Answer:![h=160.84 W/m^2-K](https://tex.z-dn.net/?f=h%3D160.84%20W%2Fm%5E2-K)
Explanation:
Given
mass flow rate=0.3 kg/s
diameter of pipe=5 cm
length of pipe=10 m
Inside temperature=22
Pipe surface =100
Temperature drop=30
specific heat of vapor(c)=2190 J/kg.k
heat supplied ![Q=mc\Delta T=0.3\times 2190\times (30)](https://tex.z-dn.net/?f=Q%3Dmc%5CDelta%20T%3D0.3%5Ctimes%202190%5Ctimes%20%2830%29)
Heat due to convection =hA(100-30)
![A=\pi d\cdot L](https://tex.z-dn.net/?f=A%3D%5Cpi%20d%5Ccdot%20L)
![A=\pi 0.05\times 10=1.571 m^2](https://tex.z-dn.net/?f=A%3D%5Cpi%200.05%5Ctimes%2010%3D1.571%20m%5E2)
![Q_{convection}=h\times 1.571\times (100-22)=122.538 h](https://tex.z-dn.net/?f=Q_%7Bconvection%7D%3Dh%5Ctimes%201.571%5Ctimes%20%28100-22%29%3D122.538%20h)
![Q=Q_{convection}](https://tex.z-dn.net/?f=Q%3DQ_%7Bconvection%7D)
19,710=122.538 h
![h=160.84 W/m^2-K](https://tex.z-dn.net/?f=h%3D160.84%20W%2Fm%5E2-K)
Answer:
we can not use the suns energy too effectively in power cells and with human power we can generate more energy
Answer:
Injections of aqueous solution of fruct
levulose of formula C, H, O,,
For
prevent dehydration such solutions are obtained
dissolving a mass m = 25g of fructose for a volume of 50
final solution.
1.1 Calculate the molecular molar mass of fructose
1.2 Determine the amount of corresponding fructose material
1.3 Calculate the molar concentration of these fructose solutions
1.4
The final velocity of the truck is found as 146.969 m/s.
Explanation:
As it is stated that the lorry was in standstill position before travelling a distance or covering a distance of 3600 m, the initial velocity is considered as zero. Then, it is stated that the lorry travels with constant acceleration. So we can use the equations of motion to determine the final velocity of the lorry when it reaches 3600 m distance.
Thus, a initial velocity (u) = 0, acceleration a = 3 m/s² and the displacement s is 3600 m. The third equation of motion should be used to determine the final velocity as below.
![2as =v^{2} - u^{2} \\\\v^{2} = 2as + u^{2}](https://tex.z-dn.net/?f=2as%20%3Dv%5E%7B2%7D%20-%20u%5E%7B2%7D%20%5C%5C%5C%5Cv%5E%7B2%7D%20%3D%202as%20%2B%20u%5E%7B2%7D)
Then, the final velocity will be
![v^{2} = 2 * 3 * 3600 + 0 = 21600\\ \\v=\sqrt{21600}=146.969 m/s](https://tex.z-dn.net/?f=v%5E%7B2%7D%20%3D%202%20%2A%203%20%2A%203600%20%2B%200%20%3D%2021600%5C%5C%20%5C%5Cv%3D%5Csqrt%7B21600%7D%3D146.969%20m%2Fs)
Thus, the final velocity of the truck is found as 146.969 m/s.
Answer:
P₂ = 1.22 kPa
Explanation:
This problem can be solved using the equation of state:
![\frac{P_1V_1}{T_1} =\frac{P_2V_2}{T_2}](https://tex.z-dn.net/?f=%5Cfrac%7BP_1V_1%7D%7BT_1%7D%20%3D%5Cfrac%7BP_2V_2%7D%7BT_2%7D)
where,
P₁ = initial pressure = 1 KPa
P₂ = final pressure = ?
V₁ = initial Volume = 1 liter
V₂ = final volume = 1.1 liter
T₁ = initial temperature = 290 k
T₂ = final temperature = 390 k
Therefore,
![\frac{(1\ kPa)(1\ liter)}{290\ k} =\frac{(P_2)(1.1\ liter)}{390\ k}\\\\P_2= \frac{(1\ kPa)(1\ liter)(390\ k)}{(290\ k)(1.1\ liter)}](https://tex.z-dn.net/?f=%5Cfrac%7B%281%5C%20kPa%29%281%5C%20liter%29%7D%7B290%5C%20k%7D%20%3D%5Cfrac%7B%28P_2%29%281.1%5C%20liter%29%7D%7B390%5C%20k%7D%5C%5C%5C%5CP_2%3D%20%5Cfrac%7B%281%5C%20kPa%29%281%5C%20liter%29%28390%5C%20k%29%7D%7B%28290%5C%20k%29%281.1%5C%20liter%29%7D)
<u>P₂ = 1.22 kPa</u>