Let current be I, charge be Q and time be t.
Here we are provided with,
I = 0.72A
t = 4s / 60s / 180s / 7s / 0.5s
We know,
I = Q/t
Case I
---------
When, t = 4s
0.72 = Q/4
Q = 0.72 * 4 = 2.88C
Case II
----------
When, t = 60s
0.72 = Q/60
Q = 0.72 * 60 = 43.2C
Case III
-----------
When, t = 180s
0.72 = Q/180
Q = 0.72 * 180 = 129.6C
Case IV
-----------
When, t = 7s
0.72 = Q/7
Q = 0.72 * 7 = 5.04C
Case V
----------
When, t = 0.5s
0.72 = Q/0.5
Q = 0.72 * 0.5 = 0.36C
Answer:
a. 572Btu/s
b.0.1483Btu/s.R
Explanation:
a.Assume a steady state operation, KE and PE are both neglected and fluids properties are constant.
From table A-3E, the specific heat of water is 
, and the steam properties as, A-4E:
Using the energy balance for the system:
Hence, the rate of heat transfer in the heat exchanger is 572Btu/s
b. Heat gained by the water is equal to the heat lost by the condensing steam.
-The rate of steam condensation is expressed as:
Entropy generation in the heat exchanger could be defined using the entropy balance on the system:
Hence,the rate of entropy generation in the heat exchanger. is 0.1483Btu/s.R
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Answer:
what's your question I can't understand
Answer:
Explanation:
initial momentum = .36 kg.m.s⁻¹
negative impulse = force x time = .02 x 12 = .24 kg.m.s⁻¹
final momentum - initial momentum = impulse
final momentum = initial momentum + impulse
= .36 - .24
= .12 kg.m.s⁻¹
