Answer:
T=kg·m^2/s^2
Explanation:
T = kg (m^2/s^2) m^3 /m^3
Here I wrote down the unit for every dimension.
T=kg m^2 / s^2
m^3 is divided between m^3, this is equal to 1.
Result: T=kg·m^2/s^2
PD: I'm not sure if this is what you ask for. I hope it helps
Answer:
i) 43.55 kg/s
ii) 40 m/s
iii) -199.32 KW
Explanation:
To resolve the above question we have to make some assumptions :
- mass flow through the system is constant
- The only interactions that are between the system and the surrounding are work and heat
- The fluid is uniform
i) first we have to determine the mass flow rate of the air
M = 
= 
---------- (1) hence M = 43.55 kg/s
ii) using this relationship : A1V1 = A2V2 hence V1 = (0.2/0.5) * 100 = 40m/s ( inlet velocity )
input this value into equation 1
iii) Next we will determine the power required to run compressor
attached below
power required = -199.32 KW ( this value indicates that there is power supplied )
Answer:
The final temperature in the vessel after the resistor has been operating for 30 min is 111.67°C
Explanation:
given information:
mass, m = 3 kg
initial temperature, T₁ = 40°C
current, I = 10 A
voltage, V = 50 V
time, t = 30 min = 1800 s
Heat for the system because of the resistance is
Q = V I t
where
V = voltage (V)
I = current (A)
t = time (s)
Q = heat transfer to the system (J)
so,
Q = V x I x t
= 50 x 10 x 1800
= 900000
= 9 x 10⁵ J
the heat transfer in the closed system is
Q = ΔU + W
where
U = internal energy
W = work done by the system
thus,
Q = ΔU + W
9 x 10⁵ = ΔU + 0, W = 0 because the tank is a well-insulated and rigid.
ΔU = 9 x 10⁵ J = 900 kJ
then, the energy change in the system is
ΔU = m c ΔT
ΔT = ΔU / m c, c = 4.186 J/g°C
= 900 / (3 x 4.186)
= 71.67°C
so,the final temperature (T₂)
ΔT = T₂ - T₁
T₂ = ΔT + T₁
= 71.67°C + 40°C
= 111.67°C
Otis Boykin is the answer.
