Answer:
Explanation:
Given:
Steam Mass rate, ms = 1.5 kg/min
= 1.5 kg/min × 1 min/60 sec
= 0.025 kg/s
Air Mass rate, ma = 100 kg/min
= 100 kg/min × 1 min/60 sec
= 1.67 kg/s
A.
Extracting the specific enthalpy and temperature values from property table of “Saturated water – Pressure table” which corresponds to temperature at 0.07 MPa.
xf, quality = 0.9.
Tsat = 89.9°C
hf = 376.57 kJ/kg
hfg = 2283.38 kJ/kg
Using the equation for specific enthalpy,
hi = hf + (hfg × xf)
= 376.57 + (2283.38 × 0.9)
= 2431.552 kJ/kg
The specific enthalpy of the outlet, h2 = hf
= 376.57 kJ/kg
B.
Rate of enthalpy (heat exchange), Q = mass rate, ms × change in specific enthalpy
= ms × (hi - h2)
= 0.025 × (2431.552 - 376.57)
= 0.025 × 2055.042
= 51.37455 kW
= 51.38 kW.
Givens
=====
V
= 4.00 L
T
= 273oK We're assuming the temperature does not change, just the
pressure.
n
= 0.864 moles
R
= 8.314 joules / mole * oK
P
= ?????
Formula
======
PV
= n*R*T
P
= n*R*T/V
P
= 0.864 * 8.314 * 273 / 4
P
= 490 kpa
You
have to add 1.6 – 0.864 = 0.736 moles of gas.
We
have to assume that the temperature and pressure remain the same when
we add the 0.736 moles of gas. We are now looking for the volume.
PV
= n*R*T
<span>
V
= 0.736 * 8.314 * 273 / 490</span>
V
= 3.41 L Remember this is at about 4 atmospheres so we have to
convert to Standard Pressure.
Total
Volume = 3.41 + 4.00 = 4.41
V1
* P1 = V2 * P2
P1
= 490 kPa
P2
= 101 kPa
V1
= 7.41 L
V2
= ????
<span>
<span>
7.41*
490 = V2 * 101
V2
= 7.41 * 490 / 101
V2
= 35.94 L
</span>
</span>
<span>You
had 4 L now you need 31.94 more.</span>
The wave speed completely depends on the characteristics and properties of the medium . . . physical properties for mechanical waves, electrical properties for electromagnedtic waves.
So if you want to change the speed of a wave, you have to change the medium . . . shoot it through some different kind of stuff. <em>(B) </em>
Answer:
x = 0.4 m
Explanation:
When a spring is stretched from its equilibrium position. Some energy is stored in the spring. This energy is called the elastic potential energy of the spring. The formula used to calculate the magnitude of this stored energy is given as follows:
P.E = (1/2)kx²
where,
P.E = Elastic Potential Energy Stored in the spring = 45 J
k = Spring Constant = 540 N/m
x = amount of stretching = ?
Therefore,
45 J = (1/2)(540 N/m)x²
x² = (45 J)(2)/(540 N/m)
x = √(0.167 m²)
<u>x = 0.4 m</u>
