Answer:
Option B
Explanation:
An operational amplifier usually has a high open loop gain of around 10^5 which allows a wide range get of feed back levels in order to achieve the desired performance so therefore a low open loop gain reduces the range feed back level thereby reducing the performance which can cause errors in the output voltage.
The heat transferred to and the work produced by the steam during this process is 13781.618 kJ/kg
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How to calcultae the heat?</h3>
The Net Change in Enthalpy will be:
= m ( h2 - h1 ) = 11.216 ( 1755.405 - 566.78 ) = 13331.618 kJ/kg
Work Done (Area Under PV curve) = 1/2 x (P1 + P2) x ( V1 - V2)
= 1/2 x ( 75 + 225) x (5 - 2)
W = 450 KJ
From the First Law of Thermodynamics, Q = U + W
So, Heat Transfer = Change in Internal Energy + Work Done
= 13331.618 + 450
Q = 13781.618 kJ/kg
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Answer:
The tube surface temperature immediately after installation is 120.4°C and after prolonged service is 110.8°C
Explanation:
The properties of water at 100°C and 1 atm are:
pL = 957.9 kg/m³
pV = 0.596 kg/m³
ΔHL = 2257 kJ/kg
CpL = 4.217 kJ/kg K
uL = 279x10⁻⁶Ns/m²
KL = 0.68 W/m K
σ = 58.9x10³N/m
When the water boils on the surface its heat flux is:

For copper-water, the properties are:
Cfg = 0.0128
The heat flux is:
qn = 0.9 * 18703.42 = 16833.078 W/m²

The tube surface temperature immediately after installation is:
Tinst = 100 + 20.4 = 120.4°C
For rough surfaces, Cfg = 0.0068. Using the same equation:
ΔT = 10.8°C
The tube surface temperature after prolonged service is:
Tprolo = 100 + 10.8 = 110.8°C