Steam enters a radiator at 16 psia and 0.97 quality. The steam flows through the radiator, is con- densed, and leaves as liquid
1 answer:
Answer:
5.328Ibm/hr
Explanation:
Through laboratory tests, thermodynamic tables were developed, these allow to know all the thermodynamic properties of a substance (entropy, enthalpy, pressure, specific volume, internal energy etc ..)
through prior knowledge of two other properties such as pressure and temperature.
for this case we can define the following equation for mass flow using the first law of thermodynamics
where
Q=capacity of the radiator =5000btu/hr
m = mass flow
then using thermodynamic tables we found entalpy in state 1 and 2
h1(x=0.97, p=16psia)=1123btu/lbm
h2(x=0, p=16psia)=184.5btu/lbm
solving
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Answer:
A) 222.58 kJ / kg
B) 0.8897 M^3/ kg
c) 0.7737 m^3/kg
D) 746.542 k
E) 536.017 kj/kg
efficiency = 58% ( approximately )
Explanation:
Given Data :
Gas constant (R) = 0.287 kJ/ kg.K
T1 = 310 k
P1 ( Kpa ) = 100
r = 11.5 ( compression ratio )
rp = 1.95 ( pressure ratio )
A ) specific internal energy at state 1
= Cv*T1 = 0.718 * 310 = 222.58 kJ / kg
B) Relative specific volume at state 1
= P1*V1 = R*T1 ( ideal gas equation )
V1 = R*T1 / P1 = (0.287* 10^3*310 ) / 100 * 10^3
V1 = 88.97 / 100 = 0.8897 M^3/ kg
C ) relative specific volume at state 2
Applying r ( compression ratio) = V1 / V2
11.5 = 0.8897 / V2
V2 = 0.8897 / 11.5 = 0.7737 m^3/kg
D) The temperature (k) at state 2
since the process is an Isentropic process we will apply the p-v-t relation
hence T2 = = 2.4082 * 310 = 746.542 k
e) specific internal energy at state 2
= Cv*T2 = 0.718 * 746.542 = 536.017 kj/kg
efficiency = output /input = 390.3511 / 667.5448 ≈ 58%
attached is a free hand diagram of an Otto cycle is attached below
