Answer:
Mass flow rate= 7.53 kg/sec
Exit Area= 0.108m^2
Explanation:
The solution and complete explanation for the above question and mentioned conditions is given below in the attached document.i hope my explanation will help you in understanding this particular question.
Answer:
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Answer: 33.35 minutes
Explanation:
A(t) = A(o) *(.5)^[t/(t1/2)]....equ1
Where
A(t) = geiger count after time t = 100
A(o) = initial geiger count = 400
(t1/2) = the half life of decay
t = time between geiger count = 66.7 minutes
Sub into equ 1
100=400(.5)^[66.7/(t1/2)
Equ becomes
.25= (.5)^[66.7/(t1/2)]
Take log of both sides
Log 0.25 = [66.7/(t1/2)] * log 0.5
66.7/(t1/2) = 2
(t1/2) = (66.7/2 ) = 33.35 minutes
Answer:
True
Explanation:
This is a true fact because From the time you are born to around the time you turn 30, your muscles grow larger and stronger.
Answer:
the quality of the refrigerant exiting the expansion valve is 0.2337 = 23.37 %
Explanation:
given data
pressure p1 = 1.4 MPa = 14 bar
temperature t1 = 32°C
exit pressure = 0.08 MPa = 0.8 bar
to find out
the quality of the refrigerant exiting the expansion valve
solution
we know here refrigerant undergoes at throtting process so
h1 = h2
so by table A 14 at p1 = 14 bar
t1 ≤ Tsat
so we use equation here that is
h1 = hf(t1) = 332.17 kJ/kg
this value we get from table A13
so as h1 = h2
h1 = h(f2) + x(2) * h(fg2)
so
exit quality = ![\frac{h1 - h(f2)}{h(fg2)}](https://tex.z-dn.net/?f=%5Cfrac%7Bh1%20-%20h%28f2%29%7D%7Bh%28fg2%29%7D)
exit quality = ![\frac{332.17- 9.04}{1382.73)}](https://tex.z-dn.net/?f=%5Cfrac%7B332.17-%209.04%7D%7B1382.73%29%7D)
so exit quality = 0.2337 = 23.37 %
the quality of the refrigerant exiting the expansion valve is 0.2337 = 23.37 %