Universal Containers (UC) has a requirement to expose a web service to their business partners. The web service will be used to
1 answer:
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Answer:
i) 43.55 kg/s
ii) 40 m/s
iii) -199.32 KW
Explanation:
To resolve the above question we have to make some assumptions :
- mass flow through the system is constant
- The only interactions that are between the system and the surrounding are work and heat
- The fluid is uniform
i) first we have to determine the mass flow rate of the air
M =
= ---------- (1) hence M = 43.55 kg/s
ii) using this relationship : A1V1 = A2V2 hence V1 = (0.2/0.5) * 100 = 40m/s ( inlet velocity )
input this value into equation 1
iii) Next we will determine the power required to run compressor
attached below
power required = -199.32 KW ( this value indicates that there is power supplied )
Answer:
Explanation:
We are given:
m = 1.06Kg
T = 22kj
Therefore we need to find coefficient performance or the cycle
= 5
For the amount of heat absorbed:
= 5 × 22 = 110KJ
For the amount of heat rejected:
= 110 + 22 = 132KJ
[tex[ q_H = \frac{Q_L}{m} [/tex];
=
= 124.5KJ
Using refrigerant table at hfg = 124.5KJ/Kg we have 69.5°c
Convert 69.5°c to K we have 342.5K
To find the minimum temperature:
;
= 285.4K
Convert to °C we have 12.4°C
From the refrigerant R -134a table at = 12.4°c we have 442KPa