Consider the double-fluid manometer attached to an air pipe shown. If the specific gravity of one fluid is 14.514.5, determine t
he specific gravity of the other fluid if pressure of air was 81kpa81kpa. Take the atmospheric pressure to be 100 kpa.
1 answer:
Complete Question
The complete question is shown on the first uploaded image
Answer:
The value is
Explanation:
From the question we are told that
The specific gravity of the first fluid is
The pressure of air is
The atmospheric pressure is
Generally the specific gravity of the second fluid is mathematically represented as
Here is the density of water with value
=>
=>
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Answer:
The answers to the question are
(i) The rate of heat loss per-unit-length (W/m) from the pipe is 131.62 W
(ii) The temperature of the outer surface of the insulation is 49.89 °C
Explanation:
To solve the question, we note that the heat transferred is given by
Where
= Temperature at the inside of the pipe = 300 °C
= Temperature at the outside of the pipe = 20 °C
r₁ =internal radius of pipe = 4.0 cm
r₂ = Outer radius of pipe = 4.5 cm
r₃ = Outer radius of the insulation = r₂ + 2.5 = 7.0 cm
= 15 W/m·K
= 0.038 W/m·K
= 75 W/m²·K
= 10 W/m²·K
Plugging in the values in the above equation where for a unit length L = 1 m, we have
Q = 131.32 W
From which we have, for the film of air at the pipe outer boundary layer
Where
for the air film on the pipe outer surface is given by
where A =area of the outside of the pipe
= = 0.227 K/W
Therefore
131.32 W = which gives
= 49.89 °C
Heat transferred by radiation = q' = ε×σ×(T₁⁴ - T₂⁴)
Where ε = 0.9, σ, = 5.67×10⁻⁸W/m²·(K⁴)
T₁ = Surface temperature of the pipe = 49.89 °C and
T₂ = Temperature of the surrounding = 20.00 °C
Plugging in the values gives, q' = 0.307 W per m²
Total heat lost per unit length = 131.32 + 0.307 =131.62 W
Given Information:
Power = P = 100 Watts
Voltage = V = 220 Volts
Required Information:
a) Current = I = ?
b) Resistance = R = ?
Answer:
a) Current = I = 0.4545 A
b) Resistance = R = 484 Ω
Explanation:
According to the Ohm’s law, the power dissipated in the light bulb is given by
Where V is the voltage across the light bulb, I is the current flowing through the light bulb and P is the power dissipated in the light bulb.
Re-arranging the above equation for current I yields,
Therefore, 0.4545 A current is flowing through the light bulb.
According to the Ohm’s law, the voltage across the light bulb is given by
Where V is the voltage across the light bulb, I is the current flowing through the light bulb and R is the resistance of the light bulb.
Re-arranging the above equation for resistance R yields,
Therefore, the resistance of the bulb is 484 Ω