Answer:
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Explanation:
Answer:
The energy in kJ is 8558.16 kJ.
Explanation:
Data presented in the problem:
Water is heated from 70 (T1) to 200 °F (T2).
Volume (V) of the water is 1 ft3.
It is required for the specific heat of water(HW), which is 1 BTU/lb°F.
First, we need to calculate the mass of water (M) presented in the process. Water density (D) is 62. 4 lb/ft3.
M = V*D = (1ft3)*(62.4 lb/ft3) = 62.4 lb Water.
.After that, we can calculate the heat required (Q).
Q = M*HW*(T2 - T1) = (62.4 lb)*(1 BTU/lb°F)(200 °F - 70°F)
Q = 62.4 * 130 BTU = 8112 BTU.
Q is converted to kJ units using the conversion factor 1 BTU = 1.055 kJ
Q = 8112 BTU * (1.055 kJ/1BTU) = 8558.16 kJ.
Finally, the energy required is 8558.16 kJ.
Answer:
h1 = 290.16kj/kg
P = 1.2311
Prandil expression at 8
P=p1/p7×pr
=8(1.2311)
=9.85
Enthalpy state at 8 corresponding to 9.85
h1 = 526.13kj/kg
Now prandtl state at 9 that correspond to 1400k.
h9 = 1515.42kj/kg
Pr = 450.5
Prandtl expression at state 10
P= p10/p9×pr
=1/8(450.5)
=56.31
Enthalpy at state 10 corresponding to prandtl 56.31
h10 = 860.39kj/kg
At 520k
h11 = 523.63kj/kg
Answer:
Explanation:
Kindly go through the attachment for the step by step approach to the solution of this question.
Answer:
Exit Temperature= T2=563.6 °R
Exit Pressure= P2= 30.06 lbf/in^2
Mass Flow rate=1.276 lb/sec
Explanation:
Answer is explained in detailed way in the attached files.
