Answer:
Explanation:
Given that : -
The desirable limit is 500 mg / l , but
allowable upto 2000 mg / l.
The take volume is V = 160.000 m3
V = 160 , 000 x 103 l
The crainage gives 150 mg / l and lake has initialy 100 mg / l
Code of tpr frpm drawn = 150 x 60, 000 x 1000
Ci = 9000 kg / gr
Cl = 100 x 160,000 x 1000
Cl = 16, 000 kg
Since allowable limit = 2000 mg / l
Cn = ( 2000 x 160, 00 x 1000 )
= 320, 000 kg
so, each year the rate increases, by 9000 kg / yr
Read level = ( 320, 000 - 16,000 )
Li = 304, 000 kg
Tr=<u>304,000</u>
900
=33.77
Answer:
7.15
Explanation:
Firstly, the COP of such heat pump must be measured that is,
![COP_{HP}=\frac{T_H}{T_H-T_L}](https://tex.z-dn.net/?f=COP_%7BHP%7D%3D%5Cfrac%7BT_H%7D%7BT_H-T_L%7D)
Therefore, the temperature relationship, ![T_H=1.15\;T_L](https://tex.z-dn.net/?f=T_H%3D1.15%5C%3BT_L)
Then, we should apply the values in the COP.
![=\frac{1.15\;T_L}{1.15-1}](https://tex.z-dn.net/?f=%3D%5Cfrac%7B1.15%5C%3BT_L%7D%7B1.15-1%7D)
![=7.67](https://tex.z-dn.net/?f=%3D7.67)
The number of heat rejected by the heat pump must then be calculated.
![Q_H=COP_{HP}\times W_{nst}](https://tex.z-dn.net/?f=Q_H%3DCOP_%7BHP%7D%5Ctimes%20W_%7Bnst%7D)
![=7.67\times5=38.35](https://tex.z-dn.net/?f=%3D7.67%5Ctimes5%3D38.35)
We must then calculate the refrigerant mass flow rate.
![m=0.264\;kg/s](https://tex.z-dn.net/?f=m%3D0.264%5C%3Bkg%2Fs)
![q_H=\frac{Q_H}{m}](https://tex.z-dn.net/?f=q_H%3D%5Cfrac%7BQ_H%7D%7Bm%7D)
![=\frac{38.35}{0.264}=145.27](https://tex.z-dn.net/?f=%3D%5Cfrac%7B38.35%7D%7B0.264%7D%3D145.27)
The
value is 145.27 and therefore the hot reservoir temperature is 64° C.
The pressure at 64 ° C is thus 1849.36 kPa by interpolation.
And, the lowest reservoir temperature must be calculated.
![T_L=\frac{T_H}{1.15}](https://tex.z-dn.net/?f=T_L%3D%5Cfrac%7BT_H%7D%7B1.15%7D)
![=\frac{64+273}{1.15}=293.04](https://tex.z-dn.net/?f=%3D%5Cfrac%7B64%2B273%7D%7B1.15%7D%3D293.04)
![=19.89\°C](https://tex.z-dn.net/?f=%3D19.89%5C%C2%B0C)
the lowest reservoir temperature = 258.703 kpa
So, the pressure ratio should be = 7.15
Correcto no se muy bien de que se trata el tema porque está en inglés.
Sorry
Answer:
3 sec
If you are driving a 30-foot vehicle at 55 mph, how many seconds of following distance should you allow? 30ft truck. = 3 sec. Since the truck is over 40 mph.
Explanation:
Full Question
1. Correct the following code and
2. Convert the do while loop the following code to a while loop
declare integer product
declare integer number
product = 0
do while product < 100
display ""Type your number""
input number
product = number * 10
loop
display product
End While
Answer:
1. Code Correction
The errors in the code segment are:
a. The use of do while on line 4
You either use do or while product < 100
b. The use of double "" as open and end quotes for the string literal on line 5
c. The use of "loop" statement on line 7
The correction of the code segment is as follows:
declare integer product
declare integer number
product = 0
while product < 100
display "Type your number"
input number
product = number * 10
display product
End While
2. The same code segment using a do-while statement
declare integer product
declare integer number
product = 0
Do
display "Type your number"
input number
product = number * 10
display product
while product < 100