Answer:
Maximum concentration for Pb is only 41%. we get 64.3% which exceeds the maximum concentration. Therefore it is not possible.
Explanation:
let's first determine the mass fraction of the phases as follows
[tex]W_{a}=\frac{m_{a} }{m_{a}+m_{p_{g2}p_{b} } }[/tex]
Given that masses are [tex]m_{a}=7.39kg[/tex] and [tex]m_{p_{g2}p_{b} }=3.81kg[/tex]
Thus, [tex]W_{a} =\frac{7.39}{7.39+3.81}=0.659[/tex]
[tex]W_{M_{g2Pb} } =1-W_{a}=1-0.659=0.341[/tex]
Now determine the concentration
[tex]W_{a}=\frac{C_{0}-C_{a} }{C_{b}-C_{a} }[/tex]
At 81% PB and 70% Pb only point at which [tex]M_{g2Pb}[/tex] exists, we can solve
[tex]0.659=\frac{70-81}{C_{b}-81 }[/tex]
[tex]C_{b}=[/tex]64.3%
Refer text book page no. 326, figure 9.20, from figure we can say maximum concentration for Pb is only 41%. we get 64.3% which exceeds the maximum concentration.
A six-lane divided multilane highway (three lanes in each direction) has a measured free-flow speed of 50 mi/h. It is on mountainous terrain with a traffic stream consisting of 7% large trucks and buses and 3% recreational vehicles. The driver population adjustment in 0.92. One direction of the highway currently operates at maximum LOS C conditions and it is known that the highway has PHF = 0.90.
Required:
How many vehicles can be added to this highway before capacity is reached, assuming the proportion of vehicle types remains the same but the peak-hour factor increases to 0.95?
Answer:
2901 vehicles
Explanation:
We are given;
Percentage of large trucks & buses; p_t = 7% = 0.07
Percentage of recreational vehicles; p_r = 3% = 0.03
PHF = 0.90
Driver population adjustment; f_p = 0.92
First of all, let's Calculate the heavy vehicle factor from the formula;
f_hv = 1/[1 + p_t(e_t - 1) + p_r(e_r - 1)]
Where;
e_t = passenger car equivalents for trucks and buses
e_r = passenger car equivalents for recreational vehicles
From the table attached, for a mountainous terrain, e_t = 6 and e_r = 4. Thus;
f_hv = 1/[1 + 0.07(6 - 1) + 0.03(4 - 1)]
f_hv = 1.44
Let's now calculate the initial hourly volume from the formula;
v_p = V1/(PHF × N × f_hv × f_p)
Where;
v_p = 15-minute passenger-car equivalent flow rate
V1 = hourly volume
N = number of lanes in each direction
From online tables of LOS criteria for multilane freeway segments, v_p = 1300 pc/hr/ln
Thus;
1300 = V1/(0.9 × 3 × 1.44 × 0.92)
V1 = 1300 × (0.9 × 3 × 1.44 × 0.92)
V1 = 4650 veh/hr
Now, let's Calculate the final hourly volume;
From online sources, the maximum capacity of a 6 lane highway with free-flow speed of 50 mi/h is 2000 pc/hr/ln.
We are told the online peak-hour factor increases to 0.95 and so PHF = 0.95.
Thus;
2000 = V2/(0.95 × 3 × 1.44 × 0.92)
V2 = 2000(0.95 × 3 × 1.44 × 0.92)
V2 = 7551 veh/hr
Number of vehicles added to the highway = V2 - V1 = 7551 - 4650 = 2901 vehicles
If the old radiator is replaced with a new one that has longer tubes made of the same material and same thickness as those in the old unit, what should the total surface area available for heat exchange be in the new radiator to achieve the desired cooling temperature gradient
Answer: hello some parts of your question is missing attached below is the missing information
The radiator of a car is a type of heat exchanger. Hot fluid coming from the car engine, called the coolant, flows through aluminum radiator tubes of thickness d that release heat to the outside air by conduction. The average temperature gradient between the coolant and the outside air is about 130 K/mm . The term ΔT/d is called the temperature gradient which is the temperature difference ΔT between coolant inside and the air outside per unit thickness of tube
answer : Total surface area = 3/2 * area of old radiator
Explanation:
we will use this relation
K = [tex]\frac{Qd }{A* change in T }[/tex]
change in T = ΔT
therefore New Area ( A ) = 3/2 * area of old radiator
Given that the thermal conductivity is the same in the new and old radiators
A body of weight 300N is lying rough
horizontal plane having
a Coefficient of friction as 0.3
Find the magnitude of the forces which can move the
body while acting at an angle of 25 with the horizonted
Answer:
Horizontal force = 89.2 N
Explanation:
The frictional force = coefficient of friction * magnitude of the force (weight of the body) * cos theta
Substituting the given values, we get -
Frictional Force = 0.3*300 * cos 25 = 89.2 N
Horizontal force = 89.2 N
state the degree of the homogeneity (1)sin() (2) (x+y+1) ². (3)√x+y(4x+3y).
Answer:
(2) ( (x+y)⁴)³(3)+(3)x(4x+y)
Explanation:
correct me if I'm wrong^_^
A 50 mm 45 mm 20 mm cell phone charger has a surface temperature of Ts 33 C when plugged into an electrical wall outlet but not in use. The surface of the charger is of emissivity 0.92 and is subject to a free convection heat transfer coefficient of h 4.5 W/m2 K. The room air and wall temperatures are T 22 C and Tsur 20 C, respectively. If electricity costs C $0.18/kW h, determine the daily cost of leaving the charger plugged in when not in use.
Answer:
C = $0.0032 per day
Explanation:
We are given;
Dimension of cell phone; 50 mm × 45 mm × 20 mm
Temperature of charger; T1 = 33°C = 306K
Emissivity; ε = 0.92
convection heat transfer coefficient; h = 4.5 W/m².K
Room air temperature; T∞ = 22°C = 295K
Wall temperature; T2 = 20°C = 293 K
Cost of electricity; C = $0.18/kW.h
Chargers are usually in the form of a cuboid, and thus, surface Area is;
A = (50 × 45) + 2(50 × 20) + 2(45 × 20)
A = 6050 mm²
A = 6.05 × 10^(-3) m²
Formula for total heat transfer rate is;
E_t = hA(T1 - T∞) + εσA((T1)⁴ - (T2)⁴)
Where σ is Stefan Boltzmann constant with a value of; σ = 5.67 × 10^(-8) W/m².K⁴
Thus;
E_t = 4.5 × 6.05 × 10^(-3) (306 - 295) + (0.92 × 6.05 × 10^(-3) × 5.67 × 10^(-8)(306^(4) - 293^(4)))
E_t = 0.7406 W = 0.7406 × 10^(-3) KW
Now, we know C = $0.18/kW.h
Thus daily cost which has 24 hours gives;
C = 0.18 × 0.7406 × 10^(-3) × 24
C = $0.0032 per day
Forget it because I almost have it
Answer:
hahahaha
Explanation: