / 0.66 = 20.60

Attached below is the remaining answers

You might be interested in
A window‐mounted air‐conditioning unit (AC) removes energy by heat transfer from a room, and rejects energy by heat transfer to
Arada [10]

Solution :

Given :

The power of the air‐conditioning (AC) unit is , W = 0.434 kW

The coefficient of performance or the COP of the air‐conditioning (AC) unit is given by  = 6.22

Therefore he heat removed is given by , $Q_H = 6.22 \times 0.434$

                                                                     $Q_H = 2.7 \ kW $

Now if the electricity is valued at  0.10 dollar per kW hour, then the operating cost of the air conditioning unit in 24 hours is given by = 0.10 x 2.7 x 24

                                                                                            = 6.48

Therefore the operating cost = $ 6.48 for 24 hours.

3 0
3 years ago
Briefly explain why small-angle grain boundaries are not as effective in interfering with the slip process as are high-angle gra
Vlada [557]

Answer:

Explanation:

Small-angle grain boundaries are not as effective in interfering with the slip process as are high-angle grain boundaries because there is not as much crystallographic misalignment in the grain boundary region for small-angle, and therefore not as much change in slip direction.

Low angle grain boundaries (quasi-coherent) are formed by the dislocation network positioned along the geometric plane with small tilt angle differences between successive peers that is tilt boundary made up edge dislocations therefore it may only divert the slip direction of the incoming gliding dislocation with very little frictional stresses. And on the other hand, a high angle grain boundary region because of their disordered almost liquid like structure which acts as a strong barrier against dislocation slip motion and causes actually formation of dislocations file-up against it by arresting their motion unless that the stress concentration at the leading dislocation becomes high enough to go though the barrier.

5 0
3 years ago
A heat pump designer claims to have an air-source heat pump whose coefficient of performance is 1.8 when heating a building whos
Anit [1.1K]

Answer:

The claim is valid.

Explanation:

Let assume that heat pump is reversible. The coefficient of performance for the heat pump is:

COP_{HP} = \frac{T_{H}}{T_{H}-T_{L}}

COP_{HP} = \frac{300\,K}{300\,K-260\,K}

COP_{HP} = 7.5

The claim is valid as real heat pumps have lower coefficients of performance.

3 0
3 years ago
What is the angular velocity (in rad/s) of a body rotating at N r.p.m.?
Darina [25.2K]

Answer:

0.1047N

Explanation:

To solve this problem we must remember the conversion factors, remembering that 1 revolution equals 2π radians and 1min equals 60s

N\frac{rev}{min} \frac{2\pi }{1rev} \frac{1min}{60} =N\frac{2\pi }{60} =0.1047N

in conclusion, to know how many rad / s an element rotates which is expressed in Rev / min we must only multiply by 0.1047

3 0
3 years ago
while performing a running compression test how should running compression compare to static compression
algol [13]

Answer:

The idle speed of a running compression should be between 50-75 PSI and that is about half of the static compression.

Explanation:

The Running or Dynamic compression is used to determine how well the cylinder in an engine  is absorbing air, reserving it for the proper length of time, and releasing it to the exhaust. The static or cranking compression test is used to check the sealing of the cylinder. Before performing the running compression test, the static compression test is first performed to rule out other issues like bent valves.

The standard value for the static compression is given by;

Compression ratio * 14.7 = Manufacturers Specification

The running compression should always be half of the static compression.

5 0
2 years ago
Other questions:
  • Please describe a real situation in which you had to troubleshoot and fix the failure of a piece equipment/machine?
    5·1 answer
  • A simple ideal Rankine cycle with water as the working fluid operates between the pressure limits of 4 MPa in the boiler and 25
    14·2 answers
  • suppose a wheel with a 15 inch diameter is used to turn a water valve stem with a radius of .95 inches. What is the Mechanical a
    15·1 answer
  • Determine the direct runoff and streamflow given the following unit hydrograph. The rainfall is collected at 4-hour intervals an
    14·1 answer
  • After a capacitor is fully chargerd, a small amount of current will flow though it. what is this current called?
    14·1 answer
  • The following liquids are stored in a storage vessel at 1 atm and 25°C. The vessels are vented with air. Determine whether the e
    5·1 answer
  • The minimum fresh air requirement of a residential building is specified to be 0.35 air changes per hour (ASHRAE, Standard 62, 1
    10·1 answer
  • When must an Assured Equipment Grounding Conductor Program (AEGCP) be in place?
    10·1 answer
  • The hydraulic tool holder is a type of what? A.port B. clamp C. Press D. Actuator​
    11·1 answer
  • Which of the following suggestions would best help alleviate the Gulf of Mexico dead zone?
    13·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!