Answer:
In Btu:
Q=0.001390 Btu.
In Joule:
Q=1.467 J
Part B:
Temperature at midpoint=274.866 C
Explanation:
Thermal Conductivity=k=30 (Btu/hr)/(ft ⋅ °F)=
Thermal Conductivity is SI units:
Length=20 cm=0.2 m= (20*0.0328) ft=0.656 ft
Radius=4/2=2 mm =0.002 m=(0.002*3.28)ft=0.00656 ft
T_1=500 C=932 F
T_2=50 C= 122 F
Part A:
In Joules (J)
Heat Q is:
In Btu:
Heat Q is:
PArt B:
At midpoint Length=L/2=0.1 m
On rearranging:
Answer:
Explanation:
solution:
from this below equation (1)
σ/2εo
...........(1)
we obtain:
Answer:
net boiler heat = 301.94 kW
Explanation:
given data
saturated steam = 6.0 bars
temperature = 18°C
flow rate = 115 m³/h = 0.03194 m³/s
heat use by boiler = 90 %
to find out
rate of heat does the boiler output
solution
we can say saturated steam is produce at 6 bar from liquid water 18°C
we know at 6 bar from steam table
hg = 2756 kJ/kg
and
enthalpy of water at 18°C
hf = 75.64 kJ/kg
so heat required for 1 kg is
=hg - hf
= 2680.36 kJ/kg
and
from steam table specific volume of saturated steam at 6 bar is 0.315 m³/kg
so here mass flow rate is
mass flow rate =
mass flow rate m = 0.10139 kg/s
so heat required is
H = h × m
here h is heat required and m is mass flow rate
H = 2680.36 × 0.10139
H = 271.75 kJ/s = 271.75 kW
now 90 % of boiler heat is used for generate saturated stream
so net boiler heat =
net boiler heat =
net boiler heat = 301.94 kW