Answer:
the surface heat-transfer coefficient due to natural convection during the initial cooling period. = 4.93 w/m²k
Explanation:
check attachement for answer explanation
Answer:
A) approximate alkalinity = 123.361 mg/l
B) exact alkalinity = 124.708 mg/l
Explanation:
Given data :
A) determine approximate alkalinity first
Bicarbonate ion = 120 mg/l
carbonate ion = 15 mg/l
Approximate alkalinity = ( carbonate ion ) * 50/30 + ( bicarbonate ion ) * 50/61
= 15 * (50/30) + 120*( 50/61 ) = 123.361 mg/l as CaCO3
B) calculate the exact alkalinity of the water if the pH = 9.43
pH + pOH = 14
9.43 + pOH = 14. therefore pOH = 14 - 9.43 = 4.57
[OH^- ] = 10^-4.57 = 2.692*10^-5 moles/l
[ OH^- ] = 2.692*10^-5 * 179/mole * 10^3 mg/g = 0.458 mg/l
[ H^+ ] = 10^-9.43 * 1 * 10^3 = 3.7154 * 10^-7 mg/l
therefore the exact alkalinity can be calculated as
= ( approximate alkalinity ) + ( [ OH^- ] * 50/17 ) - ( [ H^+ ] * 50/1 )
= 123.361 + ( 0.458 * 50/17 ) - ( 3.7154 * 10^-7 * 50/1 )
= 124.708 mg/l
Answer:
<em>The maximum efficiency the plant will ever achieve is 75%</em>
<em>Explanation:</em>
From the question given, we recall the following:
<em>Th flames in the boiler reaches a temperature of = 1200K</em>
<em>the cooling water is = 300K</em>
<em>The maximum efficiency the plant will achieve is defined as:</em>
Let nmax = 1 - Tmin /Tmax
Where,
Tmin = Minimum Temperature in plants
Tmax = Maximum Temperature in plants
The temperature of the cooling water = Tmin = 300K
The temperature of the flames in boiler = Tmax = 1200k=K
The maximum efficiency becomes:
nmax = 1 - Tmin /Tmax
nmax = 1 - 300 /1200
nmax = 1-1/4 =0.75
nmax = 75%
