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natali 33 [55]
3 years ago
14

Review 1. The combustion efficiency of a non-weather- ized oil furnace must be at least

Engineering
1 answer:
kati45 [8]3 years ago
5 0

The combustion efficiency of a non-weatherized oil furnace must be at least 75%. It uses a secondary heat exchanger to extract heat.

<h3>Condensing furnaces</h3>

A condensing oil furnace is a heating device that needs a source of oil in order to generate heat.

This device (condensing oil furnace) may use a secondary heat exchanger to extract more heat from gases.

The condensing oil furnaces exhibit a high efficiency ranging from 75% to 90+%.

Learn more about condensing oil furnaces here:

brainly.com/question/6486936

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Ammonia enters an adiabatic compressor operating at steady state as saturated vapor at 300 kPa and exits at 1400 kPa, 140◦C. Kin
hammer [34]

Answer:

a. 149.74 KJ/KG

b. 97.9%

c. 0.81 kJ/kg K

Explanation:

8 0
4 years ago
Your boss asks you to classify some of the components in a temperature measurement system. The system consists of a thermocouple
lianna [129]

Answer:

Explanation:

1) Resolution and uncertainty of both ADC ranges

a) Resolution (for +/- 5 mV range) = 5 mV / 2n where n = number of bits of ADC

Resolution = 5 mV / 28 = 5 mV / 256 = 0.0195 mV    ..................1

uncertainty (for +/- 5 mV range) = +/- 0.5*resolution = +/- 0.5*0.0195 mV = 0.00975 mV       ..........2

b) Resolution (for +/- 5 V range) = 5 V / 2n where n = number of bits of ADC

Resolution = 5 mV / 28 = 5 V / 256 = 0.0195 V = 19.5 mV    ..................3

uncertainty (for +/- 5 mV range) = +/- 0.5*resolution = +/- 0.5*19.5 mV = 9.75 mV       ..........4

2)Thermocouple sensitivity = ( Maximum output voltage - Minimum Output voltage) / (Maximum Temperature - Minimum Temperature)

Thermocouple sensitivity = (3.649mv - 0 ) / (70 - 0) = 0.0521 mV / Deg.C           ............5

This is the required Thermocouple sensitivity

3) Water bath temperature is given as 57 deg.C

Hence voltage read by Thermocouple = Sensitivity*57 = 0.0521*57 mV = 2.9697 mV    ........6

4)We need to use ADC with a range of +/- 5 mV range as ADC with +/- 5 V range can not do measurement as it's resolution is higher than output voltage.

ADC will measure voltage as 2.9695 mV                    ......................7

8 0
3 years ago
I need answer to this question
blagie [28]

Answer:

1. Graph C

2. Friction

Explanation:

1. The line on all of the graphs shown represents velocity. The formula for velocity is v=\frac{d}{t} where d is distance and t is time. Focusing on the first lap, the starting point on the graph should be the origin and the "ending" point should be (20, 3). These requirements eliminate graph A as an answer because its "end" is not (20, 3). During the break, the student does not move, so the slope of the line should be completely horizontal. The break lasted for 5 minutes, so the correct graph should have a horizontal line between the points (20, 3) and (25, 3). This requirement eliminates graph B and D because their break is either not long enough (B) or too long (D).

2. Friction slows down the movement of objects. When an object is rough, it produces more friction which causes the object to be slowed more. When an object is smooth, friction slows it less than it would for a rough object.

6 0
3 years ago
A manufacturer has been asked to produce 100 customized metal discs with a particular pattern engraved on them. Which production
topjm [15]
I’m just here for points because I have test and I need them lol
8 0
3 years ago
Read 2 more answers
It is given that 50 kg/sec of air at 288.2k is iesntropically compressed from 1 to 12 atm. Assuming a calorically perfect gas, d
denis23 [38]

The exit temperature is 586.18K and  compressor input power is 14973.53kW

Data;

  • Mass = 50kg/s
  • T = 288.2K
  • P1 = 1atm
  • P2 = 12 atm

<h3>Exit Temperature </h3>

The exit temperature of the gas can be calculated isentropically as

\frac{T_2}{T_1} = (\frac{P_2}{P_1})^\frac{y-1}{y}\\ y = 1.4\\ C_p= 1.005 Kj/kg.K\\

Let's substitute the values into the formula

\frac{T_2}{T_1} = (\frac{P_2}{P_1})^\frac{y-1}{y} \\\frac{T_2}{288.2} = (\frac{12}{1})^\frac{1.4-1}{1.4} \\ T_2 = 586.18K

The exit temperature is 586.18K

<h3>The Compressor input power</h3>

The compressor input power is calculated as

P= mC_p(T_2-T_1)\\P = 50*1.005*(586.18-288.2)\\P= 14973.53kW

The compressor input power is 14973.53kW

Learn more on exit temperature and compressor input power here;

brainly.com/question/16699941

brainly.com/question/10121263

6 0
3 years ago
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