Answer:
a mass of water required is mw= 1273.26 gr = 1.27376 Kg
Explanation:
Assuming that the steam also gives out latent heat, the heat provided should be same for cooling the hot water than cooling the steam and condense it completely:
Q = mw * cw * ΔTw = ms * cs * ΔTw + ms * L
where m = mass , c= specific heat , ΔT=temperature change, L = latent heat of condensation
therefore
mw = ( ms * cs * ΔTw + ms * L )/ (cw * ΔTw )
replacing values
mw = [182g * 2.078 J/g°C*(118°C-100°C) + 118 g * 2260 J/g ] /[4.187 J/g°C * (90.7°C-39.4°C)] = 1273.26 gr = 1.27376 Kg
Answer:
<em>the % recovery of aluminum product is 80.5%</em>
<em>the % purity of the aluminum product is 54.7%</em>
<em></em>
Explanation:
feed rate to separator = 2500 kg/hr
in one hour, there will be 2500 kg/hr x 1 hr = 2500 kg of material is fed into the machine
of this 2500 kg, the feed is known to contain 174 kg of aluminium and 2326 kg of rejects.
After the separation, 256 kg is collected in the product stream.
of this 256 kg, 140 kg is aluminium.
% recovery of aluminium will be = mass of aluminium in material collected in the product stream ÷ mass of aluminium contained in the feed material
% recovery of aluminium = 140kg/174kg x 100% = <em>80.5%</em>
% purity of the aluminium product = mass of aluminium in final product ÷ total mass of product collected in product stream
% purity of the aluminium product = 140kg/256kg
x 100% = <em>54.7%</em>
Answer:
Q = 5.06 x 10⁻⁸ m³/s
Explanation:
Given:
v=0.00062 m² /s and ρ= 850 kg/m³
diameter = 8 mm
length of horizontal pipe = 40 m
Dynamic viscosity =
μ = ρv
=850 x 0.00062
= 0.527 kg/m·s
The pressure at the bottom of the tank is:
P₁,gauge = ρ g h = 850 x 9.8 x 4 = 33.32 kN/m²
The laminar flow rate through a horizontal pipe is:
Q = 5.06 x 10⁻⁸ m³/s
Answer:
1) 1.4(D + F)
2) 1.2(D + F + T) + 1.6(L + H) + 0.5(Lr or S or R)
3) 1.2D + 1.6(Lr or S or R) + ((0.5 or 1.0)*L or 0.8W)
4) 1.2D + 1.6W + (0.5 or 1.0)*L + 0.5(Lr or S or R)
5) 1.2D + 1.0E + (0.5 or 1.0)*L + 0.2S
6) 0.9D + 1.6W + 1.6H
7) 0.9D + 1.0E + 1.6H
Explanation:
Load and Resistance Factor Design
there are 7 basic load combination of LRFD that is
1) 1.4(D + F)
2) 1.2(D + F + T) + 1.6(L + H) + 0.5(Lr or S or R)
3) 1.2D + 1.6(Lr or S or R) + ((0.5 or 1.0)*L or 0.8W)
4) 1.2D + 1.6W + (0.5 or 1.0)*L + 0.5(Lr or S or R)
5) 1.2D + 1.0E + (0.5 or 1.0)*L + 0.2S
6) 0.9D + 1.6W + 1.6H
7) 0.9D + 1.0E + 1.6H
and
here load factor for L given ( * ) mean it is permitted = 0.5 for occupancies when live load is less than or equal to 100 psf
here
D is dead load and L is live load
E is earth quake load and S is snow load
W is wind load and R is rain load
Lr is roof live load
