Answer:
Area required = 9.5 ft²
Explanation:
Step by step explanation is given in the attached document.
An algorithm is itself a general step-by-step solution of your problem. ... The most important point here is that you must use algorithms to solve problem, one way or the other. Most of the time it's better to think about your problem before you jump to coding - this phase is often called design.
Answer:
a. 81 kj/kg
b. 420.625K
c. 101.24kj/kg
Explanation:
![\frac{t2}{t1} =[\frac{p2}{p1} ]^{\frac{y-1}{y} }](https://tex.z-dn.net/?f=%5Cfrac%7Bt2%7D%7Bt1%7D%20%3D%5B%5Cfrac%7Bp2%7D%7Bp1%7D%20%5D%5E%7B%5Cfrac%7By-1%7D%7By%7D%20%7D)
t1 = 360
p1 = 0.4mpa
p2 = 1.20
y = 1.13
substitute these values into the equation
![\frac{t2}{360} =[\frac{1.20}{0.4} ]^{\frac{1.13-1}{1.13} }](https://tex.z-dn.net/?f=%5Cfrac%7Bt2%7D%7B360%7D%20%3D%5B%5Cfrac%7B1.20%7D%7B0.4%7D%20%5D%5E%7B%5Cfrac%7B1.13-1%7D%7B1.13%7D%20%7D)
![\frac{t2}{360} =[\frac{1.2}{0.4} ]^{0.1150}\\\frac{t2}{360} =1.1347](https://tex.z-dn.net/?f=%5Cfrac%7Bt2%7D%7B360%7D%20%3D%5B%5Cfrac%7B1.2%7D%7B0.4%7D%20%5D%5E%7B0.1150%7D%5C%5C%5Cfrac%7Bt2%7D%7B360%7D%20%3D1.1347)
when we cross multiply
t2 = 360 * 1.1347
= 408.5
a. the work required in the firs compressor
w=c(t2-t1)
c=1.67x10³
t1 = 360
t2 = 408.5
w = 1670(408.5-360)
= 1670*48.5
= 80995 J
= 81KJ/kg
b. 
n = 80%
t2 = 408.5
t1 = 360
0.80 = 408.5-360 ÷ t'2-360

cross multiply to get the value of t'2
0.80(t'2-360) = 48.5
0.80t'2 - 288 = 48.5
0.8t'2 = 48.5+288
0.8t'2 = 336.5
t'2 = 336.5/0.8
= 420.625
this is the temperature at the exit of the first compressor
c. cooling requirement
w = c(t2-t1)
= 1.67x10³(420.625-360)
= 1670*60.625
= 101243.75
= 101.24kj/kg
Answer:
74.12kg/day
Explanation:
Equation of reaction for the neutralization reaction of sodium hydroxide and sulfuric acid: 2NaOH + H2SO4 yields Na2SO4 + 2H2O
Mass of sulfuric acid produced per day = 90,800kg
Percentage of sulfuric acid in wastewater = 0.1%
Mass of sulfuric acid that ends up in wastewater per day = 0.1/100 × 90,800 = 90.8kg
From the equation of reaction, 2 moles of NaOH (80kg of NaOH) is required to neutralize 1 mole of H2SO4 (98kg of H2SO4)
80kg of NaOH is required to neutralize 98kg of sulfuric acid
90.8kg of sulfuric acid would be neutralized by (90.8×80)/98kg of NaOH = 74.14kg/day of NaOH
Answer:
The heat transferred to water equals 1600 kJ
Explanation:
By the conservation of energy we have
All the kinetic energy of the moving vehicle is converted into thermal energy
We know that kinetic energy of a object of mass 'm' moving with a speed of 'v' is given by

Thus

Thus the heat transferred to water equals 