The products of combustion from a burner are routed to an industrial application through a thin-walled metallic duct of diameter
Di=1 m and length L=100 m. The gas enters the duct at atmospheric pressure and a mean temperature and velocity of Tm,i=1600 K and Um,i=10 m/s, respectively. It must exit the duct at a temperature that is no less than Tm,o=1400 K. What is the minimum thickness of an alumina-silica insulation (Kins=0.125 W/m*K) needed to meet the outlet requirement under worst case conditions for which the duct is exposed to ambient air at T[infinity]=250 K and a cross-flow velocity of V=15 m/s? The properties of the gas may be approximated as those of air, and as a first estimate, the effect of the insulation thickness on the convection coefficient and the thermal resistance associated with the cross flow may be neglected.
1 answer:
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Answer:
The head loss in Psi is 0.390625 psi.
Explanation:
Fluid looses energy in the form of head loss. Fluid looses energy in the form of head loss when passes through the valve as well.
Given:
Factor cv is 48.
Flow rate of water is 30 GPM.
GPM means gallon per minute.
Calculation:
Step1
Expression for head loss for the water is given as follows:
Here, cv is valve coefficient, Q is flow rate in GPM and h is head loss is psi.
Step2
Substitute 48 for cv and 30 for Q in above equation as follows:
h = 0.390625 psi.
Thus, the head loss in Psi is 0.390625 psi.
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