Answer:
Explanation:
We are given:
m = 1.06Kg
T = 22kj
Therefore we need to find coefficient performance or the cycle
= 5
For the amount of heat absorbed:
= 5 × 22 = 110KJ
For the amount of heat rejected:
= 110 + 22 = 132KJ
[tex[ q_H = \frac{Q_L}{m} [/tex];
= 
= 124.5KJ
Using refrigerant table at hfg = 124.5KJ/Kg we have 69.5°c
Convert 69.5°c to K we have 342.5K
To find the minimum temperature:
;
= 285.4K
Convert to °C we have 12.4°C
From the refrigerant R -134a table at 
= 12.4°c we have 442KPa
Maximum shear stress in the pole is 0.
<u>Explanation:</u>
Given-
Outer diameter = 127 mm
Outer radius,
= 127/2 = 63.5 mm
Inner diameter = 115 mm
Inner radius, 
= 115/2 = 57.5 mm
Force, q = 0
Maximum shear stress, τmax = ?
τmax 
If force, q is 0 then τmax is also equal to 0.
Therefore, maximum shear stress in the pole is 0.
Answer:
thickness1=1.4m
thickness2=2.2m
convection coefficient=0.33W/m^2K
Explanation:
you must use this equation to calculate the thickness:
L=K(T2-T1)/Q
L=thickness
T=temperature
Q=heat
L1=0.04*(0--350)/10=1.4m
L2=0.1(220-0)/10=2.2m
Then use this equation to calculate the convective coefficient
H=Q/(T2-T1)
H=10/(250-220)=0.33W/m^2K
Remote?? maybe I’m not really sure
Answer:
(a) 20 MHz
(b) 1.025 KW
(c) 3.33 ns
(d) 33 pF
Explanation:
(a) 20,000,000 Hz = 20 x 10^6 Hz = 20 Mega Hz = <u>20 MHz</u>
(b) 1025 W = 1.025 x 10^3 W = 1.025 Kilo W = <u>1.025 KW</u>
(c) 0.333 x 10^(-8) s = 3.33 x 10^(-9) s = 3.33 nano s = <u>3.33 ns</u>
(d) 33 x10^(-12)F = 33 pico F = <u>33 pF</u>
