Answer:
61257.6 kW per day
Explanation:
The rating of the refrigerator is given as : 709 w
This means the power consumption is 709 joules per second
The operating time is given as 10 hours.
Change hours to second as
1 hour = 3600 seconds
10 hours = 36000 seconds
Apply the rating as;
709 w = 1 s
? = 36000 seconds
perform cross product to get;
709 * 36000= 25524000 w
25524000 = 25524kW
For a day, 24 hours will be;
25524 kW = 10 hours
? =24 hours
={24*25524}/10 = 61257.6 kW per day
Answer:
You should do it. it is okay if it does not work because if she knows it does not exist she is using it as a example of how it don't exist
Explanation:
Answer:
Pressure = 115.6 psia
Explanation:
Given:
v=800ft/s
Air temperature = 10 psia
Air pressure = 20F
Compression pressure ratio = 8
temperature at turbine inlet = 2200F
Conversion:
1 Btu =775.5 ft lbf,
= 32.2 lbm.ft/lbf.s², 1Btu/lbm=25037ft²/s²
Air standard assumptions:
= 0.0240Btu/lbm.°R, R = 53.34ft.lbf/lbm.°R = 1717.5ft²/s².°R 0.0686Btu/lbm.°R
k= 1.4
Energy balance:
As enthalpy exerts more influence than the kinetic energy inside the engine, kinetic energy of the fluid inside the engine is negligible
hence 

= 20+460 = 480°R
= 533.25°R
Pressure at the inlet of compressor at isentropic condition

=
= 14.45 psia
Answer:
a. ε₁=-0.000317
ε₂=0.000017
θ₁= -13.28° and θ₂=76.72°
b. maximum in-plane shear strain =3.335 *10^-4
Associated average normal strain ε(avg) =150 *10^-6
θ = 31.71 or -58.29
Explanation:

ε₁=-0.000317
ε₂=0.000017
To determine the orientation of ε₁ and ε₂

θ= -13.28° and 76.72°
To determine the direction of ε₁ and ε₂

=-0.000284 -0.0000335 = -0.000317 =ε₁
Therefore θ₁= -13.28° and θ₂=76.72°
b. maximum in-plane shear strain

=3.335 *10^-4

ε(avg) =150 *10^-6
orientation of γmax

θ = 31.71 or -58.29
To determine the direction of γmax

= 1.67 *10^-4
Explanation:
A.
H = Aeσ^4
Using the stefan Boltzmann law
When we differentiate
dH/dT = 4AeσT³
dH/dT = 4(0.15)(0.9)(5.67)(10^-8)(650)³
= 8.4085
Exact error = 8.4085x20
= 168.17
H(650) = 0.15(0.9)(5.67)(10^-8)(650)⁴
= 1366.376watts
B.
Verifying values
H(T+ΔT) = 0.15(0.9)(5.67)(10)^-8(670)⁴
= 1542.468
H(T+ΔT) = 0.15(0.9)(5.67)(10^-8)(630)⁴
= 1205.8104
Error = 1542.468-1205.8104/2
= 168.329
ΔT = 40
H(T+ΔT) = 0.15(0.9)(5.67)(10)^-8(690)⁴
= 1735.05
H(T-ΔT) = 0.15(0.9)(5.67)(10^-8)(610)⁴
= 1735.05-1059.83/2
= 675.22/2
= 337.61