Answer:
hello the figure attached to your question is missing attached below is the missing diagram
answer :
i) 1.347 kW
ii) 1.6192 kW
Explanation:
Attached below is the detailed solution to the problem above
First step : Calculate for Enthalpy
h1 - hf = -3909.9 kJ/kg ( For saturated liquid nitrogen at 600 kPa )
h2- hg = -222.5 kJ/kg ( For saturated vapor nitrogen at 600 kPa )
second step : Calculate the rate of heat transfer in boiler
Q1-2 = m( h2 - h1 ) = 0.008( -222.5 -(-390.9) = 1.347 kW
step 3 : find the enthalpy of superheated Nitrogen at 600 Kpa and 280 K
from the super heated Nitrogen table
h3 = -20.1 kJ/kg
step 4 : calculate the rate of heat transfer in the super heater
Q2-3 = m ( h3 - h2 )
= 0.008 ( -20.1 -(-222.5 ) = 1.6192 kW
Answer:
The ratio that a material expands in accordance with changes in temperature is called the coefficient of thermal expansion. Because Fine Ceramics possess low coefficients of thermal expansion, their distortion values, with respect to changes in temperature, are low. The coefficients of thermal expansion depend on the bond strength between the atoms that make up the materials.
Explanation:
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Answer:
0.447 s²
Explanation:
First, convert to SI units.
(354 mg) (45 km) / (0.0356 kN)
(0.354 g) (45000 m) / (35.6 N)
One Newton is kg m/s²:
(0.354 g) (45000 m) / (35.6 kg m/s²)
(0.000354 kg) (45000 m) / (35.6 kg m/s²)
Simplify:
0.447 s²
Answer:
Highest temperature rise allowable = ΔT = 21.22°C
Highest allowable temperature = ΔT + 20 = 41.22°C
Explanation:
From literature, the coefficient of volume expansion of water between 20°C and 50°C = β = (0.377 × 10⁻³) K⁻¹
Volume expansivity is given by
ΔV = V β ΔT
ΔV = Change in volume
V = initial volume
β = Coefficient of volume expansion = (0.377 × 10⁻³) K⁻¹ = 0.000377 K⁻¹
ΔT = Change in temperature = ?
It is given in the question that maximum volume increase the tank can withstand is
(ΔV/V) × 100% = 0.8%
(ΔV/V) = 0.008
V β ΔT = ΔV
β ΔT = (ΔV/V)
β ΔT = 0.008
ΔT = (0.008/β)
ΔT = (0.008/0.000377)
ΔT = 21.22°C
Highest temperature rise allowable = ΔT = 21.22°C
Highest allowable temperature = ΔT + 20 = 41.22°C
Hope this Helps
