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White raven [17]
2 years ago
13

An electrical heater is a form of sensible heating process, and heats 0.1m/s of air from 15°C and 80% RH to 50°C? The barometric

pressure is 101.325 kPa. Using the psychrometric chart given, determine the heater wattage.
Engineering
1 answer:
lawyer [7]2 years ago
3 0

Answer:

The heater load =35 KJ/kg

Explanation:

Given that

At initial condition

Temperature= 15°C

RH=80%

At final condition

Temperature= 50°C

We know that in sensible heating process humidity ratio remain constant.

Now from chart

At temperature= 15°C and RH=80%

h_1=38 \frac{KJ}{kg},v=0.8 \frac{m^3}{kg}

At  temperature= 50°C

h_2=73 \frac{KJ}{kg}

So\ the\ heater\ load =h_2-h_1

The heater load = 73 - 38 KJ/kg

The heater load =35 KJ/kg

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Answer:

Explanation:

Given that : -

The desirable limit is 500 mg / l , but

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The take volume is V = 160.000 m3

V = 160 , 000 x 103 l

The crainage gives 150 mg / l and lake has initialy 100 mg / l

Code of tpr frpm drawn = 150 x 60, 000 x 1000

Ci = 9000 kg / gr

Cl = 100 x 160,000 x 1000

Cl = 16, 000 kg

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Cn = ( 2000 x 160, 00 x 1000 )

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5 0
3 years ago
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If a 9V battery produces a current of 3 A through a load, what is the resistance of the load
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3 ohms hope this helps :D ❤

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Which measuring tool will be used to determine the diameter of a crankshaft journal?
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Answer:

The dial bore gauge measures the inside of round holes, such as the bearing journals. This one tool can measure 2″ up to 6″ diameter holes. Both tools are needed in order to check the interior and exterior dimensions of the crankshaft, rods and engine block journals, as well as the thickness of the bearings themselves.

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3 years ago
A particle is emitted from a smoke stack with diameter of 0.05 mm. In order to determine how far downstream it travels it is imp
Nikolay [14]

Answer: downward velocity = 6.9×10^-4 cm/s

Explanation: Given that the

Diameter of the smoke = 0.05 mm = 0.05/1000 m = 5 × 10^-5 m

Where radius r = 2.5 × 10^-5 m

Density = 1200 kg/m^3

Area of a sphere = 4πr^2

A = 4 × π× (2.5 × 10^-5)^2

A = 7.8 × 10^-9 m^2

Volume V = 4/3πr^3

V = 4/3 × π × (2.5 × 10^-5)^3

V = 6.5 × 10^-14 m^3

Since density = mass/ volume

Make mass the subject of formula

Mass = density × volume

Mass = 1200 × 6.5 × 10^-14

Mass M = 7.9 × 10^-11 kg

Using the formula

V = sqrt( 2Mg/ pCA)

Where

g = 9.81 m/s^2

M = mass = 7.9 × 10^-11 kg

p = density = 1200 kg/m3

C = drag coefficient = 24

A = area = 7.8 × 10^-9m^2

V = terminal velocity

Substitute all the parameters into the formula

V = sqrt[( 2 × 7.9×10^-11 × 9.8)/(1200 × 24 × 7.8×10^-9)]

V = sqrt[ 1.54 × 10^-9/2.25×10-4]

V = 6.9×10^-6 m/s

V = 6.9 × 10^-4 cm/s

6 0
3 years ago
For this problem, calculate the following by hand and show the procedure for how you obtained the results. Subsequently, solve p
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