A heat exchanger is designed to is to heat 2,500 kg/h of water from 15 to 80°C by engine oil. The configuration of the heat exch
anger is in shell-and tube and the water isflowing through the tubes. The oil makes a single shell pass, entering at 160°C and leaving at 90°C, with an averagedheat transfer coefficient from the oil to the outer wall of the tubes equals 400 W/m^2*K. The water flows through 11 brass tubes of 30 mm diameter with each tube making four passes through the shell.
1. Assuming fully developed flow for the water inside the tubes, determine the tube length per pass to achieve these specified output temperatures.You may assume the tube wall thickness is negligible in the determination of overall heat transfer coefficient.
1. Assuming fully developed flow for the water inside the tubes, determine the tube length per pass to achieve these specified output temperatures.You may assume the tube wall thickness is negligible in the determination of overall heat transfer coefficient.
1 answer:
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Answer:
Qx = 9.10 m³/s
Explanation:
given data
diameter = 85 mm
length = 2 m
depth = 9mm
N = 60 rev/min
pressure p = 11 × Pa
viscosity n = 100 Pas
angle = 18°
so Qd will be
Qd = 0.5 × π² ×D²×dc × sinA × cosA ..............1
put here value and we get
Qd = 0.5 × π² × ( 85 )²× 9
× sin18 × cos18
Qd = 94.305 × m³/s
and
Qb = p × π × D × dc³ × sin²A ÷ 12 × n × L ............2
Qb = 11 × × π × 85
× ( 9
)³ × sin²18 ÷ 12 × 100 × 2
Qb = 85.2 × m³/s
so here
volume flow rate Qx = Qd - Qb ..............3
Qx = 94.305 × - 85.2 ×
Qx = 9.10 m³/s
Answer:
The time required is 10.078 hours or 605 min
Explanation:
The formula to apply here is ;
K=(d²-d²₀ )/t
where t is time in hours
d is grain diameter to be achieved after heating in mm
d₀ is the grain diameter before heating in mm
Given
d=5.5 × 10^-2 mm
d₀=2.4 × 10^-2 mm
t₁= 500 min = 500/60 =25/3 hrs
t₂=?
n=2.2
First find K
K=(d²-d²₀ )/t₁
K={ (5.1 × 10^-2 mm)²-(2.4 × 10−2 mm)² }/ 25/3
K=(0.051²-0.024²) ÷25/2
K=0.000243 mm²/h
Re-arrange equation for K ,to get the equation for d as;
d=√(d₀²+ Kt) where now t=t₂