Answer:
A combination square is a multi-use measuring instrument which is primarily used for ensuring the integrity of a 90° angle, measuring a 45° angle, measuring the center of a circular object, find depth, and simple distance measurements. It can also be used to determine level and plumb using its spirit level vial.
Explanation:
Explanation:
The obtained data from water properties tables are:
Point 1 (condenser exit) @ 8 KPa, saturated fluid
![h_{f} = 173.358 \\h_{fg} = 2402.522](https://tex.z-dn.net/?f=h_%7Bf%7D%20%3D%20173.358%20%5C%5Ch_%7Bfg%7D%20%3D%202402.522)
Point 2 (Pump exit) @ 18 MPa, saturated fluid & @ 4 MPa, saturated fluid
![h_{2a} = 489.752\\h_{2b} = 313.2](https://tex.z-dn.net/?f=h_%7B2a%7D%20%3D%20%20489.752%5C%5Ch_%7B2b%7D%20%3D%20%20313.2)
Point 3 (Boiler exit) @ 18 MPa, saturated steam & @ 4 MPa, saturated steam
![h_{3a} = 2701.26 \\s_{3a} = 7.1656\\h_{3b} = 2634.14\\s_{3b} = 7.6876](https://tex.z-dn.net/?f=h_%7B3a%7D%20%3D%202701.26%20%5C%5Cs_%7B3a%7D%20%3D%207.1656%5C%5Ch_%7B3b%7D%20%3D%202634.14%5C%5Cs_%7B3b%7D%20%3D%207.6876)
Point 4 (Turbine exit) @ 8 KPa, mixed fluid
![x_{a} = 0.8608\\h_{4a} = 2241.448938\\x_{b} = 0.9291\\h_{4b} = 2405.54119](https://tex.z-dn.net/?f=x_%7Ba%7D%20%3D%200.8608%5C%5Ch_%7B4a%7D%20%3D%202241.448938%5C%5Cx_%7Bb%7D%20%3D%200.9291%5C%5Ch_%7B4b%7D%20%3D%202405.54119)
Calculate mass flow rates
Part a) @ 18 MPa
mass flow
![\frac{100*10^6 }{w_{T} - w_{P}} = \frac{100*10^3 }{(h_{3a} - h_{4a}) - (h_{2a} - h_{f})}\\\\= \frac{100*10^ 3}{(2701.26 - 2241.448938 ) - (489.752 - 173.358)}\\\\= 697.2671076 \frac{kg}{s} = 2510161.587 \frac{kg}{hr}](https://tex.z-dn.net/?f=%5Cfrac%7B100%2A10%5E6%20%7D%7Bw_%7BT%7D%20-%20w_%7BP%7D%7D%20%3D%20%5Cfrac%7B100%2A10%5E3%20%7D%7B%28h_%7B3a%7D%20%20-%20h_%7B4a%7D%29%20-%20%28h_%7B2a%7D%20%20-%20h_%7Bf%7D%29%7D%5C%5C%5C%5C%3D%20%5Cfrac%7B100%2A10%5E%203%7D%7B%282701.26%20%20-%202241.448938%20%29%20-%20%28489.752%20%20-%20173.358%29%7D%5C%5C%5C%5C%3D%20697.2671076%20%5Cfrac%7Bkg%7D%7Bs%7D%20%3D%202510161.587%20%5Cfrac%7Bkg%7D%7Bhr%7D)
Heat transfer rate through boiler
![Q_{in} = mass flow * (h_{3a} - h_{2a})\\Q_{in} = (697.2671076)*(2701.26-489.752)\\\\Q_{in} = 1542011.787 W](https://tex.z-dn.net/?f=Q_%7Bin%7D%20%20%3D%20mass%20flow%20%2A%20%28h_%7B3a%7D%20-%20%20h_%7B2a%7D%29%5C%5CQ_%7Bin%7D%20%3D%20%28697.2671076%29%2A%282701.26-489.752%29%5C%5C%5C%5CQ_%7Bin%7D%20%3D%201542011.787%20W)
Heat transfer rate through condenser
![Q_{out} = mass flow * (h_{4a} - h_{f})\\Q_{out} = (697.2671076)*(2241.448938-173.358)\\\\Q_{out} = 1442011.787 W](https://tex.z-dn.net/?f=Q_%7Bout%7D%20%20%3D%20mass%20flow%20%2A%20%28h_%7B4a%7D%20-%20%20h_%7Bf%7D%29%5C%5CQ_%7Bout%7D%20%3D%20%28697.2671076%29%2A%282241.448938-173.358%29%5C%5C%5C%5CQ_%7Bout%7D%20%3D%201442011.787%20W)
Thermal Efficiency
![n = \frac{W_{net} }{Q_{in} } = \frac{100*10^3}{1542011.787} \\\\n = 0.06485](https://tex.z-dn.net/?f=n%20%3D%20%5Cfrac%7BW_%7Bnet%7D%20%20%7D%7BQ_%7Bin%7D%20%7D%20%3D%20%5Cfrac%7B100%2A10%5E3%7D%7B1542011.787%7D%20%20%5C%5C%5C%5Cn%20%3D%200.06485)
Part b) @ 4 MPa
mass flow
![\frac{100*10^6 }{w_{T} - w_{P}} = \frac{100*10^3 }{(h_{3b} - h_{4b}) - (h_{2b} - h_{f})}\\\\= \frac{100*10^ 3}{(2634.14 - 2405.54119 ) - (313.12 - 173.358)}\\\\= 1125 \frac{kg}{s} = 4052374.235 \frac{kg}{hr}](https://tex.z-dn.net/?f=%5Cfrac%7B100%2A10%5E6%20%7D%7Bw_%7BT%7D%20-%20w_%7BP%7D%7D%20%3D%20%5Cfrac%7B100%2A10%5E3%20%7D%7B%28h_%7B3b%7D%20%20-%20h_%7B4b%7D%29%20-%20%28h_%7B2b%7D%20%20-%20h_%7Bf%7D%29%7D%5C%5C%5C%5C%3D%20%5Cfrac%7B100%2A10%5E%203%7D%7B%282634.14%20%20-%202405.54119%20%29%20-%20%28313.12%20%20-%20173.358%29%7D%5C%5C%5C%5C%3D%201125%20%5Cfrac%7Bkg%7D%7Bs%7D%20%3D%204052374.235%20%5Cfrac%7Bkg%7D%7Bhr%7D)
Heat transfer rate through boiler
![Q_{in} = mass flow * (h_{3b} - h_{2b})\\Q_{in} = (1125.65951)*(2634.14-313.12)\\\\Q_{in} = 2612678.236 W](https://tex.z-dn.net/?f=Q_%7Bin%7D%20%20%3D%20mass%20flow%20%2A%20%28h_%7B3b%7D%20-%20%20h_%7B2b%7D%29%5C%5CQ_%7Bin%7D%20%3D%20%281125.65951%29%2A%282634.14-313.12%29%5C%5C%5C%5CQ_%7Bin%7D%20%3D%202612678.236%20W)
Heat transfer rate through condenser
![Q_{out} = mass flow * (h_{4b} - h_{f})\\Q_{out} = (1125)*(2405.54119-173.358)\\\\Q_{out} = 2511206.089 W](https://tex.z-dn.net/?f=Q_%7Bout%7D%20%20%3D%20mass%20flow%20%2A%20%28h_%7B4b%7D%20-%20%20h_%7Bf%7D%29%5C%5CQ_%7Bout%7D%20%3D%20%281125%29%2A%282405.54119-173.358%29%5C%5C%5C%5CQ_%7Bout%7D%20%3D%202511206.089%20W)
Thermal Efficiency
![n = \frac{W_{net} }{Q_{in} } = \frac{100*10^3}{1542011.787} \\\\n = 0.038275](https://tex.z-dn.net/?f=n%20%3D%20%5Cfrac%7BW_%7Bnet%7D%20%20%7D%7BQ_%7Bin%7D%20%7D%20%3D%20%5Cfrac%7B100%2A10%5E3%7D%7B1542011.787%7D%20%20%5C%5C%5C%5Cn%20%3D%200.038275)
Answer:
(a) 3.455
(b) 21.143
(c) 16.36L/min
Explanation:
In this question, we’d be providing solution to the working process of a refrigerator given the data in the question.
Please check attachment for complete solution and step by step explanation
I don’t thick it is write but I thick it is option (c)
Answer:
4.6 mm
Explanation:
Given data includes:
thin-walled pipe diameter = 100-mm =0.1 m
Temperature of pipe
= -15° C = (-15 +273)K =258 K
Temperature of water
= 3° C = (3 + 273)K = 276 K
Temperature of ice
= 0° C = (0 +273)K =273 K
Thermal conductivity (k) from the ice table = 1.94 W/m.K ; R = 0.05
convection coefficient
=2000 W/m².K
The energy balance can be expressed as:
![q_{conduction} =q_{convention}](https://tex.z-dn.net/?f=q_%7Bconduction%7D%20%3Dq_%7Bconvention%7D)
where;
------------- equation (1)
------------ equation(2)
Equating both equation (1) and (2); we have;
![= \pi DLh_l(T_w-T_i)](https://tex.z-dn.net/?f=%3D%20%5Cpi%20DLh_l%28T_w-T_i%29)
Replacing the given data; we have:
![= \pi (0.1)*2000(276-273)](https://tex.z-dn.net/?f=%3D%20%5Cpi%20%280.1%29%2A2000%28276-273%29)
![\frac{182.84}{In(\frac{0.05}{r}) } = 1884.96](https://tex.z-dn.net/?f=%5Cfrac%7B182.84%7D%7BIn%28%5Cfrac%7B0.05%7D%7Br%7D%29%20%7D%20%3D%201884.96)
![In(\frac{0.05}{r})*1884.96 = 182.84](https://tex.z-dn.net/?f=In%28%5Cfrac%7B0.05%7D%7Br%7D%29%2A1884.96%20%3D%20182.84)
![In(\frac{0.05}{r}) = \frac{182.84}{1884.96}](https://tex.z-dn.net/?f=In%28%5Cfrac%7B0.05%7D%7Br%7D%29%20%3D%20%5Cfrac%7B182.84%7D%7B1884.96%7D)
![In(\frac{0.05}{r}) =0.0970](https://tex.z-dn.net/?f=In%28%5Cfrac%7B0.05%7D%7Br%7D%29%20%3D0.0970)
![\frac {0.05}{r} =e^{0.0970}](https://tex.z-dn.net/?f=%5Cfrac%20%7B0.05%7D%7Br%7D%20%3De%5E%7B0.0970%7D)
![\frac {0.05}{r} =1.102](https://tex.z-dn.net/?f=%5Cfrac%20%7B0.05%7D%7Br%7D%20%3D1.102)
![r=\frac{0.05}{1.102}](https://tex.z-dn.net/?f=r%3D%5Cfrac%7B0.05%7D%7B1.102%7D)
r = 0.0454
The thickness (t) of the ice layer can now be calculated as:
t = (R - r)
t = (0.05 - 0.0454)
t = 0.0046 m
t = 4.6 mm