Answer:
h = 6.35 W/m².k
Explanation:
In order to solve this problem, we will use energy balance, taking the thin hot plate as a system. According to energy balance, the rate of heat transfer to surrounding through convection must be equal to the energy stored in the plate:
Rate of Heat Transfer Through Convection = Energy Stored in Plate
- h A (Ts - T₀) = m C dT/dt
where,
h = convection heat transfer coefficient = ?
A = Surface area of plate through which heat transfer takes place = 2 x 0.3 m x 0.3 m (2 is multiplied for two sides of thin plate) = 0.18 m²
Ts = Surface Temperature of hot thin plate = 225⁰C
T₀ = Ambient Temperature = 25°C
m = mass of plate = 3.75 kg
C = Specific Heat = 2770 J/kg. k
dT/dt = rate of change in plate temperature = - 0.022 K/s
Therefore,
- h (0.18 m²)(225 - 25) k = (3.75 kg)(2770 J/kg.k)(- 0.022 k/s)
h = (- 228.525 W)/(- 36 m².k)
<u>h = 6.35 W/m².k</u>
Answer:
a)Bulk deformation process
Explanation:
<u>Rolling</u>
Rolling is a metal forming process.In rolling work piece passes through two moving rollers and get compressed.in rolling thickness of work piece will reduces and length of work piece will increase for maintaining the constant area.Due to compression bulk deformation takes place.
<u>Shearing</u>
In shearing one surface slides on another surface and deformation take place.shearing is a machining process.This is also a bilk motion deformation process.
So from above we can say that option a is right.
Answer:
D = 0.060732 in
Explanation:
given data
sp. wt. = 500 lb/ft³
diameter = 0.036 in
solution
we get here maximum diameter of rod that is express as
D = 
......................1
here 
surface tension of water at 60⁰f = 5.03 × 
lb/ft and y = 500 lb/ft³
so put here value and we will get
D = 
D = 0.005061 ft
D = 0.060732 in
Answer:
the lost work per kilogram of water for this everyday household happening = 0.413 kJ/kg
Explanation:
Given that:
Initial Temperature 
= 15°C
Initial Pressure 
= 5 atm
Final Pressure 
= 1 atm
Data obtain from steam tables of saturated water at 15°C are as follows:
Specific volume v = 1.001 cm³/gm
The change in temperature = 2°C
Specific heat of water = 4.19 J/gm.K
volume expansivity β = 1.5 × 10⁻⁴ K⁻¹
The expression to determine the change in temperature can be given as :
Δ T = 0.093 K
Now; we can calculate the lost work bt the formula:
where ;
is the temperature of the surrounding. = 20°C = (20+273.15)K = 293.15 K
From above the change in entropy is:
Thus, the lost work per kilogram of water for this everyday household happening = 0.413 kJ/kg
