Answer:
We conclude that the approximate series solution (with only one eigein value) provides systematically high results but by less than 1.5%, for the biot number range from 0.11 to 10. See attached image.
Explanation:
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Answer:
Rate of Entropy =210.14 J/K-s
Explanation:
given data:
power delivered to input = 350 hp
power delivered to output = 250 hp
temperature of surface = 180°F
rate of entropy is given as
T = 180°F = 82°C = 355 K
Rate of heat = (350 - 250) hp = 100 hp = 74600 W
Rate of Entropy
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
Answer:
Outside temperature =88.03°C
Explanation:
Conductivity of air-soil from standard table
K=0.60 W/m-k
To find temperature we need to balance energy
Heat generation=Heat dissipation
Now find the value
We know that for sphere
Given that q=500 W
so
By solving that equation we get
=88.03°C
So outside temperature =88.03°C
