Consider laminar, fully developed flow in a channel of constant surface temperature Ts. For a given mass flow rate and channel l
ength, determine which rectangular channel, b/a = 1.0, 1.43, or 2.0, will provide the highest heat transfer rate. Is this heat transfer rate greater than, equal to, or less than the heat transfer rate associated with a circular tube? answer: b/a=2
1 answer:
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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
Answer: power required to maintain level flight=82.20hp
Explanation:
Given
Area = 200 ft^2
Weight = 2000 lb
Cl( Lift coefficient)= 0.39
Cd( Drag coefficient) = 0.06
The density ρ of air at standard atmospheric pressure = 2.38 X 10^-3 slugs/ft^3
For Equilibrium to be maintained during flight conditions, the lift force must be balanced by the weight of the aircraft such that
Lift force = Weight of aircraft
(1/2)ρAU²Cl= W
1/2X 2.38 X 10^-3 X 200 X U² X 0.39 = 2000
U²= 2000 X 2 / 2.38 X 10^-3 X 200 X 0.39
U=
Velocity, U= 146.7892ft/s
Drag force of the velocity can be deduced from the formulae
Cd= Drag force(D) /1/2 ρU²A
Drag force=1/2 ρU²ACd
D=1/2 x (2.38 X 10^-3 slugs/ft^3) x (146.7892ft/s)² x 200 ft^2 x 0.06
D=307.69
Drag force= 308lb
power required to maintain level flight is given as
P = Drag force x Velocity = D x U
=308lb X 146.7892ft/s
=45,211.0736lb.ft/s
Changing to hp we have that
1 Horsepower, hp = 550 ft lbf/s
??=45,211.0736lb.ft/s
45,211.0736lb.ft/s/ 550 ft lbf/s= 82.20hp