Answer:
if I am not wrong the volumetric flow rate into the finance if the year inter 868 1.00 pm
Technician A is incorrect so that means Technician B is correct.
Answer:
Given:
high temperature reservoir 
low temperature reservoir 
thermal efficiency 
The engines are said to operate on Carnot cycle which is totally reversible.
To find the intermediate temperature between the two engines, The thermal efficiency of the first heat engine can be defined as

The thermal efficiency of second heat engine can be written as

The temperature of intermediate reservoir can be defined as

Given acceleration a = 5-3t, and its velocity is 7 at time t = 2, the value of s2 - s1 = 7
<h3>How to solve for the value of s2 - s1</h3>
We have
= 


v2 = 5x2 - 3x2 + c
= 10-6+c
= 4+c

S2 - S1

= 6 + 6+c - 2+3+c
12+c-5+c = 0
7 = c
Read more on acceleration here: brainly.com/question/605631
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 = 
exit quality = 
so exit quality = 0.2337 = 23.37 %
the quality of the refrigerant exiting the expansion valve is 0.2337 = 23.37 %