Answer:
mevaporation=˙Qhfg=1. 8 kJ /s2269. 6 kJ /kg=0 . 793×10−3kg/ s=2. 855 kg /h
Explanation:
The properties of water at 1 atm and thus at the saturation temperature of 100C are hfg =2256.4 kJ/kg (Table A-4). The net rate of heat transfer to the water is ˙Q=0 . 60×3 kW=1 . 8 kWNoting that it takes 2256.4 kJ of energy to vaporize 1 kg of saturated liquid water, therate of evaporation of water is determined to be mevaporation=˙Qhfg=1. 8 kJ /s2269. 6 kJ /kg=0 . 793×10−3kg/ s=2. 855 kg /h
Answer:
The exact time when the sample was taken is = 0.4167337 hr
Explanation:
The diagram of a sketch of the tank is shown on the first uploaded image
Let A denote the first inlet
Let B denote the second inlet
Let C denote the single outflow from the tank
From the question we are given that the diameter of A is = 1 cm = 0.01 m
Area of A is =
=
Velocity of liquid through A = 0.2 m/s
The rate at which the liquid would flow through the first inlet in terms of volume = =
i.e is
=
The rate at which the liquid would flow through the first inlet in terms of mass of the liquid = mass of liquid × the rate of flow in terms of volume
=
= 0.016324
From the question the diameter of B = 2 cm = 0.02 m
Area of B =
Velocity of liquid through B = 0.01 m/s
The rate at which the liquid would flow through the first inlet in terms of volume = =
i.e is
=
The rate at which the liquid would flow through the second inlet in terms of mass of the liquid = mass of liquid × the rate of flow in terms of volume
=
= 0.00330642
From the question The flow rate in term of volume of the outflow at the time of measurement is given as = 0.5 L/s
And also from the question the mass of potassium chloride at the time of measurement is given as 13 g/L
So The rate at which the liquid would flow through the outflow in terms of mass of the liquid = mass of liquid × the rate of flow in terms of volume
=
= Note (1 Kg = 1000 g)
= 0.0065 kg/s
Considering potassium chloride
Let denote the rate at which liquid flows in terms of mass as as i.e change in mass with respect to time hence
Input(in terms of mass flow ) - output(in terms of mass flow ) = Accumulation in the Tank(in terms of mass flow )
(0.016324 + 0.00330642) - 0.0065 =
=> 0.01313122 t = (m - )
From the question (m - ) is given as = 19.7 Kg
Hence the time when the sample was taken is given as
0.01313122 t = 19.7 Kg
=> t = 1500.2414 sec
t = .4167337 hours (1 hour = 3600 seconds)
Answer:
(a) See attachment
(b) The two planes are parallel because the intercepts for plane [220] are X = 0,5 and Y = 0,5 and for plane [110] are X = 1 and Y = 1. When the planes are drawn, they keep the same slope in a 2D plane.
(c)
Explanation:
(a) To determine the intercepts for an specific set of Miller indices, the reciprocal intercepts are taken as follows:
For [110]
For [220]
The drawn of the planes is shown in the attachments.
(b) Considering the planes as two sets of 2D straight lines with no intersection to Z axis, then the slope for these two sets are:
For (1,1):
For (0.5, 0.5):
As shown above, the slopes are exactly equal, then, the two straight lines are considered parallel and for instance, the two planes are parallel also.
(c) To calculate the d-spacing between these two planes, the distance is calculated as follows:
The Miller indices are already given in the statement. Then, the distance is:
